a) x=0

b) x=-2, y=-1

Thấy : \(\sqrt{x}\ge0\)

\(\Rightarrow P=\dfrac{\sqrt{x}+2}{2\sqrt{x}+1}>0\)

\(\Rightarrow\left|P\right|=P\)

Ta có : \(\left|P\right|=P\ge P\)

=> P = P .

Vậy \(\forall x>0\) TMYC đè bài

Ơ câu này giống câu ở dưới thế ?_? Lặp câu hỏi à bạn :v

Ta có: \(\left|P\right|\ge P\)

\(\Leftrightarrow\dfrac{\sqrt{x}+2}{2\sqrt{x}+1}\ge0\)

\(\Leftrightarrow x>0\)

+++++\(x+5x^2=0\)    \(\Leftrightarrow x.\left(1+5x\right)=0\) \(\Leftrightarrow\hept{\begin{cases}x=0\\1+5x=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=\frac{-1}{5}\end{cases}}}\)

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

\(\Leftrightarrow x.\left(x+1\right)=0\)\(\Leftrightarrow\hept{\begin{cases}x=0\\x+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=-1\end{cases}}}\)

++++\(5x.\left(x-1\right)=x-1\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x-1=0\\5x-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}}\)

a)  x + 5x^2 = 0

=> x(1+5x) = 0

=> \(\orbr{\begin{cases}x=0\\1+5x=0\end{cases}}\)   => \(\orbr{\begin{cases}x=0\\5x=-1\end{cases}}\)   => \(\orbr{\begin{cases}x=0\\x=\frac{-1}{5}\end{cases}}\)

Vậy: x=0 hoặc x=-1/5

b)  (x+1)^2 = x+1

=> (x+1)(x+1) = x+1

=>        x+1   = (x+1) : (x+1)

=>        x+1   =   1

=>         x      =   0

Vậy: x = 0

c)  5x(x-1) = x-1

=> 5x       = (x-1) : (x-1)

=> 5x       =    1

=>  x        =   1/5

Vậy: x = 1/5

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

\(\Leftrightarrow\left(2x+3\right)\left(x^2+2x+1-x^2-3x-2+x^2+4x+4\right)=\left(2x+3\right)^3\)

\(\Leftrightarrow\left(2x+3\right)\left(x^2+3x+3\right)-\left(2x+3\right)^3=0\)

\(\Leftrightarrow\left(2x+3\right)\left(4x^2+12x+9-x^2-3x-3\right)=0\)

\(\Leftrightarrow\left(2x+3\right)\left(3x^2+9x+6\right)=0\)

\(\Leftrightarrow\left(2x+3\right)\left(x+1\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=-1\\x=-2\end{matrix}\right.\)

\(\left(\dfrac{3}{2}x-1\right)\left(x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{2}x-1=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-4\end{matrix}\right.\)

Bài 1:

Để biểu thức nhận giá trị nguyên thì \(3\sqrt{x}+1⋮2\sqrt{x}-1\)

\(\Leftrightarrow6\sqrt{x}+2⋮2\sqrt{x}-1\)

\(\Leftrightarrow2\sqrt{x}-1\in\left\{1;-1;5\right\}\)

\(\Leftrightarrow2\sqrt{x}\in\left\{2;0;6\right\}\)

hay \(x\in\left\{4;0;36\right\}\)