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\(\Leftrightarrow x\left(4x-3\right)-\left(x-2\right)\left(3x+2\right)=x^2-5\)
\(\Leftrightarrow4x^2-3x-3x^2-2x+6x+4=x^2-5\)
\(\Leftrightarrow x^2+x+4=x^2-5\)
=>x+4=-5
hay x=-9(nhận)
\(\dfrac{2x-1}{3}+\dfrac{3x-2}{4}=\dfrac{4x-3}{5}\)
\(\Leftrightarrow\dfrac{20\left(2x-1\right)}{60}+\dfrac{15\left(3x-2\right)}{60}=\dfrac{12\left(4x-3\right)}{60}\)
`<=> 20(2x-1) +15(3x-2) =12(4x-3)`
`<=> 40x - 20 + 45x - 30 = 48x - 36`
`<=> 85x -50 = 48x - 36`
`<=> 85x-48x = -36+50`
`<=> 37x =14`
`<=> x= 14/37`
Vậy phương trình có nghiệm `x=14/37`
__
\(\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-6}{x^2-9}\)
\(\Leftrightarrow\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-6}{\left(x-3\right)\left(x+3\right)}\)
ĐKXĐ : \(\left\{{}\begin{matrix}x-3\ne0\\x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)
Ta có : \(\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-6}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{4\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-6}{\left(x-3\right)\left(x+3\right)}\)
`=> 5x + 15 + 4x -12=x-6`
`<=> 9x + 3=x-6`
`<=> 9x-x=-6-3`
`<=> 8x = -9`
`<=>x=-9/8(tm)`
Vậy phương trình có nghiệm `x=-9/8`
` @ yngoc`
a) Ta có: \(2\left(3x+1\right)-4\left(5-2x\right)>2\left(4x-3\right)-6\)
\(\Leftrightarrow6x+2-20+8x>8x-6-6\)
\(\Leftrightarrow14x-18-8x+12>0\)
\(\Leftrightarrow6x-6>0\)
\(\Leftrightarrow6x>6\)
hay x>1
Vậy: S={x|x>1}
b) Ta có: \(9x^2-3\left(10x-1\right)< \left(3x-5\right)^2-21\)
\(\Leftrightarrow9x^2-30x+3< 9x^2-30x+25-21\)
\(\Leftrightarrow9x^2-30x+3-9x^2+30x-4< 0\)
\(\Leftrightarrow-1< 0\)(luôn đúng)
Vậy: S={x|\(x\in R\)}
\(a,x-5\left(x-2\right)=6x\\ \Leftrightarrow x-5x+10-6x=0\\ \Leftrightarrow-10x+10=0\\ \Leftrightarrow x=1\\ b,2^3+3x^2-32x=48\\ \Leftrightarrow3x^2-32x+8=48\\ \Leftrightarrow3x^2-32x-40=0\)
Nghiệm xấu lắm bn
\(c,\left(3x+1\right)\left(x-3\right)^2=\left(3x+1\right)\left(2x-5\right)^2\\ \Leftrightarrow c,\left(3x+1\right)\left[\left(2x-5\right)^2-\left(x-3\right)^2\right]\\ \Leftrightarrow\left(3x+1\right)\left(2x-5-x+3\right)\left(2x-5+x-3\right)=0\\ \Leftrightarrow\left(3x+1\right)\left(x-2\right)\left(3x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=2\\x=\dfrac{8}{3}\end{matrix}\right.\)
\(d,9x^2-1=\left(3x+1\right)\left(4x+1\right)\\ \Leftrightarrow\left(3x+1\right)\left(4x+1\right)-\left(3x-1\right)\left(3x+1\right)=0\\ \Leftrightarrow\left(3x+1\right)\left(4x+1-3x+1\right)=0\\ \Leftrightarrow\left(3x+1\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-2\end{matrix}\right.\)
\(b,2x^3+3x^2-32x-48=0\\ \Leftrightarrow\left(2x^3-8x^2\right)+\left(11x^2-44x\right)+\left(12x-48\right)=0\\ \Leftrightarrow2x^2\left(x-4\right)+11x\left(x-4\right)+12\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(2x^2+11x+12\right)=0\\ \Leftrightarrow\left(x-4\right)\left[\left(2x^2+8x\right)+\left(3x+12\right)\right]=0\\ \Leftrightarrow\left(x-4\right)\left[2x\left(x+4\right)+3\left(x+4\right)\right]=0\\ \Leftrightarrow\left(x-4\right)\left(2x+3\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{3}{2}\\x=-4\end{matrix}\right.\)
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 \(x=\dfrac{7}{5}\)
a) \(2x-6=0\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=\dfrac{6}{2}=3\)
b) \(x^2-4x=0\)
\(\Leftrightarrow x\left(x-4\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)