Khai triển đẳng thức :
a, \(\left(x+y+z\right)^2\)
b, \(\left(x+y-z\right)^2\)
c, \(\left(x-y-z\right)^2\)
Những câu hỏi liên quan
khai triển các biểu thức sau:
\(a.\left(2x+3y\right)^2\)
\(b.2\left(\dfrac{1}{2}x^2+y\right)\left(x^2-2y\right)\)
\(c.\left(x+y+z\right)^2\)
a. (2x+3y)2= (2x)2+2.2x.3y+(3y)2
=4x2+12xy+9y2
b. 2(\(\dfrac{1}{2}\)x2+y)(x2-2y)
=(x2+2y)(x2-2y)
=x4-4y2
c, (x+y+z)2= [(x+y)+z]2
=(x+y)2+2(x+y)z+z2
=x2+2xy+y2+2xz+2yz+z2
=x2+y2+z2+2xy+2yz+2xz
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Chứng minh đẳng thức:
a) dfrac{y}{left(x-yright)left(y-zright)}+dfrac{z}{left(y-zright)left(z-xright)}+dfrac{x}{left(z-xright)left(x-yright)0}
b) dfrac{x^2}{left(x-yright)left(y-zright)}+dfrac{y^2}{left(y-zright)left(y-xright)}+dfrac{z^2}{left(z-xright)left(z-yright)1}
c) dfrac{1}{xleft(x-yright)left(x-zright)}+dfrac{1}{yleft(y-zright)left(y-xright)}+dfrac{1}{zleft(z-xright)left(z-yright)}dfrac{1}{xyz}
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Chứng minh đẳng thức:
a) \(\dfrac{y}{\left(x-y\right)\left(y-z\right)}+\dfrac{z}{\left(y-z\right)\left(z-x\right)}+\dfrac{x}{\left(z-x\right)\left(x-y\right)=0}\)
b) \(\dfrac{x^2}{\left(x-y\right)\left(y-z\right)}+\dfrac{y^2}{\left(y-z\right)\left(y-x\right)}+\dfrac{z^2}{\left(z-x\right)\left(z-y\right)=1}\)
c) \(\dfrac{1}{x\left(x-y\right)\left(x-z\right)}+\dfrac{1}{y\left(y-z\right)\left(y-x\right)}+\dfrac{1}{z\left(z-x\right)\left(z-y\right)}=\dfrac{1}{xyz}\)
a: \(\dfrac{y}{\left(x-y\right)\left(y-z\right)}-\dfrac{z}{\left(y-z\right)\left(x-z\right)}-\dfrac{x}{\left(x-y\right)\left(x-z\right)}\)
\(=\dfrac{xy-yz-xz+yz-xy+xz}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
=0
c: \(=\dfrac{1}{x\left(x-y\right)\left(x-z\right)}-\dfrac{1}{y\left(y-z\right)\left(x-y\right)}+\dfrac{1}{z\left(x-z\right)\left(y-z\right)}\)
\(=\dfrac{zy\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{zy^2-z^2y-x^2z+xz^2+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{1}{xyz}\)
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Chứng minh đẳng thức:\(\dfrac{x^2y^2z^2}{a^2b^2}+\dfrac{(x^2-a^2)\left(y^2-a^2\right)\left(z^2-a^2\right)}{a^2\left(a^2-b^2\right)}+\dfrac{(x^2-b^2)\left(y^2-b^2\right)\left(z^2-b^2\right)}{b^2\left(b^2-a^2\right)}=x^2+y^2+z^2-a^2-b^2\)
Gợi ý trong sách: Khai triển hạng tử thứ 2 và thứ 3 ở vế trái (Có nhiu đó thui)
Help me!!!!!!!!!
1.Tính:x:frac{x-1}{2}-frac{left(x-1right)left(x^2+4x+1right)}{2x^2+2x}.frac{-4x}{left(x-1right)^2}-frac{4x^2}{x^2-1}2.Chứng minh đẳng thức sau( giả sử đẳng thức có nghĩa):frac{y-z}{left(x-yright)left(x-zright)}+frac{z-x}{left(y-zright)left(y-xright)}+frac{x-y}{left(z-xright)left(z-yright)}frac{2}{x-y}+frac{2}{y-z}+frac{2}{z-x} Các bạn giúp mình với!
