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\(\frac{-1}{2}+\frac{1}{3}+\frac{2}{4}=\frac{-6}{12}+\frac{4}{12}+\frac{6}{12}\)
= \(\frac{4}{12}\)
a) \(\frac{9x^2}{11y^2}:\frac{6x}{11y}=\frac{9x^2}{11y^2}\cdot\frac{11y}{6x}=\frac{3xy}{2}\)
b) \(\frac{x^2-49}{x-7}+x-2=\frac{\left(x-7\right)\left(x+7\right)}{x-7}+x-2=x+7+x-2=2x+5\)
c) \(\frac{3}{x+3}+\frac{1}{x-3}-\frac{18}{9-x^2}\)
= \(\frac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{1\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{18}{\left(3-x\right)\left(x+3\right)}\)
= \(\frac{3x-9}{\left(x-3\right)\left(x+3\right)}+\frac{x+3}{\left(x-3\right)\left(x+3\right)}+\frac{18}{\left(x-3\right)\left(x+3\right)}\)
= \(\frac{3x-9+x+3+18}{\left(x-3\right)\left(x+3\right)}\)
= \(\frac{4x+12}{\left(x-3\right)\left(x+3\right)}\)
= \(\frac{4\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{4}{x-3}\)(đk: \(x-3\ne0\)=> \(x\ne3\))
1. Ta có:
\(\frac{1}{x}+\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+2013\right)\left(x+2014\right)}\)
\(=\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+2013}-\frac{1}{x+2014}\)
\(=\frac{2}{x}-\frac{1}{x+2014}\)
\(=\frac{2\left(x+2014\right)}{x\left(x+2014\right)}-\frac{x}{x\left(x+2014\right)}\)
\(=\frac{2x+4028-x}{x\left(x+2014\right)}=\frac{x+4028}{x\left(x+2014\right)}\)
2a) ĐKXĐ: x \(\ne\)1 và x \(\ne\)-1
b) Ta có: A = \(\frac{x^2-2x+1}{x-1}+\frac{x^2+2x+1}{x+1}-3\)
A = \(\frac{\left(x-1\right)^2}{x-1}+\frac{\left(x+1\right)^2}{x+1}-3\)
A = \(x-1+x+1-3\)
A = \(2x-3\)
c) Với x = 3 => A = 2.3 - 3 = 3
c) Ta có: A = -2
=> 2x - 3 = -2
=> 2x = -2 + 3 = 1
=> x= 1/2
a) \(\frac{x+2}{2002}\)+\(\frac{x+5}{1999}\)+\(\frac{x+201}{1803}\)=-3
⇔\(\frac{x+2}{2002}\)+\(\frac{x+5}{1999}\)+\(\frac{x+201}{1803}\)+3=0
⇔\(\frac{x+2}{2002}\)+1+\(\frac{x+5}{1999}\)+1+\(\frac{x+201}{1803}\)+1=0
⇔\(\frac{x+2004}{2002}\)+\(\frac{x+2004}{1999}\)+\(\frac{x+2004}{1803}\)=0
⇔(x+2004)(\(\frac{1}{2002}\)+\(\frac{1}{1999}\)+\(\frac{1}{1803}\))=0
Mà (\(\frac{1}{2002}\)+\(\frac{1}{1999}\)+\(\frac{1}{1803}\))≠0
⇒x+2004=0
⇔x=-2004
Vậy tập nghiệm của phương trình đã cho là:S={-2004}
\(a,\frac{1}{3x-2}-\frac{1}{3x+2}-\frac{3x-6}{4-9x^2}\)
\(=\frac{1}{3x-2}-\frac{1}{3x+2}+\frac{3\left(x-2\right)}{\left(3x+2\right)\left(3x-2\right)}\)
\(=\frac{3x+2-\left(3x-2\right)+3\left(x-2\right)}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\frac{3x+2-3x+2+3x-6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\frac{3x-2}{\left(3x-2\right)\left(3x+2\right)}=\frac{1}{3x+2}\)
\(b,\frac{18}{\left(x-3\right)\left(x^2-9\right)}-\frac{3}{x^2-6x+9}-\frac{x}{x^2-9}\)
\(=\frac{18}{\left(x-3\right)\left(x-3\right)\left(x+3\right)}-\frac{3}{\left(x-3\right)\left(x-3\right)}-\frac{x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{18-3\left(x+3\right)-x\left(x-3\right)}{\left(x-3\right)\left(x-3\right)\left(x+3\right)}\)
\(=\frac{18-3x-9-x^2+3x}{\left(x-3\right)\left(x-3\right)\left(x+3\right)}\)
\(=\frac{-x^2+9}{\left(x-3\right)\left(x-3\right)\left(x+3\right)}\)
\(=\frac{-\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x-3\right)\left(x+3\right)}=-\frac{1}{x-3}\)
Bài 1:
a) Ta có: \(2,3x-2\left(0,7+2x\right)=3,6-1,7x\)
\(\Leftrightarrow2,3x-1,4-4x-3,6+1,7x=0\)
\(\Leftrightarrow-5=0\)(vl)
Vậy: \(x\in\varnothing\)
b) Ta có: \(\frac{4}{3}x-\frac{5}{6}=\frac{1}{2}\)
\(\Leftrightarrow\frac{4}{3}x=\frac{1}{2}+\frac{5}{6}=\frac{8}{6}=\frac{4}{3}\)
hay x=1
Vậy: x=1
c) Ta có: \(\frac{x}{10}-\left(\frac{x}{30}+\frac{2x}{45}\right)=\frac{4}{5}\)
\(\Leftrightarrow\frac{9x}{90}-\frac{3x}{90}-\frac{4x}{90}-\frac{72}{90}=0\)
\(\Leftrightarrow2x-72=0\)
\(\Leftrightarrow2\left(x-36\right)=0\)
