a: HB=4,5(cm)

BC=12,5(cm)

b: \(\widehat{B}=37^0\)

Ta có: \(AB^2+HC^2=\left(AA'^2+A'B^2\right)+\left(A'H^2+A'C^2\right)\)

\(=\left(AA'^2+A'C^2\right)+\left(A'B^2+A'H^2\right)=AC^2+HB^2\)

Lại có: \(BC^2+HA^2=\left(BB'^2+B'C^2\right)+\left(B'H^2+B'A^2\right)\)

\(=\left(BB'^2+B'A^2\right)+\left(B'C^2+B'H^2\right)=AB^2+HC^2\)

\(\Rightarrow AB^2+HC^2=AC^2+HB^2=BC^2+HA^2\)

\(\Leftrightarrow16x^4-4x^2-4xy+y^2+1=0\)

\(\Leftrightarrow\left(16x^4-8x^2+1\right)+\left(4x^2-4xy+y^2\right)=0\)

\(\Leftrightarrow\left(4x^2-1\right)^2+\left(2x-y\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x^2-1=0\\2x-y=0\end{matrix}\right.\)

\(\Leftrightarrow\left(x;y\right)=\left(-\dfrac{1}{2};-1\right);\left(\dfrac{1}{2};1\right)\)

Đăng tách ra đi bạn

Bài 5: 

Theo đề, ta có:

\(\left(2x+5\right)^2-4x^2-12x=41\)

\(\Leftrightarrow20x-12x=41+25=66\)

hay \(x=8.25\left(m\right)\)

Chu vi là:

\(\left[\left(2\cdot8.25+5\right)^2+\left(4\cdot8.25^2+12\cdot8.25\right)\right]\cdot2=1667\left(m\right)\)

Bài 3: 

a: Ta có: \(A=\left(2a+3\right)^2-13\left(a+1\right)\left(a-1\right)+\left(3a-2\right)^2\)

\(=4a^2+12a+9+9a^2-12a+4-13\left(a^2-1\right)\)

\(=13a^2+13-13a^2+13\)

=26

b: Ta có: \(B=\left(4x-1\right)^3-\left(4x-3\right)\left(16x^2+3\right)\)

\(=64x^3-48x^2+12x-1-64x^3-12x+48x^2+9\)

=8

\(P=\dfrac{2\sqrt{x}+1+\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}:\dfrac{\sqrt{x}+2}{\sqrt{x}}\)

\(=\dfrac{2\sqrt{x}+1+1-x}{\sqrt{x}\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}}{\sqrt{x}+2}=\dfrac{-x+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}=\dfrac{-x+2\sqrt{x}+2}{x+3\sqrt{x}+2}\)