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\(6x^2-\left(2x+5\right)\left(3x-2\right)\)
= \(6x^2-6x^2+4x-15x+10\)
= \(-11x+10\)
(3x+2)2-(3x-2)2=5x+38
⇒(3x+2-3x+2)(3x+2+3x-2)=5x+38
⇒4.6x=5x+38
⇒19x=38
⇒x=2
Vậy...
\(\left(3x+2\right)^2-\left(3x-2\right)^2=5x+38\)
\(\Leftrightarrow\left(9x^2+12x+4\right)-\left(9x^2-12x+4\right)=5x+38\)
\(\Leftrightarrow24x=5x+38\Leftrightarrow19x=38\Leftrightarrow x=\frac{38}{19}=2\)
Vậy $x=2$
\(x\left(x-1\right)+3< \left(x-1\right)\left(x+2\right)-3x\)
\(\Leftrightarrow x^2-x+3< x^2+x-2-3x\)
\(\Leftrightarrow x^2-x-x^2-x+3x< -3-2\)
\(\Leftrightarrow x< -5\)
KL: .........................................
Lời giải:
PT $\Leftrightarrow (3x)^2-1-(x^2-1)-(8x^2+2x-3)=20$
$\Leftrightarrow 9x^2-1-x^2+1-8x^2-2x+3=20$
$\Leftrightarrow -2x=17$
$\Leftrightarrow x=\frac{-17}{2}$
a) \(3x-\left(x+5\right)=2\left(x+1\right)\)
\(\Leftrightarrow2x-5=2x+2\) ( Pt vô nghiệm )
b) \(\left(2x-6\right)\left(x+5\right)-4x=2x^2-\left(x+1\right)\)
\(\Leftrightarrow2x^2-30=2x^2-x-1\)
\(\Leftrightarrow x=29\)
c) \(x+3\left(x+5\right)-6x=7\left(x+2\right)\)
\(\Leftrightarrow4x+15-6x=7x+14\)
\(\Leftrightarrow9x=1\)
\(\Leftrightarrow x=\frac{1}{9}\)
\(6x^2-\left(2x+5\right)\left(3x-2\right)=7\)
\(\Leftrightarrow6x^2-6x^2+4x-15x+10=7\)
\(\Leftrightarrow-11x+10=7\)
\(\Leftrightarrow x=\frac{3}{11}\)
Vậy .........