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\(A=17\frac{2}{31}-\left(\frac{15}{17}+6\frac{2}{31}\right)=\left(17\frac{2}{31}-6\frac{2}{31}\right)-\frac{15}{17}=11-\frac{15}{17}=10+\left(1-\frac{15}{17}\right)=10\frac{2}{17}\)
\(B=\left(31\frac{6}{13}-36\frac{6}{13}\right)+5\frac{9}{41}=-5+5\frac{9}{41}=\frac{9}{41}\)
C=\(\left(27\frac{51}{59}-7\frac{51}{59}\right)+\frac{1}{3}=20+\frac{1}{3}=20\frac{1}{3}\)
\(D=\left(13\frac{29}{31}-2\frac{28}{31}\right)+\left(4-3\frac{7}{8}\right)=11\frac{1}{31}+\frac{1}{8}=11\frac{8+31}{31.8}=11\frac{39}{248}\)
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)=\(\sqrt{SI^2+d\left(I,DM\right)^2}=\frac{a\sqrt{22}}{4}\)
\(2-\frac{13}{9}:\frac{5}{14}-\frac{5}{9}.\frac{14}{5}\)
\(=2-\frac{13}{9}.\frac{14}{5}-\frac{5}{9}.\frac{14}{5}\)
\(=2-\frac{14}{5}.\left(\frac{13}{9}-\frac{5}{9}\right)\)
\(=2-\frac{14}{5}.\frac{8}{9}\)
\(=2-\frac{112}{45}=\frac{90}{45}-\frac{112}{45}=\frac{-22}{45}\)
a. \(2^{2\log_25+\log_{\frac{1}{2}}9}\) và \(\frac{\sqrt{626}}{6}\)
Ta có : \(2^{2\log_25+\log_{\frac{1}{2}}9}=2^{\log_225-\log_29}=2^{\log_2\frac{25}{9}}=\frac{25}{9}=\frac{\sqrt{625}}{9}< \frac{\sqrt{626}}{6}\)
\(\Rightarrow2^{2\log_25+\log_{\frac{1}{2}}9}< \frac{\sqrt{626}}{6}\)
b. \(3^{\log_61,1}\) và \(7^{\log_60,99}\)
Ta có : \(\begin{cases}\log_61,1>0\Rightarrow3^{\log_61,1}>3^0=1\\\log_60,99< 0\Rightarrow7^{\log_60,99}< 7^0=1\end{cases}\)
\(\Rightarrow3^{\log_61,1}>7^{\log_60,99}\)
c. \(\log_{\frac{1}{3}}\frac{1}{80}\) và \(\log_{\frac{1}{2}}\frac{1}{15+\sqrt{2}}\)
Ta có : \(\begin{cases}\log_{\frac{1}{2}}\frac{1}{80}=\log_{3^{-1}}80^{-1}=\log_380< \log_381=4\\\log_{\frac{1}{2}}\frac{1}{15+\sqrt{2}}=\log_{2^{-1}}\left(15+\sqrt{2}\right)^{-1}=\log_2\left(15+\sqrt{2}\right)>\log_216=4\end{cases}\)
\(\Rightarrow\log_{\frac{1}{3}}\frac{1}{80}< \log_{\frac{1}{2}}\frac{1}{15+\sqrt{2}}\)
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 \(D=\log_{\frac{1}{3}}6^2-\log_{\frac{1}{3}}400^{\frac{1}{2}}+\log_{\frac{1}{3}}\left(\sqrt[3]{45}\right)\)
\(=\log_{\frac{1}{3}}36-\log_{\frac{1}{3}}20+\log_{\frac{1}{3}}45\)
\(=\log_{\frac{1}{3}}\frac{36.45}{20}=\log_{3^{-1}}81=-\log_33^4=-4\)
\(I_1=\int cos\left(\frac{\pi x}{2}\right)dx-\int\frac{2}{6x+5}dx=\frac{2}{\pi}\int cos\left(\frac{\pi x}{2}\right)d\left(\frac{\pi x}{2}\right)-\frac{1}{3}\int\frac{d\left(6x+5\right)}{6x+5}\)
\(=\frac{2}{\pi}sin\left(\frac{\pi x}{2}\right)-\frac{1}{3}ln\left|6x+5\right|+C\)
\(I_2=-\frac{1}{2}\int\left(4-x^4\right)^{\frac{1}{2}}d\left(4-x^4\right)=-\frac{1}{2}.\frac{\left(4-x^4\right)^{\frac{3}{2}}}{\frac{3}{2}}+C=\frac{-\sqrt{\left(4-x^4\right)^3}}{3}+C\)
\(I_3=2\int e^{\frac{1}{2}\left(4+x^2\right)}d\left(\frac{1}{2}\left(4+x^2\right)\right)=2e^{\frac{1}{2}\left(4+x^2\right)}+C=2\sqrt{e^{4+x^2}}+C\)
\(I_4=-\frac{1}{2}\int\left(1-x^2\right)^{\frac{1}{3}}d\left(1-x^2\right)=-\frac{1}{2}.\frac{\left(1-x^2\right)^{\frac{4}{3}}}{\frac{4}{3}}+C=-\frac{3}{8}\sqrt[3]{\left(1-x^2\right)^4}+C\)
\(I_5=\int e^{sinx}d\left(sinx\right)=e^{sinx}+C\)
\(I_6=\int\frac{d\left(1+sinx\right)}{1+sinx}=ln\left(1+sinx\right)+C\)
\(I_7=\int\left(x+1\right)\sqrt{x-1}dx\)
Đặt \(\sqrt{x-1}=t\Rightarrow x=t^2+1\Rightarrow dx=2tdt\)
\(\Rightarrow I_7=\int\left(t^2+2\right).t.2t.dt=\int\left(2t^4+4t^2\right)dt=\frac{2}{5}t^5+\frac{4}{3}t^3+C\)
\(=\frac{2}{5}\sqrt{\left(1-x\right)^5}+\frac{4}{3}\sqrt{\left(1-x\right)^3}+C\)
\(I_8=\int\left(2x+1\right)^{20}dx\)
Đặt \(2x+1=t\Rightarrow2dx=dt\Rightarrow dx=\frac{1}{2}dt\)
\(\Rightarrow I_8=\frac{1}{2}\int t^{20}dt=\frac{1}{42}t^{21}+C=\frac{1}{42}\left(2x+1\right)^{21}+C\)
\(I_9=-3\int\left(1-x^3\right)^{-\frac{1}{2}}d\left(1-x^3\right)=-3.\frac{\left(1-x^3\right)^{\frac{1}{2}}}{\frac{1}{2}}+C=-6\sqrt{1-x^3}+C\)
\(I_{10}=\int\frac{x}{\sqrt{2x+3}}dx\)
Đặt \(\sqrt{2x+3}=t\Rightarrow x=\frac{1}{2}t^2-\frac{3}{2}\Rightarrow dx=t.dt\)
\(\Rightarrow I_{10}=\int\frac{\frac{1}{2}t^2-\frac{3}{2}}{t}.t.dt=\frac{1}{2}\int\left(t^2-3\right)dt=\frac{2}{3}t^3-\frac{3}{2}t+C\)
\(=\frac{2}{3}\sqrt{\left(2x+3\right)^3}-\frac{3}{2}\sqrt{2x+3}+C\)
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