`3xx(x-1/6)-2xx(x-1/3)=2/3`

`3x-1/2-2x+2/3=2/3`

`x=2/3-2/3+1/2`

`x=1/2`

`3x(x-1/6)-2x(x-1/3)=2/3`

`3x^2 - 1/2 x- 2x^2 + 2/3x = 2/3`

`x^2+1/6 x =2/3`

`x=(-1 \pm \sqrt97)/12`

Áp dụng \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:

\(\left|2x-1\right|+\left|2x-5\right|=\left|2x-1\right|+\left|5-2x\right|\ge\left|2x-1+5-2x\right|=\left|4\right|=4\)

Để \(\left|2x-1\right|+\left|2x-5\right|=4\) thì \(\left(2x-1\right)\left(5-2x\right)\ge0\)

TH1:\(2x-1\le0;5-2x\le0\)\(\Rightarrow\hept{\begin{cases}x\le\frac{1}{2}\\x\ge\frac{5}{2}\end{cases}}\)(loại)

TH2:\(2x-1\ge0;5-2x\ge0\)\(\Rightarrow\hept{\begin{cases}x\ge\frac{1}{2}\\x\le\frac{5}{2}\end{cases}}\)(thỏa mãn)

Áp dụng bất đẳng thức giá trị tuyệt đối ta có : 

\(\left|2x-1\right|+ \left|2x-5\right|=\left|2x-1\right|+\left|5-2x\right|=\left|2x-1+5-2x\right|=\left|4\right|=4\)

Dấu "=" xảy ra khi \(\left(2x-1\right)\left(5-2x\right)\ge0\)

Trường hợp 1 : 

\(2x-1\ge0\)\(\Rightarrow\)\(x\ge\frac{1}{2}\)\(;\)\(5-2x\ge0\)\(\Rightarrow\)\(x\le\frac{5}{2}\) 

\(\Leftrightarrow\)\(\frac{1}{2}\le x\le\frac{5}{2}\) ( thoã mãn )

Trường hợp 2 : 

\(2x-1\le0\)\(\Rightarrow\)\(x\le\frac{1}{2}\)\(;\)\(5-2x\le0\)\(\Rightarrow\)\(x\ge\frac{5}{2}\) ( loại )

Vậy \(\frac{1}{2}\le x\le\frac{5}{2}\)

Cả hai đều đúng hết ó giỏi ghê 

k ai đây ?

hello

ủa dễ mà.

???what

A+6/×^2-2×+1-3×^2/×^2+×2+1

Là sao

\(\frac{7^{x+2}+7^{x+1}+7^x}{57}=\frac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}\)

<=>\(\frac{7^x\left(7^2+7+1\right)}{57}=\frac{5^{2x}.\left(1+5+5^3\right)}{131}\)

<=>\(\frac{7^x.57}{57}=\frac{5^{2x}.131}{131}\)

<=>\(7^x=5^{2x}\)<=>\(7^x=10^x\)<=>x=0

Vậy x=0

1.Ta có: 3x(x - y) - (y - x)² = 3x(x - y) - (x - y)² = (3x - x + y)(x - y) = (2x + y)(x - y)

2. Ta có: 2x(1 - 2x) + (2x - 3)(2x + 3) = 5

=> 2x - 4x² + 4x² - 9 = 5

=> 2x = 5 + 9

=> 2x = 14

=> x = 14 : 2 = 7

\(3^{x+4}=9^{2x-1}\)

\(\Rightarrow3^{x+4}=3^{4x-2}\)

\(\Rightarrow x+4=4x-2\)

\(\Rightarrow3x=6\Rightarrow x=2\)

3x+4=92x−13x+4=92x−1

⇒3x+4=34x−2⇒3x+4=34x−2

1,(x-1)(y+5)=101

th1:x-1=101

<=>x=102

th2:y+5=101

<=>y=96

2,(x-2)(-y+5)=12

th1:x-2=12

<=>x=14

th2:-y+5=12

<=>-y=7

<=>y=-7

3,(x-5)(2x+6)<0

th1:x-5<0

<=>x<5

th2:2x+6<0

<=>2x<-6

<=>x<-3