Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(yz\left(y+z\right)+zx\left(z-x\right)-xy\left(x+y\right)\)
\(=yz\left(y+z\right)+zx\left(z-x\right)-xy\left[\left(y+z\right)-\left(z-x\right)\right]\)
\(=yz\left(y+z\right)+zx\left(z-x\right)-xy\left(y+z\right)+xy\left(z-x\right)\)
\(=y\left(y+z\right)\left(z-x\right)+x\left(z-x\right)\left(z-y\right)\)
\(=\left(z-x\right)\left(yz-xy+xz-xy\right)\)
Bài 3:
a) \(4x^2+4x+1=\left(2x+1\right)^2\)
b) \(9x^2-12x+4=\left(3x-2\right)^2\)
c) \(ab^2+\dfrac{1}{4}a^2b^4+1=\left(\dfrac{1}{2}ab^2+1\right)^2\)
Phân tích đa thức thành nhân tử
a) 125x3 + y6
= (5x)3 + (y2)3
= (5x + y2)(25x2 - 5xy2 + y4)
b) x2 - 2xy + y2 - xz + yz
= (x - y)2 - z(x - y)
= (x - y)(x - y - z)
c) x4 - x3 - x2 + 1
= (x4 - x3) - (x2 - 1)
= x3(x - 1) - (x - 1)(x + 1)
= (x - 1)(x3 - x + 1)
a) \(g\left(x,y\right)=x^2-10xy+9y^2=x^2-xy-9xy+9y^2\)
\(=x\left(x-y\right)-9y\left(x-y\right)=\left(x-y\right)\left(x-9y\right)\).
b )\(f\left(x,y\right)=x^6+x^4+x^2y^2+y^4-y^6\)
\(=x^6-y^6+x^4+x^2y^2+y^4\)
\(=\left(x^3\right)^2-\left(y^3\right)^2+\left(x^4+2x^2y^2+y^4\right)-x^2y^2\)
\(=\left(x^3-y^3\right)\left(x^3+y^3\right)+\left(x^2+y^2\right)^2-\left(xy\right)^2\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)+\left(x^2+y^2-xy\right)\left(x^2+y^2+xy\right)\)
\(=\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)\left[\left(x-y\right)\left(x+y\right)+1\right]\)
\(=\left(x^2+xy+y^2\right)\left(x^2-2y+y^2\right)\left(x^2-y^2+1\right)\)
Vậy \(f\left(x,y\right)=\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)\left(x^2-y^2+1\right)\)
1, Ta có: \(\left(x-1\right)^2\ge0\Leftrightarrow x^2-2x+1\ge0\Leftrightarrow x^2+1\ge2x\) (1)\(\left(y-1\right)^2\ge0\Leftrightarrow y^2-2y+1\ge0\Leftrightarrow y^2+1\ge2y\) (2)\(\left(z-1\right)^2\ge0\Leftrightarrow z^2-2z+1\ge0\Leftrightarrow z^2+1\ge2z\) (3)
Từ (1), (2) và (3) suy ra:
\(x^2+1+y^2+1+z^2+1\ge2x+2y+2z\)
<=> \(x^2+y^2+z^2+3\ge2\left(x+y+z\right)\) \(\xrightarrow[]{}\) đpcm
5. a, Ta có: \(\left(x-1\right)^2\ge0\Leftrightarrow x^2-2x+1\ge0\Leftrightarrow x^2+1\ge2x\) (1)
\(\left(y-1\right)^2\ge0\Leftrightarrow y^2-2y+1\ge0\Leftrightarrow y^2+1\ge2y\) (2)
\(\left(z-1\right)^2\ge0\Leftrightarrow z^2-2z+1\ge0\Leftrightarrow z^2+1\ge2z\) (3)
Từ (1),(2) và (3) suy ra:
\(x^2+1+y^2+1+z^2+1\ge2x+2y+2z\)
<=> \(x^2+y^2+z^2+3\ge2\left(x+y+z\right)\)
mà x+y+z=3
=>\(x^2+y^2+z^2+3\ge2.3=6\)
<=> \(x^2+y^2+z^2\ge6-3=3\)
<=> \(A\ge3\)
Dấu "=" xảy ra khi x=y=z=1
Vậy GTNN của A=x2+y2+z2 là 3 khi x=y=z=1
b, Ta có: x+y+z=3
=> \(\left(x+y+z\right)^2=9\)
<=> \(x^2+y^2+z^2+2xy+2yz+2xz=9\)
<=> \(x^2+y^2+z^2=9-2xy-2yz-2xz\)
mà \(x^2+y^2+z^2\ge3\) (theo a)
=> \(9-2xy-2yz-2xz\ge3\)
<=> \(-2\left(xy+yz+xz\right)\ge3-9=-6\)
<=> \(xy+yz+xz\le\dfrac{-6}{-2}=3\)
