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Bài 1:
a: \(x\left(x+y\right)+5y-x^2\)
\(=x^2+xy+5y-x^2\)
=xy+5y
b: \(\left(x-2\right)\left(y+1\right)-xy+4\)
\(=xy+x-2y-2-xy+4\)
=-2y+x+2
c: \(\dfrac{\left(4x^2y+12xy^2-8xy\right)}{2xy}\)
\(=\dfrac{2xy\cdot2x+2xy\cdot6y-2xy\cdot4}{2xy}\)
=2x+6y-4
d: \(\left(x-4\right)^2+8x-7\)
\(=x^2-8x+16+8x-7\)
\(=x^2+9\)
\(\begin{array}{l}T + H = 3{x^2}y - 2x{y^2} + xy + \left( { - 2{x^2}y + 3x{y^2} + 1} \right)\\ = 3{x^2}y - 2x{y^2} + xy - 2{x^2}y + 3x{y^2} + 1\\ = \left( {3{x^2}y - 2{x^2}y} \right) + \left( { - 2x{y^2} + 3x{y^2}} \right) + xy + 1\\ = {x^2}y + x{y^2} + xy + 1\\T - H = 3{x^2}y - 2x{y^2} + xy - \left( { - 2{x^2}y + 3x{y^2} + 1} \right)\\ = 3{x^2}y - 2x{y^2} + xy + 2{x^2}y - 3x{y^2} - 1\\ = \left( {3{x^2}y + 2{x^2}y} \right) + \left( { - 2x{y^2} - 3x{y^2}} \right) + xy - 1\\ = 5{x^2}y - 5x{y^2} + xy - 1\end{array}\)
Chọn B.
Bài 3:
a: \(x^2-16=\left(x-4\right)\cdot\left(x+4\right)\)
b: \(x^2+2x+1-y^2=\left(x+1+y\right)\left(x+1-y\right)\)
c: \(=\left(x-y\right)^2-4=\left(x-y-2\right)\left(x-y+2\right)\)
Ta có HPT:
\(\left\{{}\begin{matrix}x-y=5\\xy=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}xy-y^2=5y\\xy=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y^2=-6-5y\\xy=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-2\\x=3\end{matrix}\right.\)
Thay x = -2, y = 3 vào, ta được:
A = (-2)3 - 33 - (-2)2 + 2.(-2).3 - 32
A = -8 - 27 - 4 + (-12) - 9
A = -60
Sửa:
Ta có HPT:
\(\left\{{}\begin{matrix}x-y=-5\\xy=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}xy-y^2=-5y\\xy=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y^2=-6-\left(-5y\right)\\xy=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=-3\end{matrix}\right.\)
Thay x = -3, y = 2 vào, ta được:
A = (-3)3 - 23 - (-3)2 + 2.(-3).2 - 22
A = -27 - 8 - 9 + (-12) - 4
A = -60
b, ( 5/2 - x ) ^2
=25/4-4/5x+x^2
c,( xy/2 - x/3 ) ( xy/2 + x/3)
=(xy/2)^2-(x/3)^2
c: \(\left(\dfrac{xy}{2}-\dfrac{x}{3}\right)\left(\dfrac{xy}{2}+\dfrac{x}{3}\right)=\dfrac{x^2y^2}{4}-\dfrac{x^2}{9}\)
e: \(\left(2x+3y\right)^2=4x^2+12xy+9y^2\)
\(a)\left(x+3y\right)\left(x-2y\right)\\ =x^3-2xy+3xy-6y^2\\ =x^2+xy-6y^2\\ b)\left(2x-y\right)\left(y-5x\right)\\ = 2xy-10x^2-y^2+5xy\\ =7xy-10x^2-y^2\\ c)\left(2x-5y\right)\left(y^2-2xy\right)\\ =2xy^2-4x^2y-5y^3+10xy^2\\ =12xy^2-4x^2y-5y^2\\ d)\left(x-y\right)\left(x^2-xy-y^2\right)\\ =x^3-x^2y-xy^2-x^2y+xy^2+y^3\\ =x^3-2x^2y+y^3\)
\(B=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2+\left(xy+\frac{1}{xy}\right)^2\)
\(-\left(x+\frac{1}{x}\right)\left(y+\frac{1}{y}\right)\left(xy+\frac{1}{xy}\right)\)
\(\Rightarrow B=x^2+2+\frac{1}{x^2}+y^2+2+\frac{1}{y^2}+x^2y^2+2+\frac{1}{x^2y^2}-x^2y^2\)
\(-2-x^2-y^2-\frac{1}{y^2}-\frac{1}{x^2}-\frac{1}{x^2y^2}\)
\(\Rightarrow B=x^2y^2-x^2y^2+x^2-x^2+1.\frac{1}{x^2}+1.\frac{1}{x^2y^2}-1.\frac{1}{x^2}-1\)
\(.\frac{1}{x^2y^2}+1.\frac{1}{y^2}-1.\frac{1}{y^2}+y^2-y^2+2+2+2-2\)
\(\Rightarrow B=4\)
a: \(=x-\dfrac{3}{2}+2y\)
b: \(=\dfrac{1}{x\left(y-x\right)}-\dfrac{1}{y\left(y-x\right)}=\dfrac{y-x}{xy\left(y-x\right)}=\dfrac{1}{xy}\)
a: A+B
=x^2y+xyz+7y^2-25xy-xyz+x^2y-7y^2+xy
=-24xy+2x^y
A-B=x^2y+xyz+7y^2-25xy+xzy-x^2y+7y^2-xy
=2xyz+14y^2-26xy
b: Bậc của A là 3
bậc của B là 3
c: Khi x=-3;y=-1/2;z=0 thì:
A=9*(-1/2)+0+7*(-1/2)^2-25*(-3)*(-1/2)
=-9/2+7/4-75/2
=-42+7/4=-161/4
B=(-3)*(-1)*(-1/2)*0+(-3)^2*(-1/2)-7*1/4+(-3)*(-1/2)
=-9/2-7/4+3/2
=-3-7/4=-19/4
\(2x\left(x^2-7x-3\right)=2x^3-14x-6x\)
\(4xy^2\left(-2x^3+y^2-7xy\right)=-8x^4y^2+4xy^5-28x^2y^3\)
Áp dụng hằng đẳng thức a2 - b2 = ( a - b ) ( a + b) ta đc:
a)\(\left(x^2+x+1\right)\left(x^2-x-1\right)\)
\(=\left(x^2\right)^2-\left[\left(x+1\right)\right]^2\)
\(=x^4-\left(x^2+2x+1\right)\)
\(=x^4-x^2-2x-1\)
b)MK sửa đề nha\(\left(x^2+xy+y^2\right)\left(x^2-xy-y^2\right)\)
\(=\left(x^2\right)^2-\left[\left(xy+y^2\right)\right]^2\)
\(=x^4-x^2y^2-2xy^3-y^4\)