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Nội dung: <blockquote data-quote="uocmo_kchodoi" data-source="post: 139712" data-attributes="member: 165510">Câu 1. 1b, Đạo hàm y=<img src="https://codecogs.izyba.com/png.latex?4mx^{3}-2\left(m+1 \right)x" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" />Để hàm số có 3 cực trị thì đạo hàm y=0 phải có 3 nghiệm phân biệt. <img src="https://codecogs.izyba.com/png.latex?\Leftrightarrow m>0" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" /> hoặc <img src="https://codecogs.izyba.com/png.latex?m<-1" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" />Mặt khác theo đề ra đồ thị hàm số có 1 CĐ,2 CT nên <img src="https://codecogs.izyba.com/png.latex?m>0" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" />.Khi này gọi B là điểm CĐ. <img src="https://codecogs.izyba.com/png.latex?B\left(0,1 \right)" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" />C,D là 2 điểm CT.  <img src="https://codecogs.izyba.com/png.latex?C\left(\sqrt{\frac{m+1}{2m}},-\frac{\left(m-1 \right)^{2}}{4m} \right)" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" /> <img src="https://codecogs.izyba.com/png.latex?D\left(\sqrt{-\frac{m+1}{2m}},-\frac{\left(m-1 \right)^{2}}{4m} \right)" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" /> Vì B là trọng tâm tam giác có đỉnh  <img src="https://codecogs.izyba.com/png.latex?A\left(0,3 \right)" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" /> nên ta có.<img src="https://codecogs.izyba.com/png.latex?\begin{cases}x_{A}+x_{C}+x_{D}=3x_{B}\\y_{A}+y_{C}+y_{D}=3y_{B}\end{cases}" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" />...Gỉa ra m=1.thỏa mãn.Kết luận yêu cầu bài toán tương đương m=1.Câu 2.1ĐK.......Với đk trên ta có phương trình tương đương. <img src="https://codecogs.izyba.com/png.latex?\left(tan3x-tanx \right)+\left(cosx-cos3x \right)=0" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" /><img src="https://codecogs.izyba.com/png.latex?\Leftrightarrow \frac{sin2x}{cosx.cos3x}+2sinx.sin2x=0" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" /><img src="https://codecogs.izyba.com/png.latex?\Leftrightarrow sin2x.\left(\frac{1}{cosx.cos3x}+2sinx \right)=0" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" />TH1.  sin2x=0 => xTH2.  <img src="https://codecogs.izyba.com/png.latex?\frac{1}{cosx.cos3x}+2sinx=0" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" /><img src="https://codecogs.izyba.com/png.latex?\Leftrightarrow sin2x.cos3x=1" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" /><img src="https://codecogs.izyba.com/png.latex?\Leftrightarrow sin2x=cos3x=1" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" /> hoặc \[sin2x=cos3x=-1\] --> vn.Câu 2.2.ĐK. x#4 và x#2.Từ đề bài => bpt sẽ có nghiệm x>4 hoặc x<2.Xét các TH.TH1.  x>4 TH2. -2<x<2TH3. x=<-2Câu 4. Vẽ hình...Gọi O là giao 2 đường chéo của đáy. Từ bài ra => H nằm trên AC và CH=2HA.Kẻ MH vuông góc AB. M thuộc AB.<img src="https://codecogs.izyba.com/png.latex?\begin{cases}AB\perp MH\\AB\perp SH\end{cases}" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" /><img src="https://codecogs.izyba.com/png.latex?\Rightarrow AB\perp \left(SMH \right)\Rightarrow AB\perp SM" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" /><img src="https://codecogs.izyba.com/png.latex?\Rightarrow gocSMH=60^{0}" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" />Ta có. <img src="https://codecogs.izyba.com/png.latex?OB=\frac{a}{2},OA=\frac{a\sqrt{3}}{2}" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" />,<img src="https://codecogs.izyba.com/png.latex?OH=\frac{OA}{3}=\frac{a}{2\sqrt{3}}" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" /><img src="https://codecogs.izyba.com/png.latex?\Rightarrow HB=\frac{a}{\sqrt{3}}" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" /><img src="https://codecogs.izyba.com/png.latex?tan\left(OBH \right)=\frac{OH}{OB}=\frac{1}{\sqrt{3}}" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" />=> góc OBH=30* => góc ABH=30*.=> <img src="https://codecogs.izyba.com/png.latex?MH=HB.sin\left(ABH \right)=\frac{a}{2\sqrt{3}}" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" /><img src="https://codecogs.izyba.com/png.latex?