(x-1)2+(x+3)2=2(x-2)(x+2)
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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
1.
$(x-2)(x-5)=(x-3)(x-4)$
$\Leftrightarrow x^2-7x+10=x^2-7x+12$
$\Leftrightarrow 10=12$ (vô lý)
Vậy pt vô nghiệm.
2.
$(x-7)(x+7)+x^2-2=2(x^2+5)$
$\Leftrightarrow x^2-49+x^2-2=2x^2+10$
$\Leftrightarrow 2x^2-51=2x^2+10$
$\Leftrightarrow -51=10$ (vô lý)
Vậy pt vô nghiệm.
3.
$(x-1)^2+(x+3)^2=2(x-2)(x+2)$
$\Leftrightarrow (x^2-2x+1)+(x^2+6x+9)=2(x^2-4)$
$\Leftrightarrow 2x^2+4x+10=2x^2-8$
$\Leftrightarrow 4x+10=-8$
$\Leftrightarrow 4x=-18$
$\Leftrightarrow x=-4,5$
4.
$(x+1)^2=(x+3)(x-2)$
$\Leftrightarrow x^2+2x+1=x^2+x-6$
$\Leftrightarrow x=-7$
a. (x - 2)(x + 2) - (x - 3)2 = 9
<=> x2 - 22 - (x - 3)2 = 32
<=> x - 2 - (x - 3) = 3
<=> x - 2 - x + 3 = 3
<=> x - x = 3 - 3 + 2
<=> 0 = 2 (Vô lí)
Vậy nghiệm của PT là S = \(\varnothing\)
b: Ta có: \(\left(x-1\right)\left(x^2+1\right)-\left(x+1\right)\left(x^2-x+1\right)=x\left(2-x\right)\)
\(\Leftrightarrow x^3+x-x^2-1-x^3-1=2x-x^2\)
\(\Leftrightarrow-x^2+x-2-2x+x^2=0\)
\(\Leftrightarrow-x=2\)
hay x=-2
Bạn nên viết lại đề bài cho sáng sủa, rõ ràng để người đọc dễ hiểu hơn.
f: =>4(x^2+4x-5)-x^2-7x-10=3(x^2+x-2)
=>4x^2+16x-20-x^2-7x-10-3x^2-3x+6=0
=>6x-24=0
=>x=4
e: =>8x+16-5x^2-10x+4(x^2-x-2)=4-x^2
=>-5x^2-2x+16+4x^2-4x-8=4-x^2
=>-6x+8=4
=>-6x=-4
=>x=2/3
d: =>2x^2+3x^2-3=5x^2+5x
=>5x=-3
=>x=-3/5
b: =>2x^2-8x+3x-12+x^2-7x+10=3x^2-12x-5x+20
=>-12x-2=-17x+20
=>5x=22
=>x=22/5
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: \(S=\left\{-\dfrac{21}{8}\right\}\)