a) 7x - 35 = 0

<=> 7x = 0 + 35

<=> 7x = 35

<=> x = 5

b) 4x - x - 18 = 0

<=> 3x - 18 = 0

<=> 3x = 0 + 18

<=> 3x = 18

<=> x = 5

c) x - 6 = 8 - x

<=> x - 6 + x = 8

<=> 2x - 6 = 8

<=> 2x = 8 + 6

<=> 2x = 14

<=> x = 7

d) 48 - 5x = 39 - 2x

<=> 48 - 5x + 2x = 39

<=> 48 - 3x = 39

<=> -3x = 39 - 48

<=> -3x = -9

<=> x = 3

có bị viết nhầm thì thông cảm nha!

\(ĐKXĐ:x\ne-1;x\ne-\frac{1}{2}\)

\(PT:\Leftrightarrow\frac{x^2-4x+1}{x+1}+1+\frac{x^2-5x+1}{2x+1}=0\)

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

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

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

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(3x+2\right)=0\)

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

\(x-2=0\Leftrightarrow x=2\)

\(3x+2=0\Leftrightarrow3x=-2\Leftrightarrow x=-\frac{2}{3}\)

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

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

\(\Leftrightarrow\left(x^2-4x+1\right)\left(x+1\right)+2\left(x+1\right)\left(2x+1\right)=-\left(x^2-5x+1\right)\left(x+1\right)\)

\(\Leftrightarrow2x^3-3x^2+4x+3=-x^3+4x^2+4x-1\)

\(\Leftrightarrow2x^3-3x^2+3+x^2-4x+1=0\)

\(\Leftrightarrow3x^2-7x^2+4=0\)

\(\Leftrightarrow\left(3x^2-4x-4\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(3x^2+2x-6x-4\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[x\left(3x+2\right)-2\left(3x+2\right)\right]\left(x-1\right)=0\)

\(\Leftrightarrow\left(3x+2\right)\left(x-2\right)\left(x-1\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}3x+2=0\\x-2=0\\x-1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{2}{3}\\x=2\\x=1\end{cases}}\)

vậy:...

$\dfrac{4}{-25x^2+20x-3}=\dfrac{3}{5x-1}-\dfrac{2}{5x-3}$

Điều kiện: $5x-1\ne0,\ 5x-3\ne0$

$\Leftrightarrow x\ne\dfrac{1}{5},\ x\ne\dfrac{3}{5}$

Ta có:

$-25x^2+20x-3=-(5x-1)(5x-3)$

$\Rightarrow \dfrac{-4}{(5x-1)(5x-3)}=\dfrac{3(5x-3)-2(5x-1)}{(5x-1)(5x-3)}$

$\Leftrightarrow \dfrac{-4}{(5x-1)(5x-3)}=\dfrac{5x-7}{(5x-1)(5x-3)}$

Vì $x\ne\dfrac{1}{5},\dfrac{3}{5}$ nên:

$-4=5x-7$

$\Leftrightarrow 5x=3$

$\Leftrightarrow x=\dfrac{3}{5}$

Nhưng $x=\dfrac{3}{5}$ không thỏa mãn điều kiện.

Vậy: phương trình vô nghiệm.

$\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}-\dfrac{2}{x^2-4x+3}=0$

Điều kiện:

$x^2-3x+2=(x-1)(x-2)\ne0$

$x^2-5x+6=(x-2)(x-3)\ne0$

$x^2-4x+3=(x-1)(x-3)\ne0$

$\Rightarrow x\ne1,\ x\ne2,\ x\ne3$

Ta có:

$\dfrac{1}{(x-1)(x-2)}+\dfrac{1}{(x-2)(x-3)}-\dfrac{2}{(x-1)(x-3)}=0$

Quy đồng:

$\dfrac{x-3+x-1-2(x-2)}{(x-1)(x-2)(x-3)}=0$

$\Leftrightarrow \dfrac{x-3+x-1-2x+4}{(x-1)(x-2)(x-3)}=0$

$\Leftrightarrow \dfrac{0}{(x-1)(x-2)(x-3)}=0$

Phương trình đúng với mọi $x$ thỏa mãn điều kiện.

Vậy: $x\in\mathbb{R}\setminus{1;2;3}$.

