a) \(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)

\(\Leftrightarrow\left(25x^2+10x+1\right)-25x^2+9=30\)

\(\Leftrightarrow10x=20\Leftrightarrow x=2\)

b) \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)

\(\Leftrightarrow x^3-1-x^3+4x-5=0\)

\(\Leftrightarrow4x=6\Leftrightarrow x=\frac{3}{2}\)

\(\Rightarrow A< B\)

Ta có : \(4p\left(p-a\right)=2\left(a+b+c\right)\left(\frac{a+b+c}{2}-a\right)=2.\left(a+b+c\right).\frac{b+c-a}{2}\)

\(=\left(a+b+c\right)\left(b+c-a\right)=\left[\left(b+c\right)+a\right]\left[\left(b+c\right)-a\right]=\left(b+c\right)^2-a^2\)

\(=b^2+c^2-a^2+2bc\)

Bài cuối bạn sửa 2ab thành 2bc nhé ^^

\(a^2+b^2=a^2-2ab+b^2+2ab=\left(a-b\right)^2+2ab\)

Vì  \(\left(a-b\right)^2\ge0\Rightarrow\left(a-b\right)^2+2ab\ge2ab\left(dpcm\right)\)

\(2bc+b^2+c^2-a^2\)

\(=\left(b+c\right)^2-a^2\)

\(=\left(b+c+a\right)\cdot\left(b+c-a\right)\)

\(=2p\cdot\left(2p-a-a\right)\)

\(=4p\left(p-a\right)\)

Sửa đề : CMR : \(2bc+b^2+c^2-a^2=4p\left(p-a\right)\)

Ta có : \(VT=2bc+b^2+c^2-a^2\)

\(=\left(b+c\right)^2-a^2\)

\(=\left(b+c-a\right)\left(b+c+a\right)\)

\(=\left(b+c+a-2a\right).2p\)

\(=\left(2p-2a\right).2p\)

\(=4p^2-4ap\)

\(=4p\left(p-a\right)=VP\left(đpcm\right)\)

TC:a+b+cd=2p=>b+c=2p-a

=>(b+c)²=(2p-a)²

=>b²+2bc+c²=4p²-4pa+a²

=>b²+2bc+c²-a²=4p²-4pa

=>2bc+b²+c²-a²=4p(p-a) ĐPCM

\(2bc+b^2+c^2-a^2\)

\(=\left(b+c\right)^2-a^2\)

\(=\left(a+b+c\right)\left(b+c-a\right)\)

\(=2p\left(a+b+c-2a\right)\)

\(=2p\left(2p-2a\right)=4p\left(p-a\right)\)

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)

a+b+c = 2p => 4p = 2(a+b+c); p=(a+b+c)/2

VP = 4p(p-a) = 2(a+b+c)(\(\frac{a+b+c}{2}-a\))

= \(2\left(a+b+c\right)\left(\frac{a+b+c-2a}{2}\right)\)

=\(2\left(a+b+c\right)\cdot\frac{b+c-a}{2}=\left(a+b+c\right)\left(b+c-a\right)\)

\(=ab+ac-a^2+b^2+bc-ab+bc+c^2-ac\)

\(=2bc+b^2+c^2-a^2\) = VT (đpcm)