a. ĐKXĐ : x>1.

b. \(A=\left(\dfrac{4}{x-\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\dfrac{1}{\sqrt{x}-1}=\left[\dfrac{4}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right].\left(\sqrt{x}-1\right)=\dfrac{4+\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\left(\sqrt{x}-1\right)=\dfrac{4+x}{\sqrt{x}}\)

c. Thay \(x=4-2\sqrt{3}\) vào A, ta có:

\(A=\dfrac{4+4-2\sqrt{3}}{\sqrt{4-2\sqrt{3}}}=\dfrac{8-2\sqrt{3}}{\sqrt{\left(\sqrt{3}-1\right)^2}}=\dfrac{8-2\sqrt{3}}{\sqrt{3}-1}=\dfrac{\left(8-2\sqrt{3}\right)\left(\sqrt{3}+1\right)}{3-1}=\dfrac{8\sqrt{3}+8-6-2\sqrt{3}}{2}=\dfrac{2+6\sqrt{3}}{2}=\dfrac{2\left(1+3\sqrt{3}\right)}{2}=1+3\sqrt{3}\)

Vậy giá trị của A tại \(x=4-2\sqrt{3}\) là \(1+3\sqrt{3}\).

Lần sau ghi dấu ra xíu nhé :v

a) Đặt \(\sqrt{x}=a\Rightarrow B=\left(\dfrac{a}{a+4}+\dfrac{4}{a-4}\right):\dfrac{a^2+16}{a+2}\)

Quy đồng,rút gọn : \(B=\dfrac{a+2}{a^2-16}\Rightarrow B=\dfrac{\sqrt{x}+2}{x-16}\)

b) \(B\left(A-1\right)=\dfrac{\sqrt{x}+2}{x-16}\left(\dfrac{\sqrt{x}+4}{\sqrt{x}+2}-1\right)=\dfrac{2}{x-16}\)

x - 16 là ước của 2 => \(x\in\left\{14;15;17;18\right\}\)

mới làm quen toán 9 ;v có gì k rõ ae chỉ bảo nhé :))

dung ko the ban, sao ngan the ?

a)ĐKXĐ:x>0

P=\(\left(\frac{3}{x-1}-\frac{1}{\sqrt{x}+1}\right):\frac{1}{\sqrt{x}+1}\left(vớix>0\right)\)

=\(\left[\frac{3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}+1}\right]:\frac{1}{\sqrt{x}+1}\)

=\(\left[\frac{3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]:\frac{1}{\sqrt{x}+1}\)

= \(\left[\frac{3-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]:\frac{1}{\sqrt{x}+1}\)

=\(\frac{4-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{\sqrt{x}+1}{1}\)

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

b)Để P=\(\frac{5}{4}\left(vớix>0\right)\)

\(\Leftrightarrow\frac{4-\sqrt{x}}{\sqrt{x}-1}=\frac{5}{4}\)

\(\Leftrightarrow\frac{4-\sqrt{x}}{\sqrt{x}-1}-\frac{5}{4}=0\)

\(\Leftrightarrow\frac{4\left(4-\sqrt{x}\right)}{4\left(\sqrt{x}-1\right)}-\frac{5\left(\sqrt{x}-1\right)}{4\left(\sqrt{x}-1\right)}=0\)

\(\Rightarrow16-4\sqrt{x}-5\sqrt{x}+5=0\)

\(\Leftrightarrow21-9\sqrt{x}=0\)

\(\Leftrightarrow-9\sqrt{x}=-21\)

\(\Leftrightarrow\sqrt{x}=\frac{7}{3}\)

\(\Leftrightarrow x=\frac{21}{9}\)

Vậy:Để P=\(\frac{5}{4}\)thì x=\(\frac{21}{9}\)

c)Còn phần c thì mik chịu

...

b \(P=\dfrac{\sqrt{x}+\sqrt{x}+2}{x-4}\cdot\dfrac{\sqrt{x}-2}{2}=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)

a: Khi x=64 thì \(P=\dfrac{8+1}{8+2}=\dfrac{9}{10}\)

b: \(P=\dfrac{\sqrt{x}+\sqrt{x}+2}{x-4}\cdot\dfrac{\sqrt{x}-2}{2}=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)

a: Khi x=64 thì \(P=\dfrac{8+1}{8+2}=\dfrac{9}{10}\)

A có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}\ne0\\\sqrt{x}-1\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\\sqrt{x}\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne1\end{matrix}\right.\)

Ta có:

A = \(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

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

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

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

Kết luận: ...

ĐK của nó còn là: x ≥ 0 nữa dung doan nhé, mình viết thiếu...

a) Với x = 25 thì \(N=\frac{\sqrt{25}+1}{\sqrt{25}}=\frac{6}{5}\)

b) Ta có   \(M=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2.\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2.\left(\sqrt{x}-1\right)}\)

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

Suy ra \(S=M.N=\frac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

a. A có nghĩa khi \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-1\ne\\\frac{x+\sqrt{x}}{\sqrt{x}+1}\ne0\end{matrix}\right.0\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

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

b. \(x=7+4\sqrt{3}\Rightarrow\)A = \(\frac{\sqrt{7+4\sqrt{3}}+1}{\sqrt{7+4\sqrt{3}}}=\frac{\sqrt{\left(2+\sqrt{3}\right)^2}+1}{\sqrt{\left(2+\sqrt{3}\right)^2}}=\frac{3+\sqrt{3}}{2+\sqrt{3}}\)

cam on bn