a: ĐKXĐ: x∈R

\(\frac{5}{x^2-2x+2}-\frac{8}{x^2-2x+5}=3\)

=>\(\frac{5\left(x^2-2x+5\right)-8\left(x^2-2x+2\right)}{\left(x^2-2x+2\right)\left(x^2-2x+5\right)}=3\)

=>\(3\left(x^2-2x+2\right)\left(x^2-2x+5\right)=5x^2-10x+25-8x^2+16x-16=-3x^2+6x+9\)

=>\(3\left\lbrack\left(x^2-2x\right)^2+7\left(x^2-2x\right)+10\right\rbrack=-3\left(x^2-2x\right)+9\)

=>\(\left(x^2-2x\right)^2+7\left(x^2-2x\right)+10=-\left(x^2-2x\right)+3\)

=>\(\left(x^2-2x\right)^2+8\left(x^2-2x\right)+7=0\)

=>\(\left(x^2-2x+1\right)\left(x^2-2x+7\right)=0\)

=>\(x^2-2x+1=0\)

=>\(\left(x-1\right)^2=0\)

=>x-1=0

=>x=1(nhận)

b: ĐKXĐ: x<>0

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

=>\(\frac{x^2+3}{2x}-2+\frac{12x}{x^2+3}+1=4\)

=>\(\frac{x^2+3}{2x}+\frac{12x}{x^2+3}=4+2-1=6-1=5\)

=>\(\frac{\left(x^2+3\right)^2+24x^2}{2x\left(x^2+3\right)}=5\)

=>\(\left(x^2+3\right)^2+24x^2-10x\left(x^2+3\right)=0\)

=>\(\left(x^2+3\right)^2-4x\left(x^2+3\right)-6x\left(x^2+3\right)+24x^2=0\)

=>\(\left(x^2+3\right)\left(x^2+3-4x\right)-6x\left(x^2+3-4x\right)=0\)

=>\(\left(x^2-6x+3\right)\left(x^2-4x+3\right)=0\)

TH1: \(x^2-6x+3=0\)

=>\(x^2-6x+9-6=0\)

=>\(\left(x-3\right)^2=6\)

=>\(\left[\begin{array}{l}x-3=\sqrt6\\ x-3=-\sqrt6\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\sqrt6+3\left(nhận\right)\\ x=-\sqrt6+3\left(nhận\right)\end{array}\right.\)

TH2: \(x^2-4x+3=0\)

=>\(x^2-x-3x+3=0\)

=>(x-1)(x-3)=0

=>x=1(nhận) hoặc x=3(nhận)

b: \(\Leftrightarrow\dfrac{20}{x}-\dfrac{20}{x+20}=\dfrac{1}{6}\)

=>\(\dfrac{20x+400-20x}{x\left(x+20\right)}=\dfrac{1}{6}\)

=>x*(x+20)=400*6=2400

=>x^2+20x-2400=0

=>(x+60)(x-40)=0

=>x=-60 hoặc x=40

c: \(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}=\dfrac{8}{4x^2-1}\)

=>(2x+1)^2-(2x-1)^2=8

=>4x^2+4x+1-4x^2+4x-1=8

=>8x=8

=>x=1(nhận)

câu b sai đề rồi anh ơi và câu a đâu rồi ạ

a, 3x - 7 = 0

<=> 3x = 7

<=> x = 7/3

b, 8 - 5x = 0

<=> -5x = -8

<=> x = 8/5

c, 3x - 2 = 5x + 8

<=> -2x = 10

<=> x = -5

e) Ta có: \(\left(5x+1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-1\\x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=3\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{1}{5};3\right\}\)

1: \(\Leftrightarrow x^2-6x=x^2-7x+10\)

hay x=10

ĐKXĐ: \(x\ne-1\)

\(\dfrac{6x^2+4x+8}{x+1}=5\sqrt{2x^2+3}\)

\(\Rightarrow6x^2+4x+8=5\left(x+1\right)\sqrt{2x^2+3}\)

\(\Leftrightarrow2\left(2x^2+3\right)-5\left(x+1\right)\sqrt{2x^2+3}+2\left(x+1\right)^2=0\)

Đặt \(\left\{{}\begin{matrix}\sqrt{2x^2+3}=a\\x+1=b\end{matrix}\right.\)

\(\Rightarrow2a^2-5ab+2b^2=0\)

\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{2x^2+3}=2\left(x+1\right)\\2\sqrt{2x^2+3}=x+1\end{matrix}\right.\) (\(x\ge-1\))

\(\Rightarrow\left[{}\begin{matrix}2x^2+3=4\left(x+1\right)^2\\4\left(2x^2+3\right)=\left(x+1\right)^2\end{matrix}\right.\) (\(x\ge-1\))

\(\Leftrightarrow...\)

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

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

b: \(\Leftrightarrow\dfrac{2}{\left(x+7\right)\left(x-3\right)}=\dfrac{3x+21}{\left(x-3\right)\left(x+7\right)}\)

=>3x+21=2

=>x=-19/3

d: \(\Leftrightarrow\left(2x+1\right)^2-\left(2x-1\right)^2=8\)

\(\Leftrightarrow4x^2+4x+1-4x^2+4x-1=8\)

=>8x=8

hay x=1

1: \(\Leftrightarrow x^2-6x=x^2-7x+10\)

hay x=10

sao câu 1 hoài v ạ.Còn câu 2,3 nữa á.

điều kiện xác định \(x\ne0\)

ta có : \(\dfrac{x+1}{x^2+2x+4}-\dfrac{x-2}{x^2-2x+4}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x^2-2x+4\right)-\left(x-2\right)\left(x^2+2x+4\right)}{\left(x^2+2x+4\right)\left(x^2-2x+4\right)}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\)

\(\Leftrightarrow-x^2+2x+12=\dfrac{6}{x}\Leftrightarrow x\left(-x^2+2x+12\right)=6\)

\(\Leftrightarrow-x^3+2x^2+12x=6\Leftrightarrow-x^3+2x^2+12x-6=0\)

tới đây bn bấm máy tính nha

câu b lm tương tự nha