1.

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

\(MC:12\)

Quy đồng :

\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)

\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)

\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)

\(\Leftrightarrow6x+9-3x=-4-9+16\)

\(\Leftrightarrow-7x=3\)

\(\Leftrightarrow x=\frac{-3}{7}\)

2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)

\(MC:20\)

Quy đồng :

\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)

\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)

\(\Leftrightarrow30x+15-20=15x-2\)

\(\Leftrightarrow15x=3\)

\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)

Đây là lớp 8 nha các b giúp mk với

Do mk viết nhầm

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) ĐKXĐ: x khác +2

\(\frac{x-2}{2+x}-\frac{3}{x-2}-\frac{2\left(x-11\right)}{x^2-4}\)

<=> \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}\)

<=> (x - 2)^2 - 3(2 + x) = 2(x - 11)

<=> x^2 - 4x + 4 - 6 - 3x = 2x - 22

<=> x^2 - 7x - 2 = 2x - 22

<=> x^2 - 7x - 2 - 2x + 22 = 0

<=> x^2 - 9x + 20 = 0

<=> (x - 4)(x - 5) = 0

<=> x - 4 = 0 hoặc x - 5 = 0

<=> x = 4 hoặc x = 5

làm nốt đi 

1.

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

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

Vậy:

\(f\left(x\right)\) ko xác định tại \(x=\left\{0;3\right\}\)

\(f\left(x\right)=0\Rightarrow x=\left\{-1;1;2;4\right\}\)

\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< -1\\0< x< 1\\2< x< 3\\x>4\end{matrix}\right.\)

\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}-1< x< 0\\1< x< 2\\3< x< 4\end{matrix}\right.\)

2.

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

Vậy:

\(f\left(x\right)\) ko xác định tại \(x=\left\{-1;0\right\}\)

\(f\left(x\right)>0\Rightarrow-1< x< 0\)

\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< -1\\x>0\end{matrix}\right.\)

3.

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

Vậy:

\(f\left(x\right)\) ko xác định tại \(x=\frac{3}{2}\)

\(f\left(x\right)=0\Rightarrow x=\left\{0;1\right\}\)

\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}0< x< 1\\x>\frac{3}{2}\end{matrix}\right.\)

\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< 0\\1< x< \frac{3}{2}\end{matrix}\right.\)

4.

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

Vậy:

\(f\left(x\right)\) ko xác định tại \(x=\left\{\pm\sqrt{3};-\frac{4}{3};2\right\}\)

\(f\left(x\right)=0\Rightarrow x=\pm1\)

\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}-\sqrt{3}< x< -\frac{4}{3}\\-1< x< 1\\\sqrt{3}< x< 2\end{matrix}\right.\)

\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< -\sqrt{3}\\-\frac{4}{3}< x< -1\\1< x< \sqrt{3}\\x>2\end{matrix}\right.\)

5.

\(f\left(x\right)=x^4-x^3-x^2+x^3-x^2-x-3x^2+3x+3\)

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

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

Vậy:

\(f\left(x\right)=0\Rightarrow\left[{}\begin{matrix}x=\frac{-1\pm\sqrt{13}}{2}\\x=\frac{1\pm\sqrt{5}}{2}\end{matrix}\right.\)

\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< \frac{-1-\sqrt{13}}{2}\\\frac{1-\sqrt{5}}{2}< x< \frac{1+\sqrt{5}}{2}\\x>\frac{-1+\sqrt{13}}{2}\end{matrix}\right.\)

\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}\frac{-1-\sqrt{13}}{2}< x< \frac{1-\sqrt{5}}{2}\\\frac{1+\sqrt{5}}{2}< x< \frac{-1+\sqrt{13}}{2}\end{matrix}\right.\)

6.

\(f\left(x\right)=\frac{x^2+4x+15-\left(x-3\right)\left(x-1\right)+\left(x-2\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{x^2+7x+10}{\left(x-1\right)\left(x+1\right)}=\frac{\left(x+5\right)\left(x+2\right)}{\left(x-1\right)\left(x+1\right)}\)

Vậy:

\(f\left(x\right)\) ko xác định khi \(x=\pm1\)

\(f\left(x\right)=0\Rightarrow x=\left\{-2;-5\right\}\)

\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< -5\\-2< x< -1\\x>1\end{matrix}\right.\)

\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}-5< x< -2\\-1< x< 1\end{matrix}\right.\)

Bài 2:

a) ĐK: $x\geq \pm \frac{1}{2}; x\neq 0$

\(\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}=\frac{(2x+1)^2-(2x-1)^2}{(2x-1)(2x+1)}.\frac{10x-5}{4x}\)

\(\frac{4x^2+4x+1-(4x^2-4x+1)}{(2x-1)(2x+1)}.\frac{5(2x-1)}{4x}=\frac{8x}{(2x-1)(2x+1)}.\frac{5(2x-1)}{4x}\)

\(=\frac{10}{2x+1}\)

b) ĐK : $x\neq 0;-1$

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

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

Bài 3:
a) ĐKXĐ: \(x\neq \pm 1\)

b)

\(A=\left(\frac{x+1}{2x-2}-\frac{3}{1-x^2}-\frac{x+3}{2x+2}\right).\frac{4x^2-4}{5}\)

\(=\left[\frac{(x+1)^2}{2(x-1)(x+1)}+\frac{6}{2(x-1)(x+1)}-\frac{(x+3)(x-1)}{2(x+1)(x-1)}\right].\frac{4(x^2-1)}{5}\)

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

\(=\frac{10}{2(x-1)(x+1)}.\frac{4(x-1)(x+1)}{5}=4\)

Bài 4:

a) ĐKXĐ: \(x\neq \pm 2\)

b)

\(A=\left(\frac{x}{(x-2)(x+2)}-\frac{2(x+2)}{(x-2)(x+2)}+\frac{x-2}{(x+2)(x-2)}\right):\frac{(x-2)(x+2)+10-x^2}{x+2}\)

\(=\frac{x-2(x+2)+(x-2)}{(x-2)(x+2)}:\frac{x^2-4+10-x^2}{x+2}\)

\(=\frac{-6}{(x-2)(x+2)}:\frac{6}{x+2}=\frac{-6}{(x-2)(x+2)}.\frac{x+2}{6}=\frac{-1}{x-2}\)

b)

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

Nếu $x=\frac{1}{2}$ thì $A=\frac{-1}{\frac{1}{2}-2}=\frac{2}{3}$

Nếu $x=\frac{-1}{2}$ thì $A=\frac{-1}{\frac{-1}{2}-2}=\frac{2}{5}$

c)

Để \(A< 0\Leftrightarrow \frac{-1}{x-2}< 0\Leftrightarrow x-2>0\Leftrightarrow x>2\)

Kết hợp với ĐKXĐ ta suy ra $x>2$