Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A-B
\(=2x^3-4x^2y+131xy^2-y^4+1-\left(-2x^3-121x^2y-y^4-3\right)\)
\(=2x^3-4x^2y+131xy^2-y^4+1+2x^3+121x^2y+y^4+3\)
\(=4x^3+117x^2y+131xy^2+4\)
a) \(4x^2-6x=2x\left(2x-3\right)\)
b) \(9x^4y^3+3x^2y^4=3x^2y^3\left(3x^2+y\right)\)
c) \(3\left(x-y\right)-5x\left(y-x\right)=3\left(x-y\right)+5x\left(x-y\right)\)
\(=\left(5x+3\right)\left(x-y\right)\)
d) \(x^3-2x^2+5x=x\left(x^2-2x+5\right)\)
e) \(5\left(x+3y\right)-15x\left(x+3y\right)=\left(5-15x\right)\left(x+3y\right)\)
\(=5\left(1-3x\right)\left(x+3y\right)\)
f) \(2x^2\left(x+1\right)-4\left(x+1\right)=\left(2x^2-4\right)\left(x+1\right)\)
\(=\left(\sqrt{2}x-2\right)\left(\sqrt{2}x+2\right)\left(x+1\right)\)
3/
a/ \(A=\left(x-y\right)^2+\left(x+y\right)^2.\)
\(A=\left(x^2-2xy+y^2\right)+\left(x^2+2xy+y^2\right)\)
\(A=x^2-2xy+y^2+x^2+2xy+y^2\)
\(A=2x^2+2y^2\)
b/ \(B=\left(2a+b\right)^2-\left(2a-b\right)^2\)
\(B=\left(4a^2+4ab+b^2\right)-\left(4a^2-4ab+b^2\right)\)
\(B=4a^2+4ab+b^2-4a^2+4ab-b^2\)
\(B=8ab\)
c/ \(C=\left(x+y\right)^2-\left(x-y\right)^2\)
\(C=\left(x^2+2xy+y^2\right)-\left(x^2-2xy+y^2\right)\)
\(C=x^2+2xy+y^2-x^2+2xy-y^2\)
\(C=4xy\)
d/ \(D=\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(D=\left(4x^2-4x+1\right)-2\left(4x^2-12x+9\right)+4\)
\(D=4x^2-4x+1-8x^2+24x-18+4\)
\(D=-4x^2+20x-13\)
- Toán lớp 8
Khóa học Toán lớp 8
Gửi câu trả lời của bạn
\(\text{a)}x^3-6x^2+12x-8\)
\(=x^3-2x^2-4x^2+8x+4x-8\)
\(=\left(x^3-2x^2\right)-\left(4x^2-8x\right)+\left(4x-8\right)\)
\(=x^2\left(x-2\right)+4x\left(x-2\right)+4\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+4x+4\right)\)
\(=\left(x-2\right)\left(x+2\right)^2\)
\(\text{b)}8x^2+12x^2y+6xy^2+y^3=\left(2x+y\right)^3\)
Bài 2:
\(\text{a) }x^7+1=\left(x^{\frac{7}{3}}\right)^3+1^3=\left(x^{\frac{7}{3}}+1\right)\left[\left(x^{\frac{7}{3}}\right)^2-x^{\frac{7}{3}}+1\right]=\left(x^{\frac{7}{3}}+1\right)\left(x^{\frac{14}{3}}-x^{\frac{7}{3}}+1\right)\)
\(\text{b) }x^{10}-1=\left(x^5\right)^2-1^2=\left(x^5-1\right)\left(x^5+1\right)\)
Bài 3:
\(\text{a) }69^2-31^2=\left(69-31\right)\left(69+31\right)=38.100=3800\)
\(\text{b) }1023^2-23^2=\left(1023-23\right)\left(1023+23\right)=1000.1046=1046000\)
Bài 1
\(a,5x^2-10xy+5y^2\)
\(=5\cdot\left(x^2-2xy+y^2\right)\)
\(=5\cdot\left(x-y\right)^2\)
\(b,x^2-y^2+6y-9\)
\(=x^2-\left(y^2-6y+9\right)\)
\(=x^2-\left(y-3\right)^2\)
\(=\left(x-y+3\right)\cdot\left(x+y-3\right)\)
\(c,3x^4-75x^2y^2\)
\(=3x^2\cdot\left(x^2-25y^2\right)\)
\(=3x^2\cdot\left(x-5y\right)\cdot\left(x+5y\right)\)
\(d,x^4y+xy^4\)
\(=xy\left(x^3+y^3\right)\)
\(=xy\cdot\left(x+y\right)\cdot\left(x^2-xy+y^2\right)\)
Bài 2:
a) \(x^2-y^2+3x-3y=\left(x^2-y^2\right)+\left(3x-3y\right)\)
\(=\left(x-y\right)\left(x+y\right)+3\left(x-y\right)=\left(x-y\right)\left(x+y+3\right)\)
b) \(5x-5y+x^2-2xy+y^2=\left(5x-5y\right)+\left(x^2-2xy+y^2\right)\)
\(=5\left(x-y\right)+\left(x-y\right)^2=\left(x-y\right)\left(x-y+5\right)\)
c) \(x^2-5x+4=x^2-x-4x+4=\left(x^2-x\right)-\left(4x-4\right)\)
\(=x\left(x-1\right)-4\left(x-1\right)=\left(x-1\right)\left(x-4\right)\)
a. \(8x\left(x-2017\right)-2x+4034=0\)
\(8x\left(x-2017\right)-2\left(x-2017\right)=0\)
\(\left(8x-2\right)\left(x-2017\right)=0\)
\(\Rightarrow TH1:8x-2=0\)
\(8x=2\)
\(x=\frac{1}{4}\)
\(TH2:x-2017=0\)
\(x=2017\)
Vậy \(x\in\left\{\frac{1}{4};2017\right\}\)
Bài 1
a) \(8x\left(x-2017\right)-2x+4034=0\)
\(\Rightarrow8x\left(x-2017\right)-2\left(x-2017\right)=0\)
\(\Rightarrow\left(x-2017\right)\left(4x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=2017\\x=\frac{1}{4}\end{cases}}\)
a) \(x^4+2x^3-4x-4=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)\)
\(=\left(x^2+x\right)^2-\left(x+2\right)^2=\left(x^2+x-x-2\right)\left(x^2+x+x+2\right)\)
\(=\left(x^2-2\right)\left(x^2+2x+2\right)\)
a) Ta có: \(x^4+2x^3-4x-4\)
\(=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)\)
\(=\left(x^2+x\right)^2-\left(x+2\right)^2\)
\(=\left(x^2+x-x-2\right)\left(x^2+x+x+2\right)\)
\(=\left(x^2-2\right)\cdot\left(x^2+2x+2\right)\)
d) Ta có: \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
c) Ta có: \(x^2+y^2-x^2y^2+xy-x-y\)
\(=\left(x^2-x\right)+\left(y^2-y\right)+\left(-x^2y^2+xy\right)\)
\(=x\left(x-1\right)+y\left(y-1\right)+xy\left(-xy+1\right)\)
\(=\left(x+y+xy\right)\left(x-1+y-1+xy+1\right)\)