Đáp án : 2500           

Giải thích các bước giải

\(P=18a+30b+7a-5b\)

\(P=\left(18a+7a\right)+\left(30b-5b\right)\)

\(P=25a.25b=25\left(a+b\right)\)

Thêo đề bài \(a+b=100\)

\(--->25\left(a+b\right)=25.100=2500\)

\(P=25a+25b=25\left(a+b\right)=25.100=2500\)

\(P=18a+30b+7a-5b\)

\(P=a\left(18+7\right)+b\left(30-5\right)\)

\(P=a25+b25\)

\(P=25.\left(a+b\right)\)

Lời giải:

$P=4a^2+b^2+c^2+4ab+4ac+2bc=(2a+b+c)^2=(-1)^2=1$

a)  \(A=7a+7b=7\left(a+b\right)=7.11=77\)   
b)  \(B=13a+19b+4a-2b=17a+17b=17\left(a+b\right)=17.100=1700\)
c)  \(C=5a-4b+7a-8b=12a-12b=12\left(a-b\right)=12.8=96\)

P= \(\dfrac{1}{b^2+c^2-a^2}+\dfrac{1}{a^2+c^2-b^2}+\dfrac{1}{a^2+b^2-c^2}\)

=
\(\dfrac{a+b+c}{\left(b^2+c^2-a^2\right)\left(a+b+c\right)}+\dfrac{a+b+c}{\left(a^2+c^2-b^2\right)\left(a+b+c\right)}+\dfrac{a+b+c}{\left(a^2+b^2-c^2\right)\left(a+b+c\right)}\)
= 0+0+0 = 0
Vậy P= 0 
Ngu vãi ko bt đúng không nx

\(P=\dfrac{1}{b^2+c^2-a^2}+\dfrac{1}{a^2+c^2-b^2}+\dfrac{1}{a^2+b^2-c^2}\)

\(=\dfrac{1}{b^2+c^2-\left(-b-c\right)^2}+\dfrac{1}{a^2+c^2-\left(-c-a\right)^2}+\dfrac{1}{a^2+b^2-\left(-a-b\right)^2}\)

\(=\dfrac{1}{b^2+c^2-\left(b+c\right)^2}+\dfrac{1}{a^2+c^2-\left(c+a\right)^2}+\dfrac{1}{a^2+b^2-\left(a+b\right)^2}\)

\(=\dfrac{1}{b^2+c^2-b^2-2bc-c^2}+\dfrac{1}{a^2+c^2-a^2-2ac-c^2}+\dfrac{1}{a^2+b^2-a^2-2ab-b^2}\)

\(=\dfrac{1}{-2bc}+\dfrac{1}{-2ac}+\dfrac{1}{-2ab}\)

\(=\dfrac{a}{-2bca}+\dfrac{b}{-2acb}+\dfrac{c}{-2abc}\)

\(=\dfrac{a+b+c}{-2abc}=\dfrac{0}{-2abc}=0\)