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a) \(6.8^{x-1}+8^{x+1}=6.8^{19}+8^{21}\)
\(\Rightarrow x-1+x+1=19+21\)
\(=2x=40\)
\(\Rightarrow x=20\)
b) \(4.3^{x-1}+2.3^{x+2}=4.3^6+2.3^9\)
\(\Rightarrow x-1+x+2=6+9\)
\(\Rightarrow2x+1=15\)
\(\Rightarrow2x=14\)
\(\Rightarrow x=7\)
Bài 1:
a; \(\dfrac{7}{8}\) + \(x\) = \(\dfrac{4}{7}\)
\(x\) = \(\dfrac{4}{7}\) - \(\dfrac{7}{8}\)
\(x\) = \(\dfrac{32}{56}\) - \(\dfrac{49}{56}\)
\(x=-\) \(\dfrac{49}{56}\)
Vậy \(x=-\dfrac{49}{56}\)
b; 6 - \(x\) = - \(\dfrac{3}{4}\)
\(x\) = 6 + \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{24}{4}+\dfrac{3}{4}\)
\(x=\dfrac{27}{4}\)
Vậy \(x=\dfrac{27}{4}\)
c; \(\dfrac{1}{-5}\) + \(x\) = \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{3}{4}\) + \(\dfrac{1}{5}\)
\(x=\dfrac{15}{20}\) + \(\dfrac{4}{20}\)
\(x=\dfrac{19}{20}\)
Vậy \(x=\dfrac{19}{20}\)
Bài 1:
d; - 6 - \(x\) = - \(\dfrac{3}{5}\)
\(x\) = - 6 + \(\dfrac{3}{5}\)
\(x=-\dfrac{30}{5}\) + \(\dfrac{3}{5}\)
\(x=-\dfrac{27}{5}\)
Vậy \(x=-\dfrac{27}{5}\)
e; - \(\dfrac{2}{6}\) + \(x\) = \(\dfrac{5}{7}\)
\(x\) = \(\dfrac{5}{7}\) + \(\dfrac{2}{6}\)
\(x\) = \(\dfrac{15}{21}\) + \(\dfrac{1}{3}\)
\(x=\dfrac{15}{21}\) + \(\dfrac{7}{21}\)
\(x=\dfrac{22}{21}\)
Vậy \(x=\dfrac{22}{21}\)
f; - 8 - \(x\) = - \(\dfrac{5}{3}\)
\(x\) = \(-\dfrac{5}{3}\) + 8
\(x\) = \(\dfrac{-5}{3}\) + \(\dfrac{24}{3}\)
\(x\) = \(\dfrac{-19}{3}\)
Vậy \(x=-\dfrac{19}{3}\)
4, Q = |x+\(\frac{1}{5}\) | -x +\(\frac{4}{7}\)
xét x \(\ge\) \(-\frac{1}{5}\)
Ta Có Q = |x+\(\frac{1}{5}\) | -x + \(\frac{4}{7}\) = x+\(\frac{1}{5}\) - x +\(\frac{4}{7}\) = \(\frac{27}{35}\) (1)
xét x \(< -\frac{1}{5}\)
Ta có Q = | x +\(\frac{1}{5}\) | - x + \(\frac{4}{7}\) = -x - \(\frac{1}{5}\) - x + \(\frac{4}{7}\) = -2x + \(\frac{13}{35}\)
với x \(< -\frac{1}{5}\)
=> -2x \(>\) \(\frac{2}{5}\)
=> -2x + \(\frac{13}{35}\) \(>\frac{27}{35}\) (2)
Từ (1) và (2) => MinQ = \(\frac{27}{35}\) khi \(x\ge-\frac{1}{5}\)
5 , D = |x| + |8-x|
D = |x| + |8-x| \(\ge\) |x+8-x| = |8| = 8
Dấu ''='' xảy ra khi x(8-x) \(\ge\) 0 <=> 0\(\le\)x\(\le\) 8
Vậy MinD = 8 khi \(0\le x\le8\)
6,L= |x - 2012| + |2011 - x|
L = |x-2012| + |2011-x| \(\ge\) | x-2012 + 2011 - x | = |-1| = 1
Dấu ''= '' xảy ra khi ( x-2012)(2011-x) \(\ge\) 0
làm nốt câu 6 nãy ấn nhầm
<=> 2011\(\le\) x \(\le\) 2012
Vậy MinL = 1 khi \(2011\le x\le2012\)
7 , E = | x- \(\frac{2006}{2007}\) | + |x-1|
Ta có :
E = |x-\(\frac{2006}{2007}\) | + |1-x|
E = | x - \(\frac{2006}{2007}\) | + |1-x| \(\ge\) | x - \(\frac{2006}{2007}\) + 1 - x | = \(\frac{1}{2007}\)
