a/ \(\frac{1}{2-\sqrt{3}}+\frac{3+\sqrt{3}}{\sqrt{3}}-\frac{4}{\sqrt{3}-1}\)

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

\(=1\)

b/ \(\sqrt{3x+40}-4=x\)

\(\sqrt{3x+40}=x+4\)

Điều kiện: \(\hept{\begin{cases}3x+40\ge0\\x+4\ge0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ge-\frac{40}{3}\\x\ge-4\end{cases}}\)

\(\Leftrightarrow x\ge-\frac{40}{3}\)

Ta có: \(3x+40=x^2+8x+16\)

\(\Leftrightarrow x^2+5x-24=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-8\left(l\right)\\x=3\end{cases}}\)

\(\frac{\sqrt{a}}{a+\sqrt{a}}+\frac{\sqrt{a}-1}{2\sqrt{a}}.\left(\frac{1}{a-\sqrt{a}}+\frac{1}{a+\sqrt{a}}\right)\)

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

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

\(=\frac{1}{\sqrt{a}+1}+\frac{1}{\sqrt{a}\left(\sqrt{a}+1\right)}\)

\(=\frac{1}{\sqrt{a}}=\frac{\sqrt{a}}{a}\)

Giả sử

\(\frac{30-2\sqrt{45}}{4}>\sqrt{17}\)

\(\Leftrightarrow15>2\sqrt{17}+\sqrt{45}\)

\(\Leftrightarrow225>113+4\sqrt{765}\)

\(\Leftrightarrow28>\sqrt{765}\)

\(\Leftrightarrow784>765\) (đúng)

Vậy \(\frac{30-2\sqrt{45}}{4}>\sqrt{17}\)

Giả sử:

\(\frac{30-2\sqrt{45}}{4}>\sqrt{17}\)

\(\Leftrightarrow15>2\sqrt{17}+\sqrt{45}\)

\(\Leftrightarrow225>113+4\sqrt{765}\)

\(\Leftrightarrow28>\sqrt{765}\)

\(\Leftrightarrow784>765\)(đúng)

Vậy \(\frac{30-2\sqrt{45}}{4}>\sqrt{17}\)

ĐK \(\hept{\begin{cases}x\ge0\\x\ne4;x\ne9\end{cases}}\)

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

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

\(=\frac{\sqrt{x}+2+x-9-x+4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}+1}{\sqrt{x}+2}=\frac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}+1}{\sqrt{x}+2}\)

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

b. Với \(x=4-2\sqrt{3}\Rightarrow P=\frac{\sqrt{4-2\sqrt{3}}+1}{4-2\sqrt{3}-4}=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}+1}{-2\sqrt{3}}\)

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

c. Để \(\frac{1}{P}\le\frac{-5}{2}\Leftrightarrow\frac{x-4}{\sqrt{x}+1}+\frac{5}{2}\le0\Leftrightarrow\frac{2x-8+5\sqrt{x}+5}{2\left(\sqrt{x}+1\right)}\le0\)

\(\Leftrightarrow\frac{2x+5\sqrt{x}-3}{2\left(\sqrt{x}+1\right)}\le0\Leftrightarrow2x+5\sqrt{x}-3\le0\)vì \(2\left(\sqrt{x}+1\right)\ge0\forall x\ge0\)

\(\Leftrightarrow\left(\sqrt{x}+3\right)\left(2\sqrt{x}-1\right)\le0\Leftrightarrow2\sqrt{x}-1\le0\Leftrightarrow0\le x\le\frac{1}{4}\left(tm\right)\)

Vậy với \(0\le x\le\frac{1}{4}\)thì \(\frac{1}{P}\le-\frac{5}{2}\)

d. Ta có \(B=P\left(\sqrt{x}-2\right)=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}+1}{\sqrt{x}+2}=1-\frac{1}{\sqrt{x}+2}\)

Gỉa sử \(B\in Z\Leftrightarrow\sqrt{x}+2\inƯ\left(1\right)\Leftrightarrow\sqrt{x}+2\in\left\{-1;1\right\}\Leftrightarrow x\in\left\{\phi\right\}\)

Vậy B không nhận giá trị nguyên với mọi x để P có nghĩa

=>\(\sqrt{\left(x+3\right)^2}\)+ \(\sqrt{\left(x+4\right)^2}\)+\(\sqrt{\left(x+5\right)^2}\)=9x

=> x + 3 + x + 4 + x + 5 = 9x

=> - 6x = - 12

=> x=2

Ủa sao phá đc trị tuyệt đối hay v bạn? (căn a^2 = trị tuyệt đối của a ) 

Điều kiện: 

\(3-2x\ge0\)

\(\Leftrightarrow x\le\frac{3}{2}\)

Ta có: \(3-2x>3\)

\(\Leftrightarrow x< 0\)