$\dfrac{x-1}{2x^2-4x}-\dfrac{7}{8x}=\dfrac{5-x}{4x^2-8x}-\dfrac{1}{8x-16}$

Điều kiện: $x\ne0,\ x\ne2$

Ta có:

$\dfrac{x-1}{2x(x-2)}-\dfrac{7}{8x}=\dfrac{5-x}{4x(x-2)}-\dfrac{1}{8(x-2)}$

Quy đồng mẫu $8x(x-2)$:

$\dfrac{4(x-1)-7(x-2)}{8x(x-2)}=\dfrac{2(5-x)-x}{8x(x-2)}$

$\Leftrightarrow \dfrac{4x-4-7x+14}{8x(x-2)}=\dfrac{10-2x-x}{8x(x-2)}$

$\Leftrightarrow \dfrac{-3x+10}{8x(x-2)}=\dfrac{-3x+10}{8x(x-2)}$

Phương trình đúng với mọi $x$ thỏa mãn điều kiện.

Vậy: $x\in\mathbb{R}\setminus{0;2}$.

$\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}$

Điều kiện:

$x^2+9x+20=(x+4)(x+5)\ne0$

$x^2+11x+30=(x+5)(x+6)\ne0$

$x^2+13x+42=(x+6)(x+7)\ne0$

$\Rightarrow x\ne-4,-5,-6,-7$

Ta có:

$\dfrac{1}{(x+4)(x+5)}+\dfrac{1}{(x+5)(x+6)}+\dfrac{1}{(x+6)(x+7)}=\dfrac{1}{18}$

Quy đồng:

$\dfrac{(x+6)(x+7)+(x+4)(x+7)+(x+4)(x+5)}{(x+4)(x+5)(x+6)(x+7)}=\dfrac{1}{18}$

$\Leftrightarrow 18[(x+6)(x+7)+(x+4)(x+7)+(x+4)(x+5)]=(x+4)(x+5)(x+6)(x+7)$

Khai triển và rút gọn:

$\Leftrightarrow x^2+11x-26=0$

$\Leftrightarrow (x-2)(x+13)=0$

$\Leftrightarrow x=2$ hoặc $x=-13$

Cả hai giá trị đều thỏa mãn điều kiện.

Vậy: $x\in{-13;2}$.

ĐKXĐ: \(x\ne\left\{-1;-\frac{1}{2}\right\}\)

\(\Leftrightarrow\left(\frac{x^2-4x+1}{x+1}+1\right)+\left(\frac{x^2-5x+1}{2x+1}+1\right)=0\)

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

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

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

Tập nghiệm: \(S=\left\{1;2;-\frac{2}{3}\right\}\)

ko hiểu

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

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

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

\(\Rightarrow2x^3-7x^2-2x+1+4x^2+2x+4x+2=x^3-4x^2-4x+1\)

\(\Leftrightarrow2x^3-3x^2+4x+3-x^3+4x^2+4x-1=0\)

\(\Leftrightarrow x^3+x^2+8x-2=0\)

Bài làm

a) \(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x-4}\)

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

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

\(\Rightarrow\left(3x+2\right)^2-\left(18x-12\right)=9x^2\)

\(\Leftrightarrow9x^2+12x+4-18x+12x-9x^2=0\)

\(\Leftrightarrow6x+4=0\)

\(\Leftrightarrow x=-\frac{4}{6}\)

\(\Leftrightarrow x=-\frac{2}{3}\)

Vậy x = -2/3 là nghiệm.

@Tao Ngu :))@ 9x-4 không tách thành (3x+4)(3x-4) được đâu bạn. Chỗ đó phải là: 9x²-4

Bài thiếu đkxđ của x \(\hept{\begin{cases}3x-2\ne0\\2+3x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}3x\ne2\\3x\ne-2\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne\frac{2}{3}\\x\ne\frac{-2}{3}\end{cases}\Leftrightarrow}x\ne\pm\frac{2}{3}}\)

$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$###############################@@@@@@@@@@@@@@@@@@@@@@@$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$###############################@@@@@@@@@@@@@@@@@@@@@@@

\(x^2-4x+\frac{1}{x+1}+2=-x^2-5x+\frac{1}{2x+1}\left(ĐK:x\ne-1;-\frac{1}{2}\right)\)

\(< =>x^2-4x+\frac{1}{x+1}+2+x^2+5x-\frac{1}{2x+1}=0\)

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

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

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

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

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

\(< =>\left(2x^2+x\right)^2+\left(2x+1\right)^2x=x-4x^2+2\)

\(< =>4x^4+8x^3+9x^2-2=0\)

nhờ bạn nào đó giải giúp ạ