Bạn có thể thấy 2x-1 là a , 3x+2 là b thì 2.(2x-1)(3x+2)=2ab

nên phương trình trên có thể dùng bình phương 1 tổng 

\(\left(2x-1\right)^2+\left(3x+2\right)^2-2.\left(2x-1\right).\left(3x+2\right)=\left[\left(2x-1\right)-\left(3x+2\right)\right]^2\)

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

cái cuối là  (x+3)² nhé, không phải (x+3²) đâu

Ta có :\(\left(2x-1\right)^2+\left(3x+2\right)^2-2\left(2x-1\right)\left(3x+2\right)\) \(=\left(3x+2-2x+1\right)^2\) \(=\left(x+3\right)^2\)

Ko cần đâu bn à mk mong bn đấy

a)\(\left(3x-1\right)\left(5-\frac{1}{2}x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-1=0\\5-\frac{1}{2}x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}\)

b)\(2\left|\frac{1}{2}x-\frac{1}{3}\right|-\frac{3}{2}=\frac{1}{4}\)

    \(2\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{4}\)

    \(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{8}\)

\(\Rightarrow\hept{\begin{cases}\frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\\frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{29}{12}\\x=-\frac{13}{12}\end{cases}}\)

a)\(\left(3x-1\right)\left(\frac{-1}{2}x+5\right)=0\)
\(\Leftrightarrow\)3x - 1 = 0      hay      \(\frac{-1}{2}\)x + 5 = 0
\(\Leftrightarrow\)3x     = 1         I\(\Leftrightarrow\)\(\frac{-1}{2}\)x     = -5
\(\Leftrightarrow\)  x     = \(\frac{1}{3}\)  I\(\Leftrightarrow\)            x     = 10

b) 2 I \(\frac{1}{2}x-\frac{1}{3}\)I - \(\frac{3}{2}\)=\(\frac{1}{4}\)
\(\Leftrightarrow\) 2 I\(\frac{1}{2}x-\frac{1}{3}\)I = \(\frac{7}{4}\)
\(\Leftrightarrow\)    I\(\frac{1}{2}x-\frac{1}{3}\)I = \(\frac{7}{8}\)
\(\Leftrightarrow\)\(\frac{1}{2}x-\frac{1}{3}\)= \(\frac{7}{8}\)          hay     \(\frac{1}{2}x-\frac{1}{3}\)= \(\frac{-7}{8}\)
\(\Leftrightarrow\)\(\frac{1}{2}x\)           = \(\frac{29}{24}\)        I\(\Leftrightarrow\)\(\frac{1}{2}x\)           = \(\frac{-13}{24}\)
\(\Leftrightarrow\)      x              = \(\frac{29}{12}\)        I\(\Leftrightarrow\)      x              = \(\frac{-13}{12}\)

c) (2x +\(\frac{3}{5}\))² - \(\frac{9}{25}\)= 0
\(\Leftrightarrow\)(2x +\(\frac{3}{5}\))²       = \(\frac{9}{25}\)
\(\Leftrightarrow\) 2x +\(\frac{3}{5}\)         = \(\frac{3}{5}\)    hay      2x +\(\frac{3}{5}\)= \(\frac{-3}{5}\)
\(\Leftrightarrow\) 2x                    = 0           I \(\Leftrightarrow\)2x           = \(\frac{-6}{5}\)
\(\Leftrightarrow\)   x                    = 0           I \(\Leftrightarrow\) x           = \(\frac{-3}{5}\)

d) 3(x -\(\frac{1}{2}\)) - 5(x +\(\frac{3}{5}\)) = -x + \(\frac{1}{5}\)
\(\Leftrightarrow\)3x - \(\frac{3}{2}\)- 5x - 3 = -x + \(\frac{1}{5}\)
\(\Leftrightarrow\)-2x + x - \(\frac{9}{2}\)- \(\frac{1}{5}\)= 0
\(\Leftrightarrow\)-x = \(\frac{-47}{10}\)
\(\Leftrightarrow\) x = \(\frac{47}{10}\)

1) \(x\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)\)

\(=x\left(x^2-16\right)\)

\(=x^3-16x-\left(x^2+1\right)\left(x^2-1\right)\)

\(=x^3-16x-x^4+1\)

b) \(7x\left(4y-x\right)+4y\left(y-7x\right)-2\left(2y^2-3.5x\right)\)

\(=28xy-7x^2+4y\left(y-7x\right)-2\left(2y^2-3.5x\right)\)

\(=28xy-7x^2+4y^2-28xy-4y^2+7x\)

