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1 tháng 10 2016
Phân tích đa thức thành nhân tử:
a) \(xy+y^2-x-y=y\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(y-1\right)\)
b) \(25-x^2+4xy-4y^2=25-\left(x^2-4xy+4y^2\right)=25-\left(x-2y\right)^2\)
\(=\left(5-x+2y\right)\left(5+x-2y\right)\)
Rút gọn biểu thức;
\(A=\left(6x+1\right)^2+\left(3x-1\right)^2-2\left(3x-1\right)\left(6x+1\right)\)
\(=\left[\left(6x+1\right)-\left(3x-1\right)\right]^2=\left(6x+1-3x+1\right)=\left(3x+2\right)^2\)
Tìm a để đa thức.. Bạn chia cột dọ thì da
KT
1 tháng 10 2016
\(xy+y^2-x-y=\left(xy+y^2\right)-\left(x+y\right)=y\left(x+y\right)-\left(x+y\right)=\left(y-1\right)\left(x+y\right)\)b)\(25-\left(x^2-4xy+4y^2\right)=5^2-\left(x-2y\right)^2=\left(x-2y+5\right)\left(5-x+2y\right)\)
PN
22 tháng 10 2017
x4 + 4
= x4 + 4 + 4x² - 4x² = ( x² + 2 )² - 4x²
= ( x² + 2 - 2x )( x² + 2 + 2x )
P/s: Giang Lê Trà My lần sau giải cho cẩn thận vào nhé.
BT
0
TA
9 tháng 7 2017
a) \(x^2-5x+xy-5y=\left(x^2+xy\right)-\left(5x+5y\right)=x\left(x+y\right)-5\left(x+y\right)=\left(x-5\right)\left(x+y\right)\)
b) \(4x^2-\left(x-2\right)^2=\left(2x\right)^2-\left(x-2\right)^2=\left(2x+x-2\right)\left(2x-x+2\right)=\left(3x-2\right)\left(x+2\right)\)
c) \(48x^2y^2-3y^2+6xy-3x^2=3\left(16x^2y^2-y^2+2xy-x^2\right)=3\left[\left(4xy\right)^2-\left(y^2-2xy+x^2\right)\right]\)
\(=3\left[\left(4xy\right)^2-\left(y-x\right)^2\right]=3\left(4xy+y-x\right)\left(4xy-y+x\right)\)
d) \(2x^2-5x-7=2x^2+2x-7x-7=2x\left(x+1\right)-7\left(x+1\right)=\left(2x-7\right)\left(x+1\right)\)
28 tháng 10 2018
a) \(x^3+2x^2y+xy^2-4xz^2=x\left(x^2+2xy+y^2-4z^2\right)=x\left[\left(x+y\right)^2-\left(2z\right)^2\right]\)
\(=x\left(x+y-2z\right)\left(x+y+2z\right)\)
b)\(-8x^3+12x^2y-6xy^2+y^3=y^3+3.y.\left(2x\right)^2-3.y^2.2x-\left(2x\right)^3\)\(=\left(y-2x\right)^3\)
c)\(6x^2+7x-5=2x\left(3x+5\right)-\left(3x+5\right)=\left(3x+5\right)\left(2x-1\right)\)
d)\(x^4+64y^4=\left(x^2\right)^2+2.x^2.8y^2+\left(8y^2\right)^2-16x^2y^2=\left(x^2+8y^2\right)-\left(4xy\right)^2\)
\(=\left(x^2+8y^2-4xy\right)\left(x^2+8y^2+4xy\right)\)
e)\(x\left(2-x\right)-x+2=x\left(2-x\right)+\left(2-x\right)=\left(2-x\right)\left(x+1\right)\)
f)\(2x^2+3x-2=2x\left(x+2\right)-\left(x+2\right)=\left(x+2\right)\left(2x-1\right)\)
h)\(3x^2-6xy+3y^2-12z^2=3\left(x^2-2xy+y^2-4z^2\right)=3\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
\(=3\left(x-y-2z\right)\left(x-y+2z\right)\)
g)\(x^3-3x^2-9x+27=x^2\left(x-3\right)-9\left(x-3\right)=\left(x-3\right)\left(x^2-9\right)\)\(=\left(x-3\right)^2\left(x+3\right)\)
B2: \(x^3-5x=0\Rightarrow x\left(x^2-5\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x^2-5=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^2=5\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt{5}\end{cases}}}\)\(\Rightarrow\orbr{\begin{cases}x=0\\x^2=5\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\\orbr{\begin{cases}x=\sqrt{5}\\x=-\sqrt{5}\end{cases}}\end{cases}}\)
3 tháng 9 2018
