x^2+y^2=2 chung min dang thuc 2(x+1)(y+1)=(x+y+2)
Những câu hỏi liên quan
chung minh bat dang thuc
a/ x^2+xy+y^2+1>0
b/ x^2+5y^2-4xy>10y-4
a/ x2 + xy + y2 + 1
= [x2 + 2.x.\(\dfrac{y}{2}\) + (\(\dfrac{y}{2}\) )2 ] + \(\dfrac{3y^2}{4}\) + 1
= ( x + \(\dfrac{y}{2}\) )2 + \(\dfrac{3y^2}{4}\) + 1
Vì \(\left(x+\dfrac{y}{2}\right)^2\) \(\ge\) 0 với mọi x;y
và \(\dfrac{3y^2}{4}\ge0\) với mọi x;y
=> \(\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^2}{4}\ge0\) với mọi x;y
=> \(\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^2}{4}+1>0\)
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CM dang thuc
a, ( x - y )^2 = ( x + y )2 - 4xy
b, ( x + y + z )^2 - ( xy + yz + zx ) = 1/2 ( x + y )^2 + 1/2 ( y + z )^2 + 1/2( z + x )^2
cho x,y,z la cac so thuc duong thoa man x+y+z=1 tim min A=x^3/(x^2+xy+y^2)+y^3/(y^2+yz+z^2)+z^3/(z^2+zx+x^2)
chung minh cac hang dang thuc sau luon co gia tri duong voi moi gia tri cua bien
4)D=x^2 -2x+y^2+4y+6
5)P=(15x - 1)^2+3(7x+3)(x+1) -(x^2 -73)
cho cac so thuc x va y thoa man
\(\left(x^2+\sqrt{1+x^2}\right)\left(y^2+\sqrt{1+y^2}\right)=1\)1
chung minh x+y=0
\(\sqrt{x+2+2\sqrt{x+1}}+\sqrt{x+2-2\sqrt{ }x+1}=\frac{x+5}{2}\)\(\frac{x+5}{2}\)
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Xét \(x^2+\sqrt{1+x^2}\)ta có:
\(x^2\ge0\)
nên \(1+x^2\ge1\)
\(\Rightarrow\sqrt{1+x^2}\ge\sqrt{1}=1\)
\(\Rightarrow x^2+\sqrt{1+x^2}\ge1\)
Tương tự ta có:
\(y^2+\sqrt{1+y^2}\ge1\)
Do đó: \(\left(x^2+\sqrt{1+x^2}\right)\left(y^2+\sqrt{1+y^2}\right)\ge1\)
Dấu bằng xảy ra khi \(x=0;y=0\)
Khi đó \(x+y=0\left(ĐPCM\right)\)
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( x2 + y2 ) ( 1/x3 + 1/y3 ) = 6a2 - 5a +1 / a3 voi x+y=1, xy=a , x,y,a khac 0
De bai la chung minh dang thuc, giup minh nhe !!!!
bai 1 cmm hang dang thuc
a,(a+b+c)^2 +a^2+b^2+c^2=(a+b)^2+(b+c)^2+(c+a)^2
b, x^4+x^4+(x+y)^4=2(x^2+xy+y^2)^2
giAI HO MINH NHE NHANH LEN MINH DANG GAP
aVT=.\(\left(a+b+c\right)^2+a^2+b^2+c^2\)
=\(a^2+b^2+c^2+2ab+2ac+2bc+a^2+b^2+c^2\)
=\(2a^2+2b^2+2c^2+2ab+2ac+2bc\)
VP=\(\left(a+b\right)^2+\left(b+c\right)^2+\left(a+c\right)^2\)=\(a^2+2ab+b^2+b^2+2bc+b^2+a^2+2ac+c^2\)
=\(2a^2+2b^2+2c^2+2ab+2bc+2ac\)
Vậy VT=VP
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a)\(\text{(a+b+c)^2 +a^2+b^2+c^2=(a+b)^2+(b+c)^2+(c+a)^2}\)
Ta có:
\(\left(a+b+c\right)^2+a^2+b^2+c^2=a^2+b^2+c^2+2ab+2bc+2ac+a^2+b^2+c^2\)
\(=\left(a^2+2ab+b^2\right)+\left(b^2+2bc+c^2\right)+\left(c^2+2ca+a^2\right)\)
\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\)
Vậy \(\left(a+b+c\right)^2+a^2+b^2+c^2=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\)
b) Câu b sao chỉ có một vế vậy , hằng đẳng thức thì phải có hai vế chứ
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b) \(\text{x^4+y^4+(x+y)^4=2(x^2+xy+y^2)^2}\)
Ta có:
\(x^4+y^4+\left(x+y\right)^4=x^4+y^4+x^4+4x^3y+6x^2y^2+4xy^3+y^4\)
\(2x^4+2y^{\text{4}}+4x^3y+6x^2y^2+4xy^3=2\left(x^4+y^4+2x^3y+3x^2y^2+2xy^3\right)\)
\(=2\left[\left(x^2\right)^2+\left(y^2\right)^2+\left(xy\right)^2+2x^2.y^2+2y^2.xy+2x^2.xy\right]\)
\(=2\left(x^2+xy+y^2\right)^2\)
Vậy \(x^4+y^4+\left(x+y\right)^4=2\left(x^2+xy+y^2\right)^2\)
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cho x,y,z la 3 so thuc tuy y thoa man x+y+z=0 va -1< x<1,-1<y<1,-1<z<1.chung minh rang da thuc x^2+y^4+z^6 co gia tri khong lon hon 2
Chung minh dang thuc:
A)(x+y)(x^2-xy+y^2)+(x-y)(x^2+xy+y^2) = 2x^3
B)x^3-y^3=(x-y)((x-y)^2-xy)
\(A,VT=x^3+y^3+x^3-y^3=2x^3=VP\\ B,VT=\left(x-y\right)\left(x^2+xy+y^2\right)=\left(x-y\right)\left(x^2+2xy+y^2-xy\right)\\ =\left(x-y\right)\left[\left(x+y\right)^2-xy\right]=VP\)
Sửa câu b \(cm:x^3-y^3=\left(x-y\right)\left[\left(x+y\right)^2-xy\right]\)
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