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\(2-\frac{13}{9}:\frac{5}{14}-\frac{5}{9}.\frac{14}{5}\)
\(=2-\frac{13}{9}.\frac{14}{5}-\frac{5}{9}.\frac{14}{5}\)
\(=2-\frac{14}{5}.\left(\frac{13}{9}-\frac{5}{9}\right)\)
\(=2-\frac{14}{5}.\frac{8}{9}\)
\(=2-\frac{112}{45}=\frac{90}{45}-\frac{112}{45}=\frac{-22}{45}\)
\(A=17\frac{2}{31}-\left(\frac{15}{17}+6\frac{2}{31}\right)=\left(17\frac{2}{31}-6\frac{2}{31}\right)-\frac{15}{17}=11-\frac{15}{17}=10+\left(1-\frac{15}{17}\right)=10\frac{2}{17}\)
\(B=\left(31\frac{6}{13}-36\frac{6}{13}\right)+5\frac{9}{41}=-5+5\frac{9}{41}=\frac{9}{41}\)
C=\(\left(27\frac{51}{59}-7\frac{51}{59}\right)+\frac{1}{3}=20+\frac{1}{3}=20\frac{1}{3}\)
\(D=\left(13\frac{29}{31}-2\frac{28}{31}\right)+\left(4-3\frac{7}{8}\right)=11\frac{1}{31}+\frac{1}{8}=11\frac{8+31}{31.8}=11\frac{39}{248}\)
Ta có :
\(\begin{cases}5>1;3>1\Rightarrow\log_53>0\\15>1;4>1\Rightarrow\log_{15}4>0\\0< \frac{1}{3}< 1;\frac{7}{2}>1\Rightarrow\log_{\frac{1}{3}}\frac{14}{5}< 0\\0< 0,3< 1;\frac{7}{2}>1\Rightarrow\log_{0,3}\frac{7}{2}< 0\end{cases}\)
\(\Rightarrow A=\frac{\log_53.\log_{15}4}{\log_{\frac{1}{3}}\frac{14}{5}\log_{0,3}\frac{7}{2}}>0\)
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 \(I=\frac{-1}{2\sqrt{3}}v+c\)
\(A=\frac{\frac{3}{2}+\frac{2}{5}+\frac{1}{10}}{\frac{3}{2}+\frac{2}{3}+\frac{1}{12}}\)
\(\Rightarrow A=\frac{\frac{15}{10}+\frac{4}{10}+\frac{1}{10}}{\frac{18}{12}+\frac{8}{12}+\frac{1}{12}}=\frac{\frac{20}{10}}{\frac{27}{12}}=\frac{2}{\frac{9}{4}}=2:\frac{9}{4}=2.\frac{4}{9}=\frac{8}{9}\)
! Ko bt có đúng ko nx @@@
~ Học tốt
# Chiyuki Fujito
a) \(A=\log_{5^{-2}}5^{\frac{5}{4}}=-\frac{1}{2}.\frac{5}{4}.\log_55=-\frac{5}{8}\)
b) \(B=9^{\frac{1}{2}\log_22-2\log_{27}3}=3^{\log_32-\frac{3}{4}\log_33}=\frac{2}{3^{\frac{3}{4}}}=\frac{2}{3\sqrt[3]{3}}\)
c) \(C=\log_3\log_29=\log_3\log_22^3=\log_33=1\)
d) Ta có \(D=\log_{\frac{1}{3}}6^2-\log_{\frac{1}{3}}400^{\frac{1}{2}}+\log_{\frac{1}{3}}\left(\sqrt[3]{45}\right)\)
\(=\log_{\frac{1}{3}}36-\log_{\frac{1}{3}}20+\log_{\frac{1}{3}}45\)
\(=\log_{\frac{1}{3}}\frac{36.45}{20}=\log_{3^{-1}}81=-\log_33^4=-4\)
I*AB=> SI\(\perp\)AB
SI=\(SI=\frac{AB\sqrt{3}}{2}=\frac{a\sqrt{3}}{2}\)
\(V_{k.chop}=\frac{1}{3}.\frac{a\sqrt{3}}{2}.a^2=\frac{a^3\sqrt{3}}{4}\)
b) Kẻ IK//DM(K\(\in\)AD)
Kẻ KH\(\perp\)DM(H\(\in\)DM)
=> d(I,DM)=d(K,DM0=KH
\(\Delta IAK~\Delta DCM\Rightarrow AK=\frac{1}{2}CM=\frac{a}{6}\)=> KD=5a/6
\(cos\widehat{ADM}=cos\widehat{DMC}=\frac{CM}{DM}=\frac{\frac{a}{3}}{\frac{a\sqrt{10}}{3}}=\frac{1}{\sqrt{10}}\)
=> KH=KDsin\(\widehat{ADM}\)=\(\sqrt{1-\cos\widehat{ADM}^2}=\frac{5a}{6}.\frac{3}{\sqrt{10}}=\frac{a\sqrt{10}}{4}\)
d(S,DM)=\(\sqrt{SI^2+d\left(I,DM\right)^2}=\frac{a\sqrt{22}}{4}\)
Giải:
Ta có: \(\frac{x-2}{5}=\frac{2x-3}{4}\)
\(\Rightarrow\left(x-2\right).4=5.\left(2x-3\right)\)
\(\Rightarrow4x-8=10x-15\)
\(\Rightarrow4x-10x=8-15\)
\(\Rightarrow-6x=-7\)
\(\Rightarrow x=\frac{7}{6}\)
Vậy \(x=\frac{7}{6}\)
Giải :
Ta có : \(\frac{x-2}{5}=\frac{2x-3}{4}\)
\(\Rightarrow\left(x-2\right),4=5,\left(2x-3\right)\)
\(\Rightarrow4x-8=10x-15\)
\(\Rightarrow4x-10x=8-15\)
\(\Rightarrow-6x=-7\)
\(\Rightarrow x=\frac{7}{6}\)
Vậy \(x\) là \(\frac{7}{6}\)
3
=1/2+1/2+1/2+1/2+1/2+1/2
=1/2.6
=6/2=3