ΔABC vuông tại A

=>\(AB^2+AC^2=BC^2\)

=>\(AC^2=12^2-5^2=144-25=119\)

=>\(AC=\sqrt{119}\left(cm\right)\)

a: Xét ΔABC vuông tại A có AH là đường cao

nên \(\left\{{}\begin{matrix}AB^2=BH\cdot BC\\AC^2=CH\cdot BC\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}BH=\dfrac{5^2}{12}=\dfrac{25}{12}\left(cm\right)\\CH=\dfrac{119}{12}\left(cm\right)\end{matrix}\right.\)

b: Xét ΔABC vuông tại A có AH là đường cao

nên \(AH^2=HB\cdot HC=\dfrac{25}{12}\cdot\dfrac{119}{12}=\dfrac{25}{144}\cdot119\)

=>\(AH=\sqrt{119}\cdot\sqrt{\dfrac{25}{144}}=\dfrac{5}{12}\cdot\sqrt{119}\left(cm\right)\)

Ta thấy

\(x+23⋮31\)

\(x+8⋮15\Rightarrow\left(x+8\right)+15=x+23⋮15\)

\(\Rightarrow\left(x+23\right)=BC\left(15;31\right)\) x nhỏ nhất khi \(x+23=BCNN\left(15;31\right)\)

\(\Rightarrow BCNN\left(15;31\right)=15x31=465\)

\(\Rightarrow x+23=465\Rightarrow x=442\)

A=104-100+96-92+88-84+...-12+8

=(104-100)+(96-92)+...+(16-12)+8

=4+4+...+4+8

\(=4\cdot12+8=48+8=56\)

`8 . 2^(x - 5) = 16^7`

`=> 2^3  . 2^(x - 5) = (2^4)^7`

`=> 2^(3 + x - 5) = 2^28`

`=> x - 2 = 28`

`=> x=28+2`

`=>x=30`

Vậy: `x=30`

\(8\cdot2^{x-5}=16^7\)

=>\(2^3\cdot2^{x-5}=2^{28}\)

=>\(2^{x-2}=2^{28}\)

=>x-2=28

=>x=2+28=30

a: ĐKXĐ: x>=1/2

\(\sqrt{2x-1}=5\)

=>\(2x-1=5^2=25\)

=>2x=26

=>x=13(nhận)

b: ĐKXĐ: \(x>=-\dfrac{2}{3}\)

\(\sqrt{3x+2}=\dfrac{1}{4}\)

=>\(3x+2=\left(\dfrac{1}{4}\right)^2=\dfrac{1}{16}\)

=>\(3x=\dfrac{1}{16}-2=\dfrac{1}{16}-\dfrac{32}{16}=-\dfrac{31}{16}\)

=>\(x=-\dfrac{31}{48}\left(nhận\right)\)

c: \(\sqrt{x^2+\dfrac{1}{4}}=\sqrt{\dfrac{49}{81}}\)

=>\(x^2+\dfrac{1}{4}=\dfrac{49}{81}\)

=>\(x^2=\dfrac{49}{81}-\dfrac{1}{4}=\dfrac{115}{324}\)

=>\(x=\pm\dfrac{\sqrt{115}}{18}\)

\(1\dfrac{1}{5}:\left\{\dfrac{5}{8}+\left[\dfrac{5}{3}-\left(-\dfrac{1}{4}\right)\right]\cdot\dfrac{9}{2^2}\right\}\)

\(=\dfrac{6}{5}:\left\{\dfrac{5}{8}+\left(\dfrac{5}{3}+\dfrac{1}{4}\right)\cdot\dfrac{9}{4}\right\}\)

\(=\dfrac{6}{5}:\left\{\dfrac{5}{8}+\dfrac{23}{12}\cdot\dfrac{9}{4}\right\}\)

\(=\dfrac{6}{5}:\left\{\dfrac{5}{8}+\dfrac{23\cdot3}{16}\right\}=\dfrac{6}{5}:\left(\dfrac{10}{16}+\dfrac{69}{16}\right)\)

\(=\dfrac{6}{5}\cdot\dfrac{16}{79}=\dfrac{96}{395}\)

`n(n+1)(2n+1) = n(n+1)(n+2+n-1) = n(n+1)(n+2) + (n-1)n(n+1) `

Ta có: 

`n(n+1)(n+2)` là các số liên tiếp `=> {(n(n+1)(n+2) vdots 2),(n(n+1)(n+2) vdots 3):}`

`=> n(n+1)(n+2) vdots 6`

`(n-1)n(n+1)` là các số liên tiếp `=> {((n-1)n(n+1) vdots 2),((n-1)n(n+1) vdots 3):}`

`=> (n-1)n(n+1) vdots 6`

`=>  n(n+1)(n+2) + (n-1)n(n+1)  vdots 6`

`=> n(n+1)(2n+1)  vdots 6 (đpcm)`

\(n\left(n+1\right)\left(2n+1\right)\)

\(=n\left(n+1\right)\left(n+2+n-1\right)\)

\(=n\left(n+1\right)\left(n+2\right)+\left(n-1\right)\cdot n\cdot\left(n+1\right)\)

Vì n;n+1;n+2 là ba số nguyên liên tiếp

nên \(n\left(n+1\right)\left(n+2\right)⋮3!=6\)

Vì n-1;n;n+1 là ba số nguyên liên tiếp

nên \(\left(n-1\right)\cdot n\cdot\left(n+1\right)⋮3!=6\)

Do đó: \(n\left(n+1\right)\left(n+2\right)+\left(n-1\right)\cdot n\cdot\left(n+1\right)⋮6\)

=>\(n\left(n+1\right)\left(2n+1\right)⋮6\)