Dấu ''='' xảy ra khi (x- \(\frac{2006}{2007}\) ) ( 1-x ) \(\ge0\) <=> \(\frac{2006}{2007}\le x\le1\)
Vậy MinE = \(\frac{1}{2007}\) khi \(\frac{2006}{2007}\le x\le1\)
8 ,F = | x -\(\frac{1}{4}\) | + | \(x-\frac{3}{4}\) |
Ta có :
F = | x - \(\frac{1}{4}\) | + | \(\frac{3}{4}\) - x |
F = | x - \(\frac{1}{4}\) | + | \(\frac{3}{4}\) -x | \(\ge\) | x - \(\frac{1}{4}\) + \(\frac{3}{4}\) -x | = \(\frac{1}{2}\)
Dấu ''='' xảy ra khi ( x-\(\frac{1}{4}\) ) ( \(\frac{3}{4}-x\) ) \(\ge\) 0 <=> \(\frac{1}{4}\le x\le\frac{3}{4}\)
Vậy MinF = \(\frac{1}{2}\) khi \(\frac{1}{4}\le x\le\frac{3}{4}\)
\(\frac{x}{4}=\frac{3}{2}\)
\(\Rightarrow\frac{x}{4}=\frac{6}{4}\)
\(\Rightarrow x=6\)
vậy_
b)\(\frac{2}{x}=\frac{x}{8}\)
\(\Rightarrow x^2=2\cdot8\)
\(x^2=16\Rightarrow x=4\)
c) \(\frac{x+3}{4}=\frac{5}{3}\)
\(3\left(x+3\right)=4\cdot5\)
\(3x+9=20\)
\(3x=11\)
\(x=\frac{11}{3}\)
a) 3/4 + -1/8 = 5/8
b)-5/12 + -7/24 = -9/8
c) 4/21 - -5/28 = 31/84
d) 1 + -7/28 = 3/4
e) -4/3 - 17/6= -25/6
f) 1/3 - ( 1/2 +1/8 )= -7/24
g)1/21 - ( 1/7 - 1/3 ) = 5/21
h)1/2 - 1/4 + 1/13 + 1/8= 47/104
a) x - 1/10 = 1/15
x=1/15+1/10
x=1/6 Vay x=1/6b) -4/21 - x = -3/7
x=-4/21+3/7 x=5/21 Vay x=5/21c) x + 1/2 = 3/4 - (-1/2)
x+1/2= 5/4
x= 5/4-1/2
x=3/4
Vay x=3/4
d) 4/7 - x = 1/3 - (-2/3)
x= 4/7-1/3-2/3 x= -3/7 Vay x=-3/7Có: \(A=4\cdot\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=...........................\)
\(=\frac{3^{32}-1}{2}\)
\(B=3^{32-1}\)
=> \(A< B\)
9: =>x-3=2
=>x=5
10: =>x+1/2=1/5 hoặc x+1/2=-1/5
=>x=-7/10 hoặc x=-3/10
12:
a: =>x^2=900
=>x=30 hoặc x=-30
b: =>x=1/18*27=3/2
7: =>|x-0,4|=1,1
=>x-0,4=1,1 hoặc x-0,4=-1,1
=>x=1,5 hoặc x=-0,7
\(\left(\frac{2}{5}\right)^6:\left(\frac{2}{5}\right)^4=\left(\frac{2}{5}\right)^2=\frac{4}{25}\)
\(\left(\frac{3}{16}\right)^2:\left(\frac{9}{8}\right)^2=\frac{1}{12}\)
\(\left(\frac{2}{7}-\frac{1}{2}\right)^2=\frac{9}{196}\)
\(x-\frac{3}{8}=\frac{1}{6}-\frac{1}{5}\)
=> \(x-\frac{3}{8}=\frac{5}{30}-\frac{6}{30}=-\frac{1}{30}\)
=> \(x=-\frac{1}{30}+\frac{3}{8}\)
=> \(x=\frac{41}{120}\)
\(-\frac{7}{10}\left(x+\frac{1}{3}\right)=\frac{4}{5}\)
=> \(-\frac{7}{10}x-\frac{7}{30}=\frac{4}{5}\)
=> \(-\frac{7}{10}x=\frac{4}{5}+\frac{7}{30}=\frac{31}{30}\)
=> \(x=\frac{31}{30}:\left(-\frac{7}{10}\right)=\frac{31}{30}\cdot\left(-\frac{10}{7}\right)=-\frac{31}{21}\)
\(x-\frac{4}{3}=\frac{5}{6}\Rightarrow x=\frac{5}{6}+\frac{4}{3}=\frac{5}{6}+\frac{8}{6}=\frac{13}{6}\)
Thiếu đề
\(\frac{6}{5}+\left(x-\frac{2}{3}\right)=\frac{4}{7}\)
=> \(\frac{6}{5}+x-\frac{2}{3}=\frac{4}{7}\)
=> \(\frac{6}{5}+x=\frac{4}{7}+\frac{2}{3}=\frac{26}{21}\)
=> \(x=\frac{26}{21}-\frac{6}{5}=\frac{4}{105}\)