rut gon bieu thuc
p=(x^2+2xy)^2+2(x^2+2xy)y^2+y^4
Những câu hỏi liên quan
bai1: rut gon cac bieu thuc sau
a, (2x-y).(4x^2+2xy+y^2)-(2x+y).(4x^2-2xy+y^2)
b, (3x+2y).(9x^2-6xy+4y^2)-27x^3
c,8x.(x-2y).(x+2y)+(y-2x).(x^2+2xy+4x^2)
bai2 :cmr
a, a^3+b^3=(a+b)^3-3ab.(a+b)
b.a^3-b^3=(a-b)+3ab,(a-b)
bai2 :cmr
a, a^3+b^3=(a+b)^3-3ab.(a+b)
VP= \(\left(a+b\right)^3-3ab\left(a+b\right)\)
=\(a^3+b^3+3a^2b+3ab^2-3a^2b-3ab^2=a^3+b^3\)
=VT
b.a^3-b^3=(a-b)^3+3ab,(a-b)
\(VP=\left(a-b\right)^3+3ab\left(a-b\right)\)
=\(a^3-3a^2b+ab^2.3-b^3+3a^2b-3ab^2=a^3-b^3\)
=VT
=> ĐPCM
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bài 1.
a) = 8x^3+4x^2y+2xy^2-4x^2y-2xy^2-y^3-(8x^3-4x^2y+2xy^2+4x^2y-2xy^2+y^3)
= 8x3+4x2y+2xy2-4x2y-2xy2-y3 - 8x3+4x2y-2xy2-4x2y+2xy2-y3
=-8x2y-6y3
b) = 27x3-18x2y+12xy2+18x2y-12xy2+8y3-27x3
=8y
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rut gon : \(\frac{x^2+y^2+z^2-2xy+2xz-2y^2}{x^2-2xy+y^2-z^2}\)
Trả lời:
sửa đề: \(\frac{x^2+y^2+z^2-2xy+2xz-2yz}{x^2-2xy+y^2-z^2}\)
\(=\frac{\left(x-y+z\right)^2}{\left(x-y\right)^2-z^2}=\frac{\left(x-y+z\right)^2}{\left(x-y+z\right)\left(x-y-z\right)}=\frac{x-y+z}{x-y-z}\)
Rut gon bieu thuc
1)\(\left(a-b\right).\sqrt{\frac{ab}{\left(a-b\right)^2}}\) voi a \(\ne\) b
2)\(\frac{x-y}{y}.\sqrt{\frac{y^4}{x^2-2xy+y^2}}\) voi x\(\ne\) y
1) \(\left(a-b\right)\cdot\sqrt{\frac{ab}{\left(a-b\right)^2}}=\left(a-b\right)\cdot\frac{\sqrt{ab}}{a-b}=\sqrt{ab}\)
2) \(\frac{x-y}{y}\cdot\sqrt{\frac{y^4}{x^2-2xy+y^2}}=\frac{x-y}{y}\cdot\frac{\sqrt{y^4}}{\sqrt{\left(x-y\right)^2}}=\frac{x-y}{y}\cdot\frac{y^2}{x-y}=y\)
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Rut gon bieu thuc : \(\frac{2xy^2}{3ab}\sqrt{\frac{9a^3b^4}{8xy^3}}\)
B1: rut gon bieu thuc
a, (x+y)^2-4(x-y)^2
b, 2(x-y)(x+y)+(x+y)^2+(x-y)^2
B2: tim X
a, (2X-1)^2-4(X+2)^2=9
b, 3(X-1)^2-3X(X-5)=21
B3: Cho bieu thuc
M=(x+3)^3-(x-1)^3+12x(x-1)
a, Rut gon bieu thuc tren
b, Tinh gia tri M tai x=-2/3
c, Tim x de M=16
1)a)=>x2+y2+2xy-4(x2-y2-2xy)
=>x2+y2+2xy-4.x2+4y2+8xy
=>-3.x2+5y2+10xy
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rut gon bieu thuc [(x^3+y^3)-2(x^2-y^2)+3(x+y)^2]:(x+y)
rut gon bieu thuc sau : P=2(x+y)(x-y)-(x-y)^2+(x+y)^2-4y^2
\(P=2\left(x^2-y^2\right)-x^2+2xy-y^2+x^2+2xy+y^2-4y^2\)
\(=2\left(x^2-y^2\right)-4y^2+4xy\)
\(=2x^2-2y^2-4y^2+4xy\)
=2x^2+4xy-6y^2
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rut gọn cac bieu thưc sau rồi tinh gia trị biểu thưc vơi x= -2 1/3
a. ( 2x - 3). ( 2x + 3) - ( x + 5 )2 - ( x - 1).( x + 2) vs x = -2 1/3
b. ( x + 2y ).( x2 - 2xy + 4y2 )-( x - y ). ( x2 - xy - y2 )
c. x2 . ( x+ y ) + y2 .( x+ y) + 2x2y + 2xy2
d. ( x3 + 4x2 - x - 4) : ( x + 4)
Cho A = \(\left(\dfrac{\sqrt{x}+\sqrt{y}}{1-\sqrt{xy}}+\dfrac{\sqrt{x}-\sqrt{y}}{1+\sqrt{xy}}\right):\left(1+\dfrac{x+y+2xy}{1-xy}\right)\)
a, Rut gon bieu thuc A
b, Tinh gia tri cua A khi x = \(\dfrac{1}{1+\sqrt{2}}\)
c, Tim Max A
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a: \(A=\dfrac{\sqrt{x}+x\sqrt{y}+\sqrt{y}+y\sqrt{x}+\sqrt{x}-x\sqrt{y}-\sqrt{y}+y\sqrt{x}}{1-xy}:\dfrac{1-xy+x+y+2xy}{1-xy}\)
\(=\dfrac{2\sqrt{x}+2y\sqrt{x}}{x+y+xy+1}\)
\(=\dfrac{2\sqrt{x}\left(y+1\right)}{\left(x+1\right)\left(y+1\right)}=\dfrac{2\sqrt{x}}{x+1}\)
b: \(x=\dfrac{1}{\sqrt{2}+1}=\sqrt{2}-1\)
\(A=\dfrac{2\sqrt{\sqrt{2}-1}}{\sqrt{2}-1+1}=\sqrt{2\left(\sqrt{2}-1\right)}\)
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