Question

Answer 1

1a.

ĐK: \(x\ge\dfrac{3}{4}\)

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

\(\Leftrightarrow3x+1=4x-3\)

\(\Leftrightarrow x=4\left(tm\right)\)

1b.

ĐK: \(1\le x\le3;x\le0\)

\(\sqrt{x^2-x}=\sqrt{3-x}\)

\(\Leftrightarrow x^2-x=3-\text{​​}x\)

\(\Leftrightarrow x^2=3\)

\(\Leftrightarrow x=\pm\sqrt{3}\left(tm\right)\)

Answer 2

Bài 1: 

a: Ta có: \(\sqrt{3x+1}=\sqrt{4x-3}\)

\(\Leftrightarrow3x+1=4x-3\)

\(\Leftrightarrow-x=-4\)

hay x=4(nhận)

b: Ta có: \(\sqrt{x^2-x}=\sqrt{3-x}\)

\(\Leftrightarrow x^2-x=3-x\)

\(\Leftrightarrow x^2=3\)

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

Answer 3

1c.

ĐK: \(x\in R\)

\(\sqrt{9x^2+12x+4}=4x\)

\(\Leftrightarrow\sqrt{\left(3x+2\right)^2}=4x\)

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

TH1: \(x\ge-\dfrac{2}{3}\)

\(\left(1\right)\Leftrightarrow3x+2=4x\)

\(\Leftrightarrow x=2\left(tm\right)\)

TH2: \(x< -\dfrac{2}{3}\)

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

\(\Leftrightarrow x=-\dfrac{2}{7}\left(l\right)\)

Vậy \(x=2\).

Answer 4

1d.

ĐK: \(x\in R\)

\(\sqrt{x^2-4x+4}=x+3\)

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

TH1: \(\left\{{}\begin{matrix}x-2\ge0\\x-2=x+3\end{matrix}\right.\left(vn\right)\)

TH2: \(\left\{{}\begin{matrix}x-2< 0\\2-x=x+3\end{matrix}\right.\Leftrightarrow x=-\dfrac{1}{2}\)

Answer 5

1e.

ĐK: \(x\ge-5\)

\(x-1-\sqrt{x+5}=0\)

\(\Leftrightarrow x+5-\sqrt{x+5}-6=0\)

\(\Leftrightarrow\left(\sqrt{x+5}-3\right)\left(\sqrt{x+5}+2\right)=0\)

\(\Leftrightarrow\sqrt{x+5}=3\)

\(\Leftrightarrow x+5=9\)

\(\Leftrightarrow x=4\left(tm\right)\)

Answer 6

1f.

ĐK: \(x\ge2;x=-2\)

\(\sqrt{x^2-4}-2\sqrt{x+2}=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=6\left(tm\right)\\x=-2\left(tm\right)\end{matrix}\right.\)

Answer 7

1g.

ĐK: \(x\ge5\)

\(\sqrt{4x-20}+\sqrt{x-5}=4+3\sqrt{\dfrac{x-5}{9}}\)

\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}=4+\sqrt{x-5}\)

\(\Leftrightarrow2\sqrt{x-5}=4\)

\(\Leftrightarrow x-5=4\)

\(\Leftrightarrow x=9\left(tm\right)\)

Answer 8

1h.

ĐK: \(x\ge-1\)

\(3\sqrt{4x+4}-\sqrt{9x+9}-8\sqrt{\dfrac{x+1}{16}}=5\)

\(\Leftrightarrow6\sqrt{x+1}-3\sqrt{x+1}-2\sqrt{x+1}=5\)

\(\Leftrightarrow\sqrt{x+1}=5\)

\(\Leftrightarrow x+1=25\)

\(\Leftrightarrow x=24\left(tm\right)\)