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14) Ta có: \(\dfrac{x}{2x-6}+\dfrac{x}{2x+2}=\dfrac{2x+4}{x^2-2x-3}\)

\(\Leftrightarrow\dfrac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\dfrac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\dfrac{4x+8}{2\left(x-3\right)\left(x+1\right)}\)

Suy ra: \(x^2+x+x^2-3x-4x-8=0\)

\(\Leftrightarrow2x^2-6x-8=0\)

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

a=1; b=-3; c=-4

Vì a-b+c=0 nên phương trình có hai nghiệm phân biệt là:

\(x_1=-1\left(loại\right);x_2=\dfrac{-c}{a}=4\left(nhận\right)\)

1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)

=-27x^3-18x^2+4x+10

2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27

=7x^3+37x^2+46x+33

5:

\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)

\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)

=7x^3-48x^2+8x-35

\(2x^2-7x+5=0\)

\(2x^2-2x-5x+5=0\)

\(2x\left(x-1\right)-5\left(x-1\right)=0\)

\(\left(x-1\right)\left(2x-5\right)=0\)

\(\left[\begin{array}{nghiempt}x-1=0\\2x-5=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=1\\2x=5\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=1\\x=\frac{5}{2}\end{array}\right.\)

\(x\left(2x-5\right)-4x+10=0\)

\(x\left(2x-5\right)-2\left(2x-5\right)=0\)

\(\left(2x-5\right)\left(x-2\right)=0\)

\(\left[\begin{array}{nghiempt}x-2=0\\2x-5=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=2\\2x=5\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=2\\x=\frac{5}{2}\end{array}\right.\)

\(\left(x-5\right)\left(x+5\right)-x\left(x-2\right)=15\)

\(x^2-25-x^2+2x=15\)

\(2x=15+25\)

\(2x=40\)

\(x=\frac{40}{2}\)

\(x=20\)

\(x^2\left(2x-3\right)-12+8x=0\)

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

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

\(2x-3=0\) (vì \(x^2\ge0\Rightarrow x^2+4\ge4>0\))

\(2x=3\)

\(x=\frac{3}{2}\)

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

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

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

\(\left[\begin{array}{nghiempt}x-1=0\\x+5=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=1\\x=-5\end{array}\right.\)

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

\(4x^2-12x+9-4x^2+4x=5\)

\(-8x=5-9\)

\(-8x=-4\)

\(x=\frac{4}{8}\)

\(x=\frac{1}{2}\)

\(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(5x-2x^2+2x^2-2x=13\)

\(3x=13\)

\(x=\frac{13}{3}\)

\(2\left(x+5\right)\left(2x-5\right)+\left(x-1\right)\left(5-2x\right)=0\)

\(\left(2x+10\right)\left(2x-5\right)-\left(x-1\right)\left(2x-5\right)=0\)

\(\left(2x-5\right)\left(2x+10-x+1\right)=0\)

\(\left(2x-5\right)\left(x+11\right)=0\)

\(\left[\begin{array}{nghiempt}2x-5=0\\x+11=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}2x=5\\x=-11\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-11\end{array}\right.\)

\(a,2x^2-7x+5=0\Leftrightarrow2x^2-2x-5x+5=0\Leftrightarrow2x\left(x-1\right)-5\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(2x-5\right)=0\Rightarrow\left[{}\begin{matrix}x-1=0\\2x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\2x=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=2,5\end{matrix}\right.\)\(b,x\left(2x-5\right)-4x+10=0\Rightarrow x\left(2x-5\right)-2\left(2x-5\right)=0\Leftrightarrow\left(x-2\right)\left(2x-5\right)=0\Rightarrow\left[{}\begin{matrix}x-2=0\\2x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\2x=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=2,5\end{matrix}\right.\)\(c,\left(x-5\right)\left(x+5\right)-x\left(x-2\right)=15\Leftrightarrow x^2-25-x^2+2x-15=0\Leftrightarrow2x-40=0\Rightarrow2x=40\Rightarrow x=20\)\(d,x^2\left(2x-3\right)-12+8x=0\Rightarrow2x^3-3x^2-12+8x=0\Leftrightarrow2x^3+8x-3x^2-12=0\Leftrightarrow2x\left(x^2+4\right)-2\left(x^2+4\right)=0\Leftrightarrow\left(2x-2\right)\left(x^2+4\right)=0\Rightarrow\left[{}\begin{matrix}2x-2=0\\x^2+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=2\\x^2=-4\end{matrix}\right.\Rightarrow x=1\)

