ĐKXĐ:\(\hept{\begin{cases}a,b\ne0\\x\ne b\\x\ne c\end{cases}}\)

Ta có:\(\frac{2}{a\left(b-x\right)}-\frac{2}{b\left(b-x\right)}=\frac{1}{a\left(c-x\right)}-\frac{1}{b\left(c-x\right)}\)

      \(\Leftrightarrow\frac{2}{b-x}\left(\frac{1}{a}-\frac{1}{b}\right)=\frac{1}{c-x}\left(\frac{1}{a}-\frac{1}{b}\right)\)

\(\Leftrightarrow\left(\frac{1}{a}-\frac{1}{b}\right)\left(\frac{2}{b-x}-\frac{1}{c-x}\right)=0\)

Nếu \(a=b\)thì phương trình đúng với mọi nghiệm x

Nếu \(a\ne b\)thì phương trình có nghiệm

\(\frac{2}{b-x}-\frac{1}{c-x}=0\)

\(\Leftrightarrow\frac{2\left(c-x\right)}{\left(c-x\right)\left(b-x\right)}-\frac{1\left(b-x\right)}{\left(c-x\right)\left(b-x\right)}=0\)

\(\Rightarrow2c-2x-b+x=0\)

\(\Leftrightarrow-x=b-2c\)

\(\Leftrightarrow x=2c-b\left(tmđkxđ\right)\)

Vậy ..............................................................................................

Mày nhìn cái chóa j

a) \(4\sqrt{x}+\frac{2}{\sqrt{x}}< 2x+\frac{1}{2x}+2\)

hay \(2\sqrt{x}+\frac{1}{\sqrt{x}}< x+\frac{1}{4x}+1\)

\(\Leftrightarrow0< x+\frac{1}{4x}+1-2\sqrt{x}-\frac{1}{\sqrt{x}}\)

\(\Leftrightarrow0< \left(\sqrt{x}\right)^2-2\sqrt{x}-2\sqrt{x}\cdot1+1+\frac{1}{\left(2\sqrt{x}\right)^2}-2\cdot\frac{1}{2\sqrt{x}}\)

\(\Leftrightarrow1< \left(\sqrt{x}-1\right)^2+\left(\frac{1}{2\sqrt{x}}-1\right)^2\)

\(\Rightarrow\hept{\begin{cases}x>0\\\sqrt{x}>1\\2\sqrt{x}>1\end{cases}\Rightarrow\hept{\begin{cases}x>1\\x>\frac{1}{4}\end{cases}\Rightarrow}x>1}\)

b) \(\frac{1}{1-x^2}>\frac{3}{\sqrt{1-x^2}}-1\left(1\right)\left(ĐK:-1< x< 1\right)\)

Ta có (1) <=> \(\frac{1}{1-x^2}-1-\frac{3x}{\sqrt{1-x^2}}+2>0\)\(\Leftrightarrow\frac{x^2}{1-x^2}-\frac{3x}{\sqrt{1-x^2}}+2>0\)

Đặt \(t=\frac{x}{\sqrt{1-x^2}}\)ta được

\(t^2-3t+2>0\Leftrightarrow\orbr{\begin{cases}\frac{x}{\sqrt{1-x^2}}< 1\\\frac{x}{\sqrt{1-x^2}}>2\end{cases}\Leftrightarrow\orbr{\begin{cases}\sqrt{1-x^2}>x\left(a\right)\\2\sqrt{1-x^2}< x\left(b\right)\end{cases}}}\)

(a) <=> \(\hept{\begin{cases}x< 0\\1-x^2>0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge0\\1-x^2>x^2\end{cases}}}\)

\(\Leftrightarrow-1< x< 0\)hoặc \(\hept{\begin{cases}x\ge0\\x^2< \frac{1}{2}\end{cases}}\)

\(\Leftrightarrow-1< x< 0\)hoặc \(0\le x\le\frac{\sqrt{2}}{2}\Leftrightarrow-1< x< \frac{\sqrt{2}}{2}\)

(b) \(\Leftrightarrow\hept{\begin{cases}1-x^2>0\\x>0\\4\left(1-x^2\right)< x^2\end{cases}\Leftrightarrow\hept{\begin{cases}0< x< 1\\x^2>\frac{4}{5}\end{cases}\Leftrightarrow}\frac{2}{\sqrt{5}}< x< 1}\)

ok đợi nấu ăn xong r làm cho

a) điều kiện x>0

khi đó

\(\left(a\right)\Leftrightarrow4\left(\sqrt{4}+\frac{1}{2\sqrt{x}}\right)< 2\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)^2\)

