Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
, bạn nào biết thì giúp mình sửa nhé
a, Theo bài ra ta có:
\(=x^3-x-2x+2\)
\(=x\left(x^2-1\right)-2\left(x-1\right)\)
\(=x\left(x+1\right)\left(x-1\right)-2\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x-2\right)\)
b, theo bài ra ta có:
\(=x^3-3x^2-\left(2x^2-6x\right)-\left(3x-9\right)\)
\(=x^2\left(x-3\right)-2x\left(x-3\right)-3\left(x-3\right)\)
\(=\left(x^2-2x-3\right)\left(x-3\right)\)
c,Theo bài ra ta có:
\(=x^3+5x^2+3x^2+15x+2x+10\)
\(=x^2\left(x+5\right)+3x\left(x+5\right)+2\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2+3x+2\right)\)
\(=\left(x+5\right)\left(x^2+x+2x+2\right)=\left(x+5\right)\left(x\left(x+1\right)+2\left(x+1\right)\right)\)
\(=\left(x+5\right)\left(x+1\right)\left(x+2\right)\)
CHÚC BẠN HỌC TỐT...........
a) \(x^3-3x+2\)
= \(x^3-x^2+x^2-x-2x+2\)
= \(x^2\left(x-1\right)+x\left(x-1\right)-2\left(x-1\right)\)
= \(\left(x-1\right)\left(x^2+x-2\right)\)
= \(\left(x-1\right)\left(x^2+2x-x-2\right)\)
= \(\left(x-1\right)\left[x\left(x+2\right)-\left(x+2\right)\right]\)
= \(\left(x-1\right)\left(x+2\right)\left(x-1\right)\)
= \(\left(x-1\right)^2\left(x+2\right)\)
b) \(x^3-5x^2+3x+9\)
= \(x^3+x^2-6x^2-6x+9x+9\)
= \(x^2\left(x+1\right)-6x\left(x+1\right)+9\left(x+1\right)\)
= \(\left(x+1\right)\left(x^2-6x+9\right)\)
= \(\left(x+1\right)\left(x-3\right)^2\)
c) \(x^3+8x^2+17x+10\)
= \(x^3+x^2+7x^2+7x+10x+10\)
= \(x^2\left(x+1\right)+7x\left(x+1\right)+10\left(x+1\right)\)
= \(\left(x+1\right)\left(x^2+7x+10\right)\)
= \(\left(x+1\right)\left(x^2+2x+5x+10\right)\)
= \(\left(x+1\right)\left[x\left(x+2\right)+5\left(x+2\right)\right]\)
= \(\left(x+1\right)\left(x+2\right)\left(x+5\right)\)
d) \(x^3-3x^2+6x+4\)
Câu này đúng là sai đề rồi, mình sửa + làm bên dưới:
\(x^3+3x^2+6x+4\)
= \(x^3+x^2+2x^2+2x+4x+4\)
= \(x^2\left(x+1\right)+2x\left(x+1\right)+4\left(x+1\right)\)
= \(\left(x+1\right)\left(x^2+2x+4\right)\)
Học tốt nhé :))
a) \(7x^2-28=0\Leftrightarrow7\left(x^2-4\right)=0\Leftrightarrow x^2-4=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\) vậy \(x=2;x=-2\)
b) \(\left(2x+1\right)+x\left(2x+1\right)=0\Leftrightarrow\left(x+1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\2x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\2x=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=\dfrac{-1}{2}\end{matrix}\right.\) vậy \(x=-1;x=\dfrac{-1}{2}\)
c) \(2x^3-50x=0\Leftrightarrow2x\left(x^2-25\right)=0\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=0\\x-5=0\\x+5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\) vậy \(x=0;x=5;x=-5\)
d) \(9\left(3x-2\right)=x\left(2-3x\right)\Leftrightarrow9\left(3x-2\right)=-x\left(3x-2\right)\)
\(\Leftrightarrow9\left(3x-2\right)+x\left(3x-2\right)=0\Leftrightarrow\left(9+x\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}9+x=0\\3x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-9\\3x=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-9\\x=\dfrac{2}{3}\end{matrix}\right.\) vậy \(x=-9;x=\dfrac{2}{3}\)
