Question

bài 1: vs x,y,z là các số thực dương t/m xy+yz+xz=5 tìm min

\(p=\frac{3x+3y+3z}{\sqrt{6\left(x^2+5\right)}+\sqrt{6\left(y^2+5\right)}+\sqrt{z^2+5}}\)

bài 2 gpt

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

b)\(x^3-x^2-x=\frac{1}{3}\)

c)\(x^4+2x^3-6x^2+4x-1=0\)

Answer 1

Bài 2:

b)\(x^3-x^2-x=\frac{1}{3}\)

\(\Leftrightarrow x^3=x^2+x+\frac{1}{3}\)

\(\Leftrightarrow3x^3=3\left(x^2+x+\frac{1}{3}\right)\)

\(\Leftrightarrow3x^3=3x^2+3x+1\)

\(\Leftrightarrow4x^3=x^3+3x^2+3x+1\)

\(\Leftrightarrow4x^3=\left(x+1\right)^3\)\(\Leftrightarrow\sqrt[3]{4}x=x+1\)

\(\Leftrightarrow\sqrt[3]{4}x-x=1\)\(\Leftrightarrow x\left(\sqrt[3]{4}-1\right)=1\)

\(\Leftrightarrow x=\frac{1}{\sqrt[3]{4}-1}\)

 

Answer 2

c)\(x^4+2x^3-6x^2+4x-1=0\)

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

Ok...

Answer 3

help me>>>>

Answer 4

C2:a)CHia 2 vế cho x² thử

 

 

 

Answer 5

Ta có: \(x^3=-3x^2+3x-1\)

\(\Leftrightarrow2x^3=x^3-3x^2+3x-1\)

\(\Leftrightarrow2x^3=\left(x-1\right)^3\)

\(\Leftrightarrow\sqrt[3]{2}x=x-1\)

\(\Leftrightarrow\left(1-\sqrt[3]{2}\right)x=1\)

\(\Leftrightarrow x=\dfrac{1}{1-\sqrt[3]{2}}\)