\(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
1 tháng 10 2020
\begin{array}{l} a){\left( {ab - 1} \right)^2} + {\left( {a + b} \right)^2}\\ = {a^2}{b^2} - 2ab + 1 + {a^2} + 2ab + {b^2}\\ = {a^2}{b^2} + 1 + {a^2} + {b^2}\\ = {a^2}\left( {{b^2} + 1} \right) + \left( {{b^2} + 1} \right)\\ = \left( {{a^2} + 1} \right)\left( {{b^2} + 1} \right)\\ c){x^3} - 4{x^2} + 12x - 27\\ = {x^3} - 27 + \left( { - 4{x^2} + 12x} \right)\\ = \left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right) - 4x\left( {x - 3} \right)\\ = \left( {x - 3} \right)\left( {{x^2} + 3x + 9 - 4x} \right)\\ = \left( {x - 3} \right)\left( {{x^2} - x + 9} \right)\\ b){x^3} + 2{x^2} + 2x + 1\\ = {x^3} + 2{x^2} + x + x + 1\\ = x\left( {{x^2} + 2x + 1} \right) + \left( {x + 1} \right)\\ = x{\left( {x + 1} \right)^2} + \left( {x + 1} \right)\\ = \left( {x + 1} \right)\left( {x\left( {x + 1} \right) + 1} \right)\\ = \left( {x + 1} \right)\left( {{x^2} + x + 1} \right)\\ d){x^4} - 2{x^3} + 2x - 1\\ = {x^4} - 2{x^3} + {x^2} - {x^2} + 2x - 1\\ = {x^2}\left( {{x^2} - 2x + 1} \right) - \left( {{x^2} - 2x + 1} \right)\\ = \left( {{x^2} - 2x + 1} \right)\left( {{x^2} - 1} \right)\\ = {\left( {x - 1} \right)^2}\left( {x - 1} \right)\left( {x + 1} \right)\\ = {\left( {x - 1} \right)^3}\left( {x + 1} \right)\\ e){x^4} + 2{x^3} + 2{x^2} + 2x + 1\\ = {x^4} + 2{x^3} + {x^2} + {x^2} + 2x + 1\\ = {x^2}\left( {{x^2} + 2x + 1} \right) + \left( {{x^2} + 2x + 1} \right)\\ = \left( {{x^2} + 2x + 1} \right)\left( {{x^2} + 1} \right)\\ = {\left( {x + 1} \right)^2}\left( {{x^2} + 1} \right) \end{array} |
NT
27 tháng 10 2017
a) \(=2xy^2\left(x^2+8x+15\right)\)
\(=2xy^2\left[\left(x^2+8x+16\right)-1\right]\)
\(=2xy^2\left[\left(x+4\right)^2-1\right]\)
\(=2xy^2\left(x+4+1\right)\left(x+4-1\right)\)
\(=2xy^2\left(x+5\right)\left(x-3\right)\)
mấy câu sau tự làm nha :*
29 tháng 10 2017
b,=(x^2-10x+25)-4
=(x-5)^2-2^2
=(x-5-2)(x-5+2)
=(x-7)(x-3)
15 tháng 9 2019
1
\(\left(2xy+1\right)^2-\left(2x+y\right)^2=\left(2xy+1-2x-y\right)\left(2xy+1+2x+y\right)\)
3
\(\left(x^2+y^2-z^2\right)^2-4x^2y^2=\left(x^2+y^2-z^2-2xy\right)\left(x^2+y^2-z^2+2xy\right)\)
\(=\left(x-y-z\right)\left(x-y+z\right)\left(x+y-z\right)\left(x+y+z\right)\)
4
\(9x^2+90x+225-\left(x-7\right)^2=9\left(x^2+10x+25\right)-\left(x-7\right)^2\)
\(=9\left(x+5\right)^2-\left(x+7\right)^2\)
\(=\left(3x+15-x-7\right)\left(3x+15+x+7\right)\)
Rút gọn nốt:(
\(x^4+4\)
\(=x^4+4x^2+4-4x^2\)
\(=\left[\left(x^2\right)^2+2.x^2.2+2^2\right]-4x^2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)
\(\text{a) }3x-7x+2\\ \\=2-4x\\ \\=2\left(1-2x\right)\)
\(\text{b) }x^4+4\\ \\=x^4+4+4x^2-4x^2\\ \\=\left(x^4+4x^2+4\right)-4x^2\\ \\=\left(x^2+2\right)^2-\left(2x\right)^2\\ \\=\left(x^2+2+2x\right)\left(x^2+2-2x\right)\\ \)
\(\text{c) }4x-8y=4\left(x-2y\right)\)
\(\text{d) }x^3-x^2y-x+y\\ \\=\left(x^3-x^2y\right)-\left(x-y\right)\\ \\=x^2\left(x-y\right)-\left(x-y\right)\\ =\left(x-y\right)\left(x^2-1\right)\\ \\=\left(x-y\right)\left(x-1\right)\left(x+1\right)\)