`a)`Xét tam giác IEK vuông E, đường cao EM:

`@`\(EK=\sqrt{IK^2-IE^2}=\sqrt{13^2-5^2}=12\left(cm\right)\)

`@`\(IM=\dfrac{EI^2}{IK}=\dfrac{5^2}{13}=\dfrac{25}{13}\left(cm\right)\)

`@`\(MK=IK-IM=13-\dfrac{25}{13}=\dfrac{144}{13}\left(cm\right)\)

`@`\(EM=\dfrac{EI.EK}{IK}=\dfrac{5.12}{13}=\dfrac{60}{13}\left(cm\right)\)

`b)`Xét tam giác IEK vuông E, đường cao EM:

`@`\(IK=\sqrt{IE^2+EK^2}=\sqrt{6^2+8^2}=10\left(cm\right)\)

`@`\(EM=\sqrt{\dfrac{EI^2.EK^2}{EI^2+EK^2}}=\sqrt{\dfrac{6^2.8^2}{6^2+8^2}}=4,8\left(cm\right)\)

`@`\(MK=\dfrac{EK^2}{IK}=\dfrac{8^2}{10}=6,4\left(cm\right)\)

`a)`\(\left(-\dfrac{1}{3}\right)^7.3^7=-\dfrac{1^7}{3^7}.3^7=-1^7=-1\)

`b)`\(\left(0,125\right)^3.512=\left(0,125.8\right)^3=1^3=1\)

`c)`\(\dfrac{90^2}{15^2}=\dfrac{\left(15.6\right)^2}{15^2}=\dfrac{15^2.6^2}{15^2}=6^2=36\)

`d)`\(\dfrac{790^4}{79^4}=\dfrac{\left(79.10\right)^4}{79^4}=\dfrac{79^4.10^4}{79^4}=10^4=10000\)

á lâu quá mới onl lại tay đơ quá gõ ko nhanh đc:"((

`m)`\(9x^2-\left(x-4\right)^2=0\)

\(\Leftrightarrow\left(3x\right)^2=\left(x-4\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=x-4\\3x=4-x\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)

Vậy \(S=\left\{-2;1\right\}\)

`n)`\(x^2+5x+6=0\)

\(\Leftrightarrow x^2+2x+3x+6=0\)

`<=>x(x+2)+3(x+2)=0`

`<=>(x+2)(x+3)=0`

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

Vậy \(S=\left\{-3;-2\right\}\)

`k)`\(\left(x+3\right)^2-\left(x-2\right)\left(x+2\right)=0\)

`<=>x^2+6x+9-x^2+4=0`

`<=>6x+13=0`

`<=>x=-13/6`

Vậy \(S=\left\{-\dfrac{13}{6}\right\}\)

`a)` \(\dfrac{n+20}{n+9}=\dfrac{n+9+11}{n+9}=1+\dfrac{11}{n+9}\)

Để \(\left(n+20\right)⋮\left(n+9\right)\) thì \(\dfrac{11}{n+9}\in Z\) \(\Rightarrow n+9\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)

Mà \(n\in N\Rightarrow n+9\ge9\) 

chỉ có \(n+9=11\) là thỏa mãn

\(\Rightarrow n+9=11\rightarrow n=2\)

Vậy \(n=2\) thì \(\left(n+20\right)⋮\left(n+9\right)\)

`b)`\(\dfrac{n+15}{n+5}=\dfrac{n+5+10}{n+5}=1+\dfrac{10}{n+5}\)

Để \(\left(n+15\right)⋮\left(n+5\right)\) thì \(\dfrac{10}{n+5}\in Z\)

\(\Rightarrow n+5\inƯ\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)

Mà \(n\in N\Rightarrow n+5\ge5\)

chỉ có 5 và 10 thỏa mãn

`@n+5=5->n=0`

`@n+5=10->n=5`

Vậy \(n\in\left(0;5\right)\) thì \(\left(n+15\right)⋮\left(n+5\right)\)

`a)`Xét tam giác MNP vuông M, có:

`@`\(\sin N=\dfrac{MP}{NP}\)

\(\Leftrightarrow\sin30=\dfrac{2}{NP}\)

\(\Rightarrow NP=\dfrac{2}{\sin30}=4\left(cm\right)\)

`@`\(NM=\sqrt{NP^2-MP^2}=\sqrt{4^2-2^2}=2\sqrt{3}\left(cm\right)\)

`b)` Góc M làm sao tính được bạn?