5x² + 8xy + 5y² = 72

<=> 5x² + 10xy + 5y² - 2xy = 72

<=> 5(x² + 2xy + y²) - 2xy = 72

<=> 5(x + y)² - 2xy = 72

<=> -2xy = 72 - 5(x + y)²

A = x² + y² = (x + y)² - 2xy

= (x + y)² + 72 - 5(x + y)² 

= 72 - 4(x + y)²

(x + y)² > 0 => -4(x + y)² < 0

=> A < 72

dấu "=" xảy ra khi : x +  y = 0 <=> x = -y

\(ĐKXĐ:x\ne-1;x\ne\frac{2}{3}\)

\(pt\Leftrightarrow\frac{7x-2\left(x+1\right)+\left(3x-2\right)}{\left(3x-2\right)\left(x+1\right)}=1\)

\(\Leftrightarrow7x-2\left(x+1\right)+\left(3x-2\right)=\left(3x-2\right)\left(x+1\right)\)

\(\Leftrightarrow8x-4=3x^2-2x+3x-2\)

\(\Leftrightarrow3x^2-7x+2=0\)

\(\Delta=7^2-4.3.2=25,\sqrt{\Delta}=5\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{7+5}{6}=2\\x=\frac{7-5}{6}=\frac{1}{3}\end{cases}}\)

Tự cho đkxđ nha!!!

<=> \(\frac{x+1-x}{x+1}=\frac{7x}{\left(3x-2\right)\left(x+1\right)}-\frac{2}{3x-2}\)

<=> \(\frac{3x-2}{\left(3x-2\right)\left(x+1\right)}=\frac{7x}{\left(3x-2\right)\left(x+1\right)}-\frac{2\left(x+1\right)}{\left(3x-2\right)\left(x+1\right)}\)

<=> \(\frac{7x-2x-2-3x+2}{\left(3x-2\right)\left(x+1\right)}=0\)

<=> \(\frac{2x}{\left(3x-2\right)\left(x+1\right)}=0\)

=> 2x = 0

<=> x = 0 (TM)

Vậy ...

\(ĐKXĐ:x\ne\pm3\)

\(pt\Leftrightarrow\frac{\left(x+3\right)^2-\left(x-3\right)^2}{x^2-9}=\frac{17}{x^2-9}\)

\(\Leftrightarrow\left(x+3\right)^2-\left(x-3\right)^2=17\)

Tự dừng bấm Gửi tl

\(\Leftrightarrow x^2+6x+9-x^2+6x-9=17\)

\(\Leftrightarrow12x=17\Leftrightarrow x=\frac{17}{12}\)

\(\frac{x+1}{x-3}-\frac{1}{x-1}=\frac{2}{\left(x-1\right)\left(x-3\right)}\left(x\ne1;x\ne3\right)\)

\(\Leftrightarrow\frac{x^2-1}{\left(x-1\right)\left(x-3\right)}-\frac{x-3}{\left(x-1\right)\left(x-3\right)}-\frac{2}{\left(x-1\right)\left(x-3\right)}=0\)

\(\Leftrightarrow\frac{x^2-1-x+3-2}{\left(x-1\right)\left(x-3\right)}=0\)

\(\Rightarrow x^2-x=0\)

\(\Leftrightarrow x\left(x-1\right)=0\)

<=> x=0 hoặc x=1

Vậy x=0; x=1

\(ĐKXĐ:x\ne3;x\ne1\)

\(pt\Leftrightarrow\frac{x^2-1-x+3}{\left(x-3\right)\left(x-1\right)}=\frac{2}{\left(x-1\right)\left(x-3\right)}\)

\(\Leftrightarrow x^2-1-x+3=2\)

\(\Leftrightarrow x^2-x=0\Leftrightarrow x=0\)(vì x khác 1)

Vậy x = 0

\(\frac{x+5}{x-5}+\frac{x-5}{x+5}=\frac{2\left(x^2+25\right)}{x^2-25}\left(x\ne\pm5\right)\)

\(\Leftrightarrow\frac{x+5}{x-5}+\frac{x-5}{x+5}-\frac{2\left(x^2+25\right)}{\left(x-5\right)\left(x+5\right)}=0\)

\(\Leftrightarrow\frac{\left(x+5\right)^2}{\left(x-5\right)\left(x+5\right)}+\frac{\left(x-5\right)^2}{\left(x-5\right)\left(x+5\right)}-\frac{2x^2+50}{\left(x-5\right)\left(x+5\right)}=0\)

\(\Leftrightarrow\frac{x^2+10x+25}{\left(x-5\right)\left(x+5\right)}+\frac{x^2-10x+25}{\left(x-5\right)\left(x+5\right)}-\frac{2x^2+50}{\left(x-5\right)\left(x+5\right)}=0\)

\(\Leftrightarrow\frac{x^2+10x+25+x^2-10x+25-2x^2-50}{\left(x-5\right)\left(x+5\right)}=0\)

\(\Rightarrow\frac{0}{\left(x-5\right)\left(x+5\right)}=0\)

=> PT đúng với mọi x khác \(\pm5\)

Refund QB nhìn logic :V 

\(\frac{x+5}{x-5}+\frac{x-5}{x+5}=\frac{2\left(x^2+25\right)}{x^2-25}\)

\(\frac{x+5}{x-5}+\frac{x-5}{x+5}=\frac{2\left(x^2+25\right)}{\left(x+5\right)\left(x-5\right)}\)

\(\left(x+5\right)^2-\left(x-5\right)^2=2\left(x^2+25\right)\)

\(20x=2x^2+50\)

\(20x-2x^2-50=0\)

\(2\left(10x-x^2-25\right)=0\)

\(-x^2+10x+25=0\)

\(x^2-10x+25=0\)

\(x^2-2\left(x\right)\left(5\right)+5^2=0\)

\(\left(x-5\right)^2=0\)

\(x-5=0\Leftrightarrow x=5\)

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

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

\(2+\frac{1}{x}=\frac{x^2+1}{x}\)

\(2+\frac{1}{x}=x+\frac{1}{x}\)

\(2=x\Leftrightarrow x=2\)

\(2+\frac{1}{x}=\frac{1}{x}.\left(x^2+1\right)\)

\(\Leftrightarrow\frac{2x+1}{x}=\frac{x^2+1}{x}\)

\(\Rightarrow\orbr{\begin{cases}2x+1=x^2+1=0\\2x+1=x^2+1\end{cases}}\)

Mà \(x^2+1>0\)nên \(x^2-2x=0\Leftrightarrow x=2\)(vì x khác 0)

Vậy x = 2

1/x 1/x và x ngũ 2

2 + 1x = (1x + 2)(x² + 1)

<=> 2 + x = 2x² + 2 + x³ + x

<=> x = 2x² + x³ + x

<=> 0 = 2x² + x³

<=> 2x² + x³ = 0

<=> x²(2 + x) = 0

<=> x = 0 hoặc 2 + x = 0

<=> x = 0 hoặc x = -2