Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1 , \(x^5+x^4+1=\left(x^5+x^4+x^3\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\)
= \(x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)=\(\left(x^2+x+1\right)\left(x^3-x+1\right)\)
2 , \(x\left(x+4\right)\left(x+6\right)\left(x+10\right)+128=\left(x^2+10x\right)\left(x^2+10x+24\right)+128\)(*)
Đặt x2 + 10 = a , a>0 (1)
=> (*) <=> a(a+24)+128=a2 + 24a+128=(a+8)(a+16) (**)
Thay (1) vào (**) ta được :
(*) <=> \(\left(x^2+10+8\right)\left(x^2+10+16\right)\)
a) \(2\left(x+y\right)^2-7\left(x+y\right)+5\)
\(=2\left(x+y\right)^2-2\left(x+y\right)-5\left(x+y\right)+5\)
\(=2\left(x+y\right)\left(x+y-1\right)-5\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left(2x+2y-5\right)\)
b) c) tương tự
b, \(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3\)
\(=\left(x-y\right)^2\left(x-y\right)-\left(y-z\right)^2\left[\left(x-y\right)+\left(z-x\right)\right]+\left(z-x\right)^2\left(z-x\right)\)
\(=\left(x-y\right)^2\left(x-y\right)-\left(y-z\right)^2\left(x-y\right)-\left(y-z\right)^2\left(z-x\right)+\left(z-x\right)^2\left(z-x\right)\)
\(=\left(x-y\right)\left[\left(x-y\right)^2-\left(y-z\right)^2\right]-\left(z-x\right)\left[\left(y-z\right)^2-\left(z-x\right)^2\right]\)
\(=\left(x-y\right)\left(x-y-y+z\right)\left(x-y+y-z\right)-\left(z-x\right)\left(y-z-z+x\right)\left(y-z+z-x\right)\)
\(=\left(x-y\right)\left(x-2y+z\right)\left(x-z\right)-\left(z-x\right)\left(y-2z+x\right)\left(y-x\right)\)
\(=\left(x-y\right)\left(x-2y+z\right)\left(x-z\right)-\left(x-z\right)\left(y-2z+x\right)\left(x-y\right)\)
\(=\left(x-y\right)\left(x-z\right)\left(x-2y+z-y+2z-x\right)\)
\(=\left(x-y\right)\left(x-z\right)\left(3z-3y\right)\)
\(=3\left(x-y\right)\left(x-z\right)\left(z-y\right)\)
c, \(x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-z^2x^2\left(z-x\right)\)
\(=x^2y^2\left(y-x\right)-y^2z^2\left[\left(y-x\right)-\left(z-x\right)\right]-z^2x^2\left(z-x\right)\)
\(=x^2y^2\left(y-x\right)-y^2z^2\left(y-x\right)+y^2z^2\left(z-x\right)-z^2x^2\left(z-x\right)\)
\(=\left(x^2y^2-y^2z^2\right)\left(y-x\right)+\left(y^2z^2-z^2x^2\right)\left(z-x\right)\)
\(=y^2\left(x-z\right)\left(x+z\right)\left(y-x\right)+z^2\left(y-x\right)\left(x+y\right)\left(z-x\right)\)
\(=y^2\left(x-z\right)\left(x+z\right)\left(y-x\right)-z^2\left(y-x\right)\left(x+y\right)\left(x-z\right)\)
