Giải phương trình \(\dfrac{15x}{x^2+3x-4}-1=12.\left(\dfrac{1}{x+4}+\dfrac{1}{3x-3}\right)\)
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KK
13 tháng 3 2017
b)
\(\dfrac{1}{x-1}+\dfrac{1}{x-2}=\dfrac{1}{x+2}+\dfrac{1}{x+1}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x+1}=\dfrac{1}{x+2}-\dfrac{1}{x-2}\)
\(\Leftrightarrow\dfrac{2}{\left(x-1\right)\left(x+1\right)}=\dfrac{-4}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\dfrac{2}{x^2-1}=\dfrac{-4}{x^2-4}\)
\(\Leftrightarrow2x^2-8=-4x^2+4\) ( điều kiện \(x\ne\pm1,x\ne\pm2\) )
\(\Leftrightarrow6x^2=12\)
\(\Rightarrow x=\pm\sqrt{2}\)
KK
13 tháng 3 2017
a )
\(\dfrac{15x}{x^2+3x-4}-1=12\left(\dfrac{1}{x+4}+\dfrac{1}{3x-3}\right)\)
\(\Leftrightarrow\dfrac{15x-\left(x^2+3x-4\right)}{x^2+3x-4}=\dfrac{12}{x+4}+\dfrac{12}{3x-3}\)
\(\Leftrightarrow\dfrac{12x-x^2+4}{x^2+4x-x-4}=\dfrac{48x+12}{\left(x+4\right)\left(3x-3\right)}\)
\(\Leftrightarrow\dfrac{12x-x^2+4}{x\left(x+4\right)-\left(x+4\right)}=\dfrac{48x+12}{3\left(x+4\right)\left(x-1\right)}\)
\(\Leftrightarrow\dfrac{12x-x^2+4}{\left(x+4\right)\left(x-1\right)}=\dfrac{48x+12}{3\left(x+4\right)\left(x-1\right)}\)
\(\Leftrightarrow12x-x^2+4=\dfrac{48x+12}{3}\)
\(\Leftrightarrow12x-x^2+4=16x+4\)
\(\Leftrightarrow x^2+8x=0\)
\(\Delta=b^2-4ac\)
\(\Delta=64\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-8+\sqrt{64}}{2}=0\left(nhận\right)\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-8-\sqrt{64}}{2}=-8\left(loại\right)\end{matrix}\right.\)
Do \(x=-8\) không thỏa mãn phương trình
Vậy \(x=0\)
12 tháng 4 2022
g.\(\dfrac{1-3x}{6}+x-1=\dfrac{x+2}{2}\)
\(\Leftrightarrow\dfrac{\left(1-3x\right)+6\left(x-1\right)}{6}=\dfrac{3\left(x+2\right)}{6}\)
\(\Leftrightarrow\left(1-3x\right)+6\left(x-1\right)=3\left(x+2\right)\)
\(\Leftrightarrow1-3x+6x-6=3x+6\)
\(\Leftrightarrow-5=6\left(vô.lí\right)\)
Vậy pt vô nghiệm
12 tháng 4 2022
h.\(\dfrac{3\left(2x+1\right)}{4}-5-\dfrac{3x+2}{10}=\dfrac{2\left(3x-1\right)}{5}\)
\(\Leftrightarrow\dfrac{15\left(2x+1\right)-100-2\left(3x+2\right)}{20}=\dfrac{8\left(3x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-100-2\left(3x+2\right)=8\left(3x-1\right)\)
\(\Leftrightarrow30x+15-100-6x-4=24x-8\)
\(\Leftrightarrow-89=-8\left(vô.lí\right)\)
Vậy pt vô nghiệm
HP
4 tháng 2 2022
\(i.\dfrac{\left(2x+1\right)^2}{5}-\dfrac{\left(x-1\right)^2}{3}=\dfrac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\dfrac{4x^2+4x+1}{5}-\dfrac{x^2-2x+1}{3}=\dfrac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\dfrac{12x^2+12x+3}{15}-\dfrac{5x^2-10x+5}{15}=\dfrac{7x^2-14x-5}{15}\)
\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5=7x^2-14x-5\)
