Ta có:

\(\int\dfrac{x^3+x^2}{x^2+6x+5}dx=\int\dfrac{x^2}{x+5}dx\)

\(=\int\left(x-5+\dfrac{25}{x+5}\right)dx=\dfrac{1}{2}x^2-5x+25ln\left|x+5\right|+C\)

Đặt \(x-1=t\Rightarrow dx=dt\)

Ta có:

\(\int\dfrac{x^2}{\left(x-1\right)^5}dx=\int\dfrac{\left(t+1\right)^2}{t^5}dt\)

\(=\int\left(t^{-3}+2t^{-4}+t^{-5}\right)dt\)

\(=\dfrac{t^{-2}}{-2}+2.\dfrac{t^{-3}}{-3}+\dfrac{t^{-4}}{-4}+C\)

\(=\dfrac{-1}{2\left(x-1\right)^2}-\dfrac{2}{3\left(x-1\right)^3}-\dfrac{1}{4\left(x-1\right)^4}+C\)

Thiếu hình mạch điện r bạn

Gọi thể tích khối đa diện có các đỉnh \(\text{A,B,C,M,N}\) là V

Ta có:

\(\dfrac{V_{SAMN}}{V_{SABC}}=\dfrac{S_{SMN}}{S_{SBC}}=\dfrac{1}{4}\)\(\Rightarrow V=\dfrac{3}{4}V_{SABC}=\dfrac{3}{4}.\dfrac{1}{3}.S_{ABC}.SA=\dfrac{1}{4}.S_{ABC}.SA\)

\(=\dfrac{1}{4}.\dfrac{1}{2}.SA.AB.AC.sin\widehat{BAC}\)\(=\dfrac{a^3}{8}\)

28D

HD: ĐK: \(m\left(m-2\right)\ge0\)

 \(2^{x_1}.2^{x_2}=2m\). Mà theo gt thì \(x_1+x_2=3\) nên \(2^3=2m\Leftrightarrow m=4\left(tm\right)\)

37B

HD: \(a^xb^{x^2-2}=1\Leftrightarrow a^x=b^{2-x^2}\)

Do \(a,b>1\) nên \(xlog_ba=2-x^2\)\(\Leftrightarrow log_ba=\dfrac{2-x^2}{x}\)

Do đó

\(\dfrac{2-x_1^2}{x_1}=\dfrac{2-x_2^2}{x_2}>0\)

Đặt \(x_1+x_2=t\) thì ta có:

\(\left\{{}\begin{matrix}x_1x_2=-2\\t< 0\end{matrix}\right.\)

Do đó: \(T=\dfrac{4}{t^2}-5t=\dfrac{4}{t^2}+\left(-\dfrac{5t}{2}\right)+\left(-\dfrac{5t}{2}\right)\ge3\sqrt[3]{\dfrac{4}{t^2}.\left(-\dfrac{5t}{2}\right)^2}=3\sqrt[3]{25}\)

Đẳng thức xảy ra\(\Leftrightarrow t=-\dfrac{2}{\sqrt[3]{5}}\Rightarrow\left\{{}\begin{matrix}x_1=...\\x_2=...\end{matrix}\right.\)

38. Không có đáp án đúng

HD: ĐK:\(\left[{}\begin{matrix}m\ge-6+18\sqrt{3}\\m\le-6-18\sqrt{3}\end{matrix}\right.\) 

Ta có: \(\left\{{}\begin{matrix}3^{x_1}.3^{x_2}=243\left(1\right)\\3^{x_1}+3^{x_2}=m+6\left(2\right)\end{matrix}\right.\)

Từ (1)\(\Rightarrow x_1+x_2=5\)

Kết hợp với gt, ta có:

\(x_1.x_2=7\)\(\Rightarrow\) Không có x thoả mãn, đề sai

(Đề có thể sửa lại là \(\dfrac{1}{x_1}+\dfrac{1}{x_2}+\dfrac{1}{x_1x_2}=1\), khi đó \(x_1=2;x_2=3;m=30\)

là hợp lý)

39C

HD: Gt \(\Rightarrow3^x-3^{-x}=2cos\left(nx\right)\) (*)

PT đã cho tđ vs:

 \(9^x+\dfrac{1}{9^x}=4+2.\left(2cos^2\left(nx\right)-1\right)=4cos^2\left(nx\right)+2\)

\(\Leftrightarrow\left(3^x-3^{-x}\right)^2=4cos^2\left(nx\right)\)\(\Leftrightarrow\left[{}\begin{matrix}3^x-3^{-x}=2cos\left(nx\right)\\3^x-3^{-x}=-2cos\left(nx\right)\end{matrix}\right.\)

Do pt (*) có 2023 nghiệm nên pt đã cho có 2.2023=4046 nghiệm

Ta có:

\(nu_{n+2}-\left(3n+1\right)u_{n+1}+2\left(n+1\right)u_n=3\)

\(\Leftrightarrow n\left(u_{n+2}-2u_{n+1}\right)-\left(n+1\right)\left(u_{n+1}-2u_n\right)=3\)

Đặt \(u_{n+1}-2u_n=v_n\)

\(\Rightarrow\left\{{}\begin{matrix}v_1=u_2-2u_1=-2-2.\left(-1\right)=0\\nv_{n+1}-\left(n+1\right)v_n=3\left(1\right)\end{matrix}\right.\)

Từ \(\left(1\right)\Rightarrow\dfrac{1}{n+1}v_{n+1}-\dfrac{1}{n}v_n=\dfrac{3}{n\left(n+1\right)}\)

Ta có:

\(\dfrac{1}{2}v_2-v_1=\dfrac{3}{1.2}\)

\(\dfrac{1}{3}v_3-\dfrac{1}{2}v_2=\dfrac{3}{2.3}\)

\(\dfrac{1}{4}v_4-\dfrac{1}{3}v_3=\dfrac{3}{3.4}\)

\(...\)

\(\dfrac{1}{n}v_n-\dfrac{1}{n-1}v_{n-1}=\dfrac{3}{\left(n-1\right)n}\)

\(\dfrac{1}{n+1}v_{n+1}-\dfrac{1}{n}v_n=\dfrac{3}{n\left(n+1\right)}\)

Cộng theo vế, ta có:

\(\dfrac{1}{n+1}v_{n+1}-v_1=3\left(1-\dfrac{1}{n+1}\right)\)

\(\Rightarrow v_{n+1}=3n\Leftrightarrow v_n=3\left(n-1\right)\)

\(\Rightarrow u_{n+1}-2u_n=3\left(n-1\right)\)

\(\Leftrightarrow u_{n+1}+3\left(n+1\right)=2\left(u_n+3n\right)\)

Đặt \(a_n=u_n+3n\Rightarrow\left\{{}\begin{matrix}a_1=u_1+3=2\\a_{n+1}=2a_n\end{matrix}\right.\)

\(\Rightarrow a_n=2^n\)\(\Rightarrow u_n=2^n-3n\)\(,\forall n\in N\text{*}\)