QT
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
27 tháng 9 2017
\(a,2x^2-2xt-5x+5y\)
\(=\left(2x^2-5x\right)-\left(2xy-5y\right)\)
\(=x\left(2x-5\right)-y\left(2x-5\right)\)
\(=\left(2x-5\right)\left(x-y\right)\)
\(b,8x^2+4xy-2ax-ay\)
\(=\left(8x^2-2ax\right)+\left(4xy-ay\right)\)
\(=2x\left(4x-a\right)+y\left(4x-a\right)\)
\(=\left(4x-a\right)\left(2x+y\right)\)
\(c,x^3-4x^2+4x\)
\(=x^3-2x^2-2x^2+4x\)
\(=\left(x^3-2x^2\right)-\left(2x^2-4x\right)\)
\(=x^2\left(x-2\right)-2x\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2-2x\right)=x\left(x-2\right)\left(x-2\right)\)
\(=x\left(x-2\right)^2\)
\(d,2xy-x^2-y^2+16\)
\(=-\left(x^2-2xy+y^2-16\right)\)
\(=-\left[\left(x-y\right)^2-4^2\right]\)
\(=-\left(x-y-4\right)\left(x-y+4\right)\)
\(e,x^2-y^2-2yz-z^2\)
\(=x^2-\left(y^2+2yz+z^2\right)\)
\(=x^2-\left(y+z\right)^2=\left(x-y-z\right)\left(x+y+z\right)\)
6 tháng 8 2022
a: \(=x^2-4x+4+y^2+2y+1\)
\(=\left(x-2\right)^2+\left(y+1\right)^2\)
b: \(=x^2+10x+25+x^2-2xy+y^2\)
\(=\left(x+5\right)^2+\left(x-y\right)^2\)
c: \(=a^2+2ab+b^2+4b^2+4b+1\)
\(=\left(a+b\right)^2+\left(2b+1\right)^2\)
d: \(=2\left(x^2+b^2\right)\)
3 tháng 12 2022
1: =(4x-1)^2-3(4x-1)
=(4x-1)(4x-1-3)
=4(x-1)(4x-1)
2: =-8x^4y^5(2y+3x)
3: =(a-5)^2-4b^2
=(a-5-2b)(a-5+2b)
5: =x^2-mx-nx+mn
=x(x-m)-n(x-m)
=(x-m)(x-n)
6: =(4a^2-3a-18-4a^2-3a)(4a^2-3a-18+4a^2+3a)
=(-6a-18)(8a^2-18)
=-6(2a-3)(2x+3)(a+3)
S
31 tháng 7 2019
\(A=\left(x+1\right)^2+\left(x+2\right)^2=\left(x+1\right)^2+\left(-2-x\right)^2\ge\frac{1}{2}\left(x+1-2-x\right)^2=\frac{1}{2}.1^2=\frac{1}{2}\Rightarrow A_{min}=\frac{1}{2}\Leftrightarrow x=\frac{3}{2}\)
\(B=-2x^2-4\le0-4=-4\Rightarrow B_{max}=-4\Leftrightarrow x=0\)
\(C=-5x^2+10x-7=-5x^2+10x-5-2=-5\left(x-1\right)^2-2\le0-2=-2\Rightarrow C_{min}=-2\Leftrightarrow x-1=0\Leftrightarrow x=1\)
PT
10 tháng 9 2018
a) \(x^2-2x=24\)
\(\Rightarrow x^2-2x-24=0\)
\(\Rightarrow x^2-6x+4x-24=0\)
\(\Rightarrow x\left(x-6\right)+4\left(x-6\right)=0\)
\(\Rightarrow\left(x-6\right)\left(x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-6=0\\x+4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
b) \(\left(5-2x\right)^2-16=0\)
\(\Rightarrow\left(5-2x\right)^2-4^2=0\)
\(\Rightarrow\left(5-2x-4\right)\left(5-2x+4\right)=0\)
\(\Rightarrow\left(1-2x\right)\left(9-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}1-2x=0\\9-2x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=1\\2x=9\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\)
c)Sửa đề
\(x^2-4x+4-9x^2+6x-1=0\)
\(\Rightarrow\left(x^2-4x+4\right)-\left(9x^2-6x+1\right)=0\)
\(\Rightarrow\left(x-2\right)^2-\left(3x-1\right)^2=0\)
\(\Rightarrow\left(x-2-3x+1\right)\left(x-2+3x-1\right)=0\)
\(\Rightarrow\left(-2x-1\right)\left(4x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-2x-1=0\\4x-3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-2x=1\\4x=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\)
d) \(2x^2+y^2+2xy-4x+4=0\)
\(\Rightarrow\left(x^2+2xy+y^2\right)+\left(x^2-4x+4\right)=0\)
\(\Rightarrow\left(x+y\right)^2+\left(x-2\right)^2=0\)
Vì \(\left(x+y\right)^2\ge0\) với mọi x và y
\(\left(x-2\right)^2\ge0\) với mọi x
\(\Rightarrow\left(x+y\right)^2+\left(x-2\right)^2\ge0\) với mọi x và y
Mà \(\left(x+y\right)^2+\left(x-2\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x+y=0\\x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}y=-x\\x=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}y=-2\\x=2\end{matrix}\right.\)
1) \(a^2-2ab+b^2-9=\left(a-b\right)^2-9=\left(a-b-3\right)\left(a-b+3\right)\)
2) \(x^2+2x+1-y^2=\left(x+1\right)^2-y^2=\left(x+1-y\right)\left(x+1+y\right)\)
3) \(25-x^2-2xy-y^2=25-\left(x+y\right)^2=\left(5-x-y\right)\left(5+x+y\right)\)
\(1,=\left(a-b\right)^2-9=\left(a-b-3\right)\left(a-b+3\right)\\ 2,=\left(x+1\right)^2-y^2=\left(x-y+1\right)\left(x+y+1\right)\\ 3,=25-\left(x+y\right)^2=\left(5-x-y\right)\left(5+x+y\right)\)