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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
\(\left( {3{x^2} - 5xy - 4{y^2}} \right).\left( {2{x^2} + {y^2}} \right) + \left( {2{x^4}y^2 + {x^3}{y^3} + {x^2}{y^4}} \right):\left( {\dfrac{1}{5}xy} \right)\\\)
\(= 3{x^2}.2{x^2} + 3{x^2}.{y^2} - 5xy.2{x^2} - 5xy.{y^2} - 4{y^2}.2{x^2} - 4{y^2}.{y^2} + 2{x^4}y^2:\left( {\dfrac{1}{5}xy} \right) + {x^3}{y^3}:\left( {\dfrac{1}{5}xy} \right) + {x^2}{y^4}:\left( {\dfrac{1}{5}xy} \right)\\\)
\(= 6{x^4} + 3{x^2}{y^2} - 10{x^3}y - 5x{y^3} - 8{x^2}{y^2} - 4{y^4} + 10{x^3}y + 5{x^2}{y^2} + 5x{y^3}\\\)
\(= 6{x^4} - 4{y^4}+ ( - 10{x^3}y + 10{x^3}y) + \left( { - 5x{y^3} + 5x{y^3}} \right) + \left( {3{x^2}{y^2} - 8{x^2}{y^2} + 5{x^2}{y^2}} \right)\\\)
\(= 6{x^4} - 4{y^4}\)
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.
bài 11
a) \(x^2-xy+x\\ =x\left(x-y+1\right)\)
b)
\(x^2-2xy-4+y^2\\ =\left(x^2-2xy+y^2\right)-4\\ =\left(x-y\right)^2-4\\ =\left(x-y-2\right)\left(x-y+2\right)\)
c)
\(x^3-x^2-16x+16\\ =x^2\left(x-1\right)-16\left(x-1\right)\\ =\left(x-1\right)\left(x-4\right)\left(x+4\right)\)
bài 12
\(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(2x^2-10x-3x-2x^2=26\)
\(-13x=26\\ x=-2\)
b)
\(2\left(x+5\right)-x^2-5x=0\\ 2\left(x+5\right)-x\left(x+5\right)=0\\ \left(x+5\right)\left(2-x\right)=0\\ \left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
a)
\(\begin{array}{l}\left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right)\\ = x.{x^2} + x.xy + x.{y^2} - y.{x^2} - y.xy - y.{y^2}\\ = {x^3} + {x^2}y + x{y^2} - {x^2}y - x{y^2} - {y^3}\\ = {x^3} - {y^3}\end{array}\)
b)
\(\begin{array}{l}\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right)\\ = x.{x^2} + x.\left( { - xy} \right) + x{y^2} + y.{x^2} + y.\left( { - xy} \right) + y.{y^2}\\ = {x^3} - {x^2}y + x{y^2} + {x^2}y - x{y^2} + {y^3}\\ = {x^3} + {y^3}\end{array}\)
c)
\(\begin{array}{l}\left( {4{\rm{x}} - 1} \right)\left( {6y + 1} \right) - 3{\rm{x}}\left( {8y + \dfrac{4}{3}} \right)\\ = 4{\rm{x}}.6y + 4{\rm{x}}.1 - 1.6y - 1.1 - 3{\rm{x}}.8y - 3{\rm{x}}.\dfrac{4}{3}\\ = 24{\rm{x}}y + 4{\rm{x}} - 6y - 1 - 24{\rm{x}}y - 4{\rm{x}}\\ = - 6y - 1\end{array}\)
d)
\(\begin{array}{l}\left( {x + y} \right)\left( {x - y} \right) + \left( {x{y^4} - {x^3}{y^2}} \right):\left( {x{y^2}} \right)\\ = x.x + x.\left( { - y} \right) + y.x + y.\left( { - y} \right) + \left( {x{y^4}} \right):\left( {x{y^2}} \right) + \left( { - {x^3}{y^2}} \right):\left( {x{y^2}} \right)\\ = {x^2} - xy + xy - {y^2} + {y^2} - x^2\\ = 0\end{array}\)
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) \(\left(4x^4-8x^2y^2+12x^5y\right):\left(-4x^2\right)\)
\(=4x^4:-4x^2-8x^2y^2:-4x^2+12x^4y:-4x^2\)
\(=-x^2+2y^2-3x^2y\)
b) \(x^2\left(x-y^2\right)-xy\left(1-xy\right)-x^3\)
\(=x^3-x^2y^2-xy+x^2y^2-x^3\)
\(=-xy\)