a, \(P=\left(\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}\right):\left(1-\frac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)ĐK : \(x\ge0;x\ne\frac{1}{9}\)

\(=\left(\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}+1+8\sqrt{x}}{9x-1}\right):\left(\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)

\(=\left(\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{9x-1}\right):\frac{3}{3\sqrt{x}+1}\)

\(=\frac{3x+3\sqrt{x}}{9x-1}:\frac{3}{3\sqrt{x}+1}=\frac{3\sqrt{x}\left(\sqrt{x}+1\right)\left(3\sqrt{x}+1\right)}{3\left(9x-1\right)}\)

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

Ta có : \(\frac{x+\sqrt{x}}{3\sqrt{x}-1}=\frac{6}{5}\Rightarrow5x+5\sqrt{x}=18\sqrt{x}-6\)

\(\Leftrightarrow5x-13\sqrt{x}+6=0\Leftrightarrow x=\frac{9}{25};4\)( tmđk ) 

(x - y)(x - 2y)(x - 3y)(x - 4y) + y⁴ 

= [(x - y)(x - 4y)].[(x - 2y)(x - 3)y] + y⁴ 

= (x² - 5xy + 4y²)(x² - 5xy + 6y²) + y⁴

= (x² - 5xy + 5y² - y²)(x² - 5xy + 5y² + y²) + y⁴

= (x² - 5xy + 5y²)² - y⁴ + y⁴

=  (x² - 5xy + 5y²)²

_{\(P=\sqrt{x}+\sqrt{y} \)
\(\Rightarrow P.\sqrt{2}=\sqrt{2x}+\sqrt{2y}\)}
\(\Rightarrow P\sqrt{2}=\sqrt{X\left(X+Y\right)}+\sqrt{\left(X+Y\right)Y}\)
\(\ge\sqrt{x.x}+\sqrt{y.y}=x+y=2\)
\(\Rightarrow Pmin=\frac{2}{\sqrt{2}}=\sqrt{2}\Leftrightarrow\orbr{\begin{cases}x=0,y=2\\y=0,x=2\end{cases}}\)

Giups gì zậy bạn ?

A B H C 15

Ta có : \(\frac{AB}{AC}=\frac{3}{5}\)( gt ) => \(AB=\frac{3}{5}AC\)

Xét tam giác ABC vuông tại A, đường cao AH

* Áp dụng hệ thức : 

\(\frac{1}{AH^2}=\frac{1}{AB^2}+\frac{1}{AC^2}\Rightarrow\frac{1}{15^2}=\frac{1}{\left(\frac{3}{5}AC\right)^2}+\frac{1}{AC^2}\Rightarrow AC=5\sqrt{34}\)cm 

\(\Rightarrow AB=\frac{3}{5}AC=\frac{3}{5}.5\sqrt{34}=3\sqrt{34}\)cm 

Theo định lí Pytago : \(BC^2=AB^2+AC^2=306+850=1156\Rightarrow BC=34\)cm 

* Áp dụng hệ thức : \(AB^2=BH.BC\Rightarrow BH=\frac{AB^2}{BC}=\frac{306}{34}=9\)cm 

=> CH = BC - BH = 34 - 9 = 25 cm