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2.
a) Ta có:
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)=\left(x+1\right)\left(\frac{1}{13}+\frac{1}{14}\right)\)
Vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\ne\frac{1}{13}+\frac{1}{14}\)nên \(x+1=0\Leftrightarrow x=-1\)
Vậy x = -1
b) Ta có:
\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Rightarrow\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)
\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\Rightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}\right)=\left(x+2004\right)\left(\frac{1}{2002}+\frac{1}{2003}\right)\)
Vì \(\frac{1}{2000}+\frac{1}{2001}\ne\frac{1}{2002}+\frac{1}{2003}\)nên \(x+2004=0\Leftrightarrow x=-2004\)
Vậy, x = -2004
\(PT\Leftrightarrow\frac{x+4+2000}{2000}+\frac{x+3+2001}{2001}=\frac{x+2+2002}{2002}+\frac{x+1+2003}{2003}\)
<=> \(\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
<=> \(\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
<=> x + 2004 = 0
<=> x = -2004.
\(\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}+1\right)\)
\(\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)
\(\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
\(x+2004=0\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\right)\)
\(\Rightarrow x=-2004\)
Bài 3: Tìm x:
a. \(\left(2x-1\right)^4=81\)
\(\Rightarrow\left(2x-1\right)^4=3^4\)
=> 2x - 1 = 3
=> 2x = 4
=> x = 2
b. \(\left(x-2\right)^2=1\)
\(\Rightarrow\) \(\left(x-2\right)^2=1^2\)
=> x - 2 = 1
=> x = 3
c. \(x^{2000}=x\)
=> x = 1
d. \(\left(4x-3\right)^3=-125\)
\(\Rightarrow\left(4x-3\right)^3=\left(-5\right)^3\)
=> 4x - 3 = -5
=> 4x = -2
=> x = \(\dfrac{-1}{2}\)
a)\(\left(\frac{1}{3}\right)^{-1}-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^4.2^3=3-1+\frac{1}{16}.8=3-1+\frac{1}{2}=\frac{5}{2}\\ \)
b)\(2^2.2^3.\left(\frac{2}{3}\right)^{-2}=2^5.\frac{9}{4}=72\)
c)\(\left(\frac{4}{3}\right)^{-2}.\left(\frac{3}{4}\right)^3:\left(\frac{-2}{3}\right)^{-3}=\left(\frac{3}{4}\right)^2.\left(\frac{3}{4}\right)^3:\left(\frac{-2}{3}\right)^{-3}=\left(\frac{3}{4}\right)^5:\left(\frac{3}{2}\right)^3=\frac{9}{128}\)
2)
\(3^{x+1}=9^x\Leftrightarrow3^x.3=9^x\Rightarrow3=9^x:3^x\Rightarrow3=3^x\Rightarrow x=1\)
\(\left(x-0,1\right)^2=6,25\Leftrightarrow\left(x-0,1\right)^2=2,5^2\Rightarrow\left(x-0,1\right)=2,5\Rightarrow x=2,5+0,1=2,6\)
\(3^{2x-1}=243\Leftrightarrow3^{2x-1}=3^5\Rightarrow2x-1=5\Rightarrow2x=6\Rightarrow x=3\)
\(\left(4x-3\right)^4=\left(4x-3\right)^2\Rightarrow x=1\)
\(\frac{1999.4001+2000}{2000.4001-2001}=\frac{1999.4001+2000}{1999.4001+4001-2001}\)
\(=\frac{1999.4001+2000}{1999.4001+2000}=1\)
(1 - 4x)2n + 3 = 4n + 1
(1 - 4x)2n + 2 = (22)n + 1 = 22n + 2
=>1 - 4x = 2
=> 4x = 1 - 2
=> 4x = -1
=> x = -1/4
Tìm x :
( 1 - 4x )2n+2 = 4n+1
<=> ( 1 - 4x )2n+2 = ( 22 )n+1
<=> ( 1 - 4x )2n+2 = 22n+2
=> 1 - 4x = 2
<=> 4x = 1 - 2
<=> 4x = -1
<=> x = (-1) : 4
<=> x = -1/4
Vaayj x= -1/4