<=> \(B\le3\)
Dấu "=" xảy ra khi x=y=z=1
Vậy GTLN của B=xy+yz+xz là 3 khi x=y=z=1
a) \(\dfrac{x^2-y^2}{x^2-y^2+xz-yz}=\dfrac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)\left(x-y\right)+z\left(x-y\right)}\)
\(=\dfrac{\left(x-y\right)\left(x+y\right)}{\left(x-y\right)\left(x+y+z\right)}=\dfrac{x+y}{x+y+z}\)
b) \(\dfrac{x^2+y^2-z^2+2xy}{x^2+z^2-y^2-2xz}=\dfrac{\left(x+y\right)^2-z^2}{\left(x-z\right)^2-y^2}=\dfrac{\left(x+y-z\right)\left(x+y+z\right)}{\left(x-y-z\right)\left(x-z+y\right)}\)\(=\dfrac{x+y+z}{x-y-z}\)
c) \(\dfrac{x^2\left(x-3\right)-\left(x-3\right)}{x\left(x-3\right)}=\dfrac{\left(x-3\right)\left(x^2-1\right)}{x\left(x-3\right)}=\dfrac{x^2-1}{x}\)
d) \(\dfrac{4x^2\left(x-2\right)+3\left(x-2\right)}{4x^2\left(3x+1\right)+3\left(3x+1\right)}=\dfrac{\left(x-2\right)\left(4x^2+3\right)}{\left(3x+1\right)\left(4x^2+3\right)}=\dfrac{x-2}{3x+1}\)
a) \(\left(\frac{1}{3}u+3v\right)^2=\frac{1}{9}u^2+2uv+9v^2\)
b) \(\left(\frac{1}{2}x^2-6x\right)^2=\frac{1}{4}x^4-6x^3+36x^2\)
c) \(\left(-\frac{1}{2}a+b\right)^2=\frac{1}{4}a^2-ab+b^2\)
d) \(\left(-\frac{4}{3}a-\frac{1}{3}b\right)^2=\frac{16}{9}a^2+\frac{8}{9}ab+\frac{1}{9}b^2\)
e) \(\left(\frac{2}{3}x-\frac{3}{2}y\right)\left(\frac{2}{3}x+\frac{3}{2}y\right)=\frac{4}{9}x^2-\frac{9}{4}y^2\)
a) \(\left(\frac{1}{3}u+3v\right)^2=\frac{1}{9}u^2+2uv+9v^2\)
b) \(\left(\frac{1}{2}x^2-6x\right)^2=\frac{1}{4}x^4-6x^3+36x^2\)
c) \(\left(-\frac{1}{2}a+b\right)^2=\frac{1}{4}a^2-ab+b^2\)
d) \(\left(-\frac{4}{3}a-\frac{1}{3}b\right)^2=\frac{16}{9}a^2+\frac{8}{9}ab+\frac{1}{9}b^2\)
e) \(\left(\frac{2}{3}x-\frac{3}{2}y\right)\left(\frac{2}{3}x+\frac{3}{2}y\right)=\left(\frac{2}{3}x\right)^2-\left(\frac{3}{2}y\right)^2=\frac{4}{9}x^2-\frac{9}{4}y^2\)
a) \(\left(x+y\right)^3-x^3-y^3\)
\(=\left(x+y\right)^3-\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-x^2+xy-y^2\right]\)
\(=\left(x+y\right)\left(x^2+2xy+y^2-x^2+xy-y^2\right)\)
\(=3xy\left(x+y\right)\)
b) \(x^2+y^2+2xy+yz+xz\)
\(=\left(x^2+2xy+y^2\right)+\left(yz+xz\right)\)
\(=\left(x+y\right)^2+z\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y+z\right)\)
c) \(x^2-10xy-1+25y^2\)
\(=\left(x^2-10xy+25y^2\right)-1\)
\(=\left(x-5y\right)^2-1\)
\(=\left(x-5y-1\right)\left(x-5y+1\right)\)
d) \(ax^2-ax+bx^2-bx+a+b\)
\(=(ax^2+bx^2)-(ax+bx)+(a+b)\)
\(=x^2(a+b)-x(a+b)+(a+b)\)
\(=(a+b)(x^2-x+1)\)
e)\(x^2-2y+3xz+x-2y+3z\)
\(=(x^2+x)-(2xy+2y)+(3xz+3z)\)
\(=x(x+1)-2y(x-1)+3z(x+1)\)
\(=(x+1)(x-2y+3z)\)
f) \(xyz-xy-yz-xz+x+y+z-1\)
\(=(xyz-xy)-(yz-y)-(xz-x)+(z-1)\)
\(=xy(z-1)-y(z-1)-x(z-1)+(z-1)\)
\(=(z-1)(xy-y-x+1)\)
\(=(z-1)[y(x-1)-(x-1)]\)
\(=(z-1)(x-1)(y-1)\)
_Học tốt_
a) a 2 9 + 8 3 ay + 16 y 2 . b) 1 x 2 − 6 xy + 9 y 2 .
c) x 2 4 − y 2 z 2 36 . d) x 4 − 4 25 y 2 .