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\(a)x^5+x^4+1\)
\(=x^3\left(x^2+x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^3-x+1\right)\)
\(b)x^8+x^7+1\)
\(=\left(x^8-x^2\right)+\left(x^7-x\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)
\(#Tuyết\)
`b)x^3+y^3+z^3-3xyz`
`=x^3+3xy(x+y)+z^3-3xy(x+y)-3xyz`
`=(x+y)^3+z^3-3xy(x+y+z)`
`=(x+y+z)[(x+y)^2-z(x+y)+z^2]-3xy(x+y)`
`=(x+y+z)(x^2+2xy+y^2-zx-yz-3xy+z^2)`
`=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)`
a: \(=5a\left(x-2y\right)\)
b: \(=x\left(x-y\right)+\left(x-y\right)=\left(x-y\right)\left(x+1\right)\)
c: =(x-1)(x-7)
a)\(5ax-10ay=5a\left(x-2y\right)\)
b) \(x^2-xy+x-y=x\left(x-y\right)+\left(x-y\right)=\left(x+1\right)\left(x-y\right)\)
c) \(x^2-8x+7=\left(x-7\right)\left(x-1\right)\)
Câu 1:
a: Sửa đề: \(A=\left(x+2\right)\left(x^2-2x+4\right)+x\left(1-x\right)\left(1+x\right)\)
\(=x^3+2^3+x\left(1-x^2\right)\)
\(=x^3+8+x-x^3\)
=x+8
b: Khi x=-4 thì A=-4+8=4
c: Đặt A=-2
=>x+8=-2
=>x=-10
Câu 2:
a: \(x^3-3x^2=x^2\cdot x-x^2\cdot3=x^2\left(x-3\right)\)
b: \(5x^3+10x^2+5x\)
\(=5x\cdot x^2+5x\cdot2x+5x\cdot1\)
\(=5x\left(x^2+2x+1\right)\)
\(=5x\left(x+1\right)^2\)
Bài 1:
a: \(8x^3-2x=2x\left(4x^2-1\right)=2x\left(2x-1\right)\left(2x+1\right)\)
c: \(-5m^3\left(m+1\right)+m+1=\left(m+1\right)\left(-5m^3+1\right)\)
Câu 2:
a: \(\left(x-1\right)\left(x+1\right)-x\left(x+3\right)+7=0\)
=>\(x^2-1-x^2-3x+7=0\)
=>-3x+6=0
=>-3x=-6
=>\(x=\dfrac{-6}{-3}=2\)
b: \(2x^3-22x^2+36x=0\)
=>\(2x\left(x^2-11x+18\right)=0\)
=>\(x\left(x^2-11x+18\right)=0\)
=>\(x\left(x^2-2x-9x+18\right)=0\)
=>\(x\left[x\left(x-2\right)-9\left(x-2\right)\right]=0\)
=>\(x\left(x-2\right)\left(x-9\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\x-2=0\\x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=9\end{matrix}\right.\)
Câu 4:
1: Diện tích cỏ cần thay là:
\(105\cdot68=7140\left(m^2\right)\)
Số tiền BQL sân cần trả là:
\(7140\cdot120000=856800000\left(đồng\right)\)
2:
a: Xét tứ giác ABDC có
M là trung điểm chung của AD và BC
=>ABDC là hình bình hành
Hình bình hành ABDC có \(\widehat{BAC}=90^0\)
nên ABDC là hình chữ nhật
b: Xét ΔADE có
H,M lần lượt là trung điểm của AE,AD
=>HM là đường trung bình của ΔADE
=>HM//DE
=>BC//DE
=>\(\widehat{EDB}=\widehat{DBM}\)(hai góc so le trong)(1)
Ta có: ABDC là hình chữ nhật
=>AD=BC
mà \(MD=\dfrac{AD}{2};MB=\dfrac{BC}{2}\)
nên MD=MB
=>ΔMBD cân tại M
=>\(\widehat{MDB}=\widehat{MBD}\left(2\right)\)
Từ (1) và (2) suy ra \(\widehat{MDB}=\widehat{EDB}\)
