Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(x^2+4x+3\)
\(=x^2+3x+x+3\)
\(=x\left(x+3\right)+\left(x+3\right)\)
\(=\left(x+1\right)\left(x+3\right)\)
a. Biểu thức ko thể biểu diễn dưới dạng tích của các thừa số
b. (x-1)(4x+1)
c. -(3z^2-5y^2-6xy-3x^2)
d. x(y^2-2xy+x-9)
e. -(y-x)(y-x+2)
f. y^3+xy^2+3x^2y-y+x^2-x
HỌC TỐT.
\(a,35x^2y-14xy+21xy^2=7xy\left(5x+3y-2\right)\)
\(b,x^3-4x^2+4x=x\left(x^2-4x+4\right)=x\left(x-2\right)^2\)
\(c,x^2-7x+xy-7y=x\left(x-7\right)+y\left(x-7\right)=\left(x-7\right)\left(x+y\right)\)
\(d,x^2-y^2-10x+25=\left(x-5\right)^2-y^2=\left(x-y-5\right)\left(x+y-5\right)\)
\(e,x^3y+2x^2y^2-xyz^2+xy^3=xy\left(x^2+2xy+y^2-z^2\right)\)
\(=xy\left[\left(x+y\right)^2-z^2\right]=xy\left(x+y-z\right)\left(x+y+z\right)\)
a, \(x^3-2x^2+3x-6=x\left(x^2+3\right)-2\left(x^2+3\right)=\left(x-2\right)\left(x^2+3\right)\)
b, \(x^2+2x+1-4y^2=\left(x+1\right)^2-\left(2y\right)^2=\left(x+1-2y\right)\left(x+1+2y\right)\)
a/x +b/y +c/z =0 ->ayz+bxz+cxz=0
x/a + y/b + z/c=1 ->(x/a +y/b +z/c)^2=1
x^2/a^2 + y^2/b^2 + z^2/c^2 +2(xy/ab +yz/bc +xz/ac)=1
x^2/a^2 + y^2/b^2 + z^2/c^2 =1- 2* ayz+bxz+cxz/abc=1-2*0=1-0=1 =>ĐPCM
k hộ mik nha
#)Giải :
\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\rightarrow ayz+bxz+cxy=0\)
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\rightarrow\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\)
\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1-2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)=1-2\frac{ayz+bxz+cxy}{abc}=1-2.0=1\left(đpcm\right)\)
#~Will~be~Pens~#
a) 4x2-1=(2x)2-12=(2x-1)(2x+1)
b)25x2-0.09=(5x)2-\(\left(\frac{3}{10}\right)^2\)=\(\left(5x-\frac{3}{10}\right)\left(5x+\frac{3}{10}\right)\)
c)9x2-14=(3x)2-\(\sqrt{14}^2\)=(3x-\(\sqrt{14}\))(3x+\(\sqrt{14}\))
d) (x-y)2-4=(x-y)2-22=(x-y-2)(x-y+2)
e) 9-(x-y)2=33-(x-y)2=(3-x+y)(3+x-y)
f)(x2+4)2-16x2=(x2+4)2-(4x)2=(x2+4+4x)(x2+4-4x)
Chúc hok tốt!!!
\(a,\)\(4x^2-1\)\(=\left(2x\right)^2-1^2=\left(2x+1\right)\left(2x-1\right)\)
\(b,\)\(25x^2-0,09=\left(5x\right)^2-0,3^2=\left(5x-0,3\right)\left(5x+0,3\right)\)
\(c,\)\(9x^2-14=\left(3x\right)^2-\left(\sqrt{14}\right)^2=\left(3x-\sqrt{14}\right)\left(3x+\sqrt{14}\right)\)
\(d,\)\(\left(x-y\right)^2-4=\left(x-y\right)^2-2^2=\left(x-y+2\right)\left(x-y-2\right)\)
\(e,\)\(9-\left(x-y\right)^2=3^2-\left(x-y\right)^2=\left(3-x+y\right)\left(3+x-y\right)\)
\(f,\)\(\left(x^2+4\right)^2-16x^2=\left(x^2+4\right)^2-\left(4x\right)^2=\left(x^2+4-4x\right)\left(x^2+4+4x\right)\)
\(=\left(x^2-2.2x+2^2\right)\left(x^2+2.2x+2^2\right)\)
\(=\left(x-2\right)^2\left(x+2\right)^2\)
x8+x7+1=x8+x7+x6-x6-x5-x4+x5+x4+x3-x3-x2-x+x2+x+1=x6(x2+x+1)-x4(x2+x+1)+x3(x2+x+1)-x(x2+x+1)+(x2+x+1)=(x2 +x+1)(x6-x4+x3-x)
A = 4x2 + 6x = 2x( 2x + 3 )
B = ( x + 5 )2 - y2 = ( x + 5 - y )( x + 5 + y )
C = x2 + 2xy - 2x + y2 - 2y = ( x2 + 2xy + y2 ) - ( 2x + 2y ) = ( x + y )2 - 2( x + y ) = ( x + y )( x + y - 2 )
D = x2 - 4x + 3 = x2 - x - 3x + 3 = ( x2 - x ) - ( 3x - 3 ) = x( x - 1 ) - 3( x - 1 ) = ( x - 1 )( x - 3 )
a)4x^2+6x
=2x(2x+3)
b)(x+5)^2-y^2
=(x+5-y)(x+5+y)
c)x^2+2xy-2x+y^2-2y
=(x^2+2xy+y^2)-(2x+2y)
=(x+y)^2-2(x+y)
=(x+y)(x+y-2)
d)x^2-4x+3
=(x^2-x)-(3x-3)
=x(x-1)-3(x-1)
=(x-3)(x-1)