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a: \(=3y^2-5x^2y^3-2y^2+3x^2y^3=y^2-2x^2y^3\)
b: \(=6x-y+2x^2+3y^2-2x^2+x=7x-y+3y^2\)
c: \(=x-y+4y^2-6xy+\dfrac{10x^2}{y}\)
\(a.\left(9x^2y^3-15x^4y^4\right):3x^2y-\left(2-3x^2y\right)y^2\)
\(=3y^2-5x^2y^3-2y^2+3x^2y^3\)
\(=y^2-2x^2y^3\)
\(b.\left(6x^2-xy\right):x+\left(2x^3y+3xy^2\right):xy-\left(2x-1\right)x\)
\(=6x-y+2x^2+3y-2+x\)
\(=2x^2+7x+2y-2\)
\(c.\left(x^2-xy\right):x+\left(6x^2y^5-9x^3y^4+15x^4y^3\right):\dfrac{3}{2}x^2y^3\)
\(=x-y+4y^2-6xy+10x^2\)
a: \(=\dfrac{27a^6b^3\cdot a^2b^6}{a^8b^8}=27b\)
b: \(=3y^2-5x^2y^3-2y^2+3x^2y^3\)
\(=y^2-2x^2y^3\)
c: \(=6x-y+2x^2+3y-2x^2+x\)
\(=7x+2y\)
d: \(=x-y+2y^2-6xy+\dfrac{10x^2}{y}\)
2. CMR:
a. \(\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)=x^5-y^5\)
Ta có: VT=\(\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)=x^5+x^4y+x^3y^2+x^2y^3+xy^4-x^4y-x^3y^2-x^2y^3-xy^4-y^5=x^5-y^5=VP\)=> đpcm.
b. \(\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)=x^5+y^5\)
Ta có: VT=\(\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)=x^5-x^4y+x^3y^2-x^2y^3+xy^4+x^4y-x^3y^2+x^2y^3-xy^4+y^5=x^5+y^5=VP\)
=> đpcm.
c. \(\left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab\)
\(\Leftrightarrow x^2+bx+ax+ab=x^2+ax+bx+ab\) (đúng)
=> đpcm.
a. \(\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)\)
\(\Rightarrow x^5+x^4y+x^3y^2+x^2y^3+y^5-yx^4-x^3y^2-x^2y^3-xy^4-y^5=VP\)
\(\Rightarrow dpcm\)
b. \(\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)\)
\(\Rightarrow x^5-x^4y+x^3y^2-x^2y^3+xy^4+yx^4-x^3y^2-xy^4+y^5=VP\)
\(\Rightarrow dpcm\)
c.d làm tương tự
Bài làm
a) Biến đổi vế trái, ta được:
\(VT=\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)\)
\(=x^5+x^4y+x^3y^2+x^2y^3+xy^4-x^4y-x^3y^2-x^2y^3-xy^4-y^5\)
\(=\left(x^5-y^5\right)+\left(x^4y-x^4y\right)+\left(x^3y^2-x^3y^2\right)+\left(x^2y^3-x^2y^3\right)+\left(xy^4-xy^4\right)\)
\(=x^5-y^5=VP\left(đpcm\right)\)
b) Biến đổi vế trái, ta có:
\(VT=\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)\)
\(=x^5-x^4y+x^3y^2-x^2y^3+xy^4+x^4y-x^3y^2+x^2y^3-xy^4+y^5\)
\(=\left(x^5+y^5\right)+\left(-x^4y+x^4y\right)+\left(x^3y^2-x^3y^2\right)+\left(-x^2y^3+x^2y^3\right)+\left(xy^4-xy^4\right)\)
\(=x^5+y^5=VP\left(đpcm\right)\)
c) Biến đổi vế trái, ta có:
\(VT=\left(a+b\right)\left(a^3-a^2b+ab^2-b^3\right)\)
\(=a^4-a^3b+a^2b^2-ab^3+a^3b-a^2b^2+ab^3-b^4\)
\(=\left(a^4-b^4\right)+\left(-a^3b+a^3b\right)+\left(a^2b^2-a^2b^2\right)+\left(-ab^3+ab^3\right)\)
\(=a^4-b^4=VP\left(đpcm\right)\)
d) Đây là hằng đẳng thức, như vế phải hình như bạn viết bị sai, mik sửa là vế phải nha.