\(=\left(x-z\right)\left(y-x\right)\left[y^2\left(x+z\right)-z^2\left(x+y\right)\right]\)
\(=\left(x-z\right)\left(y-x\right)\left(y^2x+y^2z-z^2x-z^2y\right)\)
\(=\left(x-z\right)\left(y-x\right)\left[x\left(y^2-z^2\right)+yz\left(y-z\right)\right]\)
\(=\left(x-z\right)\left(y-x\right)\left[x\left(y-z\right)\left(y+z\right)+yz\left(y-z\right)\right]\)
\(=\left(x-z\right)\left(y-x\right)\left(y-z\right)\left(xy+xz+yz\right)\)
d, \(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3+z^3-3xyz-3xy\left(x+y\right)\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
Bài 1:
a) \(3x^2-2x(5+1,5x)+10=3x^2-(10x+3x^2)+10\)
\(=10-10x=10(1-x)\)
b) \(7x(4y-x)+4y(y-7x)-2(2y^2-3,5x)\)
\(=28xy-7x^2+(4y^2-28xy)-(4y^2-7x)\)
\(=-7x^2+7x=7x(1-x)\)
c)
\(\left\{2x-3(x-1)-5[x-4(3-2x)+10]\right\}.(-2x)\)
\(\left\{2x-(3x-3)-5[x-(12-8x)+10]\right\}(-2x)\)
\(=\left\{3-x-5[9x-2]\right\}(-2x)\)
\(=\left\{3-x-45x+10\right\}(-2x)=(13-46x)(-2x)=2x(46x-13)\)
Bài 2:
a) \(3(2x-1)-5(x-3)+6(3x-4)=24\)
\(\Leftrightarrow (6x-3)-(5x-15)+(18x-24)=24\)
\(\Leftrightarrow 19x-12=24\Rightarrow 19x=36\Rightarrow x=\frac{36}{19}\)
b)
\(\Leftrightarrow 2x^2+3(x^2-1)-5x(x+1)=0\)
\(\Leftrightarrow 2x^2+3x^2-3-5x^2-5x=0\)
\(\Leftrightarrow -5x-3=0\Rightarrow x=-\frac{3}{5}\)
\(2x^2+3(x^2-1)=5x(x+1)\)
1. Câu hỏi của Nguyễn Mai - Toán lớp 9 - Học trực tuyến OLM
3.
\(a,A=n^3-n+7=n\left(n-1\right)\left(n+1\right)+7\)
Có \(\left(n-1\right)n\left(n+1\right)\) là tích 3 số tự nhiên lt với \(n\in N\) nên chia hết cho 6
Mà 7 ko chia hết cho 6 nên A không chia hết cho 6
\(b,B=n^3-n=n\left(n-1\right)\left(n+1\right)\)
Như câu a thì B chia hết cho 6 hay B chia hết cho 3
Ta thấy n lẻ nên \(n=2k+1\left(k\in N\right)\)
\(\Rightarrow B=n^3-n=\left(n-1\right)n\left(n+1\right)\\ =\left(2k+1-1\right)\left(2k+1\right)\left(2k+1+1\right)\\ =2k\left(2k+1\right)\left(2k+2\right)\\ =4k\left(k+1\right)\left(2k+1\right)\)
Mà k+1 và 2k+1 là 2 số tự nhiên lt nên chia hết cho 2
\(\Rightarrow B⋮4\cdot2\left(2k+1\right)=8\left(2k+1\right)⋮8\)
Vì B chia hết cho cả 3;8 và \(\left(3;8\right)=1\) nên B chia hết 24
\(c,C=n^4+6n^3+11n^2+6n=n\left(n+1\right)\left(n+2\right)\left(n+3\right)\)
Ta thấy đây là 4 số tự nhiên lt với \(n\in N\) nên chia hết cho 24
a) \(\left(x+y\right)^5-x-y=\left(x+y\right)^5-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^4-1\right]\)
