\(\left(2x-\frac{1}{x}\right)^3\)

\(=\left(2x\right)^3-3\left(2x\right)^2.\frac{1}{x}+3.2x\left(\frac{1}{x}\right)^2+\left(\frac{1}{x}\right)^3\)

\(=8x^3-6x^2+6x-\frac{1}{x^3}\)

chắc là x=0

mình tính rồi 

B=2 x 0^2 - 6 x 0 = 0

B=2x2−6x+9/2−9/2

=2(x−3/2)\(^2\)−9/2≥−9/2

Bmin=−9/2 khi x=3/2

mik chưa đến tầm ấy đâu ...

Trả lời:

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

2) \(\left(x+4\right)\left(x-2\right)-\left(x-3\right)^2=x^2-2x+4x-8-\left(x^2-6x+9\right)\)\(=x^2+2x-8-x^2+6x-9=8x-17\)

3)  \(3x\left(2x^3-3x^2+5\right)=6x^4-9x^3+15x\)

a) \(\left(x-3\right)^2+2\left(x-3\right)\left(x+2\right)+\left(x+2\right)^2\)

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

\(=\left(2x-1\right)^2\)

Hằng đẳng thức: \(\left(a+b\right)^2=a^2+2ab+b^2\).

b) \(\left(x+5\right)^2-\left(2x+10\right)\left(x-6\right)+\left(x-6\right)^2\)

\(=\left(x+5\right)^2-2\left(x+5\right)\left(x-6\right)+\left(x-6\right)^2\)

\(=\left[\left(x+5\right)-\left(x-6\right)\right]^2\)

\(=11^2=121\)

Hằng đẳng thức: \(\left(a-b\right)^2=a^2-2ab+b^2\).

a.\(\left(x-3\right)^2+2\left(x-3\right)\left(x+2\right)+\left(x+2\right)^2\)

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

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

\(=\left(2x-1\right)^2\)

b.\(\left(x+5\right)^2-\left(2x+10\right)\left(x-6\right)+\left(x-6\right)^2\)

\(=\left(x+5\right)^2-2\left(x+5\right)\left(x-6\right)+\left(x-6\right)^2\)

\(=\left[\left(x+5\right)-\left(x-6\right)\right]^2\)

\(=\left(x+5-x+6\right)^2\)

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

\(\ge2+2\sqrt{\left(x-1\right).\frac{1}{x-1}}=2+2=4\)

Dấu \(=\)khi \(x-1=\frac{1}{x-1}\Leftrightarrow x=2\)(vì \(x>1\)).

Vậy \(minC=4\)xảy khi khi \(x=2\).

Ta có: \(C=\frac{x^2}{x-1}\)

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

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

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

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

Nhận thấy \(x-1+\frac{1}{x-1}\ge2\sqrt{\left(x-1\right)\frac{1}{x-1}}=2\)

\(\Rightarrow A_{min}=4\)

Dấu "=" xảy ra khi :

\(x-1=\frac{1}{x-1}\)

\(\Leftrightarrow\left(x-1\right)^2=1\Leftrightarrow\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}}\)

Cre: mạng

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

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

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

\(=x-\sqrt{x}-2\)

\(2.\left(x+4\right)\left(x-2\right)-\left(x-3\right)^2\)

\(=x\left(x-2\right)+4\left(x-2\right)-\left(x^2-6x+9\right)\)

\(=x^2-2x+4x-8-x^2+6x-9\)

\(=8x-17\)