3+3+3+6x2=?
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a)
=(x-2)3
b)\(\left(2-x\right)^3\)
c)\(\left(x+\dfrac{1}{3}\right)^3\)
d)\(\left(\dfrac{x}{2}+y\right)^3\)
e)
\(=\left(x-1\right)^2\left(x-1-15\right)+25\left[3\left(x-1\right)-5\right]\)
\(=\left(x-1\right)^2\left(x-16\right)+25\left(3x-3-5\right)\)
\(=\left(x-1\right)^2\left(x-16\right)+25\left(3x-8\right)\)
\(8.\left(x-3\right)^3+x^3=6x^2-12x+8\)
\(\Leftrightarrow\left(2x-6\right)^3=-x^3+6x^2-12x+8\)
\(\Leftrightarrow\left(2x-6\right)^3=\left(2-x\right)^3\)
\(\Leftrightarrow2x-6=2-x\)
\(\Leftrightarrow3x=8\)
\(\Leftrightarrow x=\dfrac{8}{3}\)
Vậy pt có nghiệm x = \(\dfrac{8}{3}\)
\(8\left(x-3\right)^3+x^3=6x^2-12x+8\)
\(< =>8\left(x^3-9x^2+27x-27\right)+x^3-6x^2+12x-8=0\)
\(< =>8x^3-72x^2+216x-216+x^3-6x^2+12x-8=0\)
\(< =>9x^3-78x^2+228x-224=0\)
\(< =>\left(3x-8\right)\left(3x^2-18x+28\right)=0\)
đến đây dễ rồi bạn tự làm
\(=\left(6x^2-9x+10x-15\right):\left(2x-3\right)\\ =\left[3x\left(2x-3\right)+5\left(2x-3\right)\right]:\left(2x-3\right)\\ =3x+5\)
\(a,6x^2-5x+3=2x-3x\left(3-2x\right)\)
\(\Leftrightarrow6x^2-5x+3=2x-9x+6x^2\)
\(\Leftrightarrow6x^2-6x^2-5x-2x+9x=-3\)
\(\Leftrightarrow2x=-3\)
\(\Leftrightarrow x=-\dfrac{3}{2}\)
\(b,\left(3x-1\right)\left(4x+3\right)=2\left(3x-1\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(4x+3\right)-2\left(3x-1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(4x+3-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\4x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{1}{4}\end{matrix}\right.\)
\(6x^2-5x+3=2x-3x\left(3-2x\right)\)
\(\Leftrightarrow6x^2-5x+3=2x-9x+6x^2\)
\(\Leftrightarrow6x^2-5x+3-2x+9x-6x^2=0\)
\(\Leftrightarrow2x+3=0\)
\(\Leftrightarrow2x=-3\)
\(\Leftrightarrow x=\dfrac{-3}{2}\)
\(\text{Vậy phương trình có tập nghiệm là }S=\left\{\dfrac{-3}{2}\right\}\)
\(\left(3x-1\right)\left(4x+3\right)=2\left(3x-1\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(4x+3\right)-2\left(3x-1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(4x+3-2\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\4x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=1\\4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=\dfrac{1}{4}\end{matrix}\right.\)
\(\text{Vậy phương trình có tập nghiệm là }S=\left\{\dfrac{1}{3};\dfrac{1}{4}\right\}\)
\(6x^2-9x-4\left(2x-3\right)=0\)
\(\Leftrightarrow3x\left(2x-3\right)-4\left(2x-3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(3x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\3x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=3\\3x=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{4}{3}\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{3}{2};\dfrac{4}{3}\right\}\)
\(6x^2-\left(2x-3\right)\left(3x+2\right)-1=0\)
\(\Rightarrow6x^2-6x^2+5x+6-1=0\)
\(\Rightarrow5x=-5\Rightarrow x=-1\)
\(\Rightarrow6x^2-\left(6x^2-5x-6\right)-1=0\\ \Rightarrow5x+5=0\\ \Rightarrow x=-1\)
a: (1-2x)^3-(1+2x)^3
\(=1^3-3\cdot1^2\cdot2x+3\cdot1\cdot\left(2x\right)^2-8x^3-8x^3-12x^2-6x-1\)
\(=1-6x+12x^2-8x^3-8x^3-12x^2-6x-1\)
\(=-16x^3-12x\)
b: \(=x^3-6x^2+12x-8-x^3-x^2+8\)
\(=-7x^2+12x\)
c: \(=x^3+8-12x+6x^2-x^3+6x^2+12x\)
\(=12x^2+8\)
3+3+3+12
=21
bằng 30 nhé
HT~