=>\(\widehat{ADB}=\widehat{EDB}\)
=>DB là phân giác của góc ADE
a) Sửa đề: \(a^2x+a^2y-7x-7y\)
\(=a^2\left(x+y\right)-7\left(x+y\right)=\left(x+y\right)\left(a^2-7\right)\)
b) \(=\left(2x-3y\right)\left(2x+3y\right)+2\left(2x-3y\right)=\left(2x-3y\right)\left(2x+3y+2\right)\)
\(c,Sửa:x^2-2x+2y-y^2=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)=\left(x-y\right)\left(x+y-2\right)\\ d,=\left(4x^4+36x^2+81\right)-36x^2\\ =\left(2x^2+9\right)^2-36x^2=\left(2x^2-6x+9\right)\left(2x^2+6x+9\right)\\ e,=x^7+x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x^2+x-x+1\\ =x^5\left(x^2+x+1\right)-x^4\left(x^2+x+1\right)+x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+1\right)\)
a) \(4x^2-1\)
\(=\left(2x\right)^2-1^2\)
\(=\left(2x-1\right)\left(2x+1\right)\)
b) \(x^2-3y^2\)
\(=x^2-\left(y\sqrt{3}\right)^2\)
\(=\left(x-y\sqrt{3}\right)\left(x+y\sqrt{3}\right)\)
c) \(9x^2-\dfrac{1}{4}\)
\(=\left(3x\right)^2-\left(\dfrac{1}{2}\right)^2\)
\(=\left(3x-\dfrac{1}{2}\right)\left(3x+\dfrac{1}{2}\right)\)
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) \(\left(x^2+4\right)^2-16x^2\)
\(=\left(x^2+4\right)^2-\left(4x\right)^2\)
\(=\left(x^2-4x+4\right)\left(x^2+4x+4\right)\)
\(=\left(x-2\right)^2\left(x+2\right)^2\)
1A:
a: \(x^3+2x=x\left(x^2+2\right)\)
b: \(3x-6y=3\left(x-2y\right)\)
c: \(5\left(x+3y\right)-15x\left(x+3y\right)\)
\(=5\left(x+3y\right)\left(1-3x\right)\)
d: \(3\left(x-y\right)-5x\left(y-x\right)\)
\(=3\left(x-y\right)+5x\left(x-y\right)\)
\(=\left(x-y\right)\left(5x+3\right)\)
1A. a. x(x2+2)
b. 3(x-2y)
c. 5(x+3y)(1-3x)
d. (x-y) (3-5x)
1B. a. 2x(2x-3)
b.xy(x2-2xy+5)
c. 2x(x+1)(x+2)
d. 2x(y-1)+2y(y-1)=2(y-1)(x-y)
\(x^7+x^2+1\)
\(=x^7+x^6+x^5+x^4+x^3+x^2+x+1\)
\(=x^5\left(x^2+x+1\right)-x^4\left(x^2+x+1\right)+x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+1\right)\)
a) \(x^7+x^2+1=\left(x^7-x\right)+\left(x^2+x+1\right)\)
\(=x\left(x^6-1\right)+\left(x^2+x+1\right)=x\left(x^3-1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left[x\left(x-1\right)\left(x^3+1\right)+1\right]\)
\(=\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+1\right)\)
b) \(x^7+x^5+1=\left(x^7+x^6+x^5\right)-\left(x^6-1\right)\)
\(=x^5\left(x^2+x+1\right)-\left(x^3-1\right)\left(x^3+1\right)\)
\(=x^5\left(x^2+x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)\)
\(=\left(x^2+x+1\right)\left[x^5-\left(x-1\right)\left(x^3+1\right)\right]\)
\(=\left(x^2+x+1\right)\left(x^5-x^4+x^3-x+1\right)\)