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1:

a: \(\left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2zx+2yz\)

b: \(\left(x-y+z\right)^2=x^2+y^2+z^2-2xy+2xz-2yz\)

c: \(\left(x-y-z\right)^2=x^2+y^2+z^2-2xy-2xz+2yz\)

2 tháng 9 2021

Bài 2: tất cả đều ở dạng tích rồi mà

27 tháng 6 2017

Bài 1:

a) -16 +(x-3)2

<=> (x-3)2-16

<=> (x-3)2 -42

<=> (x-3-4)(x-3+4)

<=> (x-7)(x+1)

b) 64+16y+y2

<=> y2 + 2.8.y + 82

<=> (y+8)2

c) \(\dfrac{1}{8}-8x^3\)

\(\Leftrightarrow\left(\dfrac{1}{2}\right)^3-\left(2x\right)^3\)

\(\Leftrightarrow\left(\dfrac{1}{2}-2x\right)\left(\dfrac{1}{4}+x+4x^2\right)\)

d)\(x^2-x+\dfrac{1}{4}\)

\(\Leftrightarrow x^2-2.\dfrac{1}{2}.x+\left(\dfrac{1}{2}\right)^2\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2\)

e) x4 + 4x2 + 4

<=> (x2)2 + 2.2.x2 +22

<=> (x2 + 2)2

g)\(8x^3+60x^2y+150xy^2+125y^3\)

\(\Leftrightarrow\left(2x+5y\right)^3\)

28 tháng 6 2017

Ban giup minh bai 2 luon voi nha Hậu Trần Công

Bài 1:

a) \(\left(a-b^2\right)\left(a+b^2\right)=a^2-b^4\)

b) \(\left(a^2+2a-3\right)\left(a^2+2a+3\right)=\left(a^2+2a\right)^2-9\)

c) \(\left(a^2+2a+3\right)\left(a^2-2a-3\right)=a^2-\left(2a+3\right)^2\)

d) \(\left(a^2-2a+3\right)\left(a^2+2a+3\right)=9-\left(a^2-2a\right)^2\)

e) \(\left(-a^2-2a+3\right)\left(-a^2-2a+3\right)=\left(-a^2-2a+3\right)^2\)

g) \(\left(a^2+2a+3\right)\left(a^2-2a+3\right)=\left(a^2+3\right)^2-4a^2\)

f) \(\left(a^2+2a\right)\left(2a-a^2\right)=4a^2-a^4\)

Bài 2 :

a) \(\left(x+1\right)\left(x^2-x+1\right)=x^3+1\)

b) \(\left(x+y+z\right)^2=\left(x+y+z\right)\left(x+y+z\right)=x^2+xy+xz+yx+y^2+yz+zx+zy+z^2=x^2+2xy+2yz+2xz+y^2+z^2\)

c) \(\left(x-y+z\right)^2=\left(x-y+z\right)\left(x-y+z\right)=x^2-xy+xz-xy+y^2-yz+xz-yz+z^2=x^2+y^2+z^2-2xy+2xz-2yz\)d) \(\left(x-2y\right)\left(x^2+2xy+4y^2\right)=\left(x-2y\right)^3\)

e) \(\left(x-y-z\right)^2=\left(x-y-z\right)\left(x-y-z\right)=x^2-xy-xz-xy+y^2+yz-xz+yz+z^2=x^2-2xy-2xz+2yz+y^2+z^2\)

9 tháng 12 2021

\(1,Q=\dfrac{a^4-2a^2+a^3-2a+a^2-2}{a^4-2a^2+2a^3-4a+a^2-2}\\ Q=\dfrac{\left(a^2-2\right)\left(a^2+a+1\right)}{\left(a^2-2\right)\left(a^2+2a+1\right)}=\dfrac{a^2+a+1}{a^2+2a+1}\)

\(Q=\dfrac{x^2+x+1}{\left(x+1\right)^2}-\dfrac{3}{4}+\dfrac{3}{4}=\dfrac{x^2+x+1-\dfrac{3}{4}x^2-\dfrac{3}{2}x-\dfrac{3}{4}}{\left(x+1\right)^2}+\dfrac{3}{4}\\ Q=\dfrac{\dfrac{1}{4}x^2-\dfrac{1}{2}x+\dfrac{1}{4}}{\left(x+1\right)^2}+\dfrac{3}{4}=\dfrac{\dfrac{1}{4}\left(x-1\right)^2}{\left(x+1\right)^2}+\dfrac{3}{4}\ge\dfrac{3}{4}\\ Q_{min}=\dfrac{3}{4}\Leftrightarrow x=1\)

9 tháng 12 2021

\(2,\text{Từ GT }\Leftrightarrow\dfrac{ayz+bxz+czy}{xyz}=0\\ \Leftrightarrow ayz+bxz+czy=0\\ \text{Ta có }\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\\ \Leftrightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{zx}{ca}\right)=0\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\cdot\dfrac{cxy+ayz+bzx}{abc}=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\cdot\dfrac{0}{abc}=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)

a: \(\left(a^2+2a+3\right)\left(a^2-2a-3\right)\)

\(=\left[a^2+\left(2a+3\right)\right]\left[a^2-\left(2a+3\right)\right]\)

\(=\left(a^2\right)^2-\left(2a+3\right)^2\)

\(=a^4-\left(2a+3\right)^2\)

b: \(\left(-a^2-2a+3\right)^2\)

\(=\left(a^2+2a-3\right)^2\)

\(=a^4+4a^2+9+4a^3-18a-6a^2\)

\(=a^4+4a^3-2a^2-18a+9\)

c: \(\left(x-y-z\right)^2\)

\(=x^2-2x\left(y+z\right)+\left(y+z\right)^2\)

\(=x^2-2xy-2xz+y^2+2yz+z^2\)

d: \(\left(x+y+z\right)\left(x-y-z\right)\)

\(=x^2-\left(y+z\right)^2\)

\(=x^2-y^2-2yz-z^2\)

1 tháng 8 2017

a, \(\left(a-b\right)\left(b-a\right)y+a-b\)

\(=\left(a-b\right)\left(by-ay+1\right)\)

b, \(\left(x-y+z\right)\left(a+y-x-z\right)b-x+y-z\)

\(=\left(x-y+z\right)\left(a+y-x-z\right)b-\left(x-y+z\right)\)

\(=\left(x-y+z\right)\left(ab+yb-xb-xz-1\right)\)

c, \(\left(2a+3\right)x-\left(2a+3\right)y+2a+3\)

\(=\left(2a+3\right)\left(x-y\right)+\left(2a+3\right)\)

\(=\left(2a+3\right)\left(x-y+1\right)\)

d, \(\left(a-b\right)x+\left(b-a\right)y-a+b\)

\(=\left(a-b\right)x-\left(a-b\right)y-\left(a-b\right)\)

\(=\left(a-b\right)\left(x-y-1\right)\)

2 tháng 8 2017

a. (a-b)(b-a)y+a-b=(a-b)[(b-a)y+1]

b. (x-y+z)(a+y-x-z)b-x+y-z=(x-y+z)(a-y-x-z)b-(x-y+z)=(x-y+z)[(a-y-x-z)b-1]

c. (2a+3)x-(2a+3)y+2a+3=(2a+3)(x-y+1)

d. (a-b)x+(b-a)y-a+b=(a-b)x-(a-b)y-(a-b)=(a-b)(x-y-1)