Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=x^3-y^3-21xy\)
\(A=\left(x-y\right).\left(x^2+xy+y^2\right)-21xy\)
\(A=7.\left(x^2+xy+y^2\right)-21xy\)
\(A=7.\left(x^2+xy+y^2+3xy\right)\)
\(A=7.\left(x^2+2xy+y^2+2xy\right)\)
\(A=7.\text{[}\left(x+y\right)^2+2xy\text{]}\)
\(A=7.\left(7^2+2xy\right)\)
\(A=7^3+14xy\)
Ngáo rồi @@
\(\)
\(A=x^3-y^3-21xy\)
\(\Rightarrow A=\left(x-y\right)\left(x^2+xy+y^2\right)-21xy\)
\(\Rightarrow A=7\left(x^2+xy+y^2\right)-21xy\)
\(\Rightarrow A=7\left(x^2+xy+y^2-3xy\right)\)
\(\Rightarrow A=7\left(x^2+y^2-2xy\right)\)
\(\Rightarrow A=7\left(x-y\right)^2\)
\(\Rightarrow A=7.7^2\)
\(\Rightarrow A=7.49\)
\(\Rightarrow A=343\)
a)
\(A=\left(x+3\right)\left(x^2-3x+9\right)-\left(54+x^3\right)\)
\(=x^3-3x^2+9x+3x^2-9x+27-54-x^3\)
\(=-27\)
or
\(A=x^3+27-54-x^3=-27\)
b)
\(B=\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-8x^3+y^3=2y^3\)
c)
\(C=\left(2x+1\right)^2+\left(1-3x\right)^2+2\left(2x+1\right)\left(3x-1\right)\)
\(=\left(2x+1+3x-1\right)^2=\left(5x\right)^2=25x^2\)
d)
\(D=\left(x-2\right)\left(x^2+2x+4\right)-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)
\(=x^3-8-\left(x-1\right)^3+3\left(x-1\right)\left(x+1\right)\)
\(=6x^2-3x-10\)
143. a) \(-6x^n.y^n.\left(-\dfrac{1}{18}x^{2-n}+\dfrac{1}{72}y^{5-n}\right)\)
\(=-6.\left(-\dfrac{1}{18}\right)x^n.x^{2-n}.y^n+\left(-6\right).\dfrac{1}{27}x^n.y^n.y^{5-n}\)
\(=\dfrac{1}{3}x^{n+2-n}y^n-\dfrac{2}{9}x^n.y^{n+5-n}\)
\(=\dfrac{1}{3}x^2y^n-\dfrac{2}{9}x^ny^5\)
b) Ta có: \(\left(5x^2-2y^2-2xy\right)\left(-xy-x^2+7y^2\right)\)
\(=5x^2\left(-xy\right)+5x^2.\left(-x^2\right)+5x^2.7y^2-2y^2.\left(-xy\right)-2y^2.\left(-x^2\right)-2y^2.7y^2-2xy.\left(-xy\right)-2xy\left(-x^2\right)-2xy.7y^2\)
\(=-5x^3y-5x^4+35x^2y^2+2xy^3+2x^2y^2-14y^4+2x^2y^2+2x^3y-14xy^3\)
Rút gọn các đa thức đồng dạng, ta có kết quả:
\(-5x^4-3x^3y+39x^2y^2-12xy^3-14y^4\)
Kết quả đã được xếp theo lũy thừa giảm dần của x
1.a, VT= \(\left(x^2+y^2\right)^2-\left(2xy\right)^2=\)\(\left(x^2+y^2-2xy\right)\left(x^2+y^2+2xy\right)=\left(x-y\right)^2\left(x+y\right)^2=VP.\left(đpcm\right)\)
b, VP=\(x\left(x-3y\right)^2+y\left(y-3x\right)^2\)\(=x\left(x^2-6xy+9y^2\right)+y\left(y^2-6xy+9x^2\right)\)\(=x^3-6x^2y+9xy^2+y^3-6xy^2+9x^2y\)
\(=x^3+3x^2y+3xy^2+y^3\)\(=\left(x+y\right)^3=VT\left(đpcm\right)\)
2. VT=\(\left(a+b\right)^3-\left(a-b\right)^3\)\(=\left(a+b-a+b\right)\left(a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2\right)\)
\(2b\left(b^2+3a^2\right)\)\(=VP\left(đpcm\right)\).
a) (x2 + y2)2 - (2xy)2
= [(x2 + y2) - 2xy].[(x2 + y2) + 2xy]
= [x2 + y2 - 2xy].[(x2 + y2 + 2xy]
= (x - y)2 . (x + y)2
a: \(\Leftrightarrow x\cdot\dfrac{1}{4}=\dfrac{1}{2}+\dfrac{1}{9}=\dfrac{11}{18}\)
hay \(x=\dfrac{11}{18}:\dfrac{1}{4}=\dfrac{11}{18}\cdot4=\dfrac{44}{18}=\dfrac{22}{9}\)
d: =>x+1;x-2 khác dấu
Trường hợp 1: \(\left\{{}\begin{matrix}x+1>0\\x-2< 0\end{matrix}\right.\Leftrightarrow-1< x< 2\)
Trường hợp 2: \(\left\{{}\begin{matrix}x+1< 0\\x-2>0\end{matrix}\right.\Leftrightarrow2< x< -1\left(loại\right)\)
e: =>x-2>0 hoặc x+2/3<0
=>x>2 hoặc x<-2/3
a: \(=x^2+2x+y^2-2y-2xy+37\)
\(=\left(x-y\right)^2+2\left(x-y\right)+37\)
\(=7^2+2\cdot7+37=49+14+37=100\)
b: \(=x^3+x^2-y^3+y^2+xy-3xy\cdot\left(7+1\right)-95\)
\(=\left(x^3-y^3\right)+\left(x^2+y^2\right)+xy-24xy-95\)
\(=\left(x-y\right)^3+3xy\left(x-y\right)+\left(x-y\right)^2+2xy-23xy-95\)
\(=7^3+3xy\cdot7+49-21xy-95\)
\(=343+49-95=297\)