a)\(2+\frac{3}{x-5}=1\)

\(\Rightarrow\frac{3}{x-5}=-1\)

\(\Rightarrow3=-x+5\)

\(\Leftrightarrow x+3=5\)

\(\Rightarrow x=2\)

a, x( x - 1) = x ( x + 2)

<=> x² - x = x² + 2x

<=>  x² - x - x² - 2x = 0

<=> -3x = 0

<=> x = 0

b, tương tự câu a

c,\(\Leftrightarrow\frac{3x-3}{4}=2-\frac{x-2}{8}\)        

\(\Leftrightarrow\frac{\left(3x-3\right)2}{8}=\frac{16}{8}-\frac{x-2}{8}\)

\(\Leftrightarrow\frac{6x-6}{8}=\frac{16}{8}-\frac{x-2}{8}\)

=> 6x - 6 = 16 - x + 2

<=> 6x + x = 16 + 2 + 6

<=> 7x = 24

<=> x=\(\frac{24}{7}\)

Các câu còn lại làm tương tự

Bài 1:

a) Ta có: \(2,3x-2\left(0,7+2x\right)=3,6-1,7x\)

\(\Leftrightarrow2,3x-1,4-4x-3,6+1,7x=0\)

\(\Leftrightarrow-5=0\)(vl)

Vậy: \(x\in\varnothing\)

b) Ta có: \(\frac{4}{3}x-\frac{5}{6}=\frac{1}{2}\)

\(\Leftrightarrow\frac{4}{3}x=\frac{1}{2}+\frac{5}{6}=\frac{8}{6}=\frac{4}{3}\)

hay x=1

Vậy: x=1

c) Ta có: \(\frac{x}{10}-\left(\frac{x}{30}+\frac{2x}{45}\right)=\frac{4}{5}\)

\(\Leftrightarrow\frac{9x}{90}-\frac{3x}{90}-\frac{4x}{90}-\frac{72}{90}=0\)

\(\Leftrightarrow2x-72=0\)

\(\Leftrightarrow2\left(x-36\right)=0\)

mà 2>0

nên x-36=0

hay x=36

Vậy: x=36

d) Ta có: \(\frac{10x+3}{8}=\frac{7-8x}{12}\)

\(\Leftrightarrow12\left(10x+3\right)=8\left(7-8x\right)\)

\(\Leftrightarrow120x+36=56-64x\)

\(\Leftrightarrow120x+36-56+64x=0\)

\(\Leftrightarrow184x-20=0\)

\(\Leftrightarrow184x=20\)

hay \(x=\frac{5}{46}\)

Vậy: \(x=\frac{5}{46}\)

e) Ta có: \(\frac{10x-5}{18}+\frac{x+3}{12}=\frac{7x+3}{6}-\frac{12-x}{9}\)

\(\Leftrightarrow\frac{2\left(10x-5\right)}{36}+\frac{3\left(x+3\right)}{36}-\frac{6\left(7x+3\right)}{36}+\frac{4\left(12-x\right)}{36}=0\)

\(\Leftrightarrow2\left(10x-5\right)+3\left(x+3\right)-6\left(7x+3\right)+4\left(12-x\right)=0\)

\(\Leftrightarrow20x-10+3x+9-42x-18+48-4x=0\)

\(\Leftrightarrow-23x+29=0\)

\(\Leftrightarrow-23x=-29\)

hay \(x=\frac{29}{23}\)

Vậy: \(x=\frac{29}{23}\)

f) Ta có: \(\frac{x+4}{5}-x-5=\frac{x+3}{2}-\frac{x-2}{2}\)

\(\Leftrightarrow\frac{2\left(x+4\right)}{10}-\frac{10x}{10}-\frac{50}{10}=\frac{25}{10}\)

\(\Leftrightarrow2x+8-10x-50-25=0\)

\(\Leftrightarrow-8x-67=0\)

\(\Leftrightarrow-8x=67\)

hay \(x=\frac{-67}{8}\)

Vậy: \(x=\frac{-67}{8}\)

g) Ta có: \(\frac{2-x}{4}=\frac{2\left(x+1\right)}{5}-\frac{3\left(2x-5\right)}{10}\)

\(\Leftrightarrow5\left(2-x\right)-8\left(x+1\right)+6\left(2x-5\right)=0\)

\(\Leftrightarrow10-5x-8x-8+12x-30=0\)

\(\Leftrightarrow-x-28=0\)

\(\Leftrightarrow-x=28\)

hay x=-28

Vậy: x=-28

h) Ta có: \(\frac{x+2}{3}+\frac{3\left(2x-1\right)}{4}-\frac{5x-3}{6}=x+\frac{5}{12}\)

\(\Leftrightarrow\frac{4\left(x+2\right)}{12}+\frac{9\left(2x-1\right)}{12}-\frac{2\left(5x-3\right)}{12}-\frac{12x}{12}-\frac{5}{12}=0\)

\(\Leftrightarrow4x+8+18x-9-10x+6-12x-5=0\)

\(\Leftrightarrow0x=0\)

Vậy: \(x\in R\)

