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a) \(\left(3x+2\right)^2-\left(3x-2\right)^2=5x+38\)
\(\Leftrightarrow\left[\left(3x+2\right)-\left(3x-2\right)\right]\left[\left(3x+2\right)+\left(3x-2\right)\right]=5x+38\)
\(\Leftrightarrow\left(3x+2-3x+2\right)\left(3x+2+3x-2\right)=5x+38\)
\(\Leftrightarrow4\cdot6x=5x+38\)
\(\Leftrightarrow24x-5x=38\)
\(\Leftrightarrow19x=38\Leftrightarrow x=\dfrac{38}{19}=2\)
Vậy \(S=\left\{2\right\}\)
b) \(\left(x+1\right)\left(x^2-2x+1\right)-2x=2\left(x-1\right)\left(x+1\right)\)
\(\Leftrightarrow x^3-2x^2+x+x^2-2x+1-2x=2\left(x^2-1\right)\)
\(\Leftrightarrow x^3-2x^2+x+x^2-2x+1-2x=2x^2-2\)
\(\Leftrightarrow x^3-2x^2+x+x^2-2x+1-2x-2x^2+2=0\)
\(\Leftrightarrow x^3-3x^2-3x+3=0\)
PT vô nghiệm , không tìm được x
Vậy \(S=\varnothing\)
c) \(3\left(x-2\right)^2+9\left(x-1\right)=3\left(x^2+x-3\right)\)
\(\Leftrightarrow3\left(x^2-2x+4\right)+9\left(x-1\right)=3\left(x^2+x-3\right)\)
\(\Leftrightarrow3x^2-6x+12+9x-9=3x^2+3x-9\)
\(\Leftrightarrow3x^2-6x+12+9x-9-3x^2-3x+9=0\)
\(\Leftrightarrow0x+12=0\)
PT vô nghiệm
Vậy \(S=\varnothing\)
Câu cuối tương tự
a: =>9x^2+12x+4-9x^2+12x-4=5x+38
=>24x=5x+38
=>19x=38
=>x=2
b: =>3x^2-12x+12+9x-9=3x^2+3x-9
=>-3x+3=3x-9
=>-6x=-12
=>x=2
a: =>9x^2+12x+4-9x^2+12x-4=5x+38
=>24x=5x+38
=>19x=38
=>x=2
e: =>x^3+1-2x=x^3-x
=>-2x+1=-x
=>-x=-1
=>x=1
f: =>x^3-6x^2+12x-8+9x^2-1=x^3+3x^2+3x+1
=>12x-9=3x+1
=>9x=10
=>x=10/9
b: \(\Leftrightarrow3x^2-12x+12+9x-9=3x^2+3x-9\)
=>-3x+3=3x-9
=>-6x=-12
=>x=2
bạn viết đề lại nhé
\(\Rightarrow3\left(x-1-4x^2+4x\right)+4\left(3x^2+9x-2x-6\right)=38\)
\(\Rightarrow15x-12x^2-3+12x^2+28x-24=38\)
\(\Rightarrow43x=65\Rightarrow x=\frac{65}{43}\)
\(9x^2-6x+1=9\left(x^2-x-2\right)+38\)
\(\Leftrightarrow-6x+1=-9x+20\Leftrightarrow3x=19\Leftrightarrow x=\dfrac{19}{3}\)
`Answer:`
`(3x-1)^2=9(x-2)(x+1)+38`
`<=>9x^2-6x+1=9(x^2+x-2x-2)+38`
`<=>9x^2-6x+1=9x^2-9x-18+38`
`<=>9x^2-9x^2-6x+9x=-18+38-1`
`<=>3x=19`
`<=>x=\frac{19}{3}`