1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 5 = ?
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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. \(\dfrac{2}{2x^2+10x}.\dfrac{x+5}{3x}\)
\(=\dfrac{2x\left(x+5\right)}{2x^2+10x.3x}\)
\(=\dfrac{2\left(x+5\right)}{2x\left(x+5\right)3x}\)
\(=\dfrac{2}{6x^2}=\dfrac{1}{3x^2}\)
\(a,\frac{x+1}{x-2}-\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{2x^2+4}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow x^2+2x+x+2-\left(x^2-2x-x+2\right)=2x^2+4\)
\(\Leftrightarrow x^2+3x+2-x^2+2x+x-2=2x^2+4\)
\(\Leftrightarrow6x=2x^2+4\)
\(\Leftrightarrow2x^2+4-6x=0\)
\(\Leftrightarrow2x^2+4-6x=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=-3\end{cases}}\)
\(b,\frac{2x+1}{x-1}=\frac{5\left(x-1\right)}{x+1}\)
\(\Leftrightarrow\left(2x+1\right)\left(x+1\right)=5\left(x-1\right)\left(x-1\right)\)
\(\Leftrightarrow2x^2+2x+x+1=5\left(x^2-2x+1\right)\)
\(\Leftrightarrow2x^2+3x+1=5x^2-10x+5\)
\(\Leftrightarrow5x^2-2x^2-10x-3x+5-1=0\)
\(\Leftrightarrow3x^2-13x+4=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-\frac{1}{3}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=\frac{1}{3}\end{cases}}}\)
\(\text{Ta có: }\) \(\frac{1}{4}x+\frac{1}{8}x+\frac{1}{16}x=1\)
\(\Rightarrow\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)x=1\)
\(\Rightarrow\frac{7}{16}x=1\)
\(\Rightarrow x=1:\frac{7}{16}\)
\(\Rightarrow x=\frac{16}{7}\)
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
b,\(\frac{2}{x-1}=\frac{6}{x+1}\)
\(2x+2=6x-6\)
\(4x=8\)
\(x=2\)
\(A=2x^3+3x^2-3-5x^2-5x=2x^3-2x^2-5x-3\\ B=125-150x+60x^2-8x^3-25+9x^2=-8x^3+69x^2-150x+100\\ C=\left(3x+1-2x+1\right)\left(3x+1+2x-1\right)=5x\left(x+2\right)=5x^2+10x\\ D=\left(2x+1-3+x\right)^2=\left(3x-2\right)^2=9x^2-12x+4\\ E=x^3-6x^2+12x-8-x^3+x+6x^2-18x=-5x-8\\ F=x^3-3x^2+3x-1-3+3x^2-x^3+1-3x=-3\)
5 nha bạn.
=5 nha