ừm cái này mình ko giải đc

Sorry nha

TL:
Bạn ơi cái Squart(2/3)

nghĩa là j v?

Mình ko hiểu

\(3x^4+4x^3-3x^2-2x+1=0\)

\(\Leftrightarrow3x^4+x^3-x^2+3x^3+x^2-x-3x^2-x+1=0\)

\(\Leftrightarrow x^2\left(3x^2+x-1\right)+x\left(3x^2+x-1\right)-\left(3x^2+x-1\right)=0\)

\(\Leftrightarrow\left(x^2+x-1\right)\left(3x^2+x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+x-1=0\left(1\right)\\3x^2+x-1=0\left(2\right)\end{cases}}\)

• ​\(\Delta_{\left(1\right)}=1^2-\left(-4\left(1.1\right)\right)=5\)

\(\Leftrightarrow x_{1,2}=\frac{-1\pm\sqrt{5}}{2}\left(tm\right)\)

• \(\Delta_{\left(2\right)}=1^2-\left(-4\left(3.1\right)\right)=13\)

\(x_{1,2}=\frac{-1\pm\sqrt{13}}{6}\left(tm\right)\)

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

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

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

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

b:

\(\dfrac{x}{12}=\dfrac{\left(\sqrt{5}+2\right)\sqrt[3]{17\sqrt{5}-38}}{\sqrt{5}+\sqrt{14-6\sqrt{5}}}\)

\(\Leftrightarrow x\cdot\dfrac{1}{12}=\dfrac{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}{\sqrt{5}+3-\sqrt{5}}\)

\(\Leftrightarrow\dfrac{x}{12}=\dfrac{1}{3}\)

=>x=36

Khi x=36 thì \(A=36-6+1=37-6=31\)

c: \(B=\dfrac{2\sqrt{x}}{A}=\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}\)

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

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

\(=\dfrac{-2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}< 0\)

=>B<2

\(2\sqrt{x}>0;x-\sqrt{x}+1=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)

=>B>0

=>0<B<2