Question

Tìm x:

a.\(\sqrt{4x-1}=2\)

b.\(2\sqrt{x-1}+3=7\)

c.\(\sqrt{x-1}+\sqrt{16x+6}=9\)

Answer 1

a) Điều kiện: \(4x-1\ge0\Leftrightarrow x\ge\dfrac{1}{4}\)

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

\(4x-1=4\)

\(4x=4+1\)

\(4x=5\)

\(x=\dfrac{5}{4}\) (nhận)

Vậy \(x=\dfrac{5}{4}\)

b) Điều kiện: \(x-1\ge0\Leftrightarrow x\ge1\)

\(2\sqrt{x-1}+3=7\)

\(2\sqrt{x-1}=7-3\)

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

\(\sqrt{x-1}=2\)

\(x-1=4\)

\(x=4+1\)

\(x=5\) (nhận)

Vậy \(x=5\)

c) Xem lại đề giùm

Answer 2

`a, 4x - 1 = 4`

`4x = 5`

`x = 1,25`

`b, 2 sqrt(x-1) = 4`

`sqrt(x-1) = 2`

`x-  1 = 4`

`x = 5`

`c, -sqrt(x-1) + 4 sqrt(x-1) = 9`

`<=> 3sqrt(x-1) = 9`

`<=> x - 1 = 9`

`<=> x = 10`