= \(\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\) #áp dụng hàng đẳng thức#
c) \(x^9-x^7-x^6-x^5+x^4+x^3+x^2+1\)nhóm vào là đc
b) \(\left(x^2+y^2\right)^3+\left(z^2-x^2\right)^3+\left(y^2+z^2\right)^3\)
=\(\left(y^2+x^2\right)\left[\left(x^2+y^2\right)^2-\left(x^2+y^2\right)\left(z^2-x^2\right)+\left(z^2-x^2\right)^2\right]+\left(y^2+z^2\right)^3\)
= \(\left(y^2+z^2\right)\left[x^4+y^4+2x^2y^2-x^2z^2+x^4-y^2z^2+x^2y^2+z^4+x^4-2x^2z^2+y^4+z^4+2y^2z^2\right]\)
=\(=\left(y^2+z^2\right)\left(2x^4+2y^4+2z^4+3x^2y^2-3x^2z^2+y^2z^2\right)\)
Câu a:
Xét tử số:
\(x^2(y-z)+y^2(z-x)+z^2(x-y)\)
\(=x^2(y-z)-y^2[(y-z)+(x-y)]+z^2(x-y)\)
\(=x^2(y-z)-y^2(y-z)-y^2(x-y)+z^2(x-y)\)
\(=(x^2-y^2)(y-z)-(y^2-z^2)(x-y)\)
\(=(x-y)(y-z)[(x+y)-(y+z)]=(x-y)(y-z)(x-z)\)
Xét mẫu số:
\(x^2y-x^2z+y^2z-y^3=x^2(y-z)-y^2(y-z)=(x^2-y^2)(y-z)\)
\(=(x-y)(x+y)(y-z)\)
Do đó:
\(\frac{x^2(y-z)+y^2(z-x)+z^2(x-y)}{x^2y-x^2z+y^2z-y^3}=\frac{(x-y)(y-z)(x-z)}{(x-y)(x+y)(y-z)}=\frac{x-z}{x+y}\)
Câu b:
Xét tử số:
\(x^5+x+1=x^5-x^2+x^2+x+1=x^2(x^3-1)+x^2+x+1\)
\(=x^2(x-1)(x^2+x+1)+(x^2+x+1)\)
\(=(x^2+x+1)(x^3-x^2+1)\)
Xét mẫu số:
\(x^3+x^2+x=x(x^2+x+1)\)
Do đó: \(\frac{x^5+x+1}{x^3+x^2+1}=\frac{(x^2+x+1)(x^3-x^2+1)}{x(x^2+x+1)}=\frac{x^3-x^2+1}{x}\)
Vì bài dài nên mình sẽ tách ra nhé.
1a. Ta có:
$x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)=-2(xy+yz+xz)$
$x^3+y^3+z^3=(x+y+z)^3-3(x+y)(y+z)(x+z)=-3(x+y)(y+z)(x+z)$
$=-3(-z)(-x)(-y)=3xyz$
$\Rightarrow \text{VT}=-30xyz(xy+yz+xz)(1)$
------------------------
$x^5+y^5=(x^2+y^2)(x^3+y^3)-x^2y^2(x+y)$
$=[(x+y)^2-2xy][(x+y)^3-3xy(x+y)]-x^2y^2(x+y)$
$=(z^2-2xy)(-z^3+3xyz)+x^2y^2z$
$=-z^5+3xyz^3+2xyz^3-6x^2y^2z+x^2y^2z$
$=-z^5+5xyz^3-5x^2y^2z$
$\Rightarrow 6(x^5+y^5+z^5)=6(5xyz^3-5x^2y^2z)$
$=30xyz(z^2-xy)=30xyz[z(-x-y)-xy]=-30xyz(xy+yz+xz)(2)$
Từ $(1);(2)$ ta có đpcm.
1b.
$x^4+y^4=(x^2+y^2)^2-2x^2y^2=[(x+y)^2-2xy]^2-2x^2y^2$
$=(z^2-2xy)^2-2x^2y^2=z^4+2x^2y^2-4xyz^2$
$x^3+y^3=(x+y)^3-3xy(x+y)=-z^3+3xyz$
Do đó:
$x^7+y^7=(x^4+y^4)(x^3+y^3)-x^3y^3(x+y)$
$=(z^4+2x^2y^2-4xyz^2)(-z^3+3xyz)+x^3y^3z$
$=7x^3y^3z-14x^2y^2z^3+7xyz^5-z^7$
$\Rightarrow \text{VT}=7x^3y^3z-14x^2y^2z^3+7xyz^5$
$=7xyz(x^2y^2-2xyz^2+z^4)$
$=7xyz(xy-z^2)$
$=7xyz[xy+z(x+y)]^2=7xyz(xy+yz+xz)^2$
$=7xyz[x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)]$
$=7xyz(x^2y^2+y^2z^2+z^2x^2)$ (đpcm)