a: Xét ΔHAC vuông tại H và ΔABC vuông tại A có

\(\widehat{C}\) chung

Do đó: ΔHAC~ΔABC

b: ΔABC vuông tại A

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

=>\(BC^2=15^2+20^2=625\)

=>BC=25

Xét ΔABC vuông tại A có AH là đường cao

nên \(\left\{{}\begin{matrix}BH\cdot BC=BA^2\\AH\cdot BC=AB\cdot AC\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}BH\cdot25=15^2=225\\AH\cdot25=15\cdot20=300\end{matrix}\right.\)

=>BH=9; AH=12

\(a,ĐK:...\\ PT\Leftrightarrow x^2-6x=x^2-7x+10\\ \Leftrightarrow x=10\left(tm\right)\\ b,ĐK:...\\ PT\Leftrightarrow2x\left(4-x\right)-\left(2-2x\right)\left(8-x\right)=\left(8-x\right)\left(4-x\right)\\ \Leftrightarrow8x-2x^2+16+18x-2x^2=32-12x+x^2\\ \Leftrightarrow3x^2-38x+16=0\left(casio\right)\\ c,ĐK:...\\ PT\Leftrightarrow2x\left(x-4\right)-4x=0\\ \Leftrightarrow2x^2-12x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)

GHI RÕ DÙM MÌNH ĐK CỦA CẢ 3 CÂU LUÔN ĐC KO Á.

Bài 13:

ĐKXĐ: x∉{0;2;-2;1/2}

a: \(B=\left(\frac{x+2}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{x+2}\right):\frac{2x^2-x}{x^2-2x}\)

\(=\left(\frac{-\left(x+2\right)}{x-2}-\frac{4x^2}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{x+2}\right):\frac{x\left(2x-1\right)}{x\left(x-2\right)}\)

\(=\frac{-\left(x+2\right)^2-4x^2+\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}\cdot\frac{x-2}{2x-1}\)

\(=\frac{-x^2-4x-4-4x^2+x^2-4x+4}{x+2}\cdot\frac{1}{2x-1}=\frac{-4x^2-8x}{\left(x+2\right)\left(2x-1\right)}\)

\(=\frac{-4x\left(x+2\right)}{\left.\left(x+2\right)\left(2x-1\right)\right.}=\frac{-4x}{2x-1}\)
b: |x|=3

=>x=3 hoặc x=-3

Khi x=3 thì \(B=\frac{-4\cdot3}{2\cdot3-1}=\frac{-12}{5}\)

Khi x=-3 thì \(B=\frac{-4\cdot\left(-3\right)}{2\cdot\left(-3\right)-1}=\frac{12}{-6-1}=\frac{-12}{7}\)

c: Để B nguyên thì -4x⋮2x-1

=>-4x+2-2⋮2x-1

=>-2⋮2x-1

mà 2x-1 lẻ

nên 2x-1∈{1;-1}

=>2x∈{2;0}

=>x∈{1;0}

Kết hợp ĐKXĐ, ta được: x=1

Bài 12:

a: ĐKXĐ: a∉{1;-1;-2}

b: \(P=\left(\frac{a+1}{2a-2}+\frac{1}{2-2a^2}\right)\cdot\frac{2a+2}{a+2}\)

\(=\left(\frac{a+1}{2\left(a-1\right)}-\frac{1}{2\left(a-1\right)\left(a+1\right)}\right)\cdot\frac{2\left(a+1\right)}{a+2}\)

\(=\frac{\left(a+1\right)^2-1}{2\left(a-1\right)\left(a+1\right)}\cdot\frac{2\left(a+1\right)}{a+2}=\frac{a\left(a+2\right)}{\left(a-1\right)\left(a+2\right)}=\frac{a}{a-1}\)

c: |a|=2

=>a=2(nhận) hoặc a=-2(loại)

Khi a=2 thì \(P=\frac{2}{2-1}=\frac21=2\)

Bài 11:

a: ĐKXĐ: x∉{2;-3}

b: \(P=\frac{x+2}{x+3}-\frac{5}{x^2+3x-2x-6}+\frac{1}{2-x}\)

\(=\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{1}{x-2}\)

\(=\frac{\left(x+2\right)\left(x-2\right)-5-\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)

\(=\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}=\frac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x-4}{x-2}\)

c: \(P=\frac{-3}{4}\)

=>\(\frac{x-4}{x-2}=\frac{-3}{4}\)

=>4(x-4)=-3(x-2)

=>4x-16=-3x+6

=>7x=22

=>\(x=\frac{22}{7}\) (nhận)

d: Để P nguyên thì x-4⋮x-2

=>x-2-2⋮x-2

=>-2⋮x-2

=>x-2∈{1;-1;2;-2}

=>x∈{3;1;4;0}

e: \(x^2-9=0\)

=>\(x^2=9\)

=>x=3(nhận) hoặc x=-3(loại)

Khi x=3 thì \(P=\frac{3-4}{3-2}=-1\)