11/2. 4 5/3- 2 5/3. 11/2
= 11/2. (4 5/3- 2 5/3)
= 11/2.  2
= 22/2= 11
    Chúc bạn học tốt nhoa^^

\(\frac{11}{2}\) .  4 . \(\frac{5}{3}\) -  2 . \(\frac{5}{3}\) . \(\frac{11}{2}\)

=  \(\frac{110}{3}\)-  \(\frac{55}{3}\)

= \(\frac{55}{3}\)

cảm ơn bạn nha

Lời giải:

a) Đặt \(\left\{\begin{matrix} u=x\\ dv=\cos 2xdx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=dx\\ v=\frac{\sin 2x}{2}\end{matrix}\right.\)

\(\Rightarrow \int x\cos 2xdx=\frac{x\sin 2x}{2}-\int \frac{\sin 2x}{2}dx=\frac{x\sin 2x}{2}+\frac{\cos 2x}{4}\)

\(\Rightarrow \int ^{\frac{\pi}{2}}_{0}x\cos 2xdx=\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|\left ( \frac{x\sin 2x}{2}+\frac{\cos 2x}{4} \right )=\frac{-1}{2}\)

b) Đặt \(\left\{\begin{matrix} u=x\\ dv=e^{-2x}dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=dx\\ v=\frac{-e^{-2x}}{2}\end{matrix}\right.\)

\(\Rightarrow \int xe^{-2x}dx=\frac{-xe^{-2x}}{2}+\int \frac{e^{-2x}}{2}dx=\frac{-xe^{-2x}}{2}-\frac{e^{-2x}}{4}\)

\(\Rightarrow \int ^{\ln 2}_{0}xe^{-2x}dx=\left.\begin{matrix} \ln 2\\ 0\end{matrix}\right|\left ( \frac{-xe^{-2x}}{2}-\frac{e^{2x}}{4} \right )=\frac{3}{16}-\frac{\ln 2}{8}\)

c)

\(\int ^{1}_{0}\ln (2x+1)dx=\frac{1}{2}\int ^{1}_{0}\ln (2x+1)d(2x+1)=\frac{1}{2}\int ^{3}_{1}\ln tdt\)

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

Do đó \(\frac{1}{2}\int ^{3}_{1}\ln tdt=\left.\begin{matrix} 3\\ 1\end{matrix}\right|\left(\frac{t\ln t-t}{2}\right)=\frac{3\ln 3}{2}-1\)

d)

Ta có \(\int ^{3}_{2}(\ln (x-1)-\ln (x+1))dx=\int ^{3}_{2}\ln (x-1)d(x-1)-\int ^{3}_{2}\ln (x+1)d(x+1)\)

\(=\int ^{2}_{1}\ln tdt-\int ^{4}_{3}\ln tdt\)

Theo phần c, ta đã chỉ ra được \(\int \ln tdt=t\ln t-t\), do đó:

\(\int ^{2}_{1}\ln tdt-\int ^{4}_{3}\ln tdt=\left.\begin{matrix} 2\\ 1\end{matrix}\right|(t\ln t-t)-\left.\begin{matrix} 4\\ 3\end{matrix}\right|(t\ln t-t)=\ln \left(\frac{27}{64}\right)\)

e)

Xét \(\int (x+1-\frac{1}{x})e^{x+\frac{1}{x}}dx=\int e^{x+\frac{1}{x}}dx+\int \left (x-\frac{1}{x}\right)e^{x+\frac{1}{x}}dx\)

Đặt \(\left\{\begin{matrix} u=e^{x+\frac{1}{x}}\\ dv=dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\left(1-\frac{1}{x^2}\right)e^{x+\frac{1}{x}}dx\\ v=x\end{matrix}\right.\)

\(\Rightarrow \int e^{x+\frac{1}{x}}dx=xe^{x+\frac{1}{x}}-\int \left(x-\frac{1}{x}\right)e^{x+\frac{1}{x}}dx\)

Do đó \(\int \left(x+1-\frac{1}{x}\right)e^{x+\frac{1}{x}}dx=xe^{x+\frac{1}{x}}\)

\(\int ^{2}_{\frac{1}{2}}\left(x+1-\frac{x}{x}\right)e^{x+\frac{1}{x}}dx=\left.\begin{matrix} 2\\ \frac{1}{2}\end{matrix}\right|xe^{x+\frac{1}{x}}=\frac{3e^{\frac{5}{2}}}{2}\)

g)

Có \(\int x\cos x\sin^2xdx=\frac{1}{2}\int x\sin 2x\sin xdx=\frac{1}{4}\int x(\cos x-\cos 3x)dx\)

Xét \( \int x\cos xdx\). Đặt \(\left\{\begin{matrix} u=x\\ dv=\cos xdx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=dx\\ v=\sin x\end{matrix}\right.\)

\(\Rightarrow \int x\cos xdx=x\sin x-\int \sin xdx=x\sin x+\cos x\)

Tương tự, \(\int x\cos 3xdx=\frac{x\sin 3x}{3}+\frac{\cos 3x}{9}\)

Do đó, \(\int ^{\frac{\pi}{2}}_{0}x\cos x\sin^2xdx=\int ^{\frac{\pi}{2}}_{0}\frac{1}{4}x(\cos x-\cos 3x)dx\)

\(=\frac{1}{4}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|(x\sin x+\cos x-\frac{x\sin 3x}{3}-\frac{\cos 3x}{9})=\frac{3\pi-4}{18}\)

h)

\(\left\{\begin{matrix} u=xe^x\\ dv=\frac{1}{(x+1)^2}dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=e^x(x+1)dx\\ v=\frac{-1}{x+1}\end{matrix}\right.\)

\(\Rightarrow \int \frac{xe^x}{(x+1)^2}dx=\frac{-xe^x}{x+1}+\int e^xdx=\frac{-xe^x}{x+1}+e^x=\frac{e^x}{x+1}\)

Do đó \(\int ^{1}_{0}\frac{xe^x}{(x+1)^2}dx=\frac{e}{2}-1\)

i)

Có \(\int \frac{1+x\ln x}{x}e^xdx=\int \frac{e^x}{x}dx+\int \ln xe^xdx\)

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

\(\Rightarrow \int ^{e}_{1}\frac{1+x\ln x}{x}e^xdx=\left.\begin{matrix} e\\ 1\end{matrix}\right|\ln xe^x=e^e\)

\(5^2.7^3.11^2.x+5^3.7^2.11=0\)

\(\Leftrightarrow5^2.7^2.11\left(7.11.x+5\right)=0\)

\(\Leftrightarrow77x+5=0\)

\(\Leftrightarrow77x=-5\)

\(\Leftrightarrow x=-\frac{5}{77}\)

các bạn đâu hết rồi, làm đi để sáng mai mình còn nộp!

1:

a: =12/10-7/10=5/10=1/2

b: \(=\dfrac{4}{13}-\dfrac{4}{13}+\dfrac{-5}{11}-\dfrac{6}{11}=-\dfrac{11}{11}=-1\)

2: 

a: x+2/7=-11/7

=>x=-11/7-2/7=-13/7

b: (x+3)/4=-7/2

=>x+3=-14

=>x=-17

Bài 6: 

a: \(Q=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{3}\)

\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)