Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\Leftrightarrow x^3-6x^2+12x-8+3\left(4x^2-12x+9\right)=x^3+9x^2+27x+27-5\left(9x^2+6x+1\right)+\left(x-1\right)\left(x-3\right)\)
\(\Leftrightarrow-6x^2+12x-8+12x^2-36x+27=9x^2+27x+27-45x^2-30x-5+\left(x-1\right)\left(x-3\right)\)
\(\Leftrightarrow6x^2-24x+19=-36x^2-3x+22+\left(x-1\right)\left(x-3\right)\)
\(\Leftrightarrow42x^2-21x-3-x^2+4x-3=0\)
\(\Leftrightarrow41x^2-17x-6=0\)
\(\Delta=\left(-17\right)^2-4\cdot41\cdot\left(-6\right)=1273\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{17-\sqrt{1273}}{82}\\x_2=\dfrac{17+\sqrt{1273}}{82}\end{matrix}\right.\)
\(\Leftrightarrow4x^2+4x+1-3\left(x^2-4x+4\right)+2\left(x^2+x-2\right)=4-2+2x\)
\(\Leftrightarrow4x^2+4x+1-3x^2+12x-12+2x^2+2x-4=2x+2\)
\(\Leftrightarrow3x^2+18x-15-2x-2=0\)
\(\Leftrightarrow3x^2+16x-17=0\)
\(\text{Δ}=16^2-4\cdot3\cdot\left(-17\right)=460>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-16-2\sqrt{115}}{6}=\dfrac{-8-\sqrt{115}}{3}\\x_2=\dfrac{-8+\sqrt{115}}{3}\end{matrix}\right.\)
\(\frac{\left(2-3x\right)^2}{3}-\frac{\left(1+2x\right)^2}{2}=\frac{3}{4}-2\left(x-1\right)\left(x+2\right)+x\left(1+x\right)\)
\(\frac{2^2-12x-3x^2}{3}-\frac{1^2+4x+2x^2}{2}=\frac{3}{4}-\left(x^2+x-2\right)+3x\)
\(\frac{2.\left(4-12x-3x^2\right)}{6}-\frac{3.\left(1+4x+2x^2\right)}{6}=\frac{11}{4}-x^2+2x\)
\(\frac{8-24x-6x^2}{6}-\frac{3+12x+2x^2}{6}=\frac{11}{4}-x^2+2x\)
\(\frac{8-24x-6x^2-3-12x-2x^2}{6}=\frac{11}{4}-x^2+2x\)
\(\frac{5-36x-8x^2}{6}=\frac{11}{4}-x^2+2x\)
Chỗ đây thì mk chịu
(2x−1)3−3(x+2)(x−3)=(3+2x)3−3x(x+1)
<=>\(8x^3-12x^2+6x-1-3x^2+3x+18=9+54x+36x^2+8x^3-3x^2-3x\)
<=>\(48x^2+42x-8=0\)
<=> \(x=\frac{-21\pm5\sqrt{33}}{48}\)
(x-2)3+2.(1+2x)2=(1+x)3-3(x-2)2-(x-1)
<=>x3-6x2+12x-8+2.(1+4x+4x2)=1+3x2+3x+x3-3.(x2-4x+4)-x+1
<=>x3-6x2+12x-8+2+8x+8x2=1+3x2+3x+x3-3x2+12x-12-x+1
<=>x3+2x2+20x-6=x3+14x+2
<=>2x2+6x-8=0
<=>2x2-2x+8x-8=0
<=>2x.(x-1)+8.(x-1)=0
<=>2(x-1)(x+4)=0
<=>x-1=0 hoặc x+4=0
<=>x=1 hoặc x=-4