\(\frac{15x}{x^2+3x-4}\)-1=12\(\left(\frac{1}{x+4}+\frac{1}{3x-3}\ri...">
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1) \(\frac{3x-1}{4}+\frac{2x-3}{3}=\frac{x-1}{2}\) Mc : 12 \(\Leftrightarrow\) \(\frac{3.\left(3x-1\right)}{12}+\frac{4.\left(2x-3\right)}{12}=\frac{6.\left(x-1\right)}{12}\) \(\Leftrightarrow\) 9x - 3 + 8x - 12 = 6x - 6 \(\Leftrightarrow\) 9x + 8x - 6x = 3 + 12 - 6 \(\Leftrightarrow\) 11x = 9 \(\Leftrightarrow\) x = 0,8 Vậy S = {0,8} 2) \(\frac{x+1}{2}-\frac{x+3}{12}=3-\frac{5-3x}{3}\) Mc : 12 \(\Leftrightarrow\)...
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1) \(\frac{3x-1}{4}+\frac{2x-3}{3}=\frac{x-1}{2}\) Mc : 12

\(\Leftrightarrow\) \(\frac{3.\left(3x-1\right)}{12}+\frac{4.\left(2x-3\right)}{12}=\frac{6.\left(x-1\right)}{12}\)

\(\Leftrightarrow\) 9x - 3 + 8x - 12 = 6x - 6

\(\Leftrightarrow\) 9x + 8x - 6x = 3 + 12 - 6

\(\Leftrightarrow\) 11x = 9

\(\Leftrightarrow\) x = 0,8

Vậy S = {0,8}

2) \(\frac{x+1}{2}-\frac{x+3}{12}=3-\frac{5-3x}{3}\) Mc : 12

\(\Leftrightarrow\) \(\frac{6.\left(x+1\right)}{12}-\frac{x+3}{12}=\frac{12.3}{12}-\frac{4.\left(5-3x\right)}{12}\)

\(\Leftrightarrow\) 6x + 6 - x + 3 = 36 - 20 - 12x

\(\Leftrightarrow\) 6x - x + 12x = -6 - 3 + 36 - 20

\(\Leftrightarrow\) 17x = 7

\(\Leftrightarrow\) x = \(\frac{7}{17}\)

Vậy S = {\(\frac{7}{17}\)}

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

\(\Leftrightarrow\) \(\frac{15.x}{15}-\frac{5.\left(x+1\right)}{15}=\frac{3.\left(2x-1\right)}{15}\)

\(\Leftrightarrow\) 15x - 5x - 5 = 6x - 3

\(\Leftrightarrow\) 15x - 5x - 6x = 5 - 3

\(\Leftrightarrow\) 4x = 2

\(\Leftrightarrow\) x = \(\frac{2}{4}=\frac{1}{2}\)

Vậy S = {\(\frac{1}{2}\)}

4) \(\frac{2x+7}{3}-\frac{x-2}{4}=-2\) Mc : 12

\(\Leftrightarrow\) \(\frac{4.\left(2x+7\right)}{12}-\frac{3.\left(x-2\right)}{12}=\frac{12.\left(-2\right)}{12}\)

\(\Leftrightarrow\) 8x + 28 -3x + 6 = -24

\(\Leftrightarrow\) 8x - 3x = -28 - 6 -24

\(\Leftrightarrow\) 5x = -58

\(\Leftrightarrow\) x = -11,6

Vậy S = {-11,6}

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

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

\(\Leftrightarrow\) 6x - 9 - 16x + 20 = 10 - 2x

\(\Leftrightarrow\) 6x - 16x + 2x = 9 - 20 + 10

\(\Leftrightarrow\) -8x = -1

\(\Leftrightarrow\) x = \(\frac{1}{8}\)

Vậy S = {\(\frac{1}{8}\)}

6) \(\frac{12x+1}{4}=\frac{9x+1}{3}-\frac{3-5x}{12}\) Mc : 12

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

\(\Leftrightarrow\) 36x + 3 = 36x + 4 - 3 + 5x

\(\Leftrightarrow\) 36x - 36x - 5x = -3 + 4 - 3

\(\Leftrightarrow\) -5x = -2

\(\Leftrightarrow x=\frac{2}{5}\)

7) \(\frac{x+6}{4}\) - \(\frac{x-2}{6}-\frac{x+1}{3}=0\) Mc : 12

\(\Leftrightarrow\) \(\frac{3.\left(x+6\right)}{12}-\frac{2.\left(x-2\right)}{12}-\frac{4.\left(x+1\right)}{12}=0\)

\(\Leftrightarrow\) 3x + 18 - 2x + 4 - 4x - 4 = 0

\(\Leftrightarrow\) 3x - 2x - 4x = -18 - 4 + 4

\(\Leftrightarrow\) -3x = -18

\(\Leftrightarrow\) x = 6

Vậy S = {6}

8) x\(^2\) - x - 6 = 0

\(\Leftrightarrow\) x\(^2\) + 2x - 3x - 6 = 0

\(\Leftrightarrow\) x.(x + 2) - 3.(x + 2) = 0

\(\Leftrightarrow\) (x - 3).(x + 2) = 0

\(\Leftrightarrow\) x - 3 = 0 hoặc x + 2 = 0

\(\Leftrightarrow\) x = 3 hoặc x = -2

Vậy S = {3; -2}

0
16 tháng 3 2020

câu 1

a)\(ĐKXĐ:x^3-8\ne0=>x\ne2\)

b)\(\frac{3x^2+6x+12}{x^3-8}=\frac{3\left(x^2-2x+4\right)}{\left(x-2\right)\left(x^2-2x+4\right)}=\frac{3}{x-2}\left(#\right)\)

Thay \(x=\frac{4001}{2000}\)zô \(\left(#\right)\)ta được

\(\frac{3}{\frac{4001}{2000}-2}=\frac{3}{\frac{4001}{2000}-\frac{4000}{2000}}=\frac{3}{\frac{1}{2000}}=6000\)

16 tháng 3 2020

c) Để phân thức trên có giá trị nguyên thì :

\(3⋮x-2\)

=>\(x-2\inƯ\left(3\right)=\left(\pm1\pm3\right)\)

=>\(x\in\left\{1,3,-1,5\right\}\)

zậy ....

14 tháng 2 2020

a.=\(\frac{7x+2}{3xy^2}.\frac{x^2y}{14x+4}\)

=\(\frac{7x+2}{3y}.\frac{x^2y}{2\left(7x+2\right)}\)

=\(\frac{1}{3y}.\frac{x}{2}\)

=\(\frac{x}{6y}\)

b.=\(\frac{8xy}{3x-1}.\frac{5-15x}{12xy^3}\)

=\(\frac{2}{3x-1}.\frac{-15x+5}{3y^2}\)

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

=\(\frac{-10}{3y^2}\)

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

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

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

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

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

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

e.=\(\frac{2\left(2x+3y\right)}{x-1}.\frac{1-x^3}{4x^2+12xy+9y^2}\)

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

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

14 tháng 2 2020

Phần C thiếu x3 , chỗ (x-1)