- (x+4)2-4x=(x-3)(x+3)-11
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`(x - 2)/3 = (x + 1)/4`
`(x - 2) . 4 = (x + 1) . 3`
`<=> 4x - 8 = 3x + 3`
`<=> 4x - 3x = 3 + 8`
`<=> (4 - 3)x = 11`
`=> x = 11`
`=>` `x = 11`
`2//(5x-8)-3(4x-5)=4(3x-4)`
`<=>5x-8-12x+15=12x-16`
`<=>-19x=-23`
`<=>x=23/19` Vậy `x=23/19`
`3//2(x^3-1)-2x^2(x+2x^4)+(4x^5+4)x=6`
`<=>2x^3-2-2x^3-4x^6+4x^6+4x=6`
`<=>4x=8`
`<=>x=2` Vậy `x=2`
a: Ta có: \(5\left(4x-1\right)+2\left(1-3x\right)-6\left(x+5\right)=10\)
\(\Leftrightarrow20x-5+2-6x-6x-30=10\)
\(\Leftrightarrow8x=43\)
hay \(x=\dfrac{43}{8}\)
b: ta có: \(2x\left(x+1\right)+3\left(x-1\right)\left(x+1\right)-5x\left(x+1\right)+6x^2=0\)
\(\Leftrightarrow2x^2+2x+3x^2-3-5x^2-5x+6x^2=0\)
\(\Leftrightarrow6x^2-3x-3=0\)
\(\Leftrightarrow2x^2-x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)
<=> 4( x2 +2.x.1 + 12) + (2x)2 - 2.2x.1 + 12 - 8(x2 + x - x - 1) = 11
<=> 4x2 + 2x +1 + 4x2 - 4x +1 - 8x2 - 8x + 8x +8 = 11
Bn Chuyển vế rồi tính Denta nhé
1) Ta có: \(5\left(x-3\right)\left(x-7\right)-\left(5x+1\right)\left(x-2\right)=-8\)
\(\Leftrightarrow5\left(x^2-10x+21\right)-\left(5x^2-10x+x-2\right)=-8\)
\(\Leftrightarrow5x^2-50x+105-5x^2+9x+2+8=0\)
\(\Leftrightarrow-41x=-115\)
hay \(x=\dfrac{115}{41}\)
2) Ta có: \(x\left(x+1\right)\left(x+2\right)-\left(x+4\right)\left(3x-5\right)=84-5x\)
\(\Leftrightarrow x\left(x^2+3x+2\right)-\left(3x^2+7x-20\right)=84-5x\)
\(\Leftrightarrow x^3+3x^2+2x-3x^2-7x+20-84+5x=0\)
\(\Leftrightarrow x^3=64\)
hay x=4
3) Ta có: \(\left(9x^2-5\right)\left(x+3\right)-3x^2\left(3x+9\right)=\left(x-5\right)\left(x+4\right)-x\left(x-11\right)\)
\(\Leftrightarrow9x^3+27x^2-5x-15-9x^3-27x^2=x^2-x-20-x^2+11x\)
\(\Leftrightarrow-5x-15=10x-20\)
\(\Leftrightarrow-5x-10x=-20+15\)
\(\Leftrightarrow x=\dfrac{-5}{-15}=\dfrac{1}{3}\)
\(a,=\left(x-2\right)\left(15x-7y\right)\\ b,=x\left(x-11\right)\left(2x-1\right)\\ c,=2x\left(x-3\right)\left(2+3y\right)\\ d,=\left(x-y\right)\left(x-7y\right)\\ e,=\left(x-3\right)\left(4x-12-2x\right)\\ =\left(x-3\right)\left(2x-12\right)=2\left(x-6\right)\left(x-3\right)\)
\(\left(x+4\right)^2-4x=\left(x-3\right)\left(x+3\right)-11\)
\(\Leftrightarrow x^2+8x+16-4x=x^2-9-11\)
\(\Leftrightarrow\left(x^2-x^2\right)+4x+\left(16+9+11\right)=0\)
\(\Leftrightarrow4x=-36\)
\(\Leftrightarrow x=-9\)
Vậy \(x=-9\)
\(\Leftrightarrow x^2+8x+16-4x=x^2-9-11\)
\(\Leftrightarrow x^2-x^2+8x-4x=-9-11-16\)
\(\Leftrightarrow4x=-36\)
\(\Leftrightarrow x=-9\)