\(A=\dfrac{-\left(\sqrt{x}+1\right)\left(2+\sqrt{x}\right)-2\sqrt{x}\left(2-\sqrt{x}\right)+5\sqrt{x}+2}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(\sqrt{x}+2\right)^2}\)

\(A=\dfrac{-3\sqrt{x}-x-2-4\sqrt{x}+2x+5\sqrt{x}+2}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)

\(A=\dfrac{-x-2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)

\(A=\dfrac{-\sqrt{x}\left(\sqrt{x}+2\right)^3}{\left(\sqrt{x}+2\right)\left(2-\sqrt{x}\right)\sqrt{x}\left(3-\sqrt{x}\right)}=\dfrac{-\left(\sqrt{x}+2\right)^2}{\left(2-\sqrt{x}\right)\left(3-\sqrt{x}\right)}\)

CHÚC BẠN HỌC TỐT NHÉ

\(A=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)^2}{-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)\sqrt{x}\left(3-\sqrt{x}\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)

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

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

\(ĐKXĐ:x\ge0,x\ne1\)

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

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

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

b/ Ta có: \(x=4-2\sqrt{3}=3-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\)

Thay \(x=\left(\sqrt{3}-1\right)^2\) vào (1) ta được:

\(\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\left(\sqrt{3}-1\right)^2+\sqrt{\left(\sqrt{3}-1\right)^2}+1}\)= \(\dfrac{\sqrt{3}-1}{4-2\sqrt{3}+\sqrt{3}-1+1}=\dfrac{\sqrt{3}-1}{4-\sqrt{3}}\) = \(\dfrac{\left(\sqrt{3}-1\right)\left(4+\sqrt{3}\right)}{\left(4-\sqrt{3}\right)\left(4+\sqrt{3}\right)}=\dfrac{3\sqrt{3}-1}{13}\)

Vậy giá trị của A khi \(x=4-2\sqrt{3}\) là \(\dfrac{3\sqrt{3}-1}{13}\)

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

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

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

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

=\(\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)

học tốt nhé anh trai

a: Sửa đề; \(P=\left(\dfrac{3x+3\sqrt{x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\right)\cdot\left(\dfrac{1}{1-\sqrt{x}}-1\right)\)

\(=\dfrac{3x+3\sqrt{x}-3-x+1+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{1-1+\sqrt{x}}{1-\sqrt{x}}\)

\(=\dfrac{3x+3\sqrt{x}-6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}}{1-\sqrt{x}}=\dfrac{3\sqrt{x}}{1-\sqrt{x}}\)

b: Để \(P=\sqrt{x}\) thì \(3\sqrt{x}=\sqrt{x}-x\)

\(\Leftrightarrow x+2\sqrt{x}=0\)

hay x=0

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 :))

a/ ĐKXĐ: \(x\ge0,x\ne1\)

\(P=\left(\dfrac{3}{\sqrt{x}-1}+\dfrac{\sqrt{x}-3}{x-1}\right):\left(\dfrac{x+2}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\)

= \(\dfrac{3\left(\sqrt{x}+1\right)+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{x+2-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

= \(\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

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

= \(\dfrac{4\sqrt{x}}{\sqrt{x}+1}\)

b/ Với \(x\ge0,x\ne1\)

Để \(P=\sqrt{x}-1\Leftrightarrow\dfrac{4\sqrt{x}}{\sqrt{x}+1}=\sqrt{x}-1\)

\(\Leftrightarrow4\sqrt{x}=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)

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

\(\Leftrightarrow\left(\sqrt{x}-2+\sqrt{5}\right)\left(\sqrt{x}-2-\sqrt{5}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-2+\sqrt{5}=0\\\sqrt{x}-2-\sqrt{5}=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2-\sqrt{5}\left(ktm\right)\\\sqrt{x}=2+\sqrt{5}\left(tm\right)\end{matrix}\right.\)

\(\Leftrightarrow x=9+4\sqrt{5}\)

Vậy để \(P=\sqrt{x}-1\) thì \(x=9+4\sqrt{5}\)

ĐKXĐ: \(x\ge0;x\ne1\)

Sửa lại đề chỗ \(\dfrac{\sqrt{x-1}}{\sqrt{x}+2}\) thành \(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)

\(P=\dfrac{3\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-1}{\sqrt{x}+2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\)

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

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

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

\(P=\dfrac{2\sqrt{x}+4-\sqrt{x}+1}{\sqrt{x}+2}=\dfrac{\sqrt{x}+5}{\sqrt{x}+2}=1+\dfrac{3}{\sqrt{x}+2}\)

Để P lớn nhất \(\Rightarrow\dfrac{3}{\sqrt{x}+2}\) lớn nhất

Mà \(\sqrt{x}+2\ge2\Rightarrow\dfrac{3}{\sqrt{x}+2}\le\dfrac{3}{2}\)

\(\Rightarrow P_{max}=1+\dfrac{3}{2}=\dfrac{5}{2}\) khi \(\sqrt{x}+2=2\Leftrightarrow x=0\)

XEM CÓ SAI ĐỀ BÀI KHÔNG, MK RÚT GỌN RA TO LẮM

\(=\dfrac{x+5\sqrt{x}+6-x+5\sqrt{x}-6}{\left(\sqrt{x}+3\right)^2\cdot\left(\sqrt{x}-3\right)}\cdot\dfrac{x-9}{\sqrt{x}}\)

\(=\dfrac{10\sqrt{x}}{\sqrt{x}}\cdot\dfrac{1}{\sqrt{x}+3}=\dfrac{10}{\sqrt{x}+3}\)