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

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

b) Với \(a=3-2\sqrt{2}\)(tmđk)

\(A=\dfrac{1}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}=\dfrac{\sqrt{a}-1}{a-1}\)

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

\(a,ĐK:x\ne-3;x\ne0;y\ne0\\ b,A=\dfrac{1}{x^2\left(x+3\right)+y^2\left(x+3\right)}=\dfrac{1}{\left(x^2+y^2\right)\left(x+3\right)}\\ x=y=0\Leftrightarrow A\in\varnothing\)

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

b) Thay x=0 vào A, ta được:

\(A=\dfrac{15\cdot\sqrt{0}-11}{0+2\sqrt{0}-3}-\dfrac{3\sqrt{0}-2}{\sqrt{0}-1}-\dfrac{2\sqrt{0}+3}{\sqrt{0}+3}\)

\(=\dfrac{-11}{-3}-\dfrac{-2}{-1}-\dfrac{3}{3}\)

\(=\dfrac{11}{3}-2-1\)

\(=\dfrac{11}{3}-\dfrac{9}{3}=\dfrac{2}{3}\)

Bài 5: 

a: Thay \(x=4+2\sqrt{3}\) vào E, ta được:

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

b: Để E<1 thì E-1<0

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

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

hay x<9

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne1\end{matrix}\right.\)

c: Để E nguyên thì \(4⋮\sqrt{x}-3\)

\(\Leftrightarrow\sqrt{x}-3\in\left\{-2;1;2;4\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{4;5;7\right\}\)

hay \(x\in\left\{16;25;49\right\}\)

Câu 2:
a) Ta có \(x=4-2\sqrt{3}\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{3}-2\right)^2}=\sqrt{3}-2\)

Thay \(x=\sqrt{3}-1\) vào \(B\), ta được

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

b) Để \(B\) âm thì \(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\) mà \(\sqrt{x}+1\ge1>0\forall x\) \(\Rightarrow\sqrt{x}-2< 0\Rightarrow\sqrt{x}< 2\Rightarrow x< 4\)

c) Ta có \(B=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=1-\dfrac{3}{\sqrt{x}+1}\)

Với mọi \(x\ge0\) thì \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+1\ge1\Rightarrow\dfrac{3}{\sqrt{x}+1}\le3\Rightarrow B=1-\dfrac{3}{\sqrt{x}+1}\ge-2\)

Dấu "=" xảy ra khi \(\sqrt{x}+1=1\Leftrightarrow x=0\)

Vậy \(B_{min}=-2\) khi \(x=0\)

1) Áp dụng bđt Cauchy cho 3 số dương ta có

 \(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{x}+x^3\ge4\sqrt[4]{\dfrac{1}{x}.\dfrac{1}{x}.\dfrac{1}{x}.x^3}=4\) (1)

\(\dfrac{3}{y^2}+y^2\ge2\sqrt{\dfrac{3}{y^2}.y^2}=2\sqrt{3}\) (2)

\(\dfrac{3}{z^3}+z=\dfrac{3}{z^3}+\dfrac{z}{3}+\dfrac{z}{3}+\dfrac{z}{3}\ge4\sqrt[4]{\dfrac{3}{z^3}.\dfrac{z}{3}.\dfrac{z}{3}.\dfrac{z}{3}}=4\sqrt{3}\) (3)

Cộng (1);(2);(3) theo vế ta được

\(\left(\dfrac{3}{x}+\dfrac{3}{y^2}+\dfrac{3}{z^3}\right)+\left(x^3+y^2+z\right)\ge4+2\sqrt{3}+4\sqrt{3}\)

\(\Leftrightarrow3\left(\dfrac{1}{x}+\dfrac{1}{y^2}+\dfrac{1}{z^3}\right)\ge3+4\sqrt{3}\)

\(\Leftrightarrow P\ge\dfrac{3+4\sqrt{3}}{3}\)

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{x}=x^3\\\dfrac{3}{y^2}=y^2\\\dfrac{3}{z^3}=\dfrac{z}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\sqrt[4]{3}\\z=\sqrt{3}\end{matrix}\right.\) (thỏa mãn giả thiết ban đầu)

2) Ta có \(4\sqrt{ab}=2.\sqrt{a}.2\sqrt{b}\le a+4b\)

Dấu"=" khi a = 4b

nên \(\dfrac{8}{7a+4b+4\sqrt{ab}}\ge\dfrac{8}{7a+4b+a+4b}=\dfrac{1}{a+b}\)

Khi đó \(P\ge\dfrac{1}{a+b}-\dfrac{1}{\sqrt{a+b}}+\sqrt{a+b}\)

Đặt \(\sqrt{a+b}=t>0\) ta được

\(P\ge\dfrac{1}{t^2}-\dfrac{1}{t}+t=\left(\dfrac{1}{t^2}-\dfrac{2}{t}+1\right)+\dfrac{1}{t}+t-1\)

\(=\left(\dfrac{1}{t}-1\right)^2+\dfrac{1}{t}+t-1\)

Có \(\dfrac{1}{t}+t\ge2\sqrt{\dfrac{1}{t}.t}=2\) (BĐT Cauchy cho 2 số dương)

nên \(P=\left(\dfrac{1}{t}-1\right)^2+\dfrac{1}{t}+t-1\ge\left(\dfrac{1}{t}-1\right)^2+1\ge1\)

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{t}-1=0\\t=\dfrac{1}{t}\end{matrix}\right.\Leftrightarrow t=1\)(tm)

khi đó a + b = 1

mà a = 4b nên \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)

Vậy MinP = 1 khi \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)

a: Thay x=16 vào A, ta được:

\(A=\dfrac{2\cdot4}{4+3}=\dfrac{8}{7}\)

a, Ta có: \(\sqrt[6]{a^4}=\sqrt[3]{\sqrt{a^4}}=\sqrt[3]{\sqrt{\left(a^2\right)^2}}=\sqrt[3]{\left|a^2\right|}=\sqrt[3]{a^2}\)

Vậy \(\sqrt[6]{a^4}=\sqrt[3]{a^2}\)

b, \(\sqrt[3]{a^2}=\sqrt[9]{a^6}=\sqrt[12]{a^8}\)