\(\frac{2x+3}{2x+1}-\frac{2x+5}{2x+7}=1-\frac{6x^2+9x-9}{\left(2x+1\right)\left(2x+7\right)}\)

\(\Leftrightarrow1+\frac{2}{2x+1}-1-\frac{2}{2x+7}-1=-\frac{6x^2+9x-9}{\left(2x+1\right)\left(2x+7\right)}\)

\(\Leftrightarrow\frac{4x+14-4x-2+6x^2+9x-9}{\left(2x+1\right)\left(2x+7\right)}=1\)

\(\Leftrightarrow\frac{6x^2+9x+3}{\left(2x+1\right)\left(2x+7\right)}=1\)

\(\Leftrightarrow\frac{6x^2+6x+3x+3}{\left(2x+1\right)\left(2x+7\right)}=1\)

\(\Leftrightarrow\frac{6x\left(x+1\right)+3\left(x+1\right)}{\left(2x+1\right)\left(2x+7\right)}=1\)

\(\Leftrightarrow\frac{3\left(2x+1\right)\left(x+1\right)}{\left(2x+1\right)\left(2x+7\right)}=1\)

\(\Leftrightarrow3x+3=2x+7\)

\(\Leftrightarrow x=4\)

\(\Leftrightarrow\dfrac{\left(2x+3\right)\left(2x+7\right)-\left(2x+5\right)\left(2x+1\right)}{\left(2x+1\right)\left(2x+7\right)}=\dfrac{4x^2+16x+7-6x^2-9x+9}{\left(2x+1\right)\left(2x+7\right)}\)

\(\Leftrightarrow-2x^2+7x+16=4x^2+20x+21-4x^2-12x-5\)

\(\Leftrightarrow-2x^2+7x+16=8x+16\)

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

=>x(2x+1)=0

=>x=0(nhận) hoặc x=-1/2(loại)

ghi rõ các bược hộ em ạ 

a/ (2x + 1)(4x – 3) – 6x(x + 5) – 2x(x – 7) + 18x

=8x^2-6x+4x-3-6x^2-30x-2x^2+14x+18x

=-3

vậy...

a) Ta có: \(\left(2x+1\right)\left(4x-3\right)-6x\left(x+5\right)-2x\left(x-7\right)+18x\)

\(=8x^2-6x+4x-3-6x^2-30x-2x^2+14x+18x\)

\(=-3\)

b) Ta có: \(9x\left(2x-5\right)-\left(6x+2\right)\left(3x-2\right)+39x\)

\(=18x^2-45x-18x^2+12x-6x+4+39x\)

\(=4\)

c) Ta có: \(4x\left(2x-3\right)+x\left(x+2\right)-9x\left(x-1\right)+x-5\)

\(=8x^2-12x+x^2+2x-9x^2+9x+x-5\)

\(=-5\)

* Trả lời:

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

\(\Leftrightarrow-3+6x-4-12x=-5x+5\)

\(\Leftrightarrow6x-12x+5x=3+4+5\)

\(\Leftrightarrow x=12\)

\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)

\(\Leftrightarrow6x-15-6+24x=-3x+7\)

\(\Leftrightarrow6x+24x+3x=15+6+7\)

\(\Leftrightarrow33x=28\)

\(\Leftrightarrow x=\dfrac{28}{33}\)

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

\(\Leftrightarrow1-3x-6x+12=-4x-5\)

\(\Leftrightarrow-3x-6x+4x=-1-12-5\)

\(\Leftrightarrow-5x=-18\)

\(\Leftrightarrow x=\dfrac{18}{5}\)

\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)

\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)

\(\Leftrightarrow-x-5x=-7\)

\(\Leftrightarrow-6x=-7\)

\(\Leftrightarrow x=\dfrac{7}{6}\)

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

\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)

\(\Leftrightarrow-15x+3x=4\)

\(\Leftrightarrow-12x=4\)

\(\Leftrightarrow x=-\dfrac{1}{3}\)

a) Ta có: \(6x\left(x-5\right)+3x\left(7-2x\right)=18\)

\(\Leftrightarrow6x^2-30x+21x-6x^2=18\)

\(\Leftrightarrow-9x=18\)

hay x=-2

Vậy: S={-2}

b) Ta có: \(2x\left(3x+1\right)+\left(4-2x\right)\cdot3x=7\)

\(\Leftrightarrow6x^2+2x+12x-6x^2=7\)

\(\Leftrightarrow14x=7\)

hay \(x=\dfrac{1}{2}\)

Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)

c) Ta có: \(0.5x\left(0.4-4x\right)+\left(2x+5\right)\cdot x=-6.5\)

\(\Leftrightarrow0.2x-2x^2+2x^2+5x=-6.5\)

