a: 5-3x=6x+7

=>-3x-6x=7-5

=>-9x=2

=>\(x=-\dfrac{2}{9}\)

b: \(\dfrac{3x-2}{6}-5=3-\dfrac{2\left(x+7\right)}{4}\)

=>\(\dfrac{3x-2}{6}+\dfrac{x+7}{2}=8\)

=>\(\dfrac{3x-2+3\left(x+7\right)}{6}=8\)

=>3x-2+3x+14=48

=>6x+12=48

=>6x=36

=>\(x=\dfrac{36}{6}=6\)

c: \(\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)

=>\(\left(x-1\right)\left(5x+3\right)-\left(3x-8\right)\left(x-1\right)=0\)

=>(x-1)(5x+3-3x+8)=0

=>(x-1)(2x+11)=0

=>\(\left[{}\begin{matrix}x-1=0\\2x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{11}{2}\end{matrix}\right.\)

d: \(\left(2x-1\right)^2-\left(x+3\right)^2=0\)

=>\(\left(2x-1-x-3\right)\left(2x-1+x+3\right)=0\)

=>\(\left(x-4\right)\left(3x+2\right)=0\)

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

a, \(-4x+5+2x-1=3\Leftrightarrow-2x=-1\Leftrightarrow x=\dfrac{1}{2}\)

b, \(-2x+2=2\Leftrightarrow x=0\)

c, \(-2x-6=-8\Leftrightarrow x=1\)

b) (x+1)^3-x(x-2)^2+x-1=0

 ⇔x^3+3x^2+3x+1-(x^3-4x^2+4x)=0

⇔ x^3+3x^2+3x+1-x^3+4x^2-4x+x-1=0

⇔7x^2-2=0

⇔7x^2=2

⇔7x^2=-2⇔x=-3

⇔7x^2=2⇔x=-căn 5

2: 

a: =>-2x=10

=>x=-5

b: =>(x-3)(2x+5)=0

=>x=3 hoặc x=-5/2

a)

\(\left\{{}\begin{matrix}3x-y=3\\2x+y=7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=3x-3\\2x+y=7\end{matrix}\right.\\ \Leftrightarrow2x+3x-3=7\\ \Leftrightarrow5x-3=7\\ \Leftrightarrow5x=10\\ \Leftrightarrow x=2\\ \Leftrightarrow y=3.2-3=6-3=3\)

Vậy \(S=\left\{x;y\right\}=\left\{2;3\right\}\)

b)

\(\left\{{}\begin{matrix}3x-y=5\\2y-x=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3x-y=5\\x=2y\end{matrix}\right.\\ \Leftrightarrow3.2y-y=5\\ \Leftrightarrow5y=5\\ \Leftrightarrow y=1\\ \Leftrightarrow x=2y=2.1=2\)

Vậy \(S=\left\{x;y\right\}=\left\{1;2\right\}\)

a. 6x² - (2x + 5)(3x - 2) = 7

<=> 6x² - 6x² + 4x - 15x + 10 = 7

<=> -11x = -3

<=> \(x=\dfrac{3}{11}\)

b. (5 - x)(25 + 5x + x²) + x(x² - 7) = 25

<=> 125 - x³ + x³ - 7x = 25

<=> -7x = 25 - 125

<=> -7x = -100

<=> \(x=\dfrac{100}{7}\)

c. (7 - 2x)² + (3 + 2x)(3 - 2x) = 30

<=> 49 - 28x + 4x² + 9 - 4x² = 30

<=> 4x² - 4x² - 28x = 30 - 49 - 9

<=> -28x = -28

<=> x = 1

a: \(=\dfrac{-3}{5}\cdot\dfrac{5}{7}+\dfrac{-3}{5}\cdot\dfrac{3}{7}+\dfrac{-3}{5}\cdot\dfrac{6}{7}\)

\(=\dfrac{-3}{5}\left(\dfrac{5}{7}+\dfrac{3}{7}+\dfrac{6}{7}\right)=\dfrac{-3}{5}\cdot2=-\dfrac{6}{5}\)

b: \(=\dfrac{3}{13}\cdot\dfrac{6}{11}+\dfrac{3}{13}\cdot\dfrac{5}{11}-\dfrac{2}{13}=\dfrac{3}{13}-\dfrac{2}{13}=\dfrac{1}{13}\)

c: =>1/2x+1+3/8=7/16

=>1/2x=-15/16

=>x=-15/8

d: =>5/2x-1/3=1/6*(-9)/2=-9/12=-3/4

=>5/2x=-3/4+1/3=-9/12+4/12=-5/12

=>x=-1/6

a, \(x<2\)

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

\(x=5(\)ko \(t/m)\)

\(x>2\)

\(-x=5\)

\(x=-5(ko\) \(t/m)\)

a: |x-2|+2x=7

=>|x-2|=-2x+7

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{7}{2}\\\left(-2x+7\right)^2=\left(x-2\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{7}{2}\\\left(2x-7-x+2\right)\left(2x-7+x-2\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{7}{2}\\\left(x-5\right)\left(3x-9\right)=0\end{matrix}\right.\Leftrightarrow x=3\)

b: |x-3|-4x=5

=>|x-3|=4x+5

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

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{5}{4}\\\left(3x+8\right)\left(5x+2\right)=0\end{matrix}\right.\Leftrightarrow x=-\dfrac{2}{5}\)

c: |2x+3|+x=2x+3

=>|2x+3|=x+3

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-3\\\left(2x+3-x-3\right)\left(2x+3+x+3\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{0;-2\right\}\)