1: \(\left(2x-2\right)\left(3x+1\right)-\left(3x-2\right)\left(2x-3\right)=5\)

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

=>\(6x^2-4x-2-6x^2+13x-6=5\)

=>9x-8=5

=>9x=13

=>\(x=\frac{13}{9}\)

2: \(\left(1-3x\right)\left(3x-5\right)-\left(2x-4\right)\left(2-3x\right)=x-6\)

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

=>\(-9x^2+18x-5+6x^2-4x-12x+8=x-6\)

=>\(-3x^2+2x+3-x+6=0\)

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

=>\(3x^2-x-9=0\)

=>\(x^2-\frac13x-3=0\)

=>\(x^2-2\cdot x\cdot\frac16+\frac{1}{36}-\frac{109}{36}=0\)

=>\(\left(x-\frac16\right)^2=\frac{109}{36}\)

=>\(x-\frac16=\pm\frac{\sqrt{109}}{6}\)

=>\(x=\frac16\pm\frac{\sqrt{109}}{6}\)

3: \(\left(2x-1\right)\left(4x^2+2x+1\right)-\left(2x+1\right)\left(4x^2-2x+1\right)=5x+6\)

=>\(8x^3-1-8x^3-1=5x+6\)

=>5x+6=-2

=>5x=-8

=>\(x=-\frac85\)

Đặt bt trên là A nha

Đổi |x-1|=|1-x|

Suy ra A=|1-x|+x-2|+|x-3|

Áp dụng BĐTGTTĐ ta có

A=|1-x|+x-2|+|x-3|\(\ge\)|1-x+x-3|=2

Dấu = xảy ra khi   \(\hept{\begin{cases}x-2=0\\1< x< 3\end{cases}}\)đồng thời xảy ra

Vậy x =2

b,

\(\left|3x+\frac{1}{2}\right|\ge0\)

\(\left|3x+\frac{1}{6}\right|\ge0\)

..........

\(\left|3x+380\right|\ge0\)

Suy ra đề bài \(\ge\)0

 suy ra 58x \(\ge\)0

Suy ra \(3x+\frac{1}{2}+3x+\frac{1}{6}+......+3x+380=58x\)

Tự tính nhé hok tốt

a) |2x - 1| = |x - 2|

TH1: 2x - 1 = x - 2 => 2x - x = -2 + 1 => x = 1

TH2: 2x - 1 = -x + 2 => 2x + x = 2 + 1 => 3x = 3 => x = 1 (như trên)

=> x = 1 

b) |3x - 2| = |1 - x|

TH1: 3x - 2 = 1 - x => 3x + x = 1 + 2 => 4x = 3 => x = 3/4

TH2: 3x - 2 = -1 + x => 3x - x = -1 + 2 => 2x = 1 => x = 1/2

=> x = 3/4;1/2

c) |5x - 6| = |3x - 4|

TH1: 5x - 6 = 3x - 4 => 5x - 3x = -4 + 6 => 2x = 2 => x = 1

TH2: 5x - 6 = -3x + 4 => 5x + 3x = 4 + 6 => 8x = 10 => x = 5/4

=> x = 1;5/4 

Ta có:I2X-1 I- I X-2I=0 

TH1:I 2X-1I =0 => 2X-1=0 => 2X = 1 => X=1/2

TH2: I X-2 I =0 => X=2

Vậy x= 1/2 và 2

Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu

a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14) 

=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84

=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84) 

=> 156 -  56x = 24x - 324 

=>  24x + 56x = 324 + 156 

=> 80x = 480 

=> x = 480 : 80 =  6 

Vậy x = 6 

b

\(\left|6+x\right|\ge0;\left(3+y\right)^2\ge0\Rightarrow\left|6+x\right|+\left(3+y\right)^2\ge0\)

Suy ra \(\left|6+x\right|+\left(3+y\right)^2=0\)\(\Leftrightarrow\hept{\begin{cases}6+x=0\\3+y=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-6\\y=-3\end{cases}}\)

a

Ta có:\(\left|3x-12\right|=3x-12\Leftrightarrow3x-12\ge0\Leftrightarrow3x\ge12\Leftrightarrow x\ge4\)

\(\left|3x-12\right|=12-3x\Leftrightarrow3x-12< 0\Leftrightarrow3x< 12\Leftrightarrow x< 4\)

Với \(x\ge4\) ta có:

\(3x-12+4x=2x-2\)

\(\Rightarrow5x=10\)

\(\Rightarrow x=2\left(KTMĐK\right)\)

Với  \(x< 4\) ta có:

\(12-3x+4x=2x-2\)

\(\Rightarrow10=x\left(KTMĐK\right)\)

a) \(\text{​​}/3x-5/-\frac{1}{7}=\frac{1}{3}\)                           b)\(\left(\frac{3}{5}x-\frac{2}{3}x-x\right).\frac{1}{7}=\frac{-5}{21}\)

  \(/3x-5/=\frac{10}{21}\)                                           \([x.\left(\frac{3}{5}-\frac{2}{3}-1\right)]=\frac{-5}{21}.7\)

 \(\Rightarrow3x-5=\frac{10}{21}hay3x-5=\frac{-10}{21}\)         \(\left[x.\frac{-16}{15}\right]=\frac{-5}{3}\)

\(3x=\frac{115}{21}\)                \(3x=\frac{95}{21}\)                         \(x=\frac{25}{16}\)

\(x=\frac{115}{63}\)                  \(x=\frac{95}{63}\)                             Vậy x = \(\frac{25}{16}\)

                      Vậy x \(\in\left\{\frac{115}{63};\frac{95}{63}\right\}\)