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1.Tính:
\(x:\frac{x-1}{2}-\frac{\left(x-1\right)\left(x^2+4x+1\right)}{2x^2+2x}.\frac{-4x}{\left(x-1\right)^2}-\frac{4x^2}{x^2-1}\)
2.Chứng minh đẳng thức sau( giả sử đẳng thức có nghĩa):
\(\frac{y-z}{\left(x-y\right)\left(x-z\right)}+\frac{z-x}{\left(y-z\right)\left(y-x\right)}+\frac{x-y}{\left(z-x\right)\left(z-y\right)}=\frac{2}{x-y}+\frac{2}{y-z}+\frac{2}{z-x}\)
Các bạn giúp mình với!
\(\frac{y-z}{\left(x-y\right)\left(x-z\right)}+\frac{z-x}{\left(y-z\right)\left(y-x\right)}+\frac{x-y}{\left(z-x\right)\left(z-y\right)}=\frac{2}{x-y}+\frac{2}{y-z}+\frac{2}{z-x}\)Chứng minh 2 đẳng thức trên bằng nhau
Xem chi tiết
\(\Leftrightarrow\) \(\frac{\left(x-z\right)-\left(x-y\right)}{\left(x-y\right)\left(x-z\right)}\)\(+\frac{\left(y-x\right)-\left(y-z\right)}{\left(y-z\right)\left(y-x\right)}+\frac{\left(z-y\right)-\left(z-x\right)}{\left(z-x\right)\left(z-y\right)}=\frac{2}{x-y}+\frac{2}{y-z}+\frac{2}{z-x}\)
\(\Leftrightarrow\)\(\frac{1}{x-y}-\frac{1}{x-z}+\frac{1}{y-z}-\frac{1}{y-x}+\frac{1}{z-x}-\frac{1}{z-y}=\frac{2}{x-y}+\frac{2}{y-z}+\frac{2}{z-x}\)
\(\Leftrightarrow\)\(\frac{1}{x-y}+\frac{1}{z-x}+\frac{1}{y-z}+\frac{1}{x-y}+\frac{1}{z-x}+\frac{1}{y-z}=\frac{2}{x-y}+\frac{2}{y-z}+\frac{2}{z-x}\)
tự lm nốt ik
chứng minh đẳng thức sau
a,\(\frac{x^2+3xy}{x^2-9y^2}+\frac{2x^2-5xy-3y^2}{6xy-x^2-9y^2}=\frac{x^2+xz+xy+yz}{3yz-x^2-xz+3xy}\)
b,\(\frac{y-z}{\left(x-y\right)\left(x-z\right)}+\frac{z-x}{\left(y-z\right)\left(y-x\right)}+\frac{x-y}{\left(z-x\right)\left(z-y\right)}=\frac{2}{x-y}+\frac{2}{y-z}+\frac{2}{z-x}\)
Khai triển hằng dẳng thức
1, \(\left(a+b-c\right)^2\)
2, \(\left(a-b+c\right)^2\)
3, \(\left(x-y+z\right).\left(x-y-z\right)\)
\(\left(a+b-c\right)^2=\left(\left(a+b\right)-c\right)^2\)
\(=\left(a+b\right)^2+c^2-2\left(a+b\right)c\)
\(=a^2+b^2+2ab+c^2-2ac-2bc\)
\(=a^2+b^2+c^2+2ab-2bc-2ca\)
\(\left(a-b+c\right)^2=\left(\left(a-b\right)+c\right)^2\)
\(=\left(a-b\right)^2+c^2+2\left(a-b\right)c\)
\(=a^2+b^2-2ab+c^2+2ac-2bc\)
\(=a^2+b^2+c^2-2ab-2bc+2ca\)
\(\left(x-y+z\right)\left(x-y-z\right)=\left(\left(x-y\right)+z\right)\left(\left(x-y\right)-z\right)\)
\(=\left(x-y\right)^2-z^2\)
\(=x^2+y^2-2xy-z^2\)
( a + b - c )2 = [ ( a + b ) - c ]2
= ( a + b )2 - 2( a + b )c + c2
= a2 + b2 + c2 + 2ab - 2bc - 2ac