9) Ta có: \(\dfrac{2x+5}{x+3}+1=\dfrac{4}{x^2+2x-3}-\dfrac{3x-1}{1-x}\)

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

\(\Leftrightarrow2x^2-2x+5x-5+x^2+2x-3-4-3x^2-10x+x+3=0\)

\(\Leftrightarrow-4x=9\)

hay \(x=-\dfrac{9}{4}\)

10) Ta có: \(\dfrac{x-1}{x+3}-\dfrac{x}{x-3}=\dfrac{7x-3}{9-x^2}\)

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3-7x}{\left(x-3\right)\left(x+3\right)}\)
Suy ra: \(x^2-4x+3-x^2-3x-3+7x=0\)

\(\Leftrightarrow0x=0\)(luôn đúng)

Vậy: S={x|\(x\notin\left\{3;-3\right\}\)}

11) Ta có: \(\dfrac{5+9x}{x^2-16}=\dfrac{2x-1}{x+4}+\dfrac{3x-1}{x-4}\)

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

Suy ra: \(2x^2-9x+4+3x^2+12x-x-4-9x-5=0\)

\(\Leftrightarrow5x^2-7x=0\)

\(\Leftrightarrow x\left(5x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{7}{5}\end{matrix}\right.\)

12) Ta có: \(\dfrac{2x}{2x-1}+\dfrac{x}{2x+1}=1+\dfrac{4}{\left(2x-1\right)\left(2x+1\right)}\)

\(\Leftrightarrow\dfrac{2x\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}+\dfrac{x\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}=\dfrac{4x^2-1+4}{\left(2x-1\right)\left(2x+1\right)}\)

Suy ra: \(4x^2+2x+2x^2-x-4x^2-3=0\)

\(\Leftrightarrow2x^2+x-3=0\)

\(\Leftrightarrow2x^2+3x-2x-3=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=1\end{matrix}\right.\)

13) Ta có: \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x^2-2x}\)

\(\Leftrightarrow\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{x-2}{x\left(x-2\right)}=\dfrac{2}{x\left(x-2\right)}\)

Suy ra: \(x^2+2x-x+2-2=0\)

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

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)

Bạn cần viết đề bài bằng công thức toán để được hỗ trợ tốt hơn. 

x^2+2x-3/3+2x/4=x^2/3

1: =>x+1/2=0 hoặc 2/3-2x=0

=>x=-1/2 hoặc x=1/3

2: =>7/6x=5/2:3,75=2/3

=>x=2/3:7/6=2/3*6/7=12/21=4/7

3: =>2x-3=0 hoặc 6-2x=0

=>x=3 hoặc x=3/2

4: =>-5x-1-1/2x+1/3=3/2x-5/6

=>-11/2x-3/2x=-5/6-1/3+1

=>-7x=-1/6

=>x=1/42

1) ĐKXĐ: \(x\ge\dfrac{5}{2}\)

\(\sqrt{x^2}=2x-5\\ \Rightarrow\left|x\right|=2x-5\\ \Rightarrow\left[{}\begin{matrix}x=2x-5\\x=5-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=\dfrac{5}{3}\left(ktm\right)\end{matrix}\right.\)

2) ĐKXĐ: \(x\ge3\)

\(\sqrt{25x^2-10x+1}=2x-6\\ \Rightarrow\left|5x-1\right|=2x-6\\ \Rightarrow\left[{}\begin{matrix}5x-1=2x-6\\5x-1=6-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\left(ktm\right)\\x=1\left(tm\right)\end{matrix}\right.\)

3) ĐKXĐ: \(x\ge\dfrac{5}{2}\)

\(\sqrt{25-10x+x^2}=2x-5\\ \Rightarrow\left|x-5\right|=2x-5\\ \Rightarrow\left[{}\begin{matrix}x-5=2x-5\\x-5=5-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=\dfrac{10}{3}\left(tm\right)\end{matrix}\right.\)

4) ĐKXĐ: \(x\ge\dfrac{1}{2}\)

\(\sqrt{1-2x+x^2}=2x-1\\ \Rightarrow\left|x-1\right|=2x-1\\ \Rightarrow\left[{}\begin{matrix}x-1=2x-1\\x-1=1-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=\dfrac{2}{3}\left(tm\right)\end{matrix}\right.\)