e) \(5x\left(x-3\right)-2x+6=0\Leftrightarrow5x\left(x-3\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(5x-2\right)\left(x-3\right)=0\) \(\Leftrightarrow\left\{{}\begin{matrix}5x-2=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}5x=2\\x=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\x=3\end{matrix}\right.\) vậy \(x=\dfrac{2}{5};x=3\)
a) \(x^2-6x+3\)
\(=x^2-2.x.3+9-6\)
\(=\left(x-3\right)^2-\left(\sqrt{6}\right)^2\)
\(=\left(x-3-\sqrt{6}\right)\left(x-3+\sqrt{6}\right)\)
b) \(9x^2+6x-8\)
\(=\left(3x\right)^2+2.3x+1-9\)
\(=\left(3x+1\right)^2-3^2\)
\(=\left(3x+1-3\right)\left(3x+1+3\right)\)
\(=\left(3x-2\right)\left(3x+4\right)\)
d) \(x^3+6x^2+11x+6\)
\(=x^3+3x^2+3x^2+9x+2x+6\)
\(=x^2\left(x+3\right)+3x\left(x+3\right)+2\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2+3x+2\right)\)
\(=\left(x+3\right)\left(x^2+x+2x+2\right)\)
\(=\left(x+3\right)\left[x\left(x+1\right)+2\left(x+1\right)\right]\)
\(=\left(x+3\right)\left(x+1\right)\left(x+2\right)\)
e) \(x^3+4x^2-29x+24\)
\(=x^3+8x^2-4x^2-32x+3x+24\)
\(=x^2\left(x+8\right)-4x\left(x+8\right)+3\left(x+8\right)\)
\(=\left(x+8\right)\left(x^2-4x+3\right)\)
\(=\left(x+8\right)\left(x^2-3x-x+3\right)\)
\(=\left(x+8\right)\left[x\left(x-3\right)-\left(x-3\right)\right]\)
\(=\left(x+8\right)\left(x-3\right)\left(x-1\right)\)
a) \(x^3-\dfrac{1}{9}x=0\)
\(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\)
\(\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\\x+\dfrac{1}{3}=0\Leftrightarrow x=-\dfrac{1}{3}\end{matrix}\right.\)
b) \(x\left(x-3\right)+x-3=0\)
\(\Rightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\Rightarrow x=3\\x+1=0\Rightarrow x=-1\end{matrix}\right.\)
c) \(2x-2y-x^2+2xy-y^2=0\) (thêm đề)
\(\Rightarrow2\left(x-y\right)-\left(x-y\right)^2=0\)
\(\Rightarrow\left(x-y\right)\left(2-x+y\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-y=0\Rightarrow x=y\\2-x+y=0\Rightarrow x-y=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=y\left(1\right)\\\left(1\right)\Rightarrow x-x=2\left(loại\right)\end{matrix}\right.\)
d) \(x^2\left(x-3\right)+27-9x=0\)
\(\Rightarrow x^2\left(x-3\right)+\left(x-3\right).9=0\)
\(\Rightarrow\left(x-3\right)\left(x^2+9\right)=0\)
\(\Rightarrow x-3=0\Rightarrow x=3.\)
\(A=\left(x+1\right)^3-\left(x+3\right)^2\left(x+1\right)+4x^2+8\)
\(A=x^3+3x^2+3x+1-\left(x^2+6x+9\right)\left(x+1\right)+4x^2+8\)
\(A=x^3+3x^2+3x+1-\left(x^3+6x^2+9x+x^2+6x+9\right)+4x^2+8\)
\(A=x^3+3x^2+3x+1-x^3-6x^2-9x-x^2-6x-9+4x^2+8\)
\(A=\left(x^3-x^3\right)+\left(3x^2-6x^2-x^2+4x^2\right)+\left(3x-9x-6x\right)+\left(1-9+8\right)\)
\(A=-12x\)
\(B=\left(x-2\right)\left(x^2+2x+4\right)-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)
\(B=x^3+2x^2+4x-2x^2-4x-8-\left(x^3+3x^2+3x+1\right)+3\left(x^2-1\right)\)
\(B=x^3+2x^2+4x-2x^2-4x-8-x^3-3x^2-3x-1+3x^2-3\)
\(B=\left(x^3-x^3\right)+\left(2x^2-2x^2-3x^2+3x^2\right)+\left(4x-4x-3x\right)+\left(-8-3-1\right)\)
\(B=-3x-12\)
Câu C tương tự.