\(\Leftrightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}>2\Leftrightarrow2x-4\sqrt{x}+1>0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}< \frac{2-\sqrt{2}}{2}\\\sqrt{x}>\frac{2+\sqrt{2}}{2}\end{cases}}\)

ĐK x >0

\(PT\Leftrightarrow2x+2\sqrt{x^2-\frac{1}{x^4}}=\frac{4}{x^2}.\)

\(\Leftrightarrow2\sqrt{x^2-\frac{1}{x^4}}=\frac{4}{x^2}-2x\)

\(\Leftrightarrow x^2-\frac{1}{x^4}=\frac{4}{x^4}-\frac{4}{x}+x^2\)(chia cả 2 vế cho 2)

\(\Leftrightarrow\frac{5}{x^4}-\frac{4}{x}=0\Leftrightarrow5-4x^3=0\Leftrightarrow4x^3=5\)

\(\Leftrightarrow x^3=\frac{5}{4}\Leftrightarrow x=\sqrt[3]{\frac{5}{4}}\)

Vậy................................

bài 1: 

a) ĐKXĐ: x khác 0; x khác -1

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

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

<=> (x - 1)(x + 1) + 1 - 2x = x

<=> x^2 - 2x = x

<=> x^2 - 2x - x = 0

<=> x^2 - 3x = 0

<=> x(x - 3) = 0

<=> x = 0 hoặc x - 3 = 0

<=> x = 0 hoặc x = 0 + 3

<=> x = 0 (ktm) hoặc x = 3 (tm)

=> x = 3

b) ĐKXĐ: x khác +-3; x khác -7/2

\(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{x^2-9}\)

<=> \(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{\left(x-3\right)\left(x+3\right)}\)

<=> 13(x + 3) + (x - 3)(x + 3) = 6(2x + 7)

<=> 13x + 30 + x^2 = 12x + 42

<=> 13x + 30 + x^2 - 12x - 42 = 0

<=> x - 12 + x^2 = 0

<=> (x - 3)(x + 4) = 0

<=> x - 3 = 0 hoặc x + 4 = 0

<=> x = 0 + 3 hoặc x = 0 - 4

<=> x = 3 (ktm) hoặc x = -4 (tm)

=> x = -4

c) ĐKXĐ: x khác +-1

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

<=> x(x + 1) - 2x = 0

<=> x^2 + x - 2x = 0

<=> x^2 - x = 0

<=> x(x - 1) = 0

<=> x = 0 hoặc x - 1 = 0

<=> x = 0 hoặc x = 0 + 1

<=> x = 0 (tm) hoặc x = 1 (ktm)

=> x = 0

d) \(\frac{x^2+2x}{x^2+1}-2x=0\)

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

<=> x(x + 2) - 2x(x^2 + 1) = 0

<=> x^2 - 2x^3 = 0

<=> x^2(1 - 2x) = 0

<=> x^2 = 0 hoặc 1 - 2x = 0

<=> x = 0 hoặc -2x = 0 - 1

<=> x = 0 hoặc -2x = -1

<=> x = 0 hoặc x = 1/2

bài 2: 

(x - 1)(x^2 + 3x - 2) - (x^3 - 1) = 0

<=> x^3 + 3x^2 - 2x - x^2 - 3x + 2 - x^2 + 1 = 0

<=> 2x^2 - 2x - 3x + 3 = 0

<=> 2x(x - 1) - 3(x - 1) = 0

<=> (2x - 3)(x - 1) = 0

<=> 2x - 3 = 0 hoặc x - 1 = 0

<=> 2x = 0 + 3 hoặc x = 0 + 1

<=> 2x = 3 hoặc x = 1

<=> x = 3/2 hoặc x = 1

bài 3:

(x^3 + x^2) + (x^2 + x) = 0

<=> x^3 + x^2 + x^2 + x = 0

<=> x^3 + 2x^2 + x = 0

<=> x(x^2 + 2x + 1) = 0

<=> x(x + 1)^2 = 0

<=> x = 0 hoặc x + 1 = 0

<=> x = 0 hoặc x = 0 - 1

<=> x = 0 hoặc x = -1