a) \(\left(\frac{5}{7}x-\frac{1}{4}\right)\left(\frac{-3}{4}x+\frac{1}{2}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\frac{5}{7}x-\frac{1}{4}=0\\\frac{-3}{4}x+\frac{1}{2}=0\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{5}{7}x=\frac{1}{4}\\\frac{-3}{4}x=\frac{-1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{20}\\x=\frac{2}{3}\end{cases}}\)

Vậy \(x=\frac{7}{20}\) hoặc x=\(\frac{2}{3}\)

b) \(\left(\frac{4}{5}+x\right)\left(x-\frac{8}{13}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\frac{4}{5}+x=0\\x-\frac{8}{13}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-4}{5}\\x=\frac{8}{13}\end{cases}}\)

Vậy x=-4/5 hoặc x=8/13

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

\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{2}=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=3\end{cases}}\)

Vậy x=1/4 hoặc x=3

\(x+\frac{7}{2}x+x=\frac{1}{2}\)

\(2x+\frac{7}{2}x=\frac{1}{2}\)

\(\left(2+\frac{7}{2}\right)x=\frac{1}{2}\)

\(\frac{11}{2}x=\frac{1}{2}\)

\(x=\frac{1}{2}:\frac{11}{2}\)

\(x=\frac{1}{11}\)

ĐKXĐ:\(x\ne\pm2;x\ne-3;x\ne0\)

\(P=1+\frac{x-3}{x^2+5x+6}\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)

\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\left[\frac{8x^2}{4x^2\left(x-2\right)}-\frac{3x}{3\left(x^2-4\right)}-\frac{1}{x+2}\right]\)

\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\left(\frac{2}{x-2}-\frac{x}{x^2-4}-\frac{1}{x+2}\right)\)

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

\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\cdot\frac{2x+4-x-x+4}{\left(x-2\right)\left(x+2\right)}\)

\(=1+\frac{8\left(x-3\right)}{\left(x+2\right)^2\left(x+3\right)\left(x-2\right)}\)

Đề sai à ??

Bài 1:

a) đkxđ: \(x\ne0;x\ne\pm1\)

\(D=\left(\frac{1}{1-x}+\frac{1}{1+x}\right)\div\left(\frac{1}{1-x}-\frac{1}{1+x}\right)+\frac{1}{x+1}\)

\(D=\left[\frac{1+x+1-x}{\left(1-x\right)\left(1+x\right)}\right]\div\left[\frac{1+x-1+x}{\left(1-x\right)\left(1+x\right)}\right]+\frac{1}{x+1}\)

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

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

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

b) Ta có: \(x^2-x=0\Leftrightarrow x\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\) đều ko thỏa mãn đkxđ

c) Khi \(D=\frac{3}{2}\)

\(\Leftrightarrow\frac{2x+1}{x+1}=\frac{3}{2}\)

\(\Leftrightarrow4x+2=3x+3\Rightarrow x=1\) không thỏa mãn đkxđ

Bài 2: (Sửa đề tí nếu sai ib t lm lại nhé:)

a) đkxđ: \(x\ne\pm1\)

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

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

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

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

b) Ta có: \(x^2-9=0\Rightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)

+ Nếu: \(x=3\)

=> \(E=\frac{4.3}{\left(3+1\right)^2}=\frac{3}{4}\)

+ Nếu: \(x=-3\)

=> \(E=\frac{4.\left(-3\right)}{\left(-3+1\right)^2}=-3\)

c) Để \(E=-3\)

\(\Leftrightarrow\frac{4x}{\left(x+1\right)^2}=-3\)

\(\Leftrightarrow4x=-3x^2-6x-3\)

\(\Leftrightarrow3x^2+10x+3=0\)

\(\Leftrightarrow\left(x+3\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\3x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=-\frac{1}{3}\end{cases}}\)

d) Để \(E< 0\)

\(\Leftrightarrow\frac{4x}{\left(x+1\right)^2}< 0\) , mà \(\left(x+1\right)^2>0\left(\forall x\right)\)

=> Để E < 0 => \(4x< 0\Rightarrow x< 0\)

Vậy x < 0 thì E < 0

e) Ta có: \(E-x-3=0\)

\(\Leftrightarrow\frac{4x}{\left(x+1\right)^2}=x+3\)

\(\Leftrightarrow4x=\left(x^2+2x+1\right)\left(x+3\right)\)

\(\Leftrightarrow x^3+5x^2+7x+3-4x=0\)

\(\Leftrightarrow x^3+5x^2+3x+3=0\)

Đến đây bấm máy tính thôi, nghiệm k đc đẹp cho lắm:

\(x=-4,4798...\) ; \(x=-0,2600...+0,7759...\) ; \(x=-0,2600...-0,7759...\)

Đúng đề mà

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

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)

=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)

b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)

=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)

c) TT

a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)

\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)

=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)

=> \(\left|50x-140\right|=\left|25x+24\right|\)

=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)

c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)

