a) \(\frac{81}{16}\)

b) \(\frac{-31}{8}\)

c) \(\frac{2417}{2401}\)

Bài 31:

a) \(\left(2^{-1}+3^{-1}\right):\left(2^{-1}-3^{-1}\right)+\left(2^{-1}.2^0\right):2^3\)

\(=\left(\frac{1}{2}+\frac{1}{3}\right):\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{2}.1\right):8\)

\(=\frac{5}{6}:\frac{1}{6}+\frac{1}{2}:8\)

\(=5+\frac{1}{16}\)

\(=\frac{81}{16}.\)

c) \(\left[\left(0,1\right)^2\right]^0+\left[\left(\frac{1}{7}\right)^1\right]^2.\frac{1}{49}.\left[\left(2^3\right)^3:2^5\right]\)

\(=1+\frac{1}{49}.\frac{1}{49}.16\)

\(=1+\frac{1}{2401}.16\)

\(=1+\frac{16}{2401}\)

\(=\frac{2417}{2401}.\)

Chúc bạn học tốt!

theo bài ra ta có 
n = 8a +7=31b +28 
=> (n-7)/8 = a 
b= (n-28)/31 
a - 4b = (-n +679)/248 = (-n +183)/248 + 2 
vì a ,4b nguyên nên a-4b nguyên => (-n +183)/248 nguyên 
=> -n + 183 = 248d => n = 183 - 248d (vì n >0 => d<=0 và d nguyên ) 
=> n = 183 - 248d (với d là số nguyên <=0) 
vì n có 3 chữ số lớn nhất => n<=999 => d>= -3 => d = -3 
=> n = 927

n=927

k nha

cái này dễ mà

\(4\cdot\left(\frac{1}{4}\right)^2+25\cdot\left[\left(\frac{3}{4}\right)^3\div\left(\frac{5}{4}\right)^3\right]\div\left(\frac{3}{2}\right)^3\)

\(=4\cdot\frac{1}{16}+25\cdot\left[\left(\frac{3}{4}\div\frac{5}{4}\right)^3\right]\div\left(\frac{3}{2}\right)^3\)

\(=\frac{1}{4}+25\cdot\left(\frac{3}{5}\right)^3\div\left(\frac{3}{2}\right)^3\)

\(=\frac{1}{4}+25\cdot\left(\frac{2}{5}\right)^3\)

\(=\frac{1}{4}+25\cdot\frac{8}{125}\)

\(=\frac{1}{4}\cdot\frac{8}{5}\)

\(=\frac{2}{5}\)

\(4.\left(\frac{1}{4}\right)^2+25.\left[\left(\frac{3}{4}\right)^3:\left(\frac{5}{4}\right)^3\right]:\left(\frac{3}{2}\right)^3\)

\(=4.\frac{1}{16}+25\left[\left(\frac{3}{4}:\frac{5}{4}\right)^3:\right]:\left(\frac{3}{2}\right)^3\)

\(=\frac{1}{4}+25.\left(\frac{3}{5}\right)^3:\left(\frac{3}{2}\right)^3\)

\(=\frac{1}{4}+25.\left(\frac{2}{5}\right)^3\)

\(=\frac{1}{4}+25.\frac{8}{125}\)

\(=\frac{1}{4}+\frac{8}{5}\)

\(=\frac{2}{5}\)

Câu 2)

Đặt \(\left\{\begin{matrix} u=\ln ^2x\\ dv=x^2dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=2\frac{\ln x}{x}dx\\ v=\frac{x^3}{3}\end{matrix}\right.\Rightarrow I=\frac{x^3}{3}\ln ^2x-\frac{2}{3}\int x^2\ln xdx\)

Đặt \(\left\{\begin{matrix} k=\ln x\\ dt=x^2dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} dk=\frac{dx}{x}\\ t=\frac{x^3}{3}\end{matrix}\right.\Rightarrow \int x^2\ln xdx=\frac{x^3\ln x}{3}-\int \frac{x^2}{3}dx=\frac{x^3\ln x}{3}-\frac{x^3}{9}+c\)

Do đó \(I=\frac{x^3\ln^2x}{3}-\frac{2}{9}x^3\ln x+\frac{2}{27}x^3+c\)

Câu 3:

\(I=\int\frac{2}{\cos 2x-7}dx=-\int\frac{2}{2\sin^2x+6}dx=-\int\frac{dx}{\sin^2x+3}\)

Đặt \(t=\tan\frac{x}{2}\Rightarrow \left\{\begin{matrix} \sin x=\frac{2t}{t^2+1}\\ dx=\frac{2dt}{t^2+1}\end{matrix}\right.\)

\(\Rightarrow I=-\int \frac{2dt}{(t^2+1)\left ( \frac{4t^2}{(t^2+1)^2}+3 \right )}=-\int\frac{2(t^2+1)dt}{3t^4+10t^2+3}=-\int \frac{2d\left ( t-\frac{1}{t} \right )}{3\left ( t-\frac{1}{t} \right )^2+16}=\int\frac{2dk}{3k^2+16}\)

Đặt \(k=\frac{4}{\sqrt{3}}\tan v\). Đến đây dễ dàng suy ra \(I=\frac{-1}{2\sqrt{3}}v+c\)

Câu 6)

\(I=-\int \frac{\left ( 1-\frac{1}{x^2} \right )dx}{x^2+2+\frac{1}{x^2}}=-\int \frac{d\left ( x+\frac{1}{x} \right )}{\left ( x+\frac{1}{x} \right )^2}=-\frac{1}{x+\frac{1}{x}}+c=-\frac{x}{x^2+1}+c\)

Câu 8)

\(I=\int \ln \left(\frac{x+1}{x-1}\right)dx=\int \ln (x+1)dx-\int \ln (x-1)dx\)

\(\Leftrightarrow I=\int \ln (x+1)d(x+1)-\int \ln (x-1)d(x-1)\)

Xét \(\int \ln tdt\) ta có:

Đặt \(\left\{\begin{matrix} u=\ln t\\ dv=dt\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=t\end{matrix}\right.\Rightarrow \int \ln tdt=t\ln t-\int dt=t\ln t-t+c\)

\(\Rightarrow I=(x+1)\ln (x+1)-(x+1)-(x-1)\ln (x-1)+x-1+c\)

\(\Leftrightarrow I=(x+1)\ln(x+1)-(x-1)\ln(x-1)+c\)