Question

1. Giải phương tình vô tỉ sau:

a.\(\sqrt{7-x}+1=x\)

b.\(x-2+\sqrt{x+22}=0\)

c.\(\dfrac{\left(7-x\right)\sqrt{7-x}+\left(x+5\right)\sqrt{x-5}}{\sqrt{7-x}+\sqrt{x-5}}=2\)

d.\(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\)

e.\(\sqrt[3]{x-1}+1=x\)

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

Answer 1

a , Ta có :

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

\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\7-x=x^2-2x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x^2-x-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\\left[{}\begin{matrix}x=3\left(tm\right)\\x=-2\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)

Vậy pt có nghiệm là x = 3

b , c , d , e , f tương tự