Question

[Lê Hồng Anh]

Tìm x

a)\(\sqrt{x-1}=2\left(x\ge1\right)\)

b)\(\sqrt{3-x}=4\left(x\le3\right)\)

c)\(2.\sqrt{3-2x}=\dfrac{1}{2}\left(x\le\dfrac{3}{2}\right)\)

d)\(4-\sqrt{x-1}=\dfrac{1}{2}\left(x\ge1\right)\)

e)\(\sqrt{x-1}-3=1\)

f)\(\dfrac{1}{2}-2.\sqrt{x+2}=\dfrac{1}{4}\)

Answer 1

[Tử Nguyệt Hàn]

a) (x≤3)





 

Answer 2

[Nguyễn Lê Phước Thịnh]

a: Ta có: \(\sqrt{x-1}=2\)

\(\Leftrightarrow x-1=4\)

hay x=5

b: Ta có: \(\sqrt{3-x}=4\)

\(\Leftrightarrow3-x=16\)

hay x=-13

c: Ta có: \(2\cdot\sqrt{3-2x}=\dfrac{1}{2}\)

\(\Leftrightarrow\sqrt{3-2x}=\dfrac{1}{4}\)

\(\Leftrightarrow-2x+3=\dfrac{1}{16}\)

\(\Leftrightarrow-2x=-\dfrac{47}{16}\)

hay \(x=\dfrac{47}{32}\)

d: Ta có: \(4-\sqrt{x-1}=\dfrac{1}{2}\)

\(\Leftrightarrow\sqrt{x-1}=\dfrac{7}{2}\)

\(\Leftrightarrow x-1=\dfrac{49}{4}\)

hay \(x=\dfrac{53}{4}\)

e: Ta có: \(\sqrt{x-1}-3=1\)

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

\(\Leftrightarrow x-1=16\)

hay x=17

f:Ta có: \(\dfrac{1}{2}-2\cdot\sqrt{x+2}=\dfrac{1}{4}\)

\(\Leftrightarrow2\cdot\sqrt{x+2}=\dfrac{1}{4}\)

\(\Leftrightarrow\sqrt{x+2}=\dfrac{1}{8}\)

\(\Leftrightarrow x+2=\dfrac{1}{64}\)

hay \(x=-\dfrac{127}{64}\)