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a) Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
ĐKXĐ: \(x\notin\left\{3;\dfrac{1}{5}\right\}\)
Ta có: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
\(\Leftrightarrow\dfrac{3\left(3-x\right)}{\left(5x-1\right)\left(3-x\right)}+\dfrac{2\left(5x-1\right)}{\left(3-x\right)\left(5x-1\right)}=\dfrac{4}{\left(5x-1\right)\left(3-x\right)}\)
Suy ra: \(9-3x+10x-2=4\)
\(\Leftrightarrow7x+7=4\)
\(\Leftrightarrow7x=-3\)
hay \(x=-\dfrac{3}{7}\)
Vậy: \(S=\left\{-\dfrac{3}{7}\right\}\)
a) Ta có: \(\left(2x-3\right)^2=\left(2x-3\right)\left(x+1\right)\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x-3-x-1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=4\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{3}{2};4\right\}\)
b) Ta có: \(x\left(2x-9\right)=3x\left(x-5\right)\)
\(\Leftrightarrow x\left(2x-9\right)-3x\left(x-5\right)=0\)
\(\Leftrightarrow x\left(2x-9\right)-x\left(3x-15\right)=0\)
\(\Leftrightarrow x\left(2x-9-3x+15\right)=0\)
\(\Leftrightarrow x\left(6-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
Vậy: S={0;6}
c) Ta có: \(3x-15=2x\left(x-5\right)\)
\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{5;\dfrac{3}{2}\right\}\)
d) Ta có: \(\dfrac{5-x}{2}=\dfrac{3x-4}{6}\)
\(\Leftrightarrow6\left(5-x\right)=2\left(3x-4\right)\)
\(\Leftrightarrow30-6x=6x-8\)
\(\Leftrightarrow30-6x-6x+8=0\)
\(\Leftrightarrow-12x+38=0\)
\(\Leftrightarrow-12x=-38\)
\(\Leftrightarrow x=\dfrac{19}{6}\)
Vậy: \(S=\left\{\dfrac{19}{6}\right\}\)
e) Ta có: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)
\(\Leftrightarrow6x+4-3x-1=12x+10\)
\(\Leftrightarrow3x+3-12x-10=0\)
\(\Leftrightarrow-9x-7=0\)
\(\Leftrightarrow-9x=7\)
\(\Leftrightarrow x=-\dfrac{7}{9}\)
Vậy: \(S=\left\{-\dfrac{7}{9}\right\}\)
\(a,=\dfrac{4x+8}{x^2+2x}=\dfrac{4\left(x+2\right)}{x\left(x+2\right)}=\dfrac{4}{x}\\ b,=\dfrac{\left(2x-3\right)-\left(2x-4\right)}{x-2}=\dfrac{2x-3-2x+4}{x-2}=\dfrac{1}{x-2}\\ c,=\dfrac{2x-1-3x-2}{x+3}=\dfrac{-x-3}{x+3}=\dfrac{-\left(x+3\right)}{x+3}=-1\\ d,=\dfrac{11x-18+x}{2x-3}=\dfrac{12x-18}{2x-3}=\dfrac{6\left(2x-3\right)}{2x-3}=6\)
\(e,=\dfrac{3x-6-9x+3}{2x+1}=\dfrac{-6x-3}{2x+1}=\dfrac{-3\left(2x+1\right)}{2x+1}=-3\)
a: \(\Leftrightarrow4\left(5x^2-3\right)+5\left(3x-1\right)< 10x\left(2x+3\right)-100\)
\(\Leftrightarrow20x^2-12x+15x-5< 20x^2+30x-100\)
=>3x-5<=30x-100
=>30x-100>3x-5
=>27x>95
hay x>95/27
b: \(\Leftrightarrow4\left(5x-2\right)-6\left(2x^2-x\right)< 4x\left(1-3x\right)-15x\)
\(\Leftrightarrow20x-8-12x^2+6x< 4x-12x^2-15x\)
=>26x-8<-11x
=>37x<8
hay x<8/37
a, \(6x^2-5x+3=2x-3x\left(3-2x\right)\)
⇔ \(6x^2-5x+3=2x-9x+6x^2\)
⇔ \(6x^2-5x+3-6x^2+9x-2x=0\)
⇔ \(2x+3=0\)
⇔ \(2x=-3\)
⇔ \(x=-\dfrac{3}{2}\)
b, \(\dfrac{2\left(x-4\right)}{4}-\dfrac{3+2x}{10}=x+\dfrac{1-x}{5}\)
⇔ \(\dfrac{20\left(x-4\right)}{4.10}-\dfrac{4\left(3+2x\right)}{4.10}=\dfrac{5x}{5}+\dfrac{1-x}{5}\)
⇔ \(\dfrac{20x-80}{40}-\dfrac{12+8x}{40}=\dfrac{5x+1-x}{5}\)
⇔ \(\dfrac{20x-80-12-8x}{40}=\dfrac{4x+1}{5}\)
⇔ \(\dfrac{12x-92}{40}-\dfrac{4x+1}{5}=0\)
⇔ \(\dfrac{12x-92}{40}-\dfrac{8\left(4x+1\right)}{40}=0\)
⇔ \(12x-92-8\left(4x+1\right)=0\)
⇔ 12x - 92 - 32x - 8 = 0
⇔ -100 - 20x = 0
⇔ 20x = -100
⇔ x = -100 : 20
⇔ x = -5