\(\Leftrightarrow36x=-3\)
\(\Leftrightarrow x=-\dfrac{1}{12}\)
28 tháng 8 2021
a: Ta có: \(3x-\left(3x+2\right)=x+3\)
\(\Leftrightarrow x+3=-2\)
hay x=-5
b: Ta có: \(\dfrac{5x-1}{4}+\dfrac{2x-1}{3}=\dfrac{3x}{2}\)
\(\Leftrightarrow15x-3+8x-4=18x\)
\(\Leftrightarrow5x=7\)
hay \(x=\dfrac{7}{5}\)
S
18 tháng 3 2021
1,\(3x-1=0\Leftrightarrow3x=-1\Leftrightarrow x=-\dfrac{1}{3}\)
2,\(2-x=3x+1\Leftrightarrow2-1=3x+x\rightarrow1=4x\Rightarrow x=-\dfrac{1}{4}\)
S
18 tháng 3 2021
3,\(2\left(x-2\right)-1=5x\Leftrightarrow2x-4-1=5x\Leftrightarrow2x-5x=4+1\Rightarrow3x=5\Rightarrow x=\dfrac{5}{3}\)
4,\(\dfrac{x}{3}-\dfrac{x}{5}=4\Leftrightarrow\dfrac{5x}{15}-\dfrac{3x}{15}=\dfrac{60}{15}\Rightarrow5x-3x=60\Rightarrow2x=60\Rightarrow x=\dfrac{60}{2}=30\)
8 tháng 2 2021
1) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{1-6x}{x-2}+\dfrac{9x+4}{x+2}=\dfrac{x\left(3x-2\right)+1}{x^2-4}\)
\(\Leftrightarrow\dfrac{\left(1-6x\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(9x+4\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{3x^2-2x+1}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)=3x^2-2x+1\)
\(\Leftrightarrow x+2-6x^2-12x+9x^2-18x+4x-8-3x^2+2x-1=0\)
\(\Leftrightarrow-23x-7=0\)
\(\Leftrightarrow-23x=7\)
\(\Leftrightarrow x=-\dfrac{7}{23}\)(nhận)
Vậy: \(S=\left\{-\dfrac{7}{23}\right\}\)
2) ĐKXĐ: \(x\notin\left\{\dfrac{2}{3};-\dfrac{2}{3}\right\}\)
Ta có: \(\dfrac{3x+2}{3x-2}-\dfrac{6}{2-3x}=\dfrac{9x^2}{9x^2-4}\)
\(\Leftrightarrow\dfrac{3x+2}{3x-2}+\dfrac{6}{3x-2}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
\(\Leftrightarrow\dfrac{3x+8}{3x-2}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
\(\Leftrightarrow\dfrac{\left(3x+8\right)\left(3x+2\right)}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
Suy ra: \(9x^2+6x+24x+16=9x^2\)
\(\Leftrightarrow30x+16=0\)
\(\Leftrightarrow30x=-16\)
hay \(x=-\dfrac{8}{15}\)(nhận)
Vậy: \(S=\left\{-\dfrac{8}{15}\right\}\)
\(\dfrac{15x}{x^2+3x-4}-1=12\left(\dfrac{1}{x+4}+\dfrac{1}{3x-3}\right)\)
ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-4\\x\ne1\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{15x-x^2-3x+4}{\left(x-1\right)\left(x+4\right)}=12\left(\dfrac{3x-3+x+4}{3\left(x+4\right)\left(x-1\right)}\right)\)
\(\Leftrightarrow\dfrac{3(12x-x^2+4)}{3\left(x-1\right)\left(x+4\right)}=12\left(\dfrac{4x+1}{3\left(x+4\right)\left(x-1\right)}\right)\)
\(\Leftrightarrow-x^2+12x+4=16x+4\)
\(\Leftrightarrow x^2+4x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(n\right)\\x=-4\left(l\right)\end{matrix}\right.\)
Vậy \(S=\left\{0\right\}\)
qui đồng là xong lười làm hả