TÌM x
( 2x + 3 )2 - ( x - 1)2 = 0
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Bài 1:
a,27x3+27x2+9x+1a,27x3+27x2+9x+1
=(3x)3+3.(3x)2.1+3.3x.12+13=(3x)3+3.(3x)2.1+3.3x.12+13
=(3x+1)3=(3x+1)3
b,x3+3√2x2y+6xy2+2√2y3b,x3+32x2y+6xy2+22y3
=x3+3.x2.√2y+3.x.(√2y)2+(√2y)3=x3+3.x2.2y+3.x.(2y)2+(2y)3
=(x+√2y)3=(x+2y)3
Bài 2:
a,x3+9x2+27x+27=0a,x3+9x2+27x+27=0
⇔(x+3)3=0⇔(x+3)3=0
⇔x+3=0⇔x=−3⇔x+3=0⇔x=−3
b,(x+1)3−x(x−2)2+x−1=0b,(x+1)3−x(x−2)2+x−1=0
⇔x3+3x2+3x+1−x3−4x2+4x+x−1=0⇔x3+3x2+3x+1−x3−4x2+4x+x−1=0
⇔−x2+8x=0⇔−x2+8x=0
⇔−x(x−8)=0
27x³ - 27x² + 3x + 1
= 27x³ - 9x² - 18x² - 3x + 6x + 1
= (27x³ - 9x²) - (18x² - 6x) - (3x - 1)
= 9x² (3x - 1) - 6x (3x - 1) - (3x - 1)
= (3x - 1) (9x² - 6x - 1)
\(\left(3x-5\right)\left(5-3x\right)+9\left(x+1\right)^2=30\)
\(\Leftrightarrow-\left(3x-5\right)^2+9\left(x+1\right)^2=30\)
\(\Leftrightarrow-\left(9x^2-30x+25\right)+9\left(x^2+2x+1\right)=30\)
\(\Leftrightarrow-9x^2+9x^2+30x+18x-25+9-30=0\)
\(\Leftrightarrow48x=46\Leftrightarrow x=\frac{23}{24}\)
\(3x^2+4\left(x-1\right)\left(x+1\right)-7x\left(x-1\right)=x+12\)
\(\Rightarrow3x^2+4\left(x^2-1^2\right)-7x^2+7x=x+12\)
\(\Rightarrow3x^2+4x^2-4-7x^2+7x-x-12=0\)
\(\Rightarrow\left(3x^2+4x^2-7x^2\right)+\left(7x-x\right)+\left(-12-4\right)=0\)
\(\Rightarrow6x-16=0\)
\(\Rightarrow6x=16\)
\(\Rightarrow x=\frac{8}{3}\)
(2x+3)2−(x−1)2=0(2x+3)2−(x−1)2=0
⇔(2x+3+x−1)(2x+3−x+1)=0⇔(2x+3+x−1)(2x+3−x+1)=0
⇔(3x+2)(x+4)=0⇔(3x+2)(x+4)=0
⇔[3x+2=0x+4=0⇒⎧⎨⎩x=−23x=−