ĐKXĐ: x≠8

Ta có: \(\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{13x-102}{3x-24}\)

\(\Leftrightarrow\frac{3}{2\left(x-8\right)}+\frac{3x-20}{x-8}+\frac{1}{8}-\frac{13x-102}{3\left(x-8\right)}=0\)

\(\Leftrightarrow\frac{9}{6\left(x-8\right)}+\frac{6\left(3x-20\right)}{6\left(x-8\right)}+\frac{6\left(x-8\right)}{48\left(x-8\right)}-\frac{2\left(13x-102\right)}{6\left(x-8\right)}=0\)

\(\Leftrightarrow9+6\left(3x-20\right)+6\left(x-8\right)-2\left(13x-102\right)=0\)

\(\Leftrightarrow9+18x-120+6x-48-26x+204=0\)

\(\Leftrightarrow45-2x=0\)

\(\Leftrightarrow2x=45\)

hay \(x=\frac{45}{2}\)(tm)

Vậy: \(x=\frac{45}{2}\)

Lời giải:
a) ĐKXĐ: $x\neq \pm 3; x\neq 0$

\(A=\frac{3-x}{x+3}.\frac{(x+3)^2}{(x-3)(x+3)}.\frac{x+3}{3x^2}\)

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

b)

Với $x=-\frac{1}{2}\Rightarrow A=-\frac{-\frac{1}{2}+3}{3(\frac{-1}{2})^2}=\frac{-10}{3}$

c)

Để $A< 0\Leftrightarrow -\frac{x+3}{3x^2}< 0$

$\Rightarrow x+3>0\Rightarrow x>-3$

Vậy $x>-3; x\neq 3; x\neq 0$

Lời giải:

a) ĐK: $x\neq 8$

PT \(\Leftrightarrow \frac{3}{2(x-8)}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{13x-102}{3(x-8)}\)

\(\Leftrightarrow \frac{36}{24(x-8)}+\frac{24(3x-20)}{24(x-8)}+\frac{3(x-8)}{24(x-8)}=\frac{8(13x-102)}{24(x-8)}\)

\(\Rightarrow 36+24(3x-20)+3(x-8)=8(13x-102)\)

\(\Leftrightarrow x=12\) (t/m)

b)

ĐK: $x\neq \pm 2$

PT \(\Leftrightarrow \frac{(x-1)(x-2)}{(x+2)(x-2)}-\frac{x(x+2)}{(x-2)(x+2)}=\frac{5x-2}{(2-x)(x+2)}=\frac{2-5x}{(x-2)(x+2)}\)

\(\Rightarrow (x-1)(x-2)-x(x+2)=2-5x\)

$\Leftrightarrow 0=0$

Vậy PT có nghiệm $x\in\mathbb{R}$ và $x\neq \pm 2$

Bài 1:

d)ĐKXĐ: \(x\ne8\)

Ta có: \(\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{13x-102}{3x-24}\)

\(\Leftrightarrow\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac{1}{8}-\frac{13x-102}{3x-24}=0\)

\(\Leftrightarrow\frac{3}{2\left(x-8\right)}+\frac{3x-20}{x-8}+\frac{1}{8}-\frac{13x-102}{3\left(x-8\right)}=0\)

MTC=24(x-8)

\(\Leftrightarrow\frac{36}{24\left(x-8\right)}+\frac{72x-480}{24\left(x-8\right)}+\frac{3x-24}{24\left(x-8\right)}-\frac{104x-816}{24\left(x-8\right)}=0\)

\(\Leftrightarrow36+72x-480+3x-24-104x+816=0\)

\(\Leftrightarrow348-29x=0\)

\(\Leftrightarrow-29x+348=0\)

\(\Leftrightarrow x=\frac{-348}{-29}=12\)

Vậy: x=12

e) ĐKXĐ: \(x\ne\pm1\)

Ta có: \(\frac{6}{x^2-1}+5=\frac{8x-1}{4x+4}-\frac{12x-1}{4-4x}\)

\(\Leftrightarrow\frac{6}{\left(x-1\right)\left(x+1\right)}+5-\frac{8x-1}{4x+4}+\frac{12x-1}{4-4x}=0\)

\(\Leftrightarrow\frac{6}{\left(x-1\right)\left(x+1\right)}+5-\frac{8x-1}{4\left(x+1\right)}+\frac{12x-1}{4\left(1-x\right)}=0\)

MTC=4(x+1)(x-1)

\(\Leftrightarrow\frac{24}{4\left(x-1\right)\left(x+1\right)}+\frac{20x^2-20}{4\left(x-1\right)\left(x+1\right)}-\frac{8x^2-9x+1}{4\left(x-1\right)\left(x+1\right)}-\frac{12x^2-11x-1}{4\left(x-1\right)\left(x+1\right)}=0\)

