Question

Answer 1

a) ĐKXĐ: x\(\ge\)\(\dfrac{7}{2}\)

\(\sqrt{2x+7}=5\Leftrightarrow2x+7=25\Leftrightarrow2x=18\Leftrightarrow x=9\)(thỏa mãn)

Answer 2

2)ĐKXĐ: x\(\ge\dfrac{1}{2}\)

\(\sqrt{2x-1}-x=0\Leftrightarrow\sqrt{2x-1}=x\Leftrightarrow2x-1=x^2\)

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

Answer 3

3) ĐKXĐ: \(4\ge x\ge-\dfrac{1}{5}\)

\(\sqrt{5x+1}=\sqrt{4-x}\Leftrightarrow5x+1=4-x\Leftrightarrow6x=3\Leftrightarrow x=\dfrac{1}{2}\)(thỏa mãn)

Vayy...........

Answer 4

Answer 5

4) Ta có: \(\sqrt{x^2-5}=\sqrt{4x-9}\)

\(\Leftrightarrow x^2-5-4x+9=0\)

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

\(\Leftrightarrow x=2\)(Loại)

Vậy: \(S=\varnothing\)

Answer 6

1) Ta có: \(\sqrt{2x+7}=5\)

nên 2x+7=25

hay x=9

Vậy: x=9