Answer:

\(6x^2-\left(2x+3\right)\left(3x-2\right)=7\)

\(\Rightarrow6x^2-\left(6x^2+9x-4x-6\right)=7\)

\(\Rightarrow6x^2-\left(6x^2+5x-6\right)=7\)

\(\Rightarrow6x^2-6x^2-5x+6=7\)

\(\Rightarrow-5x+6=7\)

\(\Rightarrow-5x=1\)

\(\Rightarrow x=\frac{-1}{5}\)

\(5x\left(12+7\right)-3x\left(80x-5\right)=-100\)

\(\Rightarrow5x.19-240x^2+15x=-100\)

\(\Rightarrow95x-240x^2+15x=-100\)

\(\Rightarrow-240x^2+110x+100=0\)

\(\Rightarrow-24x^2-11x-10=0\)

\(\Rightarrow24\left(x^2-\frac{11}{24}x+\frac{121}{2304}\right)-\frac{1081}{96}=0\)

\(\Rightarrow24\left(x-\frac{11}{48}\right)^2-\frac{1081}{96}=0\)

\(\Rightarrow24\left(x-\frac{11}{48}\right)^2=\frac{1081}{2304}\)

\(\Rightarrow\left(x-\frac{11}{48}\right)^2=\left(\frac{\pm\sqrt{1081}}{48}\right)^2\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{11}{48}=\frac{\sqrt{1081}}{48}\\x-\frac{11}{48}=\frac{-\sqrt{1081}}{48}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{1081}+11}{48}\\x=\frac{11-\sqrt{1081}}{48}\end{cases}}\)

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

\(\Rightarrow\left(21x-15x^2-35+25x\right)-\left(10x-15x^2-4+6x\right)-4=0\)

\(\Rightarrow36x-15x^2-35-16x+15x^2+4-4=0\)

\(\Rightarrow\left(-15x^2+15x^2\right)+\left(36x-16x\right)+\left(-35+4-4\right)=0\)

\(\Rightarrow30x-35=0\)

\(\Rightarrow x=\frac{7}{6}\)

a: \(\Leftrightarrow6x^2-6x^2+4x-9x+6=7\)

=>-5x=1

hay x=-1/5

b: \(\Leftrightarrow5x\left(12x+7\right)-3x\left(80x-5\right)=-100\)

\(\Leftrightarrow60x^2+35x-240x^2+15x=-100\)

\(\Leftrightarrow-180x^2+50x+100=0\)

hay \(x\in\left\{\dfrac{5+\sqrt{745}}{36};\dfrac{5-\sqrt{745}}{36}\right\}\)

c: \(\Leftrightarrow21x-15x^2-35+25x-\left(10x-15x^2-4+6x\right)=4\)

\(\Leftrightarrow-15x^2+46x-35+15x^2-16x+4=4\)

=>30x-31=4

=>30x=35

hay x=7/6

6x^2-(2x+5)(3x-2)=7

6x^2 - (6x^2 + 15x -4x -10) =7

-11x = -3

x= 3/11
(3x-5)*(7-5x)-(5x+2)(2-3x)=4

21x - 35 - 15x^2 + 25x - (10x +4 - 15x^2 - 6x) =4

46x - 10x +6x  = 4 + 35 +4

42x = 43

x= 42/42

ĐKXĐ: \(\left\{{}\begin{matrix}3x^2+5x+1>=0\\3x^2+5x-7>=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=\dfrac{-5+\sqrt{13}}{6}\\x< =\dfrac{-5-\sqrt{13}}{6}\end{matrix}\right.\\\left[{}\begin{matrix}x>=\dfrac{-5+\sqrt{109}}{6}\\x< =\dfrac{-5-\sqrt{109}}{6}\end{matrix}\right.\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x< =\dfrac{-5-\sqrt{109}}{6}\\x>=\dfrac{-5+\sqrt{109}}{6}\end{matrix}\right.\)

\(\sqrt{3x^2+5x+1}-\sqrt{3x^2+5x-7}=2\)

=>\(\sqrt{3x^2+5x+1}-3-\sqrt{3x^2+5x-7}+1=0\)

=>\(\dfrac{3x^2+5x+1-9}{\sqrt{3x^2+5x+1}+3}-\dfrac{3x^2+5x-7-1}{\sqrt{3x^2+5x-7}+1}=0\)

=>\(3x^2+5x-8=0\)

=>\(\left(3x+8\right)\left(x-1\right)=0\)

=>\(\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-\dfrac{8}{3}\left(nhận\right)\end{matrix}\right.\)

Ta có: (3x - 5)(7 - 5x) + (5x + 2)(3x - 2) - 2 = 0

=> 21x - 15x² - 35 + 25x + 15x² - 10x + 6x - 4 - 2 = 0

=> 42x  - 41 = 0

=> 42x = 41

=> x = 41/42

(3x-5)(7-5x)-(5x+2)(2-3x)=4

21x - 35 - 15x^2 + 25x - (10x +4 - 15x^2 -6x) =4

46x - 10x +6x = 4 + 35 +4

42x = 43

x= 43/42

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

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

\(3x.7-3x.5x-5.7+5.5x-5x.2+5x.3x+2.2-2.3x=4\)

\(21x-15x^2-35+25x-10x+15x^2+4-6x=4\)

\(30x-31=4\)

\(\Rightarrow30x=4+31\)

\(\Rightarrow30x=35\)

\(\Rightarrow x=35\div30\)

\(\Rightarrow x=\frac{7}{6}\)

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

\(=>21x-15x^2-35+25x-\left(10x-15x^2+4-6x\right)=4\)

\(=>21x-15x^2-35+25x-10x+15x^2-4+6x=4\)

\(=>42x-39=4=>42x=4+39=43=>x=\frac{43}{42}\)

1.

<=> \(\left[{}\begin{matrix}4-3x=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)

2.

<=>\(\left[{}\begin{matrix}7-2x=0\\4+8x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

3.

<=>\(\left[{}\begin{matrix}9-7x=0\\11-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{7}\\x=\dfrac{11}{3}\end{matrix}\right.\)

4.

<=>\(\left[{}\begin{matrix}7-14x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)

5. 

<=>\(\left[{}\begin{matrix}\dfrac{7}{8}-2x=0\\3x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{16}\\x=-\dfrac{1}{9}\end{matrix}\right.\)

6,7. ko đủ điều kiện tìm

20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)

Vậy...

1) 16 - 8x = 0 ⇔ 8(2 - x) = 0⇔ 2 - x = 0 ⇔ x = 2

Vậy phương trình có nghiệm là x = 2