\(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}=\dfrac{a-2\sqrt{ab}+b+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}=\dfrac{a+2\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}=\sqrt{a}+\sqrt{b}\)

\(\left(5-x\right)^2=25-10x+x^2\)

\(2xy^2+x^2y^4=xy^2\left(2+xy^2\right)\)

a, \(\left\{{}\begin{matrix}x+my=3m\\mx-y=m^2-2\end{matrix}\right.\)

Tại m = 1 , ta có:

\(\left\{{}\begin{matrix}x+y=3\\x-y=-1\end{matrix}\right.\)

giải hệ ta được:\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

\(\frac{a}{b}=\frac{c}{d}=\frac{2c}{2d}=\frac{3c}{3d}\)

Áp dụng t/c của dãy tỉ số bằng nhau => \(\frac{a}{b}=\frac{2c}{2d}=\frac{3c}{3d}=\frac{a+2c}{b+2d}=\frac{a-3c}{b-3d}\)

Vậy.....

\(D=9y^2-12y+5=\left(9x^2-12y+4\right)+1=\left(3x-2\right)^2+1\)Với \(y=\dfrac{2}{3}\Rightarrow D=\left(3\dfrac{2}{3}-2\right)^2+1=1\)

\(\left(x+y+z\right)^2-2\left(x+y+z\right)\left(x+y\right)+\left(x+y\right)^2=\left(x+y+z-x-y\right)=z^2\)(HĐT thứ 2)

\(A=x^2-20x+100=\left(x-10\right)^2\)

Với \(x=10\Rightarrow A=\left(10-10\right)^2=0\)

\(B=4x^2-4xy+y^2=\left(2x-y\right)^2\)

Với \(x=\dfrac{1}{2};y=1\Rightarrow B=\left(2.\dfrac{1}{2}-1\right)^2=0\)

\(C=4x^2-20x+25=\left(2x-5\right)^2\)

Với \(x=\dfrac{5}{2}\Rightarrow\left(2.\dfrac{5}{2}-5\right)^2=0\)

d, ko có x you ạ

Chiều rộng của lớp học là :
4. 1,25 +1,25 = 5+ 0,25 = 5,25(m)

\(a,x^4-2x^2=x^2\left(x^2-2\right)\)

\(b,x^2+5x+4=x^2+x+4x+4=x\left(x+1\right)+4\left(x+1\right)=\left(x+4\right)\left(x+1\right)\)\(c,^2-x-6=x^2-3x+2x-6=x\left(x-3\right)+2\left(x-3\right)=\left(x+2\right)\left(x-3\right)\)\(d,x^4+4=\left(x^4+4x^2+4\right)-4x^2\)

\(=\left(x^2+2\right)^2-\left(2x\right)^2=\left(x^2+2x+2\right)\left(x^2+2x-2\right)\)