( a - b + c )2 = [ ( a- b ) + c ]2
= ( a - b )2 + 2( a - b )c + c2
= a2 + b2 + c2 - 2ab - 2bc + 2ca
( x - y + z )( x - y - z ) = [ ( x - y ) + z ][ ( x - y ) - z ]
= ( x - y )2 - z2
= x2 + y2 - z2 - 2xy
a,\(\left(a+b-c\right)^2\)
\(=\left[\left(a+b\right)-c\right]^2\)
\(=\left(a+b\right)^2-2\left(a+b\right)c+c^2\)
\(=a^2+2ab+b^2-2\left(ac+bc\right)+c^2\)
\(=a^2+b^2+c^2+2ab-2ac-2bc\)
c,\(\left(a-b+c\right)^2\)
\(=\left[\left(a-b\right)+c\right]^2\)
\(=\left(a-b\right)^2+2\left(a-b\right)c+c^2\)
\(=a^2-2ab+b^2+2\left(ac-bc\right)+c^2\)
\(=a^2+b^2+c^2-2ab+2ac-2bc\)
c,\(\left(x-y+z\right)\left(x-y-z\right)\)
\(=\left[\left(x-y\right)+z\right]\left[\left(x-y\right)-z\right]\)
\(=\left(x-y\right)^2-z^2\)
\(=x^2-2xy+y^2-z^2\)
\(=x^2+y^2-z^2-2xy\)
Phân tích các đa thức sau thành nhân tử
a.\(x\left(y^2-z^2\right)+y\left(z^2-z^2\right)+z\left(x^2-y^2\right)\)
b.\(\left(x+y\right)\left(x^2-y^2\right)+\left(y+z\right)\left(y^2-z^2\right)+\left(z+x\right)\left(z^2-x^2\right)\)
a) x(\(y^2\)-\(z^2\))+y(\(z^2-z^2\)) + (\(x^2-y^2\))
=\(xy^2-xz^2+x^2z-y^2z\)
=\(y^2\left(x-z\right)+xz\left(x-z\right)\)
= \(y^2+xz\)
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Chứng minh các bất đẳng thức sau với x, y, z > 0
a) \(x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2}\)
b) \(x^3+y^3\ge\dfrac{\left(x+y\right)^3}{4}\)
c) \(x^4+y^4\ge\dfrac{\left(x+y\right)^4}{8}\)
e) \(x^2+y^2+z^2\ge\dfrac{\left(x+y+z\right)^2}{3}\)
f) \(x^3+y^3+z^3\ge3xyz\)
a) \(x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2}\)
\(\Leftrightarrow2x^2+2y^2\ge\left(x+y\right)^2\Leftrightarrow x^2+y^2\ge2xy\)
\(\Leftrightarrow x^2-2xy+y^2\ge0\Leftrightarrow\left(x-y\right)^2\ge0\left(đúng\right)\)
b) \(x^3+y^3\ge\dfrac{\left(x+y\right)^3}{4}\)
\(\Leftrightarrow4x^3+4y^3\ge\left(x+y\right)^3\Leftrightarrow3x^3+3y^3\ge3x^2y+3xy^2\)
\(\Leftrightarrow3x^2\left(x-y\right)-3y^2\left(x-y\right)\ge0\)
\(\Leftrightarrow3\left(x-y\right)\left(x^2-y^2\right)\ge0\Leftrightarrow3\left(x-y\right)^2\left(x+y\right)\ge0\left(đúng\right)\)
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a: Ta có: \(x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2}\)
\(\Leftrightarrow2x^2+2y^2-x^2-2xy-y^2\ge0\)
\(\Leftrightarrow x^2-2xy+y^2\ge0\)
\(\Leftrightarrow\left(x-y\right)^2\ge0\)(luôn đúng)
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