25-x^2-4y^2+4xy
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AT
28 tháng 6 2017
phân tích các đa thức sau thành nhân tử
a) 4x^2 - 4xy + 4y^2
\(=\) \(\left(2x\right)^2-4xy+\left(2y\right)^2\)
\(=\left(2x-2y\right)^2\)
b) x^2 - 4xy +4y^2
\(=x^2-4xy+\left(2y\right)^2\)
\(=\left(x-2y\right)^2\)
c) x^2 + 10x + 25
\(=x^2+2.x.5+5^2\)
\(=\left(x+5\right)^2\)
d)x^2 - 10x + 25
\(=x^2-2.x.5+5^2\)
\(=\left(x-5\right)^2\)
e) 81 - (x+1)^2
\(=9^2-\left(x+1\right)^2\)
\(=\left(9-x-1\right)\left(9+x+1\right)\)
f) 16x^2 - 64 (y + 1)^2
\(=16x^2-8^2\left(y+1\right)^2\)
\(=16x^2-\left(8y+8\right)^2\)
\(=\left(16-8y-8\right)\left(16+8y+8\right)\)
p/s: ko chắc câu cuối đâu :v
2 tháng 12 2018
\(x^2-25-4xy+4y^2\)
\(=\left(x^2-4xy+4y^2\right)-25\)
\(=\left[x^2-2\cdot x\cdot2y+\left(2y\right)^2\right]-25\)
\(=\left(x-2y\right)^2-5^2\)
\(=\left(x-2y-5\right)\cdot\left(x-2y+5\right)\)
26 tháng 8 2018
a. Ta có: x2+y2-2x+4y+5=0
⇌(x-1)2+(y-2)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y-2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
b. Ta có: 4x2+y2-4x-6y+10=0
⇌ (2x-1)2+(y-3)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\y-3=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=3\end{matrix}\right.\)
c.Ta có: 5x2-4xy+y2-4x+4=0
⇌(2x-y)2+(x-2)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}2x-y=0\\x-2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=4\\x=2\end{matrix}\right.\)
d.Ta có: 2x2-4xy+4y2-10x+25=0
⇌ (x-2y)2+(x-5)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-2y=0\\x-5=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{5}{2}\\x=5\end{matrix}\right.\)
CN
1
TT
14 tháng 8 2020
Mình làm và sửa đề đúng luôn nhé !
1) \(36x^2-a^2+10a-25\)
\(=\left(6x\right)^2-\left(a^2-10a+25\right)\)
\(=\left(6x\right)^2-\left(a-5\right)^2\)
\(=\left(6x-a+5\right)\left(6x+a-5\right)\)
2) \(4x^2-4xy+y^2-25a^2+10a-1\)
\(=\left(2x-y\right)^2-\left(5a-1\right)^2\)
\(=\left(2x-y-5a+1\right)\left(2x-y+5a-1\right)\)
3) \(m^2-6m+9-x^2+4xy-4y^2\)
\(=\left(m-3\right)^2-\left(x-2y\right)^2\)
\(=\left(m-3-x+2y\right)\left(m+3-x+2y\right)\)
AD
2
15 tháng 9 2021
a) \(x^2-2xy+y^2-1=\left(x-y\right)^2-1=\left(x-y-1\right)\left(x-y+1\right)\)
b) \(9-x^2-2xy-y^2=9-\left(x^2+2xy+y^2\right)=9-\left(x+y\right)^2=\left(3-x-y\right)\left(3+x+y\right)\)
c) \(25-x^2+4xy-4y^2=25-\left(x^2-4xy+4y^2\right)=25-\left(x-2y\right)^2=\left(5-x+2y\right)\left(5+x-2y\right)\)
KN
11 tháng 10 2019
a) \(4x^2-4xy+y^2-9\)
\(=\left(2x-y\right)^2-3^2\)
\(=\left(2x-y+3\right)\left(2x-y-3\right)\)
b) \(x^2-36+4xy+4y^2\)
\(=\left(4y^2+4xy+x^2\right)-36\)
\(=\left(2y+x\right)^2-6^2\)
\(=\left(2y+x+6\right)\left(2y+x-6\right)\)
c) \(9x^2-12xy-25+4y^2\)
\(=\left(9x^2-12xy+4y^2\right)-25\)
\(=\left(3x-2y\right)^2-5^2\)
\(=\left(3x-2y+5\right)\left(3x-2y-5\right)\)
d) \(25x^2+10x-4y^2+1\)
\(=\left(25x^2+10x+1\right)-4y^2\)
\(=\left(5x+1\right)^2-\left(2y\right)^2\)
\(=\left(5x+2y+1\right)\left(5x-2y+1\right)\)
HH
2
NN
2
2 tháng 11 2016
\(25-x^2-4xy-4y^2=5^2-\left(x+2y\right)^2=\left(5-x-2y\right)\left(5+x+2y\right)\)
\(25-x^2-4y^2+4xy=25-\left(x^2-4xy+4y^2\right)=25-\left[x^2-4xy+\left(2y\right)^2\right]=5^2-\left(x-2y\right)^2=\left[5-\left(x-2y\right)\right]\left[5+\left(x-2y\right)\right]=\left(5-x-2y\right)\left(5+x-2y\right)\)
\(=25-\left(x-2y\right)^2=\left(5-x+2y\right)\left(5+x-2y\right)\)