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\(4x^2-25y^2\)
\(\left(2x\right)^2-\left(5y\right)^2\)
\(\left(2x-5y\right)\left(2x+5y\right)\)
chọn c
\(a)\)
\(4x^2-4x+1\)\(=\left(2x-1\right)^2\)
\(b)\)
\(\left(3x+2\right)\left(2-3x\right)\)\(=4-9x^2\)
\(c)\)
\(\left(x-3\right)\left(x^2+3x+9\right)\)\(=x^3-27\)
\(a,4x^2-4x+1=\left(2x\right)^2-2.2x.1+1=\left(2x-1\right)^2\)
\(b,\left(3x+2\right)\left(2-3x\right)=\left(2+3x\right)\left(2-3x\right)=2^2-\left(3x\right)^2\)
\(c,\left(x-3\right)\left(x^2+3x+9\right)=\left(x-3\right)\left(x^2+3x.1+3^2\right)=x^3-3^3\)
\(\left(x^2+\frac{2}{5}y\right)\left(x^2-\frac{2}{5}y\right)\)
\(=x^4+\frac{2}{5}x^2y-\frac{2}{5}x^2y-\left(\frac{2}{5}y\right)^2\)
\(=x^4-\frac{4}{25}y^2\)
a: \(=\dfrac{3x-x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\)
b: \(=\dfrac{1}{x\left(y-x\right)}-\dfrac{1}{y\left(y-x\right)}\)
\(=\dfrac{y-x}{xy\left(y-x\right)}=\dfrac{1}{xy}\)
c: \(=\dfrac{\left(1-2x\right)\left(1+2x\right)}{x\left(x+4\right)}\cdot\dfrac{3x}{2\left(1-2x\right)}\)
\(=\dfrac{3\left(1+2x\right)}{2\left(x+4\right)}\)
d: \(=\dfrac{12x}{8x^3}\cdot\dfrac{15y^4}{5y^3}=\dfrac{3}{2x^2}\cdot3y=\dfrac{9y}{2x^2}\)
f: \(=\dfrac{\left(x-2\right)\left(x+2\right)}{3\left(x+4\right)}\cdot\dfrac{x+4}{2\left(x-2\right)}=\dfrac{x+2}{6}\)
f: \(=\dfrac{5x-3-x+3}{4x^2y}=\dfrac{4x}{4x^2y}=\dfrac{1}{xy}\)
g: \(=\dfrac{3x+10-x-4}{x+3}=\dfrac{2x+6}{x+3}=2\)
h: \(=\dfrac{4-2+x}{x-1}=\dfrac{x+2}{x-1}\)
n: \(=\dfrac{3x-x+6}{x\left(x+3\right)}=\dfrac{2\left(x+3\right)}{x\left(x+3\right)}=\dfrac{2}{x}\)
p: \(=\dfrac{x^2-9-x^2+9}{x\left(x-3\right)}=0\)
k: \(=\dfrac{x-2x-4+x-2}{\left(x+2\right)\left(x-2\right)}=\dfrac{-6}{x^2-4}\)
m: \(=\dfrac{3x-x+6}{x\left(2x+6\right)}=\dfrac{2x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\)
a.5x2-10xy+5y2-20z2
=5(x2-2xy+y2-4z2)
=5[ (x2-2xy+y2)-(2z)2 ]
=5[ (x-y)2-(2z)2 ]
=5(x-y-2z)(x-y+2z)
b.16x-5x2-3
=15x+x-5x2-3
=(15x-3)+(x-5x2)
=3(5x-1)+x(1-5x)
=3(5x-1)-x(5x-1)
=(5x-1)(3-x)
c.x2-5x+5y-y2
=(5y-5x)+(x2-y2)
=5(y-x)+(x-y)(x+y)
=5(y-x)-(y-x)(y+x)
=(y-x)[5-(y+x)]
=(y-x)(5-y-x)
d.3x2-6xy+3y2-12z2 (câu này hình như ở trên đề bạn ghi sai nha! Mình sửa lại luôn rồi đó)
=3(x2-2xy+y2-4z2)
=3[ (x2-2xy+y2)-(2z)2 ]
=3[ (x-y)2-(2z)2 ]
=3(x-y-2z)(x-y+2z)
e.x2+4x+3
=x2+3x+x+3
=(x2+x)+(3x+3)
=x(x+1)+3(x+1)
=(x+1)(x+3)
f.(x2+1)2-4x2
=(x2+1)2-(2x)2
=(x2+1-2x)(x2+1+2x)
h.x2-4x-5
=x2-5x+x-5
=(x2+x)+(-5x-5)
=x(x+1)-5(x+1)
-(x+1)(x-5)
a/ \(\left(x-2\right)\left(2+x\right)=\left(x-2\right)\left(x+2\right)=x^2-4\)
b/ \(\left(1-3x\right)\left(3x+1\right)=\left(1-3x\right)\left(1+3x\right)=1-\left(3x\right)^2=1-9x^2\)
c/ \(\left(4x+5y\right)\left(5y-4x\right)=\left(5y+4x\right)\left(5y-4x\right)=\left(5y\right)^2-\left(4x\right)^2=25y^2-16x^2\)