\Rightarrow SH=MH.tan60^{0}=\frac{a}{2}" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" /><img src="https://codecogs.izyba.com/png.latex?V_{SBCDH}=2.V_{SBCH}=2.\frac{1}{3}.SH.S_{BCH}" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" /><img src="https://codecogs.izyba.com/png.latex?=2.\frac{1}{3}.SH.\frac{1}{2}.BH.BC=\frac{a^{3}}{6\sqrt{3}}" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" />ta có. <img src="https://codecogs.izyba.com/png.latex?d\left(E,\left(SHD \right) \right)=\frac{1}{2}d\left(B,\left(SHD \right) \right)" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" /><img src="https://codecogs.izyba.com/png.latex?OH=\frac{1}{4}CH\Rightarrow V_{SBDH}=\frac{1}{4}V_{SBCDH}=\frac{a^{3}}{24\sqrt{3}}" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" /><img src="https://codecogs.izyba.com/png.latex?S_{SHD}=\frac{1}{2}SH.HD=\frac{1}{2}SH.HB=\frac{a^{2}}{4\sqrt{3}}" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" /><img src="https://codecogs.izyba.com/png.latex?\Rightarrow d\left(E,\left(SHD \right) \right)=\frac{1}{2}.\frac{3V_{SBHD}}{S_{SHD}}" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" /><img src="https://codecogs.izyba.com/png.latex?=\frac{a}{4}" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" />Câu 6a2.<img src="https://codecogs.izyba.com/png.latex?\int \frac{\left(x-1 \right)lnx}{xlnx+1}dx" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" /><img src="https://codecogs.izyba.com/png.latex?=\int dx-\int \frac{lnx+1}{xlnx+1}dx" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" /><img src="https://codecogs.izyba.com/png.latex?=\int dx-\int \frac{1}{xlnx+1}d\left(xlnx+1 \right)" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" /><img src="https://codecogs.izyba.com/png.latex?x-ln\left(xlnx+1 \right)" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" />p/s. Không biết cái x có trị tuyệt đối không nữa.Câu 7a. Đưa về phương trình bậc 2 ẩn <img src="https://codecogs.izyba.com/png.latex?5^{x}" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" />Câu 6b2.ĐK. <img src="https://codecogs.izyba.com/png.latex?n>\frac{5}{2}" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" />Với điều kiện trên ta có. <img src="https://codecogs.izyba.com/png.latex?f\left(n \right)=log_{2}n+log_{3}\left(2n-5 \right)+log_{5}\left(n+9 \right)=9" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" />Xét hàm <img src="https://codecogs.izyba.com/png.latex?f\left(n \right)" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" /> có đạo hàm. <img src="https://codecogs.izyba.com/png.latex?\frac{1}{nln2}+\frac{2}{\left(2n-5 \right)ln3}+\frac{1}{\left(n+9 \right)ln5}>0" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" /> với mọi <img src="https://codecogs.izyba.com/png.latex?n>\frac{5}{2}" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" />.Do đó phương trình <img src="https://codecogs.izyba.com/png.latex?f\left(n \right)=9" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" /> có nhiều nhất 1 nghiệm.Mặt khác n=16 là 1 nghiệm của phương trình nên phương trình có 1nghiệm n=16.Câu 7b. Tìm được n=6.Ta có. <img src="https://codecogs.izyba.com/png.latex?\left(x^{2}+6 \right)^{6}.\left(x^{2}-x \right)^{6}" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" /><img src="https://codecogs.izyba.com/png.latex?=x^{6}.\left(x^{2}+6 \right)^{6}.\left(x-1 \right)^{6}" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" /><img src="https://codecogs.izyba.com/png.latex?=x^{6}.\Sigma C^{k}_{6}.x^{2k}.6^{6-k}\Sigma C^{m}_{6}. x^{m}.\left(-1 \right)^{6-m}" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" /> (k,m là số tự nhiên thỏa mãn 0=<k,m=<6).<img src="https://codecogs.izyba.com/png.latex?=\Sigma C^{k}_{6}.6^{6-k}.C^{m}_{6}.\left(-1 \right)^{m}.x^{2k+m+6}" alt="" class="fr-fic fr-dii fr-draggable " data-size="" style="" />Ta bài ra. 6+2k+m=14 với k,m là số tự nhiên và 0=<k,m=<6 <=> 2k+m=8 => m là số chẵn.m=0 thì k=4m=2 thì k=3m=4 thì k=2m=6 thì k=1.=> hệ số của số hạng x^14  (thay các giá trị cuả k,m rồi cộng lại} là 403596.</blockquote>

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