Question

Bài 1: P=\(\left(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\right):\frac{\sqrt{x}-1}{2}\)
a,Rút gọn P
b,Chứng minh rằng P>0

Bài 2: P=\(\left(\frac{2\sqrt{x}+x}{x\sqrt{x}-1}-\frac{1}{\sqrt{x}-1}\right):\left(1-\frac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
a,Rút gọn P
b,Tính \(\sqrt{P}\) khi x=5+\(2\sqrt{3}\)
(hiu hiu...phiền các bạn giúp mk vs ạ)...

Answer 1

Bài 1:

a) P= \(\left(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\right):\frac{\sqrt{x}-1}{2}\) (x ≥ 0; x ≠ 4)

=\(\left(\frac{x+2}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}-1\right)\cdot\sqrt{x}}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}\right)\cdot\frac{2}{\sqrt{x}-1}\)

= \(\left(\frac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}\right)\cdot\frac{2}{\sqrt{x}-1}\)

=\(\left(\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}\right)\cdot\frac{2}{\sqrt{x}-1}\)

=\(\frac{\left(\sqrt{x}-1\right)^2\cdot2}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)\cdot\left(\sqrt{x}-1\right)}\)

=\(\frac{2}{x+\sqrt{x}+1}\)

b) Ta có: x ≥ 0 ⇒ \(\sqrt{x}\) ≥ 0

⇒ \(x+\sqrt{x}+1\) ≥ 1 > 0

mà 2 > 0 ⇒ \(\frac{2}{x+\sqrt{x}+1}\) > 0 ⇒ P > 0

Bài 2:

a) P= \(\left(\frac{2\sqrt{x}+x}{x\sqrt{x}-1}-\frac{1}{\sqrt{x}-1}\right):\left(1-\frac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\) (x ≥ 0; x ≠ 1)

=\(\left(\frac{2\sqrt{x}+x}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}\right):\left(\frac{x+\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)

=\(\left(\frac{2\sqrt{x}+x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}\right):\left(\frac{x+\sqrt{x}+1-\sqrt{x}-2}{x+\sqrt{x}+1}\right)\)

=\(\left(\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}\right):\left(\frac{x-1}{x+\sqrt{x}+1}\right)\)

=\(\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}\cdot\frac{x+\sqrt{x}+1}{x-1}\)

=\(\frac{1}{x-1}\)

b) Ta có: \(\sqrt{P}=\sqrt{\frac{1}{x-1}}\)

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

x = \(5+2\sqrt{3}\) (TM)

Thay x vào \(\sqrt{P}\) ta có:

\(\sqrt{P}=\frac{1}{\sqrt{5+2\sqrt{3}-1}}\)

=\(\frac{1}{\sqrt{4+2\sqrt{3}}}\)

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

=\(\frac{1}{\sqrt{\left(\sqrt{3}+1\right)^2}}\)

=\(\frac{1}{\left|\sqrt{3}+1\right|}\)

=\(\frac{1}{\sqrt{3}+1}\)

= \(\frac{\sqrt{3}-1}{\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-1\right)}\)

=\(\frac{\sqrt{3}-1}{2}\)

Vậy \(\sqrt{P}=\frac{\sqrt{3}-1}{2}\) khi x = \(5+2\sqrt{3}\)