\(=-7x^2+7x\)

c) \(\left(3x-1\right)\left(2x-5\right)-4\left(2x^2-5x+2\right)\)

\(=6x^2-17x+5-4\left(2x^2-5x+2\right)\)

\(=6x^2-17x+5-8x^2+20x-8\)

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

a)  x(x+4)(x-4)-(x²+1)(x²-1)

=>x(x²-4²)-(x⁴-1²)

=>x³-16x-x⁴+1

=>-x⁴-x³-15x

b)  7x(4y-x)+4y(y-7x)-2(2y²-3.5x)

=>28xy-7x²+4y²-28xy-4y²+30x

=>-7x²+30x

c)  (3x+1)(2x-5)-4(2x²-5x+2)

=>6x²-15x+2x-5-8x²+20x-8

=>-2x²+7x-13

\(\left(3x^2-2x+1\right)\left(x^2+2x+3\right)-4x\left(x^2-1\right)-3x^2\left(x^2+2\right)=3x^4+6x^3+9x^2-3x^3-4x^2-6x-4x^3+4x-3x^4-6x^2=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}2x^2-3x+4=2x-x^2-1\\2x^2-3x+4=-2x+x^2+1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x^2-3x+4-2x+x^2+1=0\\2x^2-3x+4+2x-x^2-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x^2-5x+5=0\\x^2-x+3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3\left(x^2-\frac{5}{3}x+\frac{25}{9}-\frac{25}{9}+\frac{5}{3}\right)=0\\x^2-2.x.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3\left(x-\frac{5}{3}^2\right)-\frac{10}{3}=0\\\left(x-\frac{1}{2}\right)^2+\frac{11}{4}>0\left(Loai\right)\end{cases}}\)

\(\Leftrightarrow\left(x\sqrt{3}-\frac{5\sqrt{3}}{3}\right)^2-\left(\frac{\sqrt{30}}{3}\right)^2=0\)

\(\Leftrightarrow\left(x\sqrt{3}-\frac{5\sqrt{3}}{3}-\frac{\sqrt{30}}{3}\right)\left(x\sqrt{3}-\frac{5\sqrt{3}}{3}+\frac{\sqrt{30}}{3}\right)=0\)

\(\Leftrightarrow\left(x\sqrt{3}-\frac{\sqrt{30}+5\sqrt{3}}{3}\right)\left(x\sqrt{3}+\frac{\sqrt{30}-5\sqrt{3}}{3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x\sqrt{3}-\frac{\sqrt{30}+5\sqrt{3}}{3}=0\\x\sqrt{3}+\frac{\sqrt{30}-5\sqrt{3}}{3}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5+\sqrt{10}}{3}\\x=\frac{5-\sqrt{10}}{3}\end{cases}}\)

Vậy ...

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

\(\Leftrightarrow\left|2x^2-3x+4\right|=\left|2x-x^2-1\right|\)

\(\Leftrightarrow\orbr{\begin{cases}2x^2-3x+4=2x-x^2-1\\2x^2-3x+4=x^2-2x+1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x^2-5x+5=0\\x^2-x+3=0\end{cases}}\)

\(TH1:3x^2-5x+5=0\)

Ta có: \(\Delta=5^2-4.3.5=-35< 0\)(vô nghiệm)

\(TH2:x^2-x+3=0\)

Ta có: \(\Delta=1^2-4.1.3=-11< 0\)(vô nghiệm)

Vậy pt vô nghiệm

3 . ( 2x - 1 ) - 2 = 13 

3 . ( 2x - 1 ) = 12 + 3 

3 . ( 2x - 1 ) = 15 

      2x - 1 = 15 : 3 

      2x - 1 = 5 

     2x = 5 + 1 = 6 

x = 6 : 2 = 3 

Vậy x = 3

\(3\left(2x-1\right)-2=13\)

\(3\left(2x-1\right)=15\)

\(2x-1=5\)

\(2x=6\)

\(x=3\)

3.(2x - 1 ) - 2 = 13

\(\Leftrightarrow\) 3.( 2x - 1) = 15

\(\Leftrightarrow\) 2x- 1 = 15 : 3

\(\Leftrightarrow\) 2x-1 = 5

\(\Leftrightarrow\)2x = 6

\(\Leftrightarrow\) x = 3

Vậy x = 3

|2x-1| >/  0 

|2x-1| -5 >/ -5

Vậy GTNN của B là -5.

|2x-1|> hoặc bằng 0

|2x-1-5> hoặc bằng -5

dấu bằng xảy ra khi x=3

Vậy Min B là -5 khi x=3