a. 3x² - 2x - 1 = 0

<=> 3x² - 3x + x - 1 = 0

<=> 3x(x - 1) + (x - 1) = 0

<=> (3x + 1)(x - 1) = 0

<=> \(\left[{}\begin{matrix}3x+1=0\\x-1=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=1\end{matrix}\right.\)

b. \(\dfrac{x+1}{3}+\dfrac{2x+3}{5}=\dfrac{3}{4}\)

<=> \(\dfrac{20\left(x+1\right)}{60}+\dfrac{12\left(2x+3\right)}{60}=\dfrac{45}{60}\)

<=> 20x + 20 + 24x + 36 = 45

<=> 44x = -11

<=> x = \(-\dfrac{1}{4}\)

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

    \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)

b) Pt\(\Rightarrow\)\(5\cdot4\left(x+1\right)+3\cdot4\cdot\left(2x+3\right)=3\cdot3\cdot5\)

       \(\Leftrightarrow44x=-11\Rightarrow x=-\dfrac{1}{4}\)

1)

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

=> 2x² - 4x - 2x² - x = 3

=> (2x² - 2x² ) - (4x+x) = 3

=> -5x = 3

=> x = \(\dfrac{-3}{5}\)

2) (2x-1).(x-2) - (x+3).(2x-7) = 3

=> 2x² - 4x - x + 2 - 2x² + 7x - 6x + 21 = 3

=> (2x² - 2x²) - (4x + 6x + x - 7x) + 2 + 21 = 3

=> -4x = -20

=> x = -20 : (-4)

=> x = 5

3) (x - 5).(-x + 4) - (x - 1).(x + 3) = -2x²

=> Bạn tách tương tự như mấy câu 2 nhé! Nếu không làm được thì bảo mình

a)\(\left(x-2\right)^2-\left(x-3\right)\cdot\left(x+3\right)=6\)

\(\Leftrightarrow x^2-4x+4-x^2+9-6=0\)

\(\Leftrightarrow7-4x=0\)

\(\Rightarrow x=\frac{-7}{4}\)

b)\(-4\cdot\left(x-1\right)^2+\left(2x-1\right)\cdot\left(2x+1\right)=-3\)

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

\(\Leftrightarrow-4x^2+8x-4+4x^2-1+3=0\)

\(\Leftrightarrow8x-2=0\)

\(\Rightarrow x=\frac{2}{8}=\frac{1}{4}\)

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

<=> \(1+2\left(x^2+x-2\right)-2\left(4x^2-4x+1\right)=3-8x-12-3x-1\)

<=>\(1+2x^2+2x-4-8x^2+8x-2-3+8x+12+3x+1=0\)

<=> \(-6x^2+5x+5=0\)

<=> \(\left[\begin{array}{nghiempt}x=\frac{5+\sqrt{145}}{12}\\x=\frac{5-\sqrt{145}}{12}\end{array}\right.\)

hai nghiệm trên là nghiệm của pt

\(a,\Leftrightarrow x^3-4x^2+4x=0\\ \Leftrightarrow x\left(x^2-4x+4\right)=0\\ \Leftrightarrow x\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ b,\Leftrightarrow4\left(x-1\right)=3x+6\left(2x-3\right)\\ \Leftrightarrow4x-4=3x+12x-18\\ \Leftrightarrow11x=14\Leftrightarrow x=\dfrac{14}{11}\)

a/ \(x^3-4x^2=-4x\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

b/ \(\dfrac{x-1}{3}=\dfrac{x}{4}+\dfrac{2x-3}{2}\)

\(\Leftrightarrow8\left(x-1\right)=6x+12\left(2x-3\right)\)

\(\Leftrightarrow8x-8=6x+24x-36\)

\(\Leftrightarrow8x-8=30x-36\)

\(\Leftrightarrow8x-30x=8-36\)

\(\Leftrightarrow-22x=-28\)

\(\Leftrightarrow x=\dfrac{14}{11}\)

bấm máy tính là ra 1,25 bạn nhé

a) 2/3 + 1/3(2x-1) =-5

            1/3(2x-1) = -5 - 2/3

            1/3(2x-1) = -17/3

                 2x-1  = -17/3 : 1/3 

                 2x-1  = -17

                 2x     = -17 + 1 

                 2x     = -16 

                  x      = -16 : 2 

                  x      = -8  

b) 25% .x + 2x = 4 

   x .( 25%+2 ) = 4

   x . 9/4          = 4 

   x                  = 4 : 9/4 

   x                  = 16/9

1) \(\left|2x+5\right|\ge21\Rightarrow2x+5\ge21\)hoặc \(2x+5

2b) Áp dụng bất đẳng thức giá trị tuyệt đối: |a| + |b|  \(\ge\) |a + b|. Dấu "=" xảy ra khi tích a.b \(\ge\) 0 

Ta có: B = |2x - 1| + |3 - 2x| + 5  \(\ge\) |2x - 1+3 - 2x| + 5  = |2| + 5 = 7

=> Min B = 7 khi

(2x - 1)( 3 - 2x) \(\ge\) 0 => (2x - 1)(2x - 3) \(\le\) 0 

Mà 2x - 1 > 2x - 3 nên 2x - 1 \(\ge\) 0 và 2x - 3 \(\le\)  0 

=> x \(\ge\) 1/2 và x  \(\le\) 3/2

a)\(\frac{1}{4}+\frac{1}{3}:2x=-5\)

   \(\frac{1}{3}:2x=-5-\frac{1}{4}\)

   \(\frac{1}{3}:2x=-\frac{21}{3}\)

   \(2x=\frac{1}{3}:\left(\frac{-21}{3}\right)\)

   \(2x=-\frac{1}{21}\)

   \(x=\frac{-1}{42}\)

b)\(\left(3x-\frac{1}{4}\right).\left(x+\frac{1}{2}\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}3x-\frac{1}{4}=0\\x+\frac{1}{2}=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}3x=\frac{1}{4}\\x=-\frac{1}{2}\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{1}{12}\\x=-\frac{1}{2}\end{array}\right.\)

c)\(\left(2x-5\right).\left(\frac{3}{2}x+9\right).\left(0,3x-12\right)=0\)

   \(\Rightarrow\left[\begin{array}{nghiempt}2x-5=0\\\frac{3}{2}x+9=0\\0,3x-12=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}2x=5\\\frac{3}{2}x=-9\\0,3x=12\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-6\\x=40\end{array}\right.\)

a) 1/4 + 1/3 : 2x = -5

=> 1/3 : 2x = -5 - 1/4

=> 1/3 : 2x = -21/4

=> 2x = 1/3 : (-21/4) = -4/63

=> x = -4/63 : 2 = -2/63