1) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

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

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

Suy ra: \(x^2+2x+1-\left(x^2-2x+1\right)=4\)

\(\Leftrightarrow x^2+2x+1-x^2+2x-1=4\)

\(\Leftrightarrow4x=4\)

hay x=1(loại)

Vậy: \(S=\varnothing\)

2) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

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

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

Suy ra: \(x^2+4x+4+x^2-2x=2x^2-8\)

\(\Leftrightarrow2x^2+2x+4-2x^2-8=0\)

\(\Leftrightarrow2x-4=0\)

\(\Leftrightarrow2x=4\)

hay x=2(loại)

Vậy: \(S=\varnothing\)

1. \(\dfrac{1}{x-1}-\dfrac{1}{x+1}\)

\(=\dfrac{1.\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{1\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)

\(=\dfrac{x+1}{\left(x+1\right)\left(x-1\right)}-\dfrac{x-1}{\left(x+1\right)\left(x-1\right)}\)

\(=\dfrac{x+1+\left(-x+1\right)}{\left(x+1\right)\left(x-1\right)}\)

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

2. \(\dfrac{x}{x^2-1}-\dfrac{1}{x-1}\)

\(=\dfrac{x}{\left(x+1\right)\left(x-1\right)}-\dfrac{x+1}{\left(x+1\right)\left(x-1\right)}\)

\(=\dfrac{x}{\left(x+1\right)\left(x-1\right)}+\dfrac{-\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\)

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

3. \(\dfrac{1}{x\left(x-y\right)}-\dfrac{1}{x\left(x-y\right)}\)

\(=\dfrac{1}{y\left(x-y\right)}+\dfrac{-1}{x\left(x-y\right)}\)

\(=\dfrac{1x}{y\left(x-y\right)x}+\dfrac{-1y}{x\left(x-y\right)y}\)

\(=\dfrac{x}{xy\left(x-y\right)}+\dfrac{-y}{xy\left(x-y\right)}\)

\(=\dfrac{x-y}{xy\left(x-y\right)}=\dfrac{1}{xy}\)

4. \(\dfrac{1}{x}-\dfrac{1}{x-1}\)

\(=\dfrac{1\left(x-1\right)}{x\left(x-1\right)}-\dfrac{1x}{\left(x-1\right)x}\)

\(=\dfrac{x-1}{x\left(x-1\right)}+\dfrac{-x}{x\left(x-1\right)}\)

\(=\dfrac{\left(x-1\right)-x}{x\left(x-1\right)}\)

\(=\dfrac{-1}{x\left(x-1\right)}\)

5. \(\dfrac{1}{x}-\dfrac{1}{x+1}\)

\(=\dfrac{1\left(x+1\right)}{x\left(x+1\right)}-\dfrac{1x}{\left(x+1\right)x}\)

\(=\dfrac{x+1}{x\left(x+1\right)}+\dfrac{-x}{x\left(x+1\right)}\)

\(=\dfrac{\left(x+1\right)-x}{x\left(x+1\right)}\)

6. \(\dfrac{1}{2x^2-10x}-\dfrac{1}{x-5}\)

\(=\dfrac{1}{2x\left(x-5\right)}-\dfrac{1}{x-5}\)

\(=\dfrac{1}{2x\left(x-5\right)}-\dfrac{1.2x}{2x\left(x-5\right)}\)

\(=\dfrac{1}{2x\left(x-5\right)}+\dfrac{-2x}{2x\left(x-5\right)}\)

\(=\dfrac{1-2x}{2x\left(x-5\right)}\)

7. \(\dfrac{x-1}{x^2-1}.\dfrac{x+1}{x+3}\)

\(=\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x^2-1\right)\left(x+3\right)}\)

\(=\dfrac{x^2-1}{\left(x^2-1\right)\left(x+3\right)}\)

8. \(\dfrac{2}{2x^2+10x}.\dfrac{x+5}{3x}\)

\(=\dfrac{2x\left(x+5\right)}{2x^2+10x.3x}\)

