G = \(\dfrac{x^2}{x-1}\)

= \(\dfrac{x^2-4x+4+4x-4}{x-1}\)

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

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

Vì x>1 nên \(\left\{{}\begin{matrix}\left(x-2\right)^2\text{≥}0\\x-1>0\end{matrix}\right.\)

=> G ≥ 4

=> G = 4 đạt GTNN

Dấu bằng xảy ra <=> \(\left(x-2\right)^2=0\)

<=> \(x=2\)

\(Do\) \(x>2\)

\(=>\left\{{}\begin{matrix}x-2\text{ ≥0}\\2x-1>0\end{matrix}\right.\)

\(=>\left(x-2\right)\left(2x-1\right)\text{ ≥0}\)

\(< =>2x^2-5x+2\text{≥}0\)

\(< =>2x^2+2\text{≥}5x\)

\(< =>2x+\dfrac{2}{x}\text{≥}5\)

\(< =>x+\dfrac{1}{x}\text{≥}2,5\)

\(< =>H\text{≥}2,5\)

\(< =>H=5\) \(đạt\) \(GTNN\)

Dấu bằng xảy ra khi \(x-2=0< =>x=2\)

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

\(=\dfrac{53x^3}{54}+\left(\dfrac{x^2}{54}+\dfrac{1}{2x}+\dfrac{1}{2x}\right)\)

Áp dụng BĐT Cô si cho 3 số dương

\(\dfrac{x^2}{54}+\dfrac{1}{2x}+\dfrac{1}{2x}\text{≥}3.\sqrt[3]{\dfrac{x^2}{54}.\dfrac{1}{2x}.\dfrac{1}{2x}}\)\(\text{≥}\dfrac{53.9}{54}+3.\sqrt[3]{54.4}\)\(=\dfrac{28}{3}\)

Dấu bằng xảy ra khi \(\left\{{}\begin{matrix}\dfrac{x^2}{54}=\dfrac{1}{2x}=\dfrac{1}{2x}\\x=3\end{matrix}\right.\)\(< =>x=3\)

Lời giải:
a. 

$2x^3-3x-1=2.1^3-3.1-1=2-3-1=-2$
b.

$xy^2-\frac{1}{2}y-x^3=2.(-4)^2-\frac{1}{2}(-4)-2^3=26$

c.

$x=|1|=1$

$3x^2-4x-1=3.1^2-4.1-1=3-4-1=-2$

d.

$(x+1)^2+(y-2)^2=(2+1)^2+(-3-2)^2=3^2+(-5)^2=9+25=34$

a) \(\sqrt{36\left(x-5\right)^2}\left(x\ge5\right)=6\left|x-5\right|=6\left(x-5\right)=6x-30\)

b) \(\sqrt{\dfrac{1}{4}\left(1-x\right)^2}\left(x>1\right)=\dfrac{1}{2}\left|1-x\right|=\dfrac{1}{2}\left(x-1\right)=\dfrac{1}{2}x-\dfrac{1}{2}\)

c) \(\sqrt{x^2\left(2x-4\right)^2}\left(x\ge2\right)=\left|x\left(2x-4\right)\right|=x\left(2x-4\right)=2x^2-4x\)

d) \(\dfrac{1}{x}\sqrt{x^2\left(1+x\right)^2}\left(x< -1\right)=\dfrac{1}{x}\left|x\left(1+x\right)\right|=\dfrac{1}{x}x\left(1+x\right)=1+x\)

Lời giải:
1. Áp dụng BĐT Cô-si

$G=\frac{x^2}{x-1}=\frac{(x^2-1)+1}{x-1}=x+1+\frac{1}{x-1}$

$=(x-1)+\frac{1}{x-1}+2$
$\geq 2\sqrt{(x-1).\frac{1}{x-1}}+2=2+2=4$ 

Vậy $G_{\min}=4$. Giá trị này đạt tại $x-1=\frac{1}{x-1}$

$\Leftrightarrow x=0$ hoặc $x=2$

2.

Áp dụng BĐT Cô-si:

$H=x+\frac{1}{x}=(\frac{x}{4}+\frac{1}{x})+\frac{3}{4}x$

$\geq 2\sqrt{\frac{x}{4}.\frac{1}{x}}+\frac{3}{4}x$
$=1+\frac{3}{4}x\geq 1+\frac{3}{4}.2=\frac{5}{2}$ (do $x\geq 2$)

Vậy $H_{\min}=\frac{5}{2}$. Giá trị này đạt tại $x=2$
 

3.

Áp dụng BĐT Cô-si:

$K=x^2+\frac{1}{x}=(\frac{x^2}{54}+\frac{1}{2x}+\frac{1}{2x})+\frac{53}{54}x^2$

$\geq 3\sqrt[3]{\frac{x^2}{54}.\frac{1}{2x}.\frac{1}{2x}}+\frac{53}{54}x^2$
$=\frac{1}{2}+\frac{53}{54}x^2\geq \frac{1}{2}+\frac{53}{54}.3^2=\frac{28}{3}$ (do $x\geq 3$)

Vậy $K_{\min}=\frac{28}{3}$ khi $x=3$

1.

\(f\left(x\right)=\dfrac{4}{x}+\dfrac{x-1+1}{1-x}=\dfrac{2^2}{x}+\dfrac{1}{1-x}-1\ge\dfrac{\left(2+1\right)^2}{x+1-x}-1=8\)

\(f\left(x\right)_{min}=8\) khi \(x=\dfrac{2}{3}\)

2.

\(f\left(x\right)=\dfrac{1}{x}+\dfrac{1}{1-x}\ge\dfrac{4}{x+1-x}=4\)

\(f\left(x\right)_{min}=4\) khi \(x=\dfrac{1}{2}\)

x=23x=23

2.

x=12

a) A= \(3x^2 - 2x+1\) với |x| = \(\dfrac{1}{2}\)

Với |x| = \(\dfrac{1}{2}\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Khi \(x=\dfrac{1}{2}\) ⇒ \(A=3.\left(\dfrac{1}{2}\right)^2-2.\dfrac{1}{2}+1=\dfrac{3}{4}-1+1=\dfrac{3}{4}\)

Khi \(x=-\dfrac{1}{2}\) ⇒ \(A=3.\left(-\dfrac{1}{2}\right)^2-2.\left(-\dfrac{1}{2}\right)+1=\dfrac{3}{4}+1+1=\dfrac{3}{4}+2=\dfrac{11}{4}\)

Vậy...

a: Thay x=2 vào B, ta được:

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

Chọn A