Ta có:\(B=x+\dfrac{1}{x}=\left(\dfrac{x}{4}+\dfrac{1}{x}\right)+\dfrac{3x}{4}\)

Áp dụng bất đẳng thức Cô-si ta có:

   \(\dfrac{x}{4}+\dfrac{1}{x}\ge2\sqrt{\dfrac{x}{4}\cdot\dfrac{1}{x}}=1\)

Ta có: \(\dfrac{3x}{4}\ge\dfrac{3.2}{4}=\dfrac{3}{2}\)

\(\Rightarrow B=1+\dfrac{3}{2}=\dfrac{5}{2}\)

Dấu "=" xảy ra ⇔ x=2

Vậy \(MinB=\dfrac{5}{2}\Leftrightarrow x=2\)

\(B=x+\dfrac{1}{x}=\left(\dfrac{x}{4}+\dfrac{1}{x}\right)+\dfrac{3}{4}x\ge2\sqrt{\dfrac{x}{4x}}+\dfrac{3}{4}.2=1+\dfrac{3}{2}=\dfrac{5}{2}\)(do \(x\ge2\))

\(minB=\dfrac{5}{2}\Leftrightarrow x=2\)

\(A=x+\dfrac{1}{x}=x+\dfrac{1}{16x}+\dfrac{15}{16x}\ge2\sqrt{x.\dfrac{1}{16x}}+\dfrac{15}{16x}\ge\dfrac{1}{2}+\dfrac{15}{4}=\dfrac{17}{4}\)(do \(x\le\dfrac{1}{4}\Rightarrow\dfrac{15}{16x}\le\dfrac{15}{4}\))

\(minA=\dfrac{17}{4}\Leftrightarrow x=\dfrac{1}{4}\)

\(\dfrac{3}{4}:x+\dfrac{1}{2}:\dfrac{1}{4}=4\)

\(\Rightarrow\dfrac{3}{4}:x+2=4\)

\(\Rightarrow\dfrac{3}{4}:x=2\Rightarrow x=\dfrac{3}{8}\)

\(\dfrac{3}{4}:x+\dfrac{1}{2}:\dfrac{1}{4}=4\)

\(\Leftrightarrow\dfrac{3}{4}:x=2\)

hay \(x=\dfrac{3}{8}\)

Tham khảo: Tìm x, y biết x^2+y^2+1/x^2+1/y^2=4 - thanh duy

Để \(B< 0\) mà 30>0\(\Rightarrow x-2< 0\Rightarrow x< 2\)

x\(\le\)31 (\(\forall\)x)
và x\(\notin\left\{-4,4\right\}\) thì B<0

k=-4x6=-24

=>x=-24/y

\(\Leftrightarrow x=-24:\dfrac{12}{5}=-24\cdot\dfrac{5}{12}=-10\)

1) b) \(\left(x-3y\right)^2+6\left(x-3\right)+9=\left(x-3y+3\right)^2\)

    c) \(x^2+x+\dfrac{1}{4}=\left(x+\dfrac{1}{2}\right)^2\)

2) \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=11\)

\(\Rightarrow x^2+6x+9-x^2+4=11\)

\(\Rightarrow6x=-2\Rightarrow x=-\dfrac{1}{3}\)

Lời giải:

Áp dụng BĐT AM-GM

$A=\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{x}}\geq 2\sqrt[4]{\frac{1}{xy}}$

Cũng áp dụng AM-GM:

$4=x+y\geq 2\sqrt{xy}\Rightarrow xy\leq 4$

Do đó: $A\geq 2\sqrt[4]{\frac{1}{xy}}\geq 2\sqrt[4]{\frac{1}{4}}=\sqrt{2}$

Vậy $A_{\min}=\sqrt{2}$ khi $x=y=2$

\(x^2=900\Leftrightarrow x^2=30^2\Rightarrow x=30\)

Chọn A

mik nghĩ A