\(\Leftrightarrow24+20x^2-20-8x^2+9x-1-12x^2+11x+1=0\)

\(\Leftrightarrow20x+4=0\)

\(\Leftrightarrow20x=-4\)

\(\Leftrightarrow x=-\frac{4}{20}=-0,2\)(loại)

Vậy: x không có giá trị

g) Ta có: \(\frac{\frac{x+1}{x-1}-\frac{x-1}{x+1}}{1+\frac{x+1}{x-1}}=\frac{1}{2}\)

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

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

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

\(\Leftrightarrow\frac{4x\cdot\left(x-1\right)}{\left(x-1\right)\left(x+1\right)\cdot2x}-\frac{1}{2}=0\)

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

MTC=2(x+1)

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

\(\Leftrightarrow2-x+1=0\)

\(\Leftrightarrow1-x=0\)

\(\Leftrightarrow x=1\)(loại vì không thỏa mãn ĐKXĐ)

Vậy: x không có giá trị

a) \(\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{3x-102}{3x-24}\) \(ĐK:x\ne8\)

\(\Leftrightarrow\frac{3}{2\left(x-8\right)}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{3x-102}{3\left(x-8\right)}\)

\(\Leftrightarrow\frac{3.3}{6.\left(x-8\right)}+\frac{6.\left(3x-20\right)}{6\left(x-8\right)}-\frac{2\left(3x-102\right)}{6\left(x-8\right)}=\frac{-1}{8}\)

\(\Leftrightarrow\frac{9+18x-120-6x+204}{6\left(x-8\right)}=\frac{-1}{8}\)

\(\Leftrightarrow\frac{12x+93}{6\left(x-8\right)}=\frac{-1}{8}\)

\(\Leftrightarrow8\left(12x+93\right)=-6\left(x-8\right)\)

\(\Leftrightarrow96x+744=-6x+48\)

\(\Leftrightarrow102x=-696\)

\(\Leftrightarrow x=\frac{-116}{17}\) (nhận)

Vậy .....

b) \(\frac{1}{3-x}+\frac{14}{x^2-9}=\frac{x-4}{3+x}+\frac{7}{3+x}\) \(ĐK:x\ne\pm3\)

\(\Leftrightarrow\frac{1}{3-x}+\frac{14}{\left(x-3\right)\left(3+x\right)}=\frac{x-4}{3+x}+\frac{7}{3+x}\)

\(\Leftrightarrow-\frac{3+x}{\left(x-3\right)\left(3+x\right)}+\frac{14}{\left(x-3\right)\left(3+x\right)}=\frac{\left(x-4\right)\left(x-3\right)}{\left(3+x\right)\left(x-3\right)}+\frac{7\left(x-3\right)}{\left(3+x\right)\left(x-3\right)}\)

\(\Leftrightarrow\frac{-3-x+14}{\left(x-3\right)\left(x+3\right)}=\frac{\left(x-4\right)\left(x-3\right)}{\left(3+x\right)\left(x-3\right)}+\frac{7\left(x-3\right)}{\left(3+x\right)\left(x-3\right)}\)

\(\Leftrightarrow-3-x+14=x^2-3x-4x+12+7x-21\)

\(\Leftrightarrow x=-5\) (nhận)

Vậy ....

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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!

la`thu'hai nga`y 19 nhe

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

b) Bạn kiểm tra lại đề bài

c) \(\frac{3}{1-4x}=\frac{2}{4x+1}-\frac{8}{16x^2-1}\left(x\ne\pm\frac{1}{4}\right)\)

\(\Leftrightarrow\frac{3}{1-4x}-\frac{2}{4x+1}+\frac{8}{16x^2-1}=0\)

\(\Leftrightarrow\frac{-3}{4x+1}-\frac{2}{4x+1}+\frac{8}{\left(4x+1\right)\left(4x-1\right)}=0\)

\(\Leftrightarrow\frac{-3\left(4x-1\right)}{\left(4x-1\right)\left(4x+1\right)}-\frac{2\left(4x-1\right)}{\left(4x-1\right)\left(4x+1\right)}+\frac{8}{\left(4x-1\right)\left(4x+1\right)}=0\)

\(\Leftrightarrow\frac{-12x+3}{\left(4x-1\right)\left(4x+1\right)}-\frac{8x-2}{\left(4x-1\right)\left(4x+1\right)}+\frac{8}{\left(4x-1\right)\left(4x+1\right)}=0\)

\(\Leftrightarrow\frac{-12x+3-8x+2+8}{\left(4x-1\right)\left(4x+1\right)}=0\)

=> -20x+13=0

<=> -20x=-13

<=> \(x=\frac{13}{20}\left(tmđk\right)\)