tìm x biết
-21/13x+1/3= -2/3
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c, \(\frac{-21}{13}x+\frac{1}{3}=\frac{2}{3}\Rightarrow\frac{-21}{13}x=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}\Rightarrow x=\frac{1}{3}:\frac{-21}{3}=\frac{-1}{21}\)
b)\(\frac{3}{4}x-\frac{1}{8}=\frac{3}{7}\)\(\Leftrightarrow\frac{3}{4}x=\frac{3}{7}+\frac{1}{8}=\frac{31}{56}\)\(\Leftrightarrow x=\frac{31}{56}:\frac{3}{4}=\frac{31}{42}\)
c)\(-\frac{21}{13}x+\frac{1}{3}=\frac{2}{3}\Leftrightarrow-\frac{21}{13}x=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}\Leftrightarrow x=\frac{1}{3}:-\frac{21}{13}=-\frac{13}{63}\)
a \(\dfrac{1}{x-y}+\dfrac{2}{x+y}+\dfrac{3x}{y^2-x^2}\)
\(=\dfrac{x+y+2x-2y-3x}{\left(x-y\right)\left(x+y\right)}=\dfrac{-y}{\left(x-y\right)\left(x+y\right)}\)
b: \(\dfrac{1}{x-2}+\dfrac{1}{x+2}-\dfrac{4x-4}{x^2-4}\)
\(=\dfrac{x+2+x-2-4x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{-2x+4}{\left(x-2\right)\left(x+2\right)}\)
=-2/x+2
c: \(\dfrac{x+1}{x+3}-\dfrac{x-1}{3-x}+\dfrac{2x-2x^2}{x^2-9}\)
\(=\dfrac{\left(x+1\right)\left(x-3\right)+\left(x-1\right)\left(x+3\right)+2x-2x^2}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{x^2-2x-3+x^2+2x-3+2x-2x^2}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{2x-6}{\left(x+3\right)\left(x-3\right)}=\dfrac{2}{x+3}\)
1/
a/ \(D=2x\left(10x^2-5x-2\right)-5x\left(4x^2-2x-1\right)\)
\(D=2x\left[10\left(x^2-\frac{1}{2}x-\frac{1}{5}\right)\right]-5x\left[4\left(x^2-\frac{1}{2}x-\frac{1}{4}\right)\right]\)
\(D=20x\left(x^2-\frac{1}{2}x-\frac{1}{5}\right)-20x\left(x^2-\frac{1}{2}x-\frac{1}{4}\right)\)
\(D=20x^3-10x^2-4x-20x^3+10x^2+5x\)
\(D=x\)
b/ Mình xin sửa lại đề:
Tính giá trị biểu thức \(E\left(x\right)=x^5-13x^4+13x^3-13x^2+13x+2012\)
Tại x = 12
\(E\left(x\right)=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x-1\right)x+2012\)
\(E\left(x\right)=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2-x+2012\)
\(E\left(x\right)=2012-x\)
\(E\left(x\right)=2000\)
2/
a/ \(2x\left(x-5\right)-x\left(3+2x\right)=26\)
<=> \(2x^2-10x-3x-2x^2=26\)
<=> \(-13x=26\)
<=> \(x=-2\)
b/ Bạn vui lòng coi lại đề.
3a/ Ta có \(D=x\left(5x-3\right)-x^2\left(x-1\right)+x\left(x^2-6x\right)-10+3x\)
\(D=5x^2-3x-x^3+x^2+x^3-6x^2-10+3x\)
\(D=-10\)
Vậy giá trị của D không phụ thuộc vào x (đpcm)
d: ĐKXĐ: x<>-4; x<>-5; x<>-6; x<>-7
\(PT\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
=>\(\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
=>\(\dfrac{x+7-x-4}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)
=>x^2+11x+28=54
=>x^2+11x-26=0
=>(x+13)(x-2)=0
=>x=2 hoặc x=-13
e: \(\dfrac{x-241}{17}+\dfrac{x-220}{19}+\dfrac{x-195}{21}+\dfrac{x-166}{23}=10\)
\(\Leftrightarrow\left(\dfrac{x-241}{17}-1\right)+\left(\dfrac{x-220}{19}-2\right)+\left(\dfrac{x-195}{21}-3\right)+\left(\dfrac{x-166}{23}-4\right)=0\)
=>x-258=0
=>x=258
Ta có pt <=> (x-3)(3x-1)(x2 -x+1) = 0
<=> x = 3 hoặc x = \(\frac{1}{3}\)
a) \(\frac{3}{7}-x=\frac{1}{4}-\left(-\frac{3}{-5}\right)\)
\(\Rightarrow\frac{3}{7}-x=\frac{1}{4}-\frac{3}{5}\)
\(\Rightarrow\frac{3}{7}-x=\frac{-7}{20}\)
\(\Rightarrow x=\frac{3}{7}-\left(\frac{-7}{20}\right)=\frac{3}{7}+\frac{7}{20}\)
\(\Rightarrow x=\frac{109}{140}\)
b) \(\frac{-21}{-13x}+\frac{1}{3}=\frac{-2}{-3}\)
\(\Rightarrow\frac{21}{13x}+\frac{1}{3}=\frac{2}{3}\)
\(\Rightarrow\frac{21}{13x}=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}\)
\(\Rightarrow13x=\frac{3.21}{1}=63\Rightarrow x=\frac{63}{13}\)
\(\frac{-21}{13}\)x + \(\frac{1}{3}\)= \(\frac{-2}{3}\)
\(\frac{-21}{13}\)x = \(\frac{-2}{3}\)- \(\frac{1}{3}\)
\(\frac{-21}{13}\)x =\(\frac{-2}{3}\)- \(\frac{1}{3}\)
\(\frac{-21}{13}\)x = ( -1 )
x = (-1) :\(\frac{-21}{13}\)
x = \(\frac{13}{21}\)
Học tốt ^-^