Cho \(a+b+c=2m\). Chúng minh rằng: \(2bc+b^2+c^2-a^2=4m\left(m-a\right)\)
Cho \(a+b+c=2p\). Chứng minh rằng:
\(2bc+b^2+c^2-a^2=4p\left(p-a\right)\)
\(2bc+b^2+c^2-a^2\)
\(=\left(b+c\right)^2-a^2\)
\(=\left(b+c+a\right)\cdot\left(b+c-a\right)\)
\(=2p\cdot\left(2p-a-a\right)\)
\(=4p\left(p-a\right)\)
Cho a + b + c = 4m. Chứng minh rằng:
\(\left(\dfrac{a+b-c}{2}\right)^2+\left(\dfrac{a-b+c}{2}\right)^2+\left(\dfrac{-a+b+c}{2}\right)^2=a^2+b^2+c^2-4m^2\)
\(\left(\dfrac{a+b-c}{2}\right)^2+\left(\dfrac{a-b+c}{2}\right)^2+\left(\dfrac{-a+b+c}{2}\right)^2\)
\(=\left(\dfrac{4m-2c}{2}\right)^2+\left(\dfrac{4m-2b}{2}\right)^2+\left(\dfrac{4m-2a}{2}\right)^2\)
\(=\left(2m-c\right)^2+\left(2m-b\right)^2+\left(2m-a\right)^2\)
\(=4m^2-4mc+c^2+4m^2-4mb+b^2+4m^2-4ma+a^2\)
\(=a^2+b^2+c^2+12m^2-4m\left(a+b+c\right)\)
\(=a^2+b^2+c^2+12m^2-4m\cdot4m\)
\(=a^2+b^2+c^2+12m^2-16m^2\)
\(=a^2+b^2+c^2-4m^2\)
Cho a + b + c = 4m. Chứng minh rằng:
\(\left(\frac{a+b+c}{2}\right)^2+\left(\frac{a-b+c}{2}\right)^2+\left(\frac{-a+b+c}{2}\right)=a^2+b^2+c^2-4m^2.\)
Cho a+b+c=2m. Chứng minh
\(\left(\frac{a+b-c}{2}\right)^2+\left(\frac{a-b+c}{2}\right)^2+\left(\frac{-a+b=c}{2}\right)^2=a^2+b^2+c^2-4m^2\)
a+b+c=2m thì 4m (m-a)=b^2+c^2-a^2-2bc
Cho a+b+c = 2p . Chứng minh rằng đẳng thức : \(2bc+b^2+c^2-a^2=4p\left(p-a\right)\)
\(2bc+b^2+c^2-a^2\)
\(=\left(b+c\right)^2-a^2\)
\(=\left(a+b+c\right)\left(b+c-a\right)\)
\(=2p\left(a+b+c-2a\right)\)
\(=2p\left(2p-2a\right)=4p\left(p-a\right)\)
biến đổi vế phải ta được:
4p(p -a ) = 4p\(^2\)-4pa
=(2p)\(^2\)-2p.2a
=(a+b+c)\(^2\)-2a(a+b+c)
=\(a^2+b^2+c^2+2ab+2ac+2bc\)-\(2a^2-2ab-2ac\)
=\(2bc+b^2+c^2-a^2\)=vế trái (đpcm)
Chung minh dang thuc
2bc+b^2+c^2-a^2=4m (m-a)
biet a+b+c=2m
(5a-3a+8c)×(5a-3b-8c)=(3a-5b)^2 biet a^2-b^2=4c^2
1) Cho \(a+b+c=2p\). Chúng minh hằng đẳng thức
\(2bc+b^2+c^2-a^2=4p\left(p-a\right)\) )
2) Cho biểu thức
\(M=\left(x-a\right)\left(x-b\right)+\left(x-b\right)\left(x-c\right)+\left(x-c\right)\left(x-a\right)+x^2\)
Tính M theo a,b,c biết rằng \(x=\frac{1}{2}a+\frac{1}{2}b+\frac{1}{2}c\)
HELP ME!!!!!!!!!! NHANH NHANH GIÙM MK NHA
2)
M= (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)+x^2
= x^2-bx-ax+ab+x^2-cx-bx+bc+x^2-ax-cx+ac+x^2
= 4x^2-2bx-2ax-2cx+ab+bc+ac
=4x^2-2x(a+b+c)+ab+bc+ac
= 2x [ 2x-(a+b+c)2x] +ab+bc+ac (1)
Mặt khác : x=\(\frac{1}{2}\)a+\(\frac{1}{2}\)b+\(\frac{1}{2}\)c
<=> x =\(\frac{1}{2}\)(a+b+c)
<=>2x=a+b+c
=> Vế phải của (1) bằng : a+b+c (a+b+c-a-b-c)+ab+bc+ac
<=> ( a+b+c ).0 + ab+bc+ac
<=> ab+bc+ac
hay M= ab+bc+ac
Vậy M=ab+bc+ac
(b+c )2 - a2 = ( b+c -a ) ( b+c + a ) = ( a+b+c -2a ) 2p = (2p - 2a )2p = (p-a ) 4p
Cho tam giác ABC có các cạnh là a, b, c và có p là nửa chu vi. Chứng minh rằng:
a) \(a^2-b^2-c^2+2bc=4\left(p-b\right)\left(p-c\right)\)
b)\(p^2+\left(p-â\right)^2+\left(p-b\right)^2+\left(p-c\right)^2=a^2+b^2+c^2\)
p là nửa chu vi =>a+b+c=2p
a, \(a^2-b^2-c^2+2bc=a^2-\left(b^2-2bc+c^2\right)=a^2-\left(b-c\right)^2=\left(a-b+c\right)\left(a+b-c\right)\)
\(=\left(a+b+c-2b\right)\left(a+b+c-2c\right)=\left(2p-2b\right)\left(2p-2c\right)=4\left(p-b\right)\left(p-c\right)\) (đpcm)
b, \(p^2+\left(p-a\right)^2+\left(p-b\right)^2+\left(p-c\right)^2=p^2+p^2-2pa+a^2+p^2-2pb+b^2+p^2-2pc+c^2\)
\(=4p^2-2p\left(a+b+c\right)+a^2+b^2+c^2=4p^2-2p.2p+a^2+b^2+c^2=a^2+b^2+c^2\) (đpcm)