Bài 2:

a) Ta có: \(5\left(x-1\right)\left(2x-1\right)=3\left(x+8\right)\left(x-1\right)\)

\(\Leftrightarrow5\left(x-1\right)\left(2x-1\right)-3\left(x-1\right)\left(x+8\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[5\left(2x-1\right)-3\left(x+8\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(10x-5-3x-24\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(7x-29\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\7x-29=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\7x=29\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{29}{7}\end{matrix}\right.\)

Vậy: Tập nghiệm \(S=\left\{1;\frac{29}{7}\right\}\)

b) Ta có: \(\left(3x-2\right)\left(x+6\right)\left(x^2+5\right)=0\)(1)

Ta có: \(x^2\ge0\forall x\)

\(\Rightarrow x^2+5\ge5\ne0\forall x\)(2)

Từ (1) và (2) suy ra:

\(\left[{}\begin{matrix}3x-2=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=-6\end{matrix}\right.\)

Vậy: Tập nghiệm \(S=\left\{\frac{2}{3};-6\right\}\)

c) Ta có: \(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)

\(\Leftrightarrow27x^3-8-\left(27x^3-1\right)-x+4=0\)

\(\Leftrightarrow27x^3-8-27x^3+1-x+4=0\)

\(\Leftrightarrow-x-3=0\)

\(\Leftrightarrow-x=3\)

hay x=-3

Vậy: Tập nghiệm S={-3}

d) Ta có: \(x\left(x-1\right)-\left(x-3\right)\left(x+4\right)=5x\)

\(\Leftrightarrow x^2-x-\left(x^2+x-12\right)-5x=0\)

\(\Leftrightarrow x^2-x-x^2-x+12-5x=0\)

\(\Leftrightarrow12-7x=0\)

\(\Leftrightarrow7x=12\)

hay \(x=\frac{12}{7}\)

Vậy: Tập nghiệm \(S=\left\{\frac{12}{7}\right\}\)

e) Ta có: (2x+1)(2x-1)=4x(x-7)-3x

\(\Leftrightarrow4x^2-1-4x^2+28x+3x=0\)

\(\Leftrightarrow31x-1=0\)

\(\Leftrightarrow31x=1\)

hay \(x=\frac{1}{31}\)

Vậy: Tập nghiệm \(S=\left\{\frac{1}{31}\right\}\)

a) $0,25x+1,5=0$

$\mathrm{=>\; x\; =\; (0\; -\; 1,5)\; :\; 0,25\; =\; -1,5\; :\; 0,25\; =\; -6}$

Vậy x = -6.

b) $6,36-5,3x=0$

=> x = (0 + 6,36) : 5,3 = 6,36 : 5,3 =\(\dfrac{6}{5}=1,2\)
Vậy x = 1,2.

c) $\frac{4}{3}x-\frac{5}{6}=\frac{1}{2}$

Vậy x = 1.

d) $-\frac{5}{9}x+1=\frac{2}{3}x-10$

$\mathrm{=>\; \backslash (\backslash dfrac\{-5\}\{9\}x-\backslash dfrac\{2\}\{3\}x=\backslash dfrac\{-11\}\{9\}x=-10-1=-11\backslash )}$

$\mathrm{=>\; \backslash (x=-11:\backslash dfrac\{-11\}\{9\}=9\backslash )}$

Vậy x = 9.

a)\(\frac{x}{5}+\frac{2x+1}{3}=\frac{x-5}{15}\)

     \(\frac{3x}{15}+\frac{10x+5}{15}=\frac{x-5}{15}\)

       \(3x+10x+5=x-5\)

       \(13x+5-x+5=0\)

        \(12x=-10\)

        \(x=-\frac{5}{6}\)

a, 8/x-8 + 11/x-11 = 9/x-9  + 10/ x-10

b, x/x-3 - x/x-5 = x/x-4 - x/x-6

c, 4/x^2-3x+2  - 3/2x^2-6x+1   +1 = 0

d, 1/x-1 + 2/ x-2  + 3/x-3  = 6/x-6

e, 2/2x+1 - 3/2x-1 = 4/4x^2-1

f, 2x/x+1 + 18/x^2+2x-3 = 2x-5 /x+3

g, 1/x-1 + 2x^2 -5/x^3 -1  = 4/ x^2 +x+1

\(\frac{x-3}{5}-\frac{2x-1}{10}=\frac{x+1}{2}+\frac{1}{4}\)

\(< =>\frac{\left(x-3\right).4}{20}-\frac{\left(2x-1\right).2}{20}=\frac{\left(x+1\right).10}{20}+\frac{5}{20}\)

\(< =>4x-12-4x+2=10x+10+5\)

\(< =>10x=-10-10-5=-25\)

\(< =>x=-\frac{25}{10}=-\frac{5}{2}\)

\(\frac{x+3}{2}-\frac{2x-1}{3}-1=\frac{x+5}{5}\)

\(< =>\frac{\left(x+3\right).15}{30}-\frac{\left(2x-1\right).10}{30}-\frac{30}{30}=\frac{\left(x+5\right).5}{30}\)\(< =>15x+45-20x+10-30=5x+25\)

\(< =>-5x+25=5x+25< =>10x=0< =>x=0\)