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\(x^4-x^2+2x-1\)
\(=x^4-\left(x^2-2x+1\right)\)
\(=x^4-\left(x-1\right)^2\)
\(=\left(x^2-x+1\right)\left(x^2+x-1\right)\)
hk
tốt
Phân tích thành nhân tử : -4x^2+12xy-9y^2+25(sử dụng hằng đẳng thức)
giúp mink với, mink đang cần gấp
\(=-2x^2+6xy-3y^2+25\)
\(=-\left(2x^2-6xy+3y^2\right)+25\)
\(=-\left(2x^2+3y^2\right)+25\)
\(=\left(2x^2-3y^2\right)+25\)
\(=\left(2x^2+3y^2\right).\left(2x^2-3y^2\right)+25\)
a, \(\left(4x+5\right)^2=\left(4x+5\right)\left(4x+5\right)=\left[\left(4x+5\right)4x\right]+\left[\left(4x+5\right)5\right]=4x^2+20x+25\)
b, \(\left(5x-2\right)^2=\left(5x-2\right)\left(5x-2\right)=\left[\left(5x-2\right)5x-\left(5x-2\right)2\right]=5x^2-10x+25\)
b, \(8^2-12x^2=\left(8^2-12x^2\right)\left(8^2+12x^2\right)\)
đúng ko :)
@No name: Bị sai rồi nhé, a,b,c sai hết :>
a) ( 4x + 5 )2
= ( 4x )2 + 2.4x.5 + 52
= 16x2 + 40x + 25
b) ( 5x - 2 )2
= ( 5x )2 - 2.5x.2 + 22
= 25x2 - 20x + 4
c) 82 - 12x2
= 64 - 12x2
= ( V8 - V12x )( V8 + V12x )
4: \(\left(2x+3\right)^3-1\)
\(=\left(2x+3-1\right)\left(4x^2+12x+9+2x+3+1\right)\)
\(=\left(2x+2\right)\left(4x^2+14x+13\right)\)
\(=2\left(x+1\right)\left(4x^2+14x+13\right)\)
5: \(4x^2+20xy+25y^2=\left(2x+5y\right)^2\)
6: \(x^4-64xy^3\)
\(=x\left(x^3-64y^3\right)\)
\(=x\left(x-4y\right)\left(x^2+4xy+16y^2\right)\)
\(\left(x^2-2xz+z^2\right)-\left(y^2-2yt+t^2\right)=\left(x-z\right)^2-\left(y-t\right)^2=\left(x-z+y-1\right)\left(x-z-y+t\right)\)
Chúc Bạn học tốt
T I C K nha
a) \(x^6-y^6\)
\(=\left(x^3\right)^2-\left(y^3\right)^2\)
\(=\left(x^3-y^3\right)\left(x^3+y^3\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)\)
b) \(10ab+0,25a^2+100b^2\)
\(=\left(0,5a\right)^2+2\cdot0,5a\cdot10b+\left(10b\right)^2\)
\(=\left(0,5a+10b\right)^2\)
c) \(9x^2-xy+\frac{1}{36}y^2\)
\(=\left(3x\right)^2-2\cdot3x\cdot\frac{1}{6}y+\left(\frac{1}{6}y\right)^2\)
\(=\left(3x-\frac{1}{6}y\right)^2\)
\(9\left(x-3y\right)^2-25\left(2x+y\right)^2\)
\(=\left[3\left(x-3y\right)\right]^2-\left[5\left(2x+y\right)\right]^2\)
\(=\left(3x-9y\right)^2-\left(10x+5y\right)^2\)
\(=\left[3x-9y+10x+5y\right]\left[3x-9y-\left(10x+5y\right)\right]\)
\(=\left(13x-4y\right)\left(-7x-14y\right)\)
\(=-7\left(x+2y\right)\left(13x-4y\right)\)
9(x - 3y)² - 25(2x + y)²
= 3².(x - 3y)² - 5².(2x + y)²
= (3x - 9y)² - (10x + 5y)²
= (3x - 9y - 10x - 5y)(3x - 9y + 10x + 5y)
= (-7x - 14y)(13x - 4y)
= -7(x + 2y)(13x - 4y)
\(\dfrac{1}{x-y}-\dfrac{1}{x+y}+\dfrac{2x}{\left(x-y\right)\left(x+y\right)}\\ \dfrac{x+y}{\left(x-y\right)\left(x+y\right)}-\dfrac{x-y}{\left(x-y\right)\left(x+y\right)}+\dfrac{2x}{\left(x-y\right)\left(x+y\right)}\\ \dfrac{x+y-x+y+2x}{\left(x-y\right)\left(x+y\right)}\\ \dfrac{2x+2y}{\left(x-y\right)\left(x+y\right)}\\ \dfrac{2}{x-y}\)
\(\dfrac{1}{x-y}+\dfrac{1}{x+y}+\dfrac{2x}{x^2-y^2}=\dfrac{x+y}{\left(x-y\right)\left(x+y\right)}+\dfrac{x-y}{\left(x-y\right)\left(x+y\right)}+\dfrac{2x}{\left(x-y\right)\left(x+y\right)}=\dfrac{4x}{\left(x-y\right)\left(x+y\right)}\)
\(\dfrac{x^2}{2}-2x^2\)
\(=\dfrac{1}{2}x^2-2x^2\)
\(=-\dfrac{3}{2}x^2\)