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RH
18 tháng 1 2022
a) (3x + 2)2 - (3x - 2)2 = 5x + 38
<=> 6x.4 = 5x + 38 <=> 19x = 38 <=> x = 2
b) 3(x - 2)2 + 9(x - 1) = 3(x2 + x - 3)
<=> 3x2 - 12x + 12 + 9x - 9 = 3x2 + 3x - 9
<=> -6x = -12 <=> x = 2
c) (x + 3)2 - (x - 3)2 = 6x + 8
<=> 2x.6 = 6x + 8 <=> 6x = 8 <=> x = 4/3
d) (x - 1)3 - x(x + 1)2 = 5x(2 - x) - 11(x + 2)
<=> x3 - 3x2 + 3x - 1 - x3 - 2x2 - x = 10x - 5x2 - 11x - 22
<=> 3x = -21 <=> x = -7
e) (x + 1)(x2 - x + 1) - 2x = x(x - 1)(x + 1)
<=> x3 - 1 - 2x = x3 - x
<=> x = -1
11 tháng 3 2023
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
9 tháng 9 2021
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: \(2x\left(x+1\right)+3\left(x-1\right)\left(x+1\right)-5x\left(x+1\right)+6x^2=0\)
\(\Leftrightarrow2x^2+2x+3x^2-3-5x^2-5x+6x^2=0\)
\(\Leftrightarrow6x^2-3x-3=0\)
\(\Leftrightarrow2x^2-x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)
11 tháng 3 2023
c: =>x^3+9x^2+27x+27-x^2+6x-9-x^2+6x-9=6x+18
=>x^3+7x^2+39x+9-6x-18=0
=>x^3+7x^2+33x-9=0
=>\(x\simeq0.26\)
d: =>x^3-3x^2+3x-1-x^3-2x^2-x=10x-5x^2-11x-22
=>x^3-5x^2+2x-1=-5x^2-x-22
=>x^3+3x+21=0
=>\(x\simeq2.40\)
28 tháng 4 2016
\(\Leftrightarrow x^3-3x^2+3x-1-x.\left(x^2+2x+1\right)=10x-5.x^2-11x-22\)
\(\Leftrightarrow x^3-3x^2+3x-1-x^3-2x^2-x-10x+5x^2+11x+22=0\)
\(\Leftrightarrow3x+21=0\Leftrightarrow x=-7\)
PN
1
26 tháng 9 2021
a: Ta có: \(\left(x-2\right)^3-x\left(x+1\right)\left(x-1\right)+6x^2=5\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+x+6x^2=5\)
\(\Leftrightarrow13x=13\)
hay x=1
5 tháng 2 2021
a) Ta có: \(\left(x+2\right)^2+2\left(x-4\right)=\left(x-4\right)\left(x-2\right)\)
\(\Leftrightarrow x^2+4x+4+2x-8=x^2-6x+8\)
\(\Leftrightarrow x^2+6x-4-x^2+6x-8=0\)
\(\Leftrightarrow12x-12=0\)
\(\Leftrightarrow12x=12\)
hay x=1
Vậy: S={1}
b) Ta có: \(\left(x+1\right)\left(2x-3\right)-3\left(x-2\right)=2\left(x-1\right)\)
\(\Leftrightarrow2x^2-3x+2x-3-3x+6=2x-2\)
\(\Leftrightarrow2x^2-4x+3-2x+2=0\)
\(\Leftrightarrow2x^2-6x+5=0\)
\(\Leftrightarrow2\left(x^2-3x+\dfrac{5}{2}\right)=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{1}{4}=0\)
\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2+\dfrac{1}{4}=0\)(Vô lý)
Vậy: \(S=\varnothing\)
c) Ta có: \(\left(x+3\right)^2-\left(x-3\right)^2=6x+18\)
\(\Leftrightarrow x^2+6x+9-\left(x^2-6x+9\right)-6x-18=0\)
\(\Leftrightarrow x^2-9-x^2+6x-9=0\)
\(\Leftrightarrow6x-18=0\)
\(\Leftrightarrow6x=18\)
hay x=3
Vậy: S={3}
d) Ta có: \(\left(x-1\right)^3-x\left(x+1\right)^2=5x\left(2-x\right)-11\left(x+2\right)\)
\(\Leftrightarrow x^3-3x^2+3x-1-x\left(x^2+2x+1\right)=5x-5x^2-11x-22\)
\(\Leftrightarrow x^3-3x^2+3x-1-x^3-2x^2-x=-5x^2-6x-22\)
\(\Leftrightarrow-5x^2+2x-1+5x^2+6x+22=0\)
\(\Leftrightarrow8x+21=0\)
\(\Leftrightarrow8x=-21\)
hay \(x=-\dfrac{21}{8}\)
Vậy: \(S=\left\{-\dfrac{21}{8}\right\}\)
\(\left(x-1\right)^3-x\left(x+1\right)^2=5x\left(2-x\right)-11\left(x+2\right)\)
\(\Leftrightarrow x^3-3x^2+3x-1-x\left(x^2+2x+1\right)=10x-5x^2-11x-22\)
\(\Leftrightarrow x^3-3x^2+3x-1-x^3-2x^2-x=-5x^2-x-22\)
\(\Leftrightarrow-5x^2+2x-1=-5x^2-x-22\)
\(\Leftrightarrow3x=-21\)
\(\Leftrightarrow x=-7\)