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vậy trả lời câu a thôi

a: \(\Leftrightarrow x=\sqrt{2x+3}\)

=>x^2=2x+3 và x>=0

=>x^2-2x-3=0 và x>=0

=>x=3

b: \(\Leftrightarrow\left\{{}\begin{matrix}x< =8\\x^2+x+12=x^2-16x+64\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< =8\\17x=52\end{matrix}\right.\)

=>x=52/17

a) \(x-\sqrt{2x+3}=0\)

⇔ \(x=\sqrt{2x+3}\left(x\ge0\right)\)

⇔ \(x^2=2x+3\)

⇔ \(x^2-2x-3=0\)

⇔ x² + x - 3x - 3 = 0

⇔ x(x+1) - 3(x+1) = 0

⇔ (x-3)(x+1) = 0

⇔\(\left\{{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\Leftrightarrow x=3\)

a: \(\Leftrightarrow x=\sqrt{2x+3}\)

=>x^2=2x+3 và x>=0

=>x^2-2x-3=0 và x>=0

=>x=3

b: \(\Leftrightarrow\left\{{}\begin{matrix}x< =8\\x^2+x+12=x^2-16x+64\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< =8\\17x=52\end{matrix}\right.\)

=>x=52/17

a) \(x=\sqrt{2x+3}\) (đk \(x\ge-\dfrac{2}{3}\) )

\(\Leftrightarrow x^2-2x-3=0\Leftrightarrow\left[{}\begin{matrix}x=-1\left(l\right)\\x=3\left(nh\right)\end{matrix}\right.\)

b)ĐK \(x\le8\)

\(\Leftrightarrow x^2+x+12=\left(8-x\right)^2\)

\(\Leftrightarrow x^2+x+12=x^2-16x+64\)

\(\Leftrightarrow17x=52\Rightarrow x=\dfrac{52}{17}\)

c) ĐK \(x\ge1\)

\(\Leftrightarrow\left(2\sqrt{x-1}+\sqrt{x+2}\right)^2=\left(x+3\right)^2\)

\(\Leftrightarrow4\left(x-1\right)+x+2+4\sqrt{\left(x-1\right)\left(x+2\right)}=x^2+6x+9\)

\(\Leftrightarrow4\sqrt{x^2+x-2}=x^2+x+11\)

\(\Leftrightarrow=x^2+x+2-4\sqrt{x^2+x-2}+9=0\)( vô lí)

suy ra pt vô nghiệm

d) ĐK \(x\ge3\)

\(\Leftrightarrow x^2-6x+9-\left(x-3\right)-2\sqrt{x-3}+2=0\)

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

Đặt \(t=\sqrt{x-3}\)

\(\Leftrightarrow t^4-t^2-2t+2=0\)

\(\Leftrightarrow t^2\left(t^2-1\right)-2\left(t-1\right)=0\)

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

\(\Leftrightarrow\left(t-1\right)\left(t^3+t^2-2\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}t=1\\t^2+2t+2=0\left(vl\right)\end{matrix}\right.\)

\(\Leftrightarrow\sqrt{x-3}=1\Leftrightarrow x-3=1\Rightarrow x=4\)

cảm ơn bn saint suppapong udomkaewkanjana