Chúc bạn học tốt!!!
A = \(\left(x+1\right)^3-\left(x+3\right)^2.\left(x+1\right)+4x^2+8\)
A = \(\left(x+1\right)\left(x+1-x-3\right)\left(x+1+x+3\right)+4x^2+8\)
A = \(\left(x+1\right).\left(-2\right).\left(2x+4\right)+4x^2+8\)
A = \(\left(-2\right)\left(2x^2+4x+2x+4\right)+4x^2+8\)
A = \(\left(-2\right)\left(2x^2+6x+4\right)+4x^2+8\)
A = \(-4x^2-12x-8+4x^2+8=-12x\)
b) B = \(\left(x-2\right)\left(x^2+2x+4\right)-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)
B = \(x^3-8-\left(x+1\right)\left(x^2+2x+1+3x-3\right)\)
B = \(x^3-8-\left(x+1\right)\left(x^2+5x-2\right)\)
B = \(x^3-8-x^3-5x^2+2x-x^2-5x+2\)
B = \(-6x^2-3x-6\)
a: \(x^2-4x+3=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
=>x=1 hoặc x=3
b: \(x^2+x-12=0\)
=>(x+4)(x-3)=0
=>x=3 hoặc x=-4
c: \(3x^2+2x-5=0\)
\(\Leftrightarrow3x^2+5x-3x-5=0\)
=>(3x+5)(x-1)=0
=>x=1 hoặc x=-5/3
d: \(x^4-2x^2-3=0\)
\(\Leftrightarrow x^4-3x^2+x^2-3=0\)
\(\Leftrightarrow x^2-3=0\)
hay \(x\in\left\{\sqrt{3};-\sqrt{3}\right\}\)
a: \(A=\left(x-5\right)\left(2x+3\right)-2x\left(x-3\right)\)
\(=2^{X2}+3x-10x-15-2x^2+6x\)
=-x-15
b: \(B=\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)\)
\(=48x^2-12x-20x+5+3x-48x^2-7+112x\)
\(=83x-2\)
b, \(2x\left(x-2\right)-\left(2-x\right)^2=2x\left(x-2\right)+\left(x-2\right)^2=\left(x-2\right)\left(2x+x-2\right)=\left(x-2\right)\left(3x-2\right)\)
>> T sẽ ko lm hết đâu :v <<
a) \(4x^3-14x^2\)
\(2x^2.\left(2x-7\right)\)
b) \(2x\left(x-2\right)-\left(2-x\right)^2\)
\(2x\left(x-2\right)-\left(-x\left(x-2\right)\right)^2\)
\(2x\left(x-2\right)-\left(x-2\right)^2\)
\(\left(x-2\right)\left(2x-\left(x-2\right)\right)\)
\(\left(x-2\right)\left(2x-x+2\right)\)
\(\left(x-2\right)\left(x+2\right)\)
c) \(\left(x-3\right)^3+3x\)
\(x^3-9x^2+27x-27+3x\)
\(x^3-9x^2+30x-27\)