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2 tháng 8 2017
a)
\(4x^2-9y^2+6x-9y=\left(2x-3y\right)\left(2x+3\right)+3\left(2x-3y\right)\)
\(=\left(2x-3y\right)\left(2x+3y+3\right)\)
b)
\(1-2x+2yz+x^2-y^2-z^2=\left(x^2-2x+1\right)-\left(y^2-2yz+z^2\right)\) (đổi dấu)
\(=\left(x-1\right)^2-\left(y-z\right)^2\)
c)
\(x^3-1+5x^2-5+3x-3=\left(x-1\right)\left(x^2+x+1\right)+5\left(x-1\right)\left(x+1\right)+3\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1+5\left(x+1\right)+3\right)\)
\(=\left(x-1\right)\left(x^2+x+1+5x+5+3\right)\)
\(=\left(x-1\right)\left(x^2+6x+9\right)=\left(x-1\right)\left(x+3\right)^2\)
R
9 tháng 7 2019
Bài 1:tìm x ,biết:
a) (2x - 1)(3x + 2) - 6x(x + 1) = 0
\(\Leftrightarrow6x^2+x-2-6x^2-6x=0\)
\(\Leftrightarrow-5x=2\)
\(\Leftrightarrow x=\frac{-2}{5}\)
b) \(\left(4x-1\right)^2-\left(2x+1\right)\left(8x-3\right)=0\)
\(\Leftrightarrow16x^2-8x+1-16x^2-2x+3=0\)
\(\Leftrightarrow-10x=-4\)
\(\Leftrightarrow x=\frac{2}{5}\)
c) \(4x^2-1=2\left(2x+1\right)\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-1\right)-2\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{3}{2}\end{cases}}\)
2a) \(4x^2-9y^2-6y-1=4x^2-\left(3y+1\right)^2\)
\(=\left(2x-3y-1\right)\left(2x+3y+1\right)\)
b) \(4x^2-1-2x\left(2x-1\right)=\left(2x-1\right)\left(2x+1\right)-2x\left(2x-1\right)\)
\(=1.\left(2x-1\right)\)
c) \(x^2-8x-4y^2+16=\left(x-4\right)^2-4y^2\)
\(=\left(x-4-2y\right)\left(x-4+2y\right)\)
d) \(9x^2-12x-y^2+4=\left(3x-2\right)^2-y^2\)
\(=\left(3x-2-y\right)\left(3x-2+y\right)\)
e) \(4x^2+10x-5=4x^2+2.2.\frac{5}{2}x+\frac{25}{4}-\frac{25}{4}-5\)
\(=\left(2x+\frac{5}{2}\right)^2-\frac{45}{4}\)
\(=\left(2x+\frac{5+3\sqrt{5}}{2}\right)\left(2x+\frac{5-3\sqrt{5}}{2}\right)\)
6 tháng 4 2020
a) Ta có: \(4x^2+4x+1\)
\(=\left(2x\right)^2+2\cdot2x\cdot1+1\)
\(=\left(2x+1\right)^2\)
b) Sửa đề: \(x^2-6x+9-9y^2\)
Ta có: \(x^2-6x+9-9y^2\)
\(=\left(x-3\right)^2-\left(3y\right)^2\)
\(=\left(x-3-3y\right)\left(x-3+3y\right)\)
c) Ta có: \(12x-9-4x^2\)
\(=-\left(4x^2-12x+9\right)\)
\(=-\left[\left(2x\right)^2-2\cdot2x\cdot3+3^2\right]\)
\(=-\left(2x-3\right)^2\)
d) Ta có: \(1-9x+27x^2-27x^3\)
\(=1^3-3\cdot1^2\cdot3x+3\cdot1\cdot\left(3x\right)^2-\left(3x\right)^3\)
\(=\left(1-3x\right)^3\)
15 tháng 7 2018
\(x^2+4x-9y^2+4\)
\(=\left(x^2+2.2x+2^2\right)-\left(3y\right)^2\)
\(=\left(x+2\right)^2-\left(3y\right)^2\)
\(=\left(x+2-3y\right)\left(x+2+3y\right)\)
\(x^2-9y^2-6y-1\)
\(=x^2-\left[\left(3y\right)^2+2.3y+1^2\right]\)
\(=x^2-\left(3y+1\right)^2\)
\(=\left(x-3y-1\right)\left(x+3y+1\right)\)
Tham khảo nhé~
15 tháng 7 2018
a/Ta có:x2+4x-9y2+4
=x2+4x+4-(3y)2
=(x+2)2-(3y)2
=(x+2-3y)(x+2+3y)
b/Ta có:x2-9y2-6xy-1
=x2-6xy-(3y)2-1
=(x-3y-1)(x-3y+1)
S
0
17 tháng 8 2019
\(36x^2-9y^2-12x-6y\)
\(=\left(36x^2-12x+1\right)-\left(9y^2+6y+1\right)\)
\(=\left(6x-1\right)^2-\left(9y+1\right)\)
\(=\left(6x+9y\right)\left(6x-3y-2\right)\)
\(=3\left(2x+3y\right)\left(6x-3y-2\right)\)
a) \(9y^2-4x^2+6x+9y\)
\(=\left(3y-2x\right)\left(3y+2x\right)+3\left(3y+2x\right)\)
\(=\left(3y+2x\right)\left(3y-2x+3\right)\)
b) \(x^3+4x^2-12x\)
\(=x\left(x^2+4x-12\right)\)
\(=x\left(x^2-2x+6x-12\right)\)
\(=x\left(x\left(x-2\right)+6\left(x-2\right)\right)\)
\(=x\left(x-2\right)\left(x+6\right)\)
a)\(9y^2-4x^2+6x+9y=\left(9y^2-4x^2\right)+\left(6x+9y\right)=\left[\left(3y\right)^2-\left(2x\right)^2\right]+3\left(2x+3y\right)\)
\(=\left(3y-2x\right)\left(3y+2x\right)+3\left(3y+2x\right)=\left(3y+2x\right)\left[\left(3y-2x\right)+3\right]=\left(3y+2x\right)\left(3y-2x+3\right)\)
b)\(x^3+4x^2-12x=x^3-2x^2+6x^2-12x=\left(x^3-2x^2\right)+\left(6x^2-12x\right)\)
\(=x^2\left(x-2\right)+6x\left(x-2\right)=\left(x-2\right)\left(x^2+6x\right)=x\left(x-2\right)\left(x+6\right)\)