=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)

Bài 2 : a. |2x - 5| = x + 1

 TH1 : 2x - 5 = x + 1

    => 2x - 5 - x = 1

    => 2x - x - 5 = 1

    => 2x - x = 6

    => x = 6

TH2 : -2x + 5 = x + 1

   => -2x + 5 - x = 1

   => -2x - x + 5 = 1

   => -3x = -4

   => x = 4/3

Ba bài còn lại tương tự

cho mình

a/

=> x= 0 hoặc x+2 = 0

=> x = 0 hoặc x= 2

b/

Vì 1/2 +2/3 -1/4  = 11/12 khác 0

=> x-3 = 0 

=> x= 3

c/

=> x-10 = 0 hoặc x-20 = 0

=> x= 10 hoặc x= 20

d/

=> x-2 = 0 hoặc x^2 -1 = 0

=> x= 2 hoặc x^2 = 1

=>. x= 2 hoặc x= 1 hoặc x= -1

e/

=> 2.|4-x| = 8

=> |4-x| = 4

=> 4-x = 4 hoặc -4

=> x= 0 hoặc 8

a/ \(\Leftrightarrow\left(x+2\right)^2-3\left|x+2\right|=0\)

\(\Leftrightarrow\left|x+2\right|^2-3\left|x+2\right|=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left|x+2\right|=0\\\left|x+2\right|=3\end{matrix}\right.\)

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

b/

\(\Leftrightarrow\left|x+2\right|^2-3\left|x+2\right|-4=0\)

\(\Leftrightarrow\left(\left|x+2\right|+1\right)\left(\left|x+2\right|-4\right)=0\)

\(\Leftrightarrow\left|x+2\right|-4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\)

c/

\(\Leftrightarrow\left|x^2-3\right|^2-6\left|x^2-3\right|+5=0\)

\(\Leftrightarrow\left(\left|x^2-3\right|-1\right)\left(\left|x^2-3\right|-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left|x^2-3\right|=1\\\left|x^2-3\right|=5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-3=1\\x^2-3=-1\\x^2-3=5\\x^2-3=-5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2=4\\x^2=2\\x^2=8\\x^2=-2\left(l\right)\end{matrix}\right.\)

d/ ĐKXĐ: ...

\(\Leftrightarrow\frac{\left|x-2\right|^2}{\left(x-1\right)^2}+\frac{2\left|x-4\right|}{x-1}=3\)

Đặt \(\frac{\left|x-2\right|}{x-1}=a\)

\(a^2+2a-3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left|x-2\right|=x-1\\\left|x-2\right|=-3\left(x-1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left|x-2\right|=x-1\left(x\ge1\right)\\\left|x-2\right|=3-3x\left(x\le1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=x-1\left(vn\right)\\x-2=1-x\\x-2=3-3x\\x-2=3x-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=\frac{4}{5}\\x=\frac{1}{2}\end{matrix}\right.\)

e/ ĐKXĐ: ...

Đặt \(\left|\frac{2x-1}{x+2}\right|=a>0\)

\(a-\frac{2}{a}=1\Leftrightarrow a^2-a-2=0\)

\(\Rightarrow\left[{}\begin{matrix}a=-1\left(l\right)\\a=2\end{matrix}\right.\) \(\Rightarrow\left|\frac{2x-1}{x+2}\right|=2\)

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

f/ ĐKXĐ: ...

Đặt \(\left|x-\frac{1}{x}\right|=a\ge0\Rightarrow a^2=x^2+\frac{1}{x^2}-2\Rightarrow x^2+\frac{1}{x^2}=a^2+2\)

Phương trình trở thành:

\(a^2+2-10=2a\)

\(\Leftrightarrow a^2-2a-8=0\Rightarrow\left[{}\begin{matrix}a=4\\a=-2\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\left|x-\frac{1}{x}\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{1}{x}=4\\x-\frac{1}{x}=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-1=0\\x^2+4x-1=0\end{matrix}\right.\)

Bài 3: \(A=\frac{\left(2a+b+c\right)\left(a+2b+c\right)\left(a+b+2c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

Đặt a+b=x;b+c=y;c+a=z

\(A=\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz}\ge\frac{2\sqrt{xy}.2\sqrt{yz}.2\sqrt{zx}}{xyz}=\frac{8xyz}{xyz}=8\)

Dấu = xảy ra khi \(a=b=c=\frac{1}{3}\)

Bài 4: \(A=\frac{9x}{2-x}+\frac{2}{x}=\frac{9x-18}{2-x}+\frac{18}{2-x}+\frac{2}{x}\ge-9+\frac{\left(\sqrt{18}+\sqrt{2}\right)^2}{2-x+x}=-9+\frac{32}{2}=7\)

Dấu = xảy ra khi\(\frac{\sqrt{18}}{2-x}=\frac{\sqrt{2}}{x}\Rightarrow x=\frac{1}{2}\)

a)x=2

b)x=3

c)x=20