cây hỏi là gì bn ?

câu hỏi đâu bn

`Answer:`

\(\frac{2-x}{2001}-1=\frac{1-x}{2002}-\frac{x}{2003}\)

\(\Leftrightarrow\frac{2-x}{2001}+1-1-1=\frac{1-x}{2002}-\frac{x}{2003}\)

\(\Leftrightarrow\frac{2-x}{2001}+1=\frac{1-x}{2002}+1-\frac{x}{2003}+1\)

\(\Leftrightarrow\frac{2-x+2001}{2001}=\frac{1-x+2002}{2002}-\frac{x+2003}{2003}\)

\(\Leftrightarrow\frac{2003-x}{2001}=\frac{2003-x}{2002}+\frac{2003-x}{2003}\)

\(\Leftrightarrow\frac{2003-x}{2001}-\frac{2003-x}{2002}-\frac{2003-x}{2003}=0\)

\(\Leftrightarrow\left(2003-x\right)\left(\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)

\(\Leftrightarrow2003-x=0\)

\(\Leftrightarrow x=2003\)

`Answer:`

`(1+1/x)^3.(1+x^3)=16`

`<=>(x+1)^3.(1+x^3)=16x^3`

`<=>(x+1)^4.(x^2-x+1)=16x^3`

`<=>x^6+3x^5+3x^4+2x^3+3x^2+3x+1=16x^3`

`<=>x^6+3x^5+3x^4-14x^3+3x^3+3=0`

`<=>(x-1)^2.(x+1)^2.(x^2+x+1)=0`

`<=>x=1`

a; \(\Rightarrow6x-8x+3=8\Leftrightarrow-x=5\Leftrightarrow x=-5\)

b, \(\Rightarrow4x-8-6x+9=12x-12\Leftrightarrow14x=13\Leftrightarrow x=\frac{13}{14}\)

c, TH1 : x = 0 

TH2 : x + 1 = 0 <=> x = -1 

TH3 : x + 3/4 = 0 <=> x = -3/4 

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

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

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x-4\right)=0\)

TH1 : x = 1 ; TH2 : x = -2 ; TH3 : x = 4 

Câu hỏi đâu bạn

trả lời được cái gì bạn ?

con tim nha bạn

HT

TL : Con tim bn nha

ta có : 

\(\frac{x^2-3x+12}{x-1}=x-2+\frac{10}{x-1}=x-1+\frac{10}{x-1}-1\)

Ta cần điều kiện \(x-1>0\) thì biểu thức mới tồn tại giá trị nhỏ nhất. khi đó

áp dụng bất đẳng thức Cauchy ta có \(x-1+\frac{10}{x-1}-1\ge2\sqrt{10}-1\)

\(x.\left(2x-9\right)=3x.\left(x-5\right)\)

\(\Leftrightarrow2x^2-9x=3x^2-15x\)

\(\Leftrightarrow2x^2-9x-3x^2+15x=0\)

\(\Leftrightarrow-x^2+6x=0\)

\(\Leftrightarrow x.\left(-x+6\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\-x+6=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\-x=-6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

Vậy.....

$x.\left(2x-9\right)=3x.\left(x-5\right)$

$\iff 2x^2-9x=3x^2-15x$

$\iff 2x^2-9x-3x^2+15x=0$

$\iff -x^2+6x=0$

$\iff x.\left(-x+6\right)=0$

$\iff \{\begin{array}{c}x=0\\ -x+6=0\end{array}$

$\iff \{\begin{array}{c}x=0\\ -x=-6\end{array}$

$\iff \{\begin{array}{c}x=0\\ x=6\end{array}$

Vậy x=6.

HT