mà 2>0
nên x-36=0
hay x=36
Vậy: x=36
d) Ta có: \(\frac{10x+3}{8}=\frac{7-8x}{12}\)
\(\Leftrightarrow12\left(10x+3\right)=8\left(7-8x\right)\)
\(\Leftrightarrow120x+36=56-64x\)
\(\Leftrightarrow120x+36-56+64x=0\)
\(\Leftrightarrow184x-20=0\)
\(\Leftrightarrow184x=20\)
hay \(x=\frac{5}{46}\)
Vậy: \(x=\frac{5}{46}\)
e) Ta có: \(\frac{10x-5}{18}+\frac{x+3}{12}=\frac{7x+3}{6}-\frac{12-x}{9}\)
\(\Leftrightarrow\frac{2\left(10x-5\right)}{36}+\frac{3\left(x+3\right)}{36}-\frac{6\left(7x+3\right)}{36}+\frac{4\left(12-x\right)}{36}=0\)
\(\Leftrightarrow2\left(10x-5\right)+3\left(x+3\right)-6\left(7x+3\right)+4\left(12-x\right)=0\)
\(\Leftrightarrow20x-10+3x+9-42x-18+48-4x=0\)
\(\Leftrightarrow-23x+29=0\)
\(\Leftrightarrow-23x=-29\)
hay \(x=\frac{29}{23}\)
Vậy: \(x=\frac{29}{23}\)
f) Ta có: \(\frac{x+4}{5}-x-5=\frac{x+3}{2}-\frac{x-2}{2}\)
\(\Leftrightarrow\frac{2\left(x+4\right)}{10}-\frac{10x}{10}-\frac{50}{10}=\frac{25}{10}\)
\(\Leftrightarrow2x+8-10x-50-25=0\)
\(\Leftrightarrow-8x-67=0\)
\(\Leftrightarrow-8x=67\)
hay \(x=\frac{-67}{8}\)
Vậy: \(x=\frac{-67}{8}\)
g) Ta có: \(\frac{2-x}{4}=\frac{2\left(x+1\right)}{5}-\frac{3\left(2x-5\right)}{10}\)
\(\Leftrightarrow5\left(2-x\right)-8\left(x+1\right)+6\left(2x-5\right)=0\)
\(\Leftrightarrow10-5x-8x-8+12x-30=0\)
\(\Leftrightarrow-x-28=0\)
\(\Leftrightarrow-x=28\)
hay x=-28
Vậy: x=-28
h) Ta có: \(\frac{x+2}{3}+\frac{3\left(2x-1\right)}{4}-\frac{5x-3}{6}=x+\frac{5}{12}\)
\(\Leftrightarrow\frac{4\left(x+2\right)}{12}+\frac{9\left(2x-1\right)}{12}-\frac{2\left(5x-3\right)}{12}-\frac{12x}{12}-\frac{5}{12}=0\)
\(\Leftrightarrow4x+8+18x-9-10x+6-12x-5=0\)
\(\Leftrightarrow0x=0\)
Vậy: \(x\in R\)
Bài 2:
a) Ta có: \(5\left(x-1\right)\left(2x-1\right)=3\left(x+8\right)\left(x-1\right)\)
\(\Leftrightarrow5\left(x-1\right)\left(2x-1\right)-3\left(x-1\right)\left(x+8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[5\left(2x-1\right)-3\left(x+8\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(10x-5-3x-24\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(7x-29\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\7x-29=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\7x=29\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{29}{7}\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{1;\frac{29}{7}\right\}\)
b) Ta có: \(\left(3x-2\right)\left(x+6\right)\left(x^2+5\right)=0\)(1)
Ta có: \(x^2\ge0\forall x\)
\(\Rightarrow x^2+5\ge5\ne0\forall x\)(2)
Từ (1) và (2) suy ra:
\(\left[{}\begin{matrix}3x-2=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=-6\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{\frac{2}{3};-6\right\}\)
c) Ta có: \(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)
\(\Leftrightarrow27x^3-8-\left(27x^3-1\right)-x+4=0\)
\(\Leftrightarrow27x^3-8-27x^3+1-x+4=0\)
\(\Leftrightarrow-x-3=0\)
\(\Leftrightarrow-x=3\)
hay x=-3
Vậy: Tập nghiệm S={-3}
d) Ta có: \(x\left(x-1\right)-\left(x-3\right)\left(x+4\right)=5x\)
\(\Leftrightarrow x^2-x-\left(x^2+x-12\right)-5x=0\)
\(\Leftrightarrow x^2-x-x^2-x+12-5x=0\)
\(\Leftrightarrow12-7x=0\)
\(\Leftrightarrow7x=12\)
hay \(x=\frac{12}{7}\)
Vậy: Tập nghiệm \(S=\left\{\frac{12}{7}\right\}\)
e) Ta có: (2x+1)(2x-1)=4x(x-7)-3x
\(\Leftrightarrow4x^2-1-4x^2+28x+3x=0\)
\(\Leftrightarrow31x-1=0\)
\(\Leftrightarrow31x=1\)
hay \(x=\frac{1}{31}\)
Vậy: Tập nghiệm \(S=\left\{\frac{1}{31}\right\}\)
a)\(\frac{59^3-41^3}{18}+59.41=\frac{\left(59-41\right)\left(59^2+59.41+41^2\right)}{18}+59.41\)
\(=\frac{18.\left(59^2+59.41+41^2\right)}{18}+59.41=59^2+59.41+41^2+59.41=59^2+2.59.41+41^2=\left(59+41\right)^2=100^2\)
=10000