*ĐKXĐ: \(x\ge0\)

Ta có: \(x=2\sqrt{x}\Leftrightarrow x-2\sqrt{x}=0\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\sqrt{x}=2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\) (tm)

*Ta có: \(x^2=3\Leftrightarrow x^2-3=0\Leftrightarrow\left(x-\sqrt{3}\right)\left(x+\sqrt{3}=0\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\sqrt{3}=0\\x+\sqrt{3}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\)

*Ta có: \(x^2=15\Leftrightarrow x^2-15=0\Leftrightarrow\left(x-\sqrt{15}\right)\left(x+\sqrt{15}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\sqrt{15}=0\\x+\sqrt{15}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{15}\\x=-\sqrt{15}\end{matrix}\right.\)

\(\dfrac{1}{4}x^2-\dfrac{1}{3}xy+\dfrac{1}{9}y^2=\left(\dfrac{1}{2}x\right)^2-2.\left(\dfrac{1}{2}x\right).\left(\dfrac{1}{3}y\right)+\left(\dfrac{1}{3}y\right)^2=\left(\dfrac{1}{2}x-\dfrac{1}{3}y\right)^2\)

\(4-6x+\dfrac{9}{4}x^2=2^2-2.2.\dfrac{2}{3}x+\left(\dfrac{2}{3}x\right)^2=\left(2-\dfrac{2}{3}x\right)^2\)

\(x^3+3x^2+3x+1=\left(x+1\right)^3\)

\(x^6+3x^5+3x^4+x^3=x^3\left(x^3+3x^2+3x+1\right)=x^3\left(x+1\right)^3\)

a, \(8-12x+6x^2-x^3=\left(2-x\right)^3\)

b, \(\dfrac{1}{4}x^2-\dfrac{1}{3}xy+\dfrac{1}{9}y^2=\left(\dfrac{1}{2}x-\dfrac{1}{3}y\right)^2\)

d, \(x^3+3x^2+3x+1=\left(x+1\right)^3\)

e, \(x^3\left(x^3+3x^2+3x+1\right)=x^3\left(x+1\right)^3\)

4*125*25*8

=(4*25)*(125*8)

=100*1000

=100 000

100000

phân số chỉ 8 con gà là

1 - 3/5 = 2/5 (số gà còn lại sau khi bán buổi sáng)

số gà còn lại sau khi bán buổi sáng là 8 : 2/5 = 20 (con gà)

phân số chỉ 20 con gà là 1 - 1/3 = 2/3 số gà

bác ba có tất cả số gà là 20 : 2/3 = 30 (con gà)

đs.....

x : (-9/2) = -4/25

            x=-4/25 . (-9/2)

           x= 18/25

 Vậy x=18/25

x=-4*-9/

x=18/25

x=o,72

vậy...

*Ta có: \(A=15x-23y=\left(13x-26y\right)+2x+3y⋮13\)

Mà \(13x-26y⋮13\)

\(\Rightarrow2x+3y⋮13\)

\(\Rightarrow B⋮13\)

* Ta có: \(A=15x-23y=\left(13x-26y\right)+2x+3y\)

Vì \(13x-26y⋮13\) và \(B=2x+3y⋮13\)

\(\Rightarrow A⋮13\)