\(\left(a+b\right)\left(a^2-ab+b^2\right)=a^3+b^3\)
Biến đổi vế trái, ta có:
\(VT=\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(=a^3-a^2b+ab^2+a^2b-ab^2+b^3\)
\(=\left(a^3+b^3\right)+\left(-a^2b+a^2b\right)+\left(ab^2-ab^2\right)\)
\(=a^3+b^3=VP\left(đpcm\right)\)
a) \(Q=\left(x-y\right)^2-4\left(x-y\right)\left(x+2y\right)+4\left(x+2y\right)^2\)
\(Q=\left(x-y\right)^2-2\cdot\left(x-y\right)\cdot2\left(x+2y\right)+\left[2\left(x+2y\right)\right]^2\)
\(Q=\left[\left(x-y\right)-2\left(x+2y\right)\right]^2\)
\(Q=\left(x-y-2x-4y\right)^2\)
\(Q=\left(-x-5y\right)^2\)
b) \(A=\left(xy+2\right)^3-6\left(xy+2\right)^2+12\left(xy+2\right)-8\)
\(A=\left(xy+2\right)^3-3\cdot2\cdot\left(xy+2\right)^2+3\cdot2^2\cdot\left(xy+2\right)-2^3\)
\(A=\left[\left(xy+2\right)-2\right]^3\)
\(A=\left(xy+2-2\right)^3\)
\(A=\left(xy\right)^3\)
\(A=x^3y^3\)
c) \(\left(x+2\right)^3+\left(x-2\right)^3-2x\left(x^2+12\right)\)
\(=\left(x^3+6x^2+12x+8\right)+\left(x^2-6x^2+12x-8\right)-\left(2x^3+24x\right)\)
\(=x^3+6x^2+12x+8+x^2-6x^2+12x-8-2x^3-24x\)
\(=\left(x^3+x^3-2x^3\right)+\left(6x^2-6x^2\right)+\left(12x+12x-24x\right)+\left(8-8\right)\)
\(=0\)
a: =(x-y)^2-2(x-y)(2x+4y)+(2x+4y)^2
=(x-y-2x-4y)^2=(-x-5y)^2=x^2+10xy+25y^2
b: =(xy+2-2)^3=(xy)^3=x^3y^3
c: =x^3+6x^2+12x+8+x^3-6x^2+12x-8-2x(x^2+12)
=24x+2x^3-2x^3-24x
=0
\(\left(\dfrac{x^2}{x+y}+y\right).\left(\dfrac{1}{x^2-xy}-\dfrac{3y^3}{x^4-xy^3}-\dfrac{y}{x^3+x^2y+xy^2}\right)\)
\(=\left(\dfrac{x^2+xy+y^2}{x+y}\right).\left(\dfrac{1}{x\left(x-y\right)}-\dfrac{3y^2}{x\left(x^3-y^3\right)}-\dfrac{y}{x\left(x^2+xy+y^2\right)}\right)\)\(=\left(\dfrac{x^2+xy+y^2}{x+y}\right).\left(\dfrac{x^2+xy+y^2}{x\left(x^3-y^3\right)}-\dfrac{3y^2}{x\left(x^3-y^3\right)}-\dfrac{xy-y^2}{x\left(x^3-y^3\right)}\right)\)
\(=\dfrac{x\left(x^3-y^3\right)}{x^3-xy^2}.\dfrac{x^2+xy+y^2-3y^2-xy+y^2}{x\left(x^3-y^3\right)}\\ =\dfrac{x^2-y^2}{x\left(x^2-y^2\right)}=\dfrac{1}{x}\)
