<=> \(\left(\frac{x-ab}{a+b}-c\right)+\left(\frac{x-ac}{a+c}-b\right)+\left(\frac{x-bc}{b+c}-a\right)=0\)

<=>\(\frac{x-ab-ac-bc}{a+b}+\frac{x-ab-ac-bc}{a+c}+\frac{x-ab-ac-bc}{b+c}=0\)

<=>\(\left(x-ab-ac-bc\right)\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)=0\)

Vì \(a\ne-b;b\ne-c;c\ne-a\) nên tổng 3 phân số kia khác 0

=> (x-ab-ac-ca)=0

=>x=ab+ac+ca

\(\frac{x-ab}{a+b}+\frac{x-ac}{a+c}+\frac{x-bc}{b+c}=a+b+c\)

\(\frac{x-ab}{a+b}-c+\frac{x-ac}{a+c}-b+\frac{x-bc}{b+c}-a=0\)

\(\frac{x-ab-ac-bc}{a+b}+\frac{x-ac-ba-bc}{a+c}+\frac{x-bc-ab-ac}{b+c}=0\)

\(\left(x-ab-ac-bc\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)=0\)

\(x-ab-ac-bc=0\)

\(x=ab+ac+bc\)

\(B=\frac{9-x}{\sqrt{x}+3}-\frac{x-6\sqrt{x}+9}{\sqrt{x}-3}-6\)(đk: x ≥ 0 và x ≠ 9)

\(B=\frac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}{\sqrt{x}+3}-\frac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}-3}-6\)

\(B=\left(3-\sqrt{x}\right)-\left(\sqrt{x}-3\right)-6\)

\(B=3-\sqrt{x}-\sqrt{x}+3-6\)

\(B=-2\sqrt{x}\)

\(A=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}+\frac{x}{36-x}\)(đk: x ≥ 0 và x ≠ 36)

\(=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}-\frac{x}{x-36}\)

\(=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}-\frac{x}{x-36}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+6\right)-3\left(\sqrt{x-6}\right)-x}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)

\(=\frac{x+6\sqrt{x}-3\sqrt{x}+18-x}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)

\(=\frac{3\sqrt{x}+18}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)

\(=\frac{3(\sqrt{x}+6)}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)

\(=\frac{3}{\sqrt{x}-6}\)

a) Đặt A=\(\frac{x^2-1}{x^2}\)

Ta có:

\(\Rightarrow A=\frac{x^2}{x^2}-\frac{1}{x^2}\)

\(\Rightarrow A=1-\frac{1}{x^2}\)

\(\Rightarrow x\in Z\) để thỏa mãn A<0

b)\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)

=>(a^2+b^2)*cd=(c^2+d^2)*ab

a^2cd+b^2cd=abc^c+abd^2

a^2cd+b^2cd-c^2ab-d^2ab=0

(a^2cd-abd^2+(b^2cd-abc^2)=0

ad(ac-bd)-bc(ac-bd)=0

(ad-bc)(ac-bd)=0

=>ad-bc=0 hoặc ac-bd=0

ad=bc ac=bd

=>a/b=c/d hoặc a/d=b/c

c)Để PEZ thì 5 chia hết cho \(\sqrt{x}-3\)

=>\(\sqrt{x}-3\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}=>\sqrt{x}\in\left\{4;2;8;-2\right\}\)

=>xE{16;4;64}

1)

\(\Leftrightarrow\frac{x}{8}=\frac{y}{12}=\frac{z}{15}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y-z}{8+12-15}=\frac{10}{5}=2\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x}{8}=2\Rightarrow x=16\\\frac{y}{12}=2\Rightarrow x=24\\\frac{z}{15}=2\Rightarrow z=30\end{matrix}\right.\)

2)

Đặt \(\frac{x}{2}=\frac{y}{5}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=5k\end{matrix}\right.\)

xy=10 <=> 2k.5k=10

<=>10k²=10

<=> k=1

\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=5\end{matrix}\right.\)

3)

\(\frac{a}{b}=\frac{c}{d}\Leftrightarrow ad=bc\)

\(\Leftrightarrow\frac{a}{c}=\frac{b}{d}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}=\frac{a-b}{c-d}\Leftrightarrow\left(a+b\right)\left(c-d\right)=\left(c+d\right)\left(a-b\right)\)

\(\Leftrightarrow\frac{a+b}{a-b}=\frac{c+d}{c-d}\) (đpcm)