2a^2+b^2+c^2>=2a(b+c)
K
Khách
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LH
2
KB
11 tháng 9 2018
a ) \(\left(2a+b\right)^2-\left(2a+b\right)\left(2a-b\right)-2a\left(b-a\right)\)
\(=4a^2+4ab+b^2-\left(4a^2-b^2\right)-2ab+2a^2\)
\(=4a^2+4ab+b^2-4a^2+b^2-2ab+2a^2\)
\(=2a^2+2ab+2b^2\)
\(=\left(a^2+2ab+b^2\right)+a^2+b^2\)
\(=\left(a+b\right)^2+a^2+b^2\)
b ) \(\left(a+b-c\right)^2-\left(a+b\right)^2+2c\left(a+b\right)\)
\(=\left(a+b\right)^2-2\left(a+b\right)c+c^2-\left(a+b\right)^2+2c\left(a+b\right)\)
\(=\left[\left(a+b\right)^2-\left(a+b\right)^2\right]+\left[2c\left(a+b\right)-2\left(a+b\right)c\right]+c^2\)
\(=c^2\)
TV
12 tháng 9 2018
@Khôi Bùi
\(\left(2a+b\right)^2-\left(2a+b\right)\left(2a-b\right)-2a\left(b-a\right)\)
\(=\left(2a+b\right)\left[\left(2a+b\right)-\left(2a-b\right)\right]-2a\left(b-a\right)\)
\(=2b\left(2a+b\right)-2a\left(b-a\right)\)
\(=4ab+2b^2-2ab+2a^2=2\left(a^2+ab+b^2\right)\)
\(\left(a+b-c\right)^2-\left(a+b\right)^2+2c\left(a+b\right)\)
\(=\left(a+b-c+a+b\right)\left(a+b-c-a-b\right)+2c\left(a+b\right)\)
\(=-c\left(2a+2b-c\right)+2c\left(a+b\right)=\)
\(-2c\left(a+b\right)+c^2+2c\left(a+b\right)=c^2\)
4 tháng 8 2022
a: \(=a^2-b^4\)
b: \(=\left(a^2+2a\right)^2-9\)
c: \(=a^2-\left(2a+3\right)^2\)
d: \(=a^4-\left(2a-3\right)^2\)
e: \(=\left(-a^2-2a+3\right)^2\)
g: \(=4a^2-a^4\)
KT
0
1 tháng 3 2022
Áp dụng bđt cauchy, ta có
\(\dfrac{2a}{b+c}+\dfrac{b+c}{2a}\ge2.\sqrt{\dfrac{2a}{b+c}\dfrac{b+c}{2a}}=2\)(đpcm)
Dấu ''='' xảy ra <=>2a = b + c
H
0
CQ
0
AH
Akai Haruma
Giáo viên
25 tháng 10 2021
Lời giải:
$B=(a^4+b^4-2a^2b^2)+c^4-2c^2(a^2-b^2)-4b^2c^2$
$=(a^2-b^2)^2+c^4-2c^2(a^2-b^2)-(2bc)^2$
$=(a^2-b^2-c^2)^2-(2bc)^2$
$=(a^2-b^2-c^2-2bc)(a^2-b^2-c^2+2bc)$
$=[a^2-(b+c)^2][a^2-(b-c)^2]$
$=(a-b-c)(a+b+c)(a-b+c)(a+b-c)$
TD
26 tháng 6 2018
a)\(\left(a+b+c\right)^2-\left(a+b\right)^2-c^2\\ =\left(a+b\right)^2+2\left(a+b\right)c+c^2-\left(a+b\right)^2-c^2\\ =2\left(a+b\right)c\)
b)\(\left(a+b+c\right)^2-\left(b+c\right)^2-2a\left(b+c\right)\\ =a^2+2a\left(b+c\right)+\left(b+c\right)^2-\left(b+c\right)^2-2a\left(b+c\right)\\ =a^2\)
c)\(\left(3a+1\right)^2-2\left(2a+5\right)\left(3a+1\right)+\left(2a+5\right)^2\\ =\left(3a+1-2a-5\right)^2\\ =\left(a-4\right)^2\)
19 tháng 6 2019
Bài 1:
a) \(\left(a-b^2\right)\left(a+b^2\right)=a^2-b^4\)
b) \(\left(a^2+2a-3\right)\left(a^2+2a+3\right)=\left(a^2+2a\right)^2-9\)
c) \(\left(a^2+2a+3\right)\left(a^2-2a-3\right)=a^2-\left(2a+3\right)^2\)
d) \(\left(a^2-2a+3\right)\left(a^2+2a+3\right)=9-\left(a^2-2a\right)^2\)
e) \(\left(-a^2-2a+3\right)\left(-a^2-2a+3\right)=\left(-a^2-2a+3\right)^2\)
g) \(\left(a^2+2a+3\right)\left(a^2-2a+3\right)=\left(a^2+3\right)^2-4a^2\)
f) \(\left(a^2+2a\right)\left(2a-a^2\right)=4a^2-a^4\)
19 tháng 6 2019
Bài 2 :
a) \(\left(x+1\right)\left(x^2-x+1\right)=x^3+1\)
b) \(\left(x+y+z\right)^2=\left(x+y+z\right)\left(x+y+z\right)=x^2+xy+xz+yx+y^2+yz+zx+zy+z^2=x^2+2xy+2yz+2xz+y^2+z^2\)
c) \(\left(x-y+z\right)^2=\left(x-y+z\right)\left(x-y+z\right)=x^2-xy+xz-xy+y^2-yz+xz-yz+z^2=x^2+y^2+z^2-2xy+2xz-2yz\)d) \(\left(x-2y\right)\left(x^2+2xy+4y^2\right)=\left(x-2y\right)^3\)
e) \(\left(x-y-z\right)^2=\left(x-y-z\right)\left(x-y-z\right)=x^2-xy-xz-xy+y^2+yz-xz+yz+z^2=x^2-2xy-2xz+2yz+y^2+z^2\)
a^4+a^3b+ab^3+b^4>=4