ko bt tthg

= 1 nhe

\(\left(5\sqrt{8}-2\sqrt{18}+3\sqrt{50}\right)\left(2\sqrt{6}-3\sqrt{10}\right)\)

\(=\left(10\sqrt{2}-6\sqrt{2}+15\sqrt{2}\right)\left(2\sqrt{6}-3\sqrt{10}\right)\)

\(=19\sqrt{2}\left(2\sqrt{6}-3\sqrt{10}\right)=76\sqrt{3}-114\sqrt{5}\)

ĐK: \(x>0,x\ne1\).

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

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

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

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

\(x-\sqrt{x}+1=x-\sqrt{x}+\frac{1}{4}+\frac{3}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Dấu \(=\)khi \(x=\frac{1}{4}\).

\(Q=\frac{2\sqrt{x}}{x-\sqrt{x}+1}\Rightarrow Qx-Q\sqrt{x}+Q=2\sqrt{x}\)

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

Với \(Q=0\Rightarrow x=0\)không thỏa mãn. 

Với \(Q\ne0\): 

Đặt \(\sqrt{x}=t>0\).

\(Qt^2-t\left(Q+2\right)+Q=0\)

\(\Delta=\left(Q+2\right)^2-4Q^2=-3Q^2+4Q+4\)

Phương trình có nghiệm suy ra \(-3Q^2+4Q+4\ge0\Leftrightarrow-\frac{2}{3}\le Q\le2\)

mà \(Q\inℤ\)\(\Rightarrow Q\in\left\{0,1,2\right\}\).

Với từng giá trị \(Q\)ta thế trực tiếp tìm giá trị của \(x\).

\(a,\frac{x+2\sqrt{x}}{x-4}\)

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

\(\frac{\sqrt{x}}{\sqrt{x}-2}\)

\(b,\frac{x+10\sqrt{x}+25}{3x+15\sqrt{x}}\)

\(\frac{\left(\sqrt{x}+5\right)^2}{3\sqrt{x}\left(\sqrt{x}+5\right)}\)

\(\frac{\sqrt{x}+5}{3\sqrt{x}}\)

\(c,\frac{\sqrt{x}-3x}{9x-6\sqrt{x}+1}\)

\(\frac{-\sqrt{x}\left(3\sqrt{x}-1\right)}{\left(3\sqrt{x}-1\right)^2}\)

\(\frac{-\sqrt{x}}{3\sqrt{x}-1}\)

hình 9 hả bn