qua học 24 mà coi

\(3a^2+4ab+b^2=3a^2+3ab+ab+b^2=3a\left(a+b\right)+b\left(a+b\right)=\left(3a+b\right)\left(a+b\right)\)

xong AM -GM

\(a,A=\sqrt{27}+\frac{2}{\sqrt{3}-2}-\sqrt{\left(1-\sqrt{3}\right)^2}\)

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

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

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

       \(=-3\)

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

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

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

    \(=\frac{\sqrt{x}-1}{\sqrt{x}}\)

b, Ta có \(B< A\)

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

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

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

\(\Leftrightarrow\frac{4\sqrt{x}-1}{\sqrt{x}}< 0\)

\(\Leftrightarrow4\sqrt{x}-1< 0\left(Do\sqrt{x}>0\right)\)

\(\Leftrightarrow\sqrt{x}< \frac{1}{4}\)

\(\Leftrightarrow0< x< \frac{1}{2}\)(Kết hợp ĐKXĐ)

Vậy ...

a) \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\)

\(=\left[-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right]\cdot\left(\sqrt{2}-\sqrt{5}\right)\)

\(=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)

\(=-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)

\(=-\left(2-5\right)\)

\(=-\left(-3\right)\)

\(=3\)

b) Ta có:

\(x^2-x\sqrt{3}+1\) 

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

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

Mà: \(\left(x-\dfrac{\sqrt{3}}{2}\right)^2\ge0\forall x\) nên

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

Dấu "=" xảy ra:

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

\(\Leftrightarrow x=\dfrac{\sqrt{3}}{2}\)

Vậy: GTNN của biểu thức là \(\dfrac{1}{4}\) tại \(x=\dfrac{\sqrt{3}}{2}\)

a)

\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\\ =\left(-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =\left(-\sqrt{2}-\sqrt{5}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}^2-\sqrt{5}^2\right)\\ =-\left(2-5\right)\\ =-\left(-3\right)\\ =3\)

bn sử dụng bất đẳng thức cô si đi

Nguyễn Đại Nghĩa,bác nói cụ thể hơn được ko :v

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

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

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

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

Giúp mình câu b vớiiii. Câu a mình làm được A= \(\frac{\sqrt{x}}{x-\sqrt{x}+1}\)

a) \(\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\frac{2a+\sqrt{a}}{\sqrt{a}}+1\)

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

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

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

b) \(\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-2\sqrt{a}=2\)

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

\(\Leftrightarrow a^2-2\sqrt{a}.a+2a-\sqrt{a}-2a=2a-2\sqrt{a}+2\)

\(\Leftrightarrow a^2-2\sqrt{a}.a+2a-\sqrt{a}-2a=-2\sqrt{a}+2\)

\(\Leftrightarrow-2\sqrt{a}.a+2a-\sqrt{a}-2a=-2\sqrt{a}+2-a^2\)

\(\Leftrightarrow-2\sqrt{a}.a-\sqrt{a}=-2\sqrt{a}+2-a^2\)

\(\Leftrightarrow-2a\sqrt{a}+\sqrt{a}=2-a^2\)

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

\(\Leftrightarrow\left[\sqrt{a}.\left(2a+1\right)\right]^2=\left(2-a^2\right)^2\)

\(\Leftrightarrow4a^3-4a^2+a=4-4a^2+a^4\)

\(\Leftrightarrow\orbr{\begin{cases}a=4\left(\text{thỏa mãn}\right)\\a=1\left(\text{loại}\right)\end{cases}}\)

=> a = 4

Cách ngắn hơn :

\(đkxđ\Leftrightarrow x\ge0\)

\(A=\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\frac{2a+\sqrt{a}}{\sqrt{a}}+1\)

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

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

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

\(=a+\sqrt{a}-2\sqrt{a}=a-\sqrt{a}\)

\(b,A=2\Rightarrow a-\sqrt{a}=2\)

\(\Rightarrow a-\sqrt{a}-2=0\)

\(\Rightarrow a+\sqrt{a}-2\sqrt{a}-2=0\)

\(\Rightarrow\sqrt{a}\left(\sqrt{a}+1\right)-2\left(\sqrt{a}+1\right)=0\)

\(\Rightarrow\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\sqrt{a}=2\\\sqrt{a}=-1\end{cases}\Rightarrow\orbr{\begin{cases}a=4\\a\in\varnothing\end{cases}}}\)

\(\Rightarrow a=4\)

\(c,A=a-\sqrt{a}=\sqrt{a}^2-2.\sqrt{a}.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}\)

\(=\left(\sqrt{a}-\frac{1}{2}\right)^2-\frac{1}{4}\)

\(\Rightarrow A_{min}=-\frac{1}{4}\Leftrightarrow\left(\sqrt{a}-\frac{1}{2}\right)^2=0\)

\(\Rightarrow\sqrt{a}=\frac{1}{2}\Rightarrow a=\frac{1}{4}\)

Vậy với \(a=\frac{1}{4}\)thì A có giá trị nhỏ nhất là \(-\frac{1}{4}\)

Làm màu quá :))