\(=\dfrac{2\left(x+5\right)}{2x\left(x+5\right)3x}\)

\(=\dfrac{2}{6x^2}=\dfrac{1}{3x^2}\)

\(a,\frac{x+1}{x-2}-\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)

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

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

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

\(\Leftrightarrow6x=2x^2+4\)

\(\Leftrightarrow2x^2+4-6x=0\)

\(\Leftrightarrow2x^2+4-6x=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=-3\end{cases}}\)

\(b,\frac{2x+1}{x-1}=\frac{5\left(x-1\right)}{x+1}\)

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

\(\Leftrightarrow2x^2+2x+x+1=5\left(x^2-2x+1\right)\)

\(\Leftrightarrow2x^2+3x+1=5x^2-10x+5\)

\(\Leftrightarrow5x^2-2x^2-10x-3x+5-1=0\)

\(\Leftrightarrow3x^2-13x+4=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-\frac{1}{3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-\frac{1}{3}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=\frac{1}{3}\end{cases}}}\)

\(c,\frac{x-1}{x+2}-\frac{x}{x-2}=\frac{5x-2}{4-x^2}\)

\(\Leftrightarrow\frac{x-1}{x+2}-\frac{x}{x-2}=\frac{2-5x}{x^2-4}\)

\(\Leftrightarrow\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{2-5x}{\left(x-2\right)\left(x+2\right)}\)

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

\(\Leftrightarrow-5x+2=2-5x\)

\(\Leftrightarrow-5x+5x=2-2\)

\(\Leftrightarrow0=0\)

=>pt luôn có nghiệm với mọi x.

\(\text{Ta có: }\) \(\frac{1}{4}x+\frac{1}{8}x+\frac{1}{16}x=1\)

\(\Rightarrow\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)x=1\)

\(\Rightarrow\frac{7}{16}x=1\)

\(\Rightarrow x=1:\frac{7}{16}\)

\(\Rightarrow x=\frac{16}{7}\)

Tìm x :

a, 1/4 * x + 1/8 * x + 1/16 * x = 1

x * ( 1/4 + 1/8 + 1/16 ) = 1

x * 7/16 = 1

     x = 1 : 7/16

     x = 16/7

b, 1/5 + 1/3 x ( x + 1 ) = 1/4

            1/3 x ( x + 1 ) = 1/4 - 1/5

            1/3 x ( x + 1 ) = 1/20

                      x + 1   = 1/20 : 1/3

                      x + 1 = 3/20

                      x      = 1 - 3/20 

                     x      = 7/20

Tính nhanh :

1/5 x 27 + 1/5 x 33 + 1/5 x 40

= 1/5 x ( 27 + 33 + 40 )

= 1/5 x 100

= 20 

a,\(2x-5=3x+15\)

\(3x-2x=-5-15\)

\(x=-20\)

b,\(\frac{2}{x-1}=\frac{6}{x+1}\)

\(2x+2=6x-6\)

\(4x=8\)

\(x=2\)

\(\frac{2x+1}{x-1}=\frac{5.\left(x-1\right)}{x+1}\)

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

\(2x^2+3x+1=5x^2-10+5\)

\(3x^2-3x=10-5+1=6\)

\(3x.\left(x-1\right)=6\)

\(x.\left(x-1\right)=3\)

Lập bảng

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À không chắc nhé!

\(A=2x^3+3x^2-3-5x^2-5x=2x^3-2x^2-5x-3\\ B=125-150x+60x^2-8x^3-25+9x^2=-8x^3+69x^2-150x+100\\ C=\left(3x+1-2x+1\right)\left(3x+1+2x-1\right)=5x\left(x+2\right)=5x^2+10x\\ D=\left(2x+1-3+x\right)^2=\left(3x-2\right)^2=9x^2-12x+4\\ E=x^3-6x^2+12x-8-x^3+x+6x^2-18x=-5x-8\\ F=x^3-3x^2+3x-1-3+3x^2-x^3+1-3x=-3\)