Đặt \(t=\sqrt{x}\) thì \(A=\dfrac{t}{t+5};B=\dfrac{2t}{t-4}-\dfrac{t^2+12t}{t^2-16}=\dfrac{2t\left(t+4\right)-t^2-12t}{t^2-16}=\dfrac{t^2-4t}{t^2-16}=\dfrac{t}{t+4}\)

\(\dfrac{A}{B}=\dfrac{t}{t+5}:\dfrac{t}{t+4}=\dfrac{t+4}{t+5}\) (với điều kiện \(t\ne0\)\(\Leftrightarrow x>0\))

1) Khi \(x=4\) thì \(t=2,A=\dfrac{2}{7}\).

2) \(B=\dfrac{t}{t+4}=\dfrac{\sqrt{x}}{\sqrt{x}+4}\).

3) \(\dfrac{A}{B}=\dfrac{5}{6}\Leftrightarrow\dfrac{t+4}{t+5}=\dfrac{5}{6}\)\(\Leftrightarrow6t+24=5t+25\)\(\Leftrightarrow t=1\)\(\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\).

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

A=\(\dfrac{\sqrt{x}+1}{\sqrt{x^2}+2.2.\sqrt{x}+2^2}:\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+2\right)}+\dfrac{x}{\sqrt{x}+2}\right)\)

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

A\(=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)^2}.\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{x+x\sqrt{x}}\)

A\(=\dfrac{\left(\sqrt{x}+1\right)\left[\sqrt{x}\left(\sqrt{x}+2\right)\right]}{\left(\sqrt{x}+2\right)^2.\left(x+x\sqrt{x}\right)}\)

A\(=\dfrac{\left(\sqrt{x}+1\right).\sqrt{x}}{\left(\sqrt{x}+2\right).\left[x\left(\sqrt{x}+1\right)\right]}\)

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

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

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

b) \(\dfrac{1}{x+2\sqrt{x}}\ge\dfrac{1}{3\sqrt{x}}\)

\(\Leftrightarrow\dfrac{1}{x+2\sqrt{x}}-\dfrac{1}{3\sqrt{x}}\ge0\)

\(\Leftrightarrow\dfrac{3\sqrt{x}-x-2\sqrt{x}}{\left(x+2\sqrt{x}\right)\left(3\sqrt{x}\right)}\ge0\)

\(\Leftrightarrow\dfrac{\sqrt{x}-x}{3x\sqrt{x}+6x}\ge0\)

\(\Leftrightarrow\dfrac{\sqrt{x}\left(1-\sqrt{x}\right)}{\sqrt{x}\left(3x+6\sqrt{x}\right)}\ge0\)

\(\Leftrightarrow\dfrac{1-\sqrt{x}}{3x+6\sqrt{x}}\ge0\)

1) ĐKXĐ: \(\left\{{}\begin{matrix}\sqrt{x}\ge0\\x-9\ne0\\\sqrt{x}-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)\(A=\left(\dfrac{2\sqrt{x}}{x-9}+\dfrac{1}{\sqrt{x}-3}\right):\dfrac{3}{\sqrt{x}-3}=\dfrac{2\sqrt{x}+\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{3}=\dfrac{3\sqrt{x}+3}{3\left(\sqrt{x}+3\right)}=\dfrac{3\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+3\right)}\)2) Để A=\(\dfrac{5}{6}\) thì \(\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+3\right)}=\dfrac{5}{6}\Leftrightarrow\left(\sqrt{x}+1\right)6=\left(\sqrt{x}+3\right)5\Leftrightarrow6\sqrt{x}+6=5\sqrt{x}+15\Leftrightarrow\sqrt{x}=9\Leftrightarrow x=81\)

1. Ta có:

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

\(=\dfrac{2\sqrt{x}.\left(\sqrt{x}-3\right)}{3\left(x-9\right)}+\dfrac{1}{3}\)

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

\(=\dfrac{3x-6\sqrt{x}-9}{3x-27}\)

\(=\dfrac{x-2\sqrt{x}-3}{x-9}\)

a: Thay x=4 vào P, ta được:

\(P=\dfrac{1}{2-3}+\dfrac{5}{2+3}-\dfrac{10\cdot2}{4-9}=-1+1-\dfrac{20}{-5}=4\)

b: \(P=\dfrac{\sqrt{x}+3+5\sqrt{x}-15-10\sqrt{x}}{x-9}\)

\(=\dfrac{-4\sqrt{x}-12}{x-9}=\dfrac{-4}{\sqrt{x}-3}\)

Bạn nào làm được bài này thì giúp mình với ạ ! mình đang cần gấp

Bài 4:

\(AH=\sqrt{9\cdot16}=12\left(cm\right)\)

\(AB=\sqrt{9\cdot25}=15\left(cm\right)\)

AC=căn(25^2-15^2)=20(cm)

Xét ΔABC vuông tại A có sin ABC=AC/BC=4/5

nên góc ABC=53 độ

Ta chứng minh: \(\sqrt[4]{5}\) là 1 nghiệm của phương trình

\(\dfrac{2}{\sqrt{4-3a+2a^2-a^3}}=a+1\)

\(\Leftrightarrow\dfrac{2}{4-3a+2a^2-a^3}=a^2+2a+1\)

\(\Leftrightarrow a\left(a^4-5\right)=0\)

\(\Rightarrow a=\sqrt[4]{5}\)

Từ đây ta suy ra được

\(x=\dfrac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125}}}=1+\sqrt[4]{5}\)

Ta lại có:

\(Q=\dfrac{1}{x^2+x}+\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+...+\dfrac{1}{x^2+4015x+4030056}\)

\(=\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(c+3\right)}+...+\dfrac{1}{\left(x+2007\right)\left(x+2008\right)}\)

\(=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+...+\dfrac{1}{x+2007}+\dfrac{1}{x+2008}\)

\(=\dfrac{1}{x}-\dfrac{1}{x+2008}=\dfrac{2008}{x^2+2008x}\)

Thế x vô nữa là xong

\(B=8x+\dfrac{6}{x}+18y+\dfrac{7}{y}=\left(8x+\dfrac{2}{x}\right)+\left(18y+\dfrac{2}{y}\right)+\left(\dfrac{4}{x}+\dfrac{5}{y}\right)\ge8+12+23=43\)

Dấu bằng xảy ra khi \(\left(x;y\right)=\left(\dfrac{1}{2};\dfrac{1}{3}\right)\)

Vậy, \(MinB\) là \(43\) khi \(\left(x;y\right)=\left(\dfrac{1}{2};\dfrac{1}{3}\right)\)