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

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

=> \(15x^2+3x+10x+2=15x^2+21x-5x-7\)

=> \(\left(15x^2-15x^2\right)+\left(3x+5x+10x-21x\right)=-7-2\)   (chuyển vế)

=> \(-3x=-9\Rightarrow x=\frac{-9}{-3}=3\)

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

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

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

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

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

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

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

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

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

nhân chéo

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

=> 2(3x+2)=5x+7

=> 6x+4-5x-7=0

=> x-3=0

=> x=3

a) \(=x^2-49-x^2\) \(=-49\)

b) \(=25x^2-1-25x^2-1\) \(=-2\)

c) \(=16x^2-1-16x^2+8x-1\) \(=8x-2\)

d) \(=9x^2-30x+25-9x^2+25\) \(=50-30x\)

a: \(\Leftrightarrow12x^2-10x-12x^2-28x=7\)

=>-38x=7

hay x=-7/38

b: \(\Leftrightarrow-10x^2-5x+9x^2+6x+x^2-\dfrac{1}{2}x=0\)

=>1/2x=0

hay x=0

c: \(\Leftrightarrow18x^2-15x-18x^2-14x=15\)

=>-29x=15

hay x=-15/29

d: \(\Leftrightarrow x^2+2x-x-3=5\)

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

\(\text{Δ}=1^2-4\cdot1\cdot\left(-8\right)=33>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-1-\sqrt{33}}{2}\\x_2=\dfrac{-1+\sqrt{33}}{2}\end{matrix}\right.\)

e: \(\Leftrightarrow-15x^2+10x-10x^2-5x-5x=4\)

\(\Leftrightarrow-25x^2=4\)

\(\Leftrightarrow x^2=-\dfrac{4}{25}\left(loại\right)\)

a/ pt đãcho tương đương với

6x\(^2\)+ 21x -2x-7-6x+5x-6x+5= 16

<=>18x=18

=> x=1

b/ pt đã cho tương đương với

10x\(^2\)+9x-10x\(^2\)-15x+2x+3= 8

<=> -4x=5

<=.> x=-\(\frac{5}{4}\)

c/ pt đã cho tương đương với

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

<=>42x=41

<=> x= \(\frac{41}{42}\)

d/ pt đã cho tương đương với

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

<=> x\(^3\)+6x\(^2\)+x\(^2\)+6x-x\(^3\)=5x

<=> 8x\(^2\)+6x-5x=0

<=>8x\(^2\)+16x-10x-5x=0

<=> (x+2)2x-5(x+2)=0

<=> (x+2)(2x-5)=0

<=>x+2=0 hoặc 2x+5=0

=> x=-2 hoặc x= -\(\frac{5}{2}\)