a) \(9x^2-1=\left(3x-1\right)\left(5x+8\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)\left(5x+8\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1-5x-8\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(-2x-7\right)=0\)
\(TH_1:3x-1=0\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
\(TH_2:-2x-7=0\)
\(\Leftrightarrow x=-\dfrac{7}{2}\)
Vậy pt có tập nghiệm \(S=\left\{\dfrac{1}{3};-\dfrac{7}{2}\right\}\)
b) \(2x^3-5x^2+3x=0\)
\(\Leftrightarrow2x^3-2x^2-3x^2+3x=0\)
\(\Leftrightarrow2x^2\left(x-1\right)-3x\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(2x-3\right)=0\)
\(TH_1:x=0\)
\(TH_2:x-1=0\)
\(\Leftrightarrow x=1\)
\(TH_3:2x-3=0\)
\(\Leftrightarrow x=\dfrac{3}{2}\)
Vậy pt có tập nghiệm \(S=\left\{0;1;\dfrac{3}{2}\right\}\)
c) \(9x^2-16-x\left(3x+4\right)=0\)
\(\Leftrightarrow\left(9x^2-16\right)-x\left(3x+4\right)=0\)
\(\Leftrightarrow\left(3x-4\right)\left(3x+4\right)-x\left(3x+4\right)=0\)
\(\Leftrightarrow\left(3x+4\right)\left(2x-4\right)=0\)
\(TH_1:3x+4=0\)
\(\Leftrightarrow x=-\dfrac{4}{3}\)
\(TH_2:2x-4=0\)
\(\Leftrightarrow x=2\)
Vậy pt có tập nghiệm \(S=\left\{-\dfrac{4}{3};2\right\}\)
d) \(\dfrac{5x+4}{3}-1=\dfrac{3x-2}{4}\)
\(\Leftrightarrow\dfrac{20x+16}{12}-\dfrac{12}{12}=\dfrac{9x-6}{12}\)
\(\Rightarrow20x+16-12=9x-6\)
\(\Leftrightarrow20x-9x=-6-16+12\)
\(\Leftrightarrow11x=-10\)
\(\Leftrightarrow x=-\dfrac{10}{11}\)
Vậy pt có nghiệm duy nhất \(x=-\dfrac{10}{11}\)
a) Ta có: \(9x^2-1=\left(3x-1\right)\left(5x+8\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)=\left(3x-1\right)\left(5x+8\right)\)
\(\Leftrightarrow3x+1=5x+8\)
\(\Leftrightarrow3x-5x=8-1\)
\(\Leftrightarrow-2x=7\)
\(\Leftrightarrow x=\dfrac{-7}{2}\)
Vậy \(X=\dfrac{-7}{2}\)
b) Ta có: \(2x^3-5x^2+3x=0\)
\(\Leftrightarrow x\left(2x^2-5x+3\right)=0\)
\(\Leftrightarrow x\left[\left(2x^2-2x\right)-\left(3x-3\right)\right]=0\)
\(\Leftrightarrow x\left(x-1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\2x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy \(x=1\) hoặc \(x=0\) hoặc \(x=\dfrac{3}{2}\)
c) \(9x^2-16-x\left(3x+4\right)=0\)
\(\Leftrightarrow9x^2-16-3x^2-4x=0\)
\(\Leftrightarrow6x^2-4x-16=0\)
\(\Leftrightarrow2\left(3x^2-2x-8\right)=0\)
\(\Leftrightarrow3x^2-6x+4x-8=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-4}{3}\end{matrix}\right.\)
Vậy \(x=2\) hoặc \(x=\dfrac{-4}{3}\)
d) \(\dfrac{5x+4}{3}-1=\dfrac{3x-2}{4}\)
\(\Leftrightarrow\dfrac{20x+16}{12}-\dfrac{12}{12}=\dfrac{9x-6}{12}\)
\(\Leftrightarrow20x+16-12=9x-6\)
\(\Leftrightarrow20x+16-12-9x+6=0\)
\(\Leftrightarrow11x+10=0\)
\(\Leftrightarrow x=\dfrac{-10}{11}\)
Vậy \(x=\dfrac{-10}{11}\)
a: \(\Leftrightarrow7\left(7-3x\right)+12\left(5x+2\right)=84\left(x+13\right)\)
\(\Leftrightarrow49-21x+60x+24=84x+1092\)
\(\Leftrightarrow39x-84x=1092-73\)
=>-45x=1019
hay x=-1019/45
b: \(\Leftrightarrow21\left(x+3\right)-14=4\left(5x+9\right)-7\left(7x-9\right)\)
=>21x+63-14=20x+36-49x+63
=>21x+49=-29x+99
=>50x=50
hay x=1
c: \(\Leftrightarrow7\left(2x+1\right)-3\left(5x+2\right)=21x+63\)
=>14x+7-15x-6-21x-63=0
=>-22x-64=0
hay x=-32/11
d: \(\Leftrightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-17\cdot105\)
=>70x-105-30x-45=84x+63-1785
=>40x-150-84x+1722=0
=>-44x+1572=0
hay x=393/11