\(\Leftrightarrow5.2x=-6.5\)

hay \(x=-\dfrac{5}{4}\)

Vậy: \(S=\left\{-\dfrac{5}{4}\right\}\)

d) Ta có: \(\left(x+3\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)=6\)

\(\Leftrightarrow x^2+5x+6-\left(x^2+3x-10\right)=6\)

\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)

\(\Leftrightarrow2x+16=6\)

\(\Leftrightarrow2x=-10\)

hay x=-5

Vậy: S={-5}

e) Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)

\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)

\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)

\(\Leftrightarrow14x=0\)

hay x=0

Vậy: S={0}

Điều kiện: \(x \ne -\dfrac{1}{2}\) và \(x \ne -\dfrac{7}{2}\)

\(\begin{array}{l} \dfrac{{\left( {2x + 3} \right)\left( {2x + 7} \right)}}{{\left( {2x + 1} \right)\left( {2x + 7} \right)}} - \dfrac{{\left( {2x + 5} \right)\left( {2x + 1} \right)}}{{\left( {2x + 7} \right)\left( {2x + 1} \right)}} = \dfrac{{\left( {2x + 7} \right)\left( {2x + 1} \right)}}{{\left( {2x + 7} \right)\left( {2x + 1} \right)}} - \dfrac{{6{x^2} + 9x - 9}}{{\left( {2x + 7} \right)\left( {2x + 1} \right)}}\\ \Leftrightarrow \dfrac{{4{x^2} + 20x + 21 - 4{x^2} - 12x - 5}}{{\left( {2x + 7} \right)\left( {2x + 1} \right)}} = \dfrac{{4{x^2} + 16x + 7 - 6{x^2} - 9x + 9}}{{\left( {2x + 7} \right)\left( {2x + 1} \right)}}\\ \Leftrightarrow \dfrac{{8x + 16}}{{\left( {2x + 7} \right)\left( {2x + 1} \right)}} = \dfrac{{ - 2{x^2} + 7x + 16}}{{\left( {2x + 7} \right)\left( {2x + 1} \right)}}\\ \Rightarrow 8x + 16 = - 2{x^2} + 7x + 16 \Leftrightarrow 2{x^2} + x = 0 \Leftrightarrow x\left( {2x + 1} \right) = 0 \Leftrightarrow \left[ \begin{array}{l} x = 0 \text{(nhận)}\\ x = - \dfrac{1}{2} \text{(loại)} \end{array} \right. \end{array}\)

Vậy phương trình có nghiệm duy nhất $x=0$

TÌM X

a) (3x+2)(2x+9)-(6x+1)(x+2)=7

=> 6x² + 31x +18 - 6x² - 13x - 2 - 7 = 0

=> 18x + 9 = 0 => 9(2x + 1) = 0 => 2x + 1 = 0 => x = -1/2

b) (x-2)(x+5)-(x+3)(x+2)=-6

=> x² + 3x - 10 - x² - 5x -6 + 6 = 0 => -2x -10 = 0 => -2(x + 5) = 0

=> x + 5 = 0 => x = -5

c) 3(2x-1)(3x-1)-(2x-3)(9x-1)=0

=> 18x² - 15x +3 - 18x² + 29x -3 = 0 => 14x = 0 => x = 0

a) \(\left(3x+2\right)\left(2x+9\right)-\left(6x+1\right)\left(x+2\right)=7\\\Rightarrow 6x^2+31x+18-6x^2-16x-2-7=0\\ \Rightarrow18x+9=0\Rightarrow9\left(2x+1\right)=0\Rightarrow2x+1=0\Rightarrow x=-\frac{1}{2}\)

b) \(\left(x-2\right)\left(x+5\right)-\left(x+3\right)\left(x+2\right)=-6\\ \Rightarrow x^2+3x-10-x^2-5x-6+6=0\\ \Rightarrow-2x-10=0\\ \Rightarrow-2\left(x+5\right)=0\\ \Rightarrow x+5=0\\ \Rightarrow x=-5\)

c) \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\\ \Rightarrow18x^2-15x+3-18x^2+29x-3=0\\ \Rightarrow14x=0\\ \Rightarrow x=0\)

(6x+1)(2x-5)=12x²-30x+2x-5=12x²-28x-5

(2x+5)²-2x(2x+8)=4x²+20x+25-4x²-16x=4x+25

(3x-5)(2x-1)-(2x+3)(3x+7)+30x=6x²-3x-10x+5=6x²-13x+5

(X-1)²-(x+1)(x-1)=x²-2x+1-x²+1=-2x+2

(3x+2)(9x²-6x+4)-(3+x)(x-3)=27x³+8+9-x²=27x³-x²+17