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\(\Leftrightarrow x^3-6x^2+12x-8+3\left(4x^2-12x+9\right)=x^3+9x^2+27x+27-5\left(9x^2+6x+1\right)+\left(x-1\right)\left(x-3\right)\)
\(\Leftrightarrow-6x^2+12x-8+12x^2-36x+27=9x^2+27x+27-45x^2-30x-5+\left(x-1\right)\left(x-3\right)\)
\(\Leftrightarrow6x^2-24x+19=-36x^2-3x+22+\left(x-1\right)\left(x-3\right)\)
\(\Leftrightarrow42x^2-21x-3-x^2+4x-3=0\)
\(\Leftrightarrow41x^2-17x-6=0\)
\(\Delta=\left(-17\right)^2-4\cdot41\cdot\left(-6\right)=1273\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{17-\sqrt{1273}}{82}\\x_2=\dfrac{17+\sqrt{1273}}{82}\end{matrix}\right.\)
(2x−1)3−3(x+2)(x−3)=(3+2x)3−3x(x+1)
<=>\(8x^3-12x^2+6x-1-3x^2+3x+18=9+54x+36x^2+8x^3-3x^2-3x\)
<=>\(48x^2+42x-8=0\)
<=> \(x=\frac{-21\pm5\sqrt{33}}{48}\)
\(\frac{\left(2-3x\right)^2}{3}-\frac{\left(1+2x\right)^2}{2}=\frac{3}{4}-2\left(x-1\right)\left(x+2\right)+x\left(1+x\right)\)
\(\frac{2^2-12x-3x^2}{3}-\frac{1^2+4x+2x^2}{2}=\frac{3}{4}-\left(x^2+x-2\right)+3x\)
\(\frac{2.\left(4-12x-3x^2\right)}{6}-\frac{3.\left(1+4x+2x^2\right)}{6}=\frac{11}{4}-x^2+2x\)
\(\frac{8-24x-6x^2}{6}-\frac{3+12x+2x^2}{6}=\frac{11}{4}-x^2+2x\)
\(\frac{8-24x-6x^2-3-12x-2x^2}{6}=\frac{11}{4}-x^2+2x\)
\(\frac{5-36x-8x^2}{6}=\frac{11}{4}-x^2+2x\)
Chỗ đây thì mk chịu
\(\Leftrightarrow\dfrac{1}{3}\left(4x^2-4x+1\right)-\dfrac{1}{2}\left(9x^2+6x+1\right)=\dfrac{1}{3}\left(2x-3x^2-2+3x\right)\)
\(\Leftrightarrow\dfrac{4}{3}x^2-\dfrac{4}{3}x+\dfrac{4}{3}-\dfrac{9}{2}x^2-3x-\dfrac{1}{2}=\dfrac{1}{3}\left(-3x^2+5x-2\right)\)
\(\Leftrightarrow x^2\cdot\dfrac{-19}{6}-\dfrac{13}{3}x+\dfrac{5}{6}+x^2-\dfrac{5}{3}x+\dfrac{2}{3}=0\)
\(\Leftrightarrow x^2\cdot\dfrac{-13}{6}-6x+\dfrac{3}{2}=0\)
\(\text{Δ}=\left(-6\right)^2-4\cdot\left(-\dfrac{13}{6}\right)\cdot\dfrac{3}{2}=49\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{6-7}{2\cdot\dfrac{-13}{6}}=\dfrac{3}{13}\\x_2=\dfrac{6+7}{2\cdot\dfrac{-13}{6}}=-3\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{1}{2}\left(x^2-4x+4\right)-\dfrac{13}{3}\left(x^2+6x+9\right)=\dfrac{1}{4}\left(x^2-3x+2\right)-2\left(9x^2+3x-2\right)\)
\(\Leftrightarrow x^2\cdot\dfrac{1}{2}-2x+2-\dfrac{13}{3}x^2-26x-39=\dfrac{1}{4}x^2-\dfrac{3}{4}x+\dfrac{1}{2}-18x^2-6x+4\)
\(\Leftrightarrow x^2\cdot\dfrac{167}{12}-\dfrac{85}{4}x-\dfrac{83}{2}=0\)
\(\Leftrightarrow167x^2-255x-498=0\)
\(\text{Δ}=\left(-255\right)^2-4\cdot167\cdot\left(-498\right)=397689\)
Vì Δ>0 nên phương trình có 2 nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{255-\sqrt{397689}}{334}\\x_2=\dfrac{255+\sqrt{397689}}{334}\end{matrix}\right.\)
(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
a)
\(A=\left(x+3\right)\left(x^2-3x+9\right)-\left(54+x^3\right)\)
\(=x^3-3x^2+9x+3x^2-9x+27-54-x^3\)
\(=-27\)
or
\(A=x^3+27-54-x^3=-27\)
b)
\(B=\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-8x^3+y^3=2y^3\)
c)
\(C=\left(2x+1\right)^2+\left(1-3x\right)^2+2\left(2x+1\right)\left(3x-1\right)\)
\(=\left(2x+1+3x-1\right)^2=\left(5x\right)^2=25x^2\)
d)
\(D=\left(x-2\right)\left(x^2+2x+4\right)-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)
\(=x^3-8-\left(x-1\right)^3+3\left(x-1\right)\left(x+1\right)\)
\(=6x^2-3x-10\)
