=> :# -> :3 (đúng khong dậy mẹ:>?)

\(\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{3}{\sqrt{x}-3}+\dfrac{6\sqrt{x}}{x-9}\left(dkxd:x\ge0,x\ne9\right)\)

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

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

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

\(=\dfrac{x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{x-9}{x-9}=1\)

\(a,M=\left(\dfrac{x+\sqrt{x}+10}{x-9}-\dfrac{1}{\sqrt{x}-3}\right):\dfrac{1}{\sqrt{x}-3}\left(dkxd:x\ge0,x\ne9\right)\)

\(=\left(\dfrac{x+\sqrt{x}+10}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{1}{\sqrt{x}-3}\right).\left(\sqrt{x}-3\right)\)

\(=\dfrac{x+\sqrt{x}+10-\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\left(\sqrt{x}-3\right)\)

\(=\dfrac{x+7}{\sqrt{x}+3}\)

\(b,\) Sửa : Tìm giá trị của M khi x = 25

\(x=25\Rightarrow M=\dfrac{25+7}{\sqrt{25}+3}=\dfrac{32}{5+3}=4\)

\(c,\) \(M< \sqrt{x}+1\Leftrightarrow\dfrac{x+7}{\sqrt{x}+3}< \sqrt{x}+1\)

\(\Leftrightarrow\dfrac{x+7}{\sqrt{x}+3}-\left(\sqrt{x}+1\right)< 0\)

\(\Leftrightarrow\dfrac{x+7-\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}< 0\)

\(\Leftrightarrow x+7-x-3\sqrt{x}-\sqrt{x}-3< 0\)

\(\Leftrightarrow-4\sqrt{x}< -4\)

\(\Leftrightarrow\sqrt{x}>1\)

\(\Leftrightarrow x>1\)

Vậy để \(M< \sqrt{x}+1\) thì \(x>1\)

Bài 13 :

\(a,x=9\Rightarrow A=\dfrac{3}{\sqrt{9}-1}+\dfrac{\sqrt{9}+3}{9-1}=\dfrac{3}{3-1}+\dfrac{3+3}{8}=\dfrac{3}{2}+\dfrac{6}{8}=\dfrac{9}{4}\)

\(b,\) \(dkxd:x\ge0,x\ne1\)

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

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

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

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

\(=\dfrac{4\sqrt{x}+6}{x-1}\)

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

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

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

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

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

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

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

Ta có : \(M=A:B=\dfrac{4\sqrt{x}+6}{x-1}:\dfrac{\sqrt{x}+2}{x+\sqrt{x}-2}\)

\(=\dfrac{4\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{x-\sqrt{x}+2\sqrt{x}-2}{\sqrt{x}+2}\)

\(=\dfrac{4\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)}{\sqrt{x}+2}\)

\(=\dfrac{4\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+2}\)

\(=\dfrac{4\sqrt{x}+6}{\sqrt{x}+1}\)

Vậy \(M=\dfrac{4\sqrt{x}+6}{\sqrt{x}+1}\)

Bài 9 :

\(a,\) \(x=25\Rightarrow A=\dfrac{2\sqrt{25}}{\sqrt{25}+3}=\dfrac{2.5}{5+3}=\dfrac{10}{8}=\dfrac{5}{4}\)

\(b,M=A+B=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{11\sqrt{x}-3}{x-9}\left(dkxd:x\ge0,x\ne9\right)\)

\(=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)+11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}+\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{3x+9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

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

\(=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)

Bài 10 :

\(a,x=4\Rightarrow A=\dfrac{4-\sqrt{4}}{4-1}=\dfrac{4-2}{3}=\dfrac{2}{3}\)

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

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

\(c,\) \(\dfrac{\sqrt{x}}{\sqrt{x}+1}>\dfrac{\sqrt{x}-2}{\sqrt{x}}\Rightarrow A>B\)

Huhu, em cám ơn ạ.

\(a,\) Rút gọn \(Q=\dfrac{x^3+y^3}{x^2-xy+y^2}.\dfrac{x+y}{x^2-y^2}\left(dkxd:x\ne y\right)\)

\(=\dfrac{\left(x+y\right)\left(x^2-xy+y^2\right)}{x^2-xy+y^2}.\dfrac{x+y}{\left(x-y\right)\left(x+y\right)}\)

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

\(=\dfrac{x+y}{x-y}\)

Vậy \(Q=\dfrac{x+y}{x-y}\) với \(x\ne y\)

\(b,\) Thay \(x=\sqrt{7-4\sqrt{3}}=\sqrt{\left(2-\sqrt{3}\right)^2}=\left|2-\sqrt{3}\right|=2-\sqrt{3}\)

             \(y=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}=\left|\sqrt{3}-1\right|=\sqrt{3}-1\)

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

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

Vậy \(P=\dfrac{1}{3-2\sqrt{3}}\)

\(3x^2+4x-7=0\)

\(a,\) Để pt có 2 nghiệm phân biệt thì \(\Delta>0\Rightarrow4^2-4.3.\left(-7\right)=100>0\)

Vậy pt có 2 nghiệm phân biệt \(x_1,x_2\)

\(b,\)Theo Vi-ét, ta có :

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=-\dfrac{4}{3}\\x_1x_2=\dfrac{c}{a}=-\dfrac{7}{3}\end{matrix}\right.\)

Ta có : \(2x_1-\left(x_1-x_2-x_1x_2\right)\)

\(=2x_1-x_1+x_2-x_1x_2\)

\(=x_1+x_2-x_1x_2\)

\(=-\dfrac{4}{3}-\left(-\dfrac{7}{3}\right)\)

\(=-\dfrac{4}{3}+\dfrac{7}{3}\)

\(=\dfrac{3}{3}=1\)

Vậy giá trị của biểu thức là \(1\)