\(2^{x-1}+5.2^{x-2}=\frac{7}{32}\)

b, 2005 = /x - 4 | + |x -...">

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28 tháng 6 2018

a, \(2^{x-1}+5.2^{x-2}=\frac{7}{32}\)

=>\(2^{x-1}+\frac{5}{2}.2^{x-1}=\frac{7}{32}\)

=>\(2^{x-1}\left(1+\frac{5}{2}\right)=\frac{7}{32}\)

=>\(2^{x-1}\cdot\frac{7}{2}=\frac{7}{32}\)

=>\(2^{x-1}=\frac{1}{16}=\frac{1}{2^4}=2^{-4}\)

=>x-1=-4

=>x=-5

b, |x - 4| + |x - 10| + |x + 101| + |x + 990| + |x + 1000| = |4-x|+|10-x|+|x+101|+|x+990|+|x+1000|

Ta có: \(\left|4-x\right|\ge4-x;\left|10-x\right|\ge10-x;\left|x+990\right|\ge x+990;\left|x+1000\right|\ge x+1000\)

\(\Rightarrow\left|4-x\right|+\left|10-x\right|+\left|x+990\right|+\left|x+1000\right|\ge4-x+10-x+x+990+x+1000\)

\(\Rightarrow\left|4-x\right|+\left|10-x\right|+\left|x+101\right|+\left|x+990\right|+\left|x+1000\right|\ge2004+\left|x+101\right|\)

\(\Rightarrow2005\ge2004+\left|x+101\right|\)

\(\Rightarrow\left|x+1\right|\le1\)

\(\Rightarrow-1\le x+101\le1\)

\(\Rightarrow-102\le x\le-100\)

Vì \(x\in Z\)

\(\Rightarrow x\in\left\{-102;-101;-100\right\}\)

29 tháng 6 2018

bài a nhầm 2 dòng cuối

=>x-1=-4

=>x=-3

22 tháng 8 2017
bài làm
A=1.2.3+2.3.4+3.4.5+...+98.99.100
4A=1.2.3.4+2.3.4.4+3.4.5.4+...+98.99.100.4
4A=1.2.3.(4-0)+2.3.4.(5-1)+...+98.99.100.(101-97)
4A=1.2.3.4+2.3.4.5-1.2.3.4+...+98.99.100.101-97.98.99.100
4A=1.2.3.4-1.2.3.4+2.3.4.5-...-97.98.99.100+98.99.100.101
4A=98.99.100.101
4A=97990200
A=979902004979902004
A=24497550
22 tháng 8 2017

a, Vào câu hỏi tương tự nhé

b, Vì \(\hept{\begin{cases}\left|x+3\right|\ge0\\\left|x+1\right|\ge0\end{cases}\Rightarrow\left|x+3\right|+\left|x+1\right|\ge0\Rightarrow3x\ge0\Rightarrow x\ge0}\)

=> x+3+x+1=3x

=> 2x+4=3x

=>x=4

c, \(\left|x-4\right|+\left|x-10\right|+\left|x+101\right|+\left|x+990\right|+\left|x+1000\right|=\left|4-x\right|+\left|10-x\right|+\left|x+101\right|+\left|x+990\right|+\left|x+1000\right|\)

Có \(\left|4-x\right|\ge4-x;\left|10-x\right|\ge10-x;\left|x+990\right|\ge x+990;\left|x+1000\right|\ge x+1000\)

=>\(\left|4-x\right|+\left|10-x\right|+\left|x+101\right|+\left|x+990\right|+\left|x+1000\right|\)

=> \(2005\ge4-x+10-x+x+990+x+1000+\left|x+101\right|\)

=> \(2005\ge\left|x+101\right|+2004\)

=> \(\left|x+101\right|\le1\)

=> \(x+101\in\left\{-1;0;1\right\}\Rightarrow x\in\left\{-102;-101;-100\right\}\)

d, tương tự b

21 tháng 10 2016

b)\(2^{x-1}+5\cdot2^{x-2}=\frac{7}{32}\)

\(2^x:2+5\cdot2^x:2^2=\frac{7}{32}\)

\(2^x:2+2^x:\frac{4}{5}=\frac{7}{32}\)

\(2^x\cdot\left(\frac{1}{2}+\frac{5}{4}\right)=\frac{7}{32}\)

\(2^x\cdot\frac{7}{4}=\frac{7}{32}\)

\(2^x=\frac{7}{32}:\frac{7}{4}=\frac{1}{8}\)

\(2^x=\frac{2^0}{2^3}=2^{-3}\)

\(\Rightarrow x=-3\)

21 tháng 10 2016

a) \(4^x+4^{x+3}=4160\)

\(\Rightarrow4^x+4^x.4^3=4160\)

\(\Rightarrow4^x.\left(1+4^3\right)=4160\)

\(\Rightarrow4^x.65=4160\)

\(\Rightarrow4^x=64\)

\(\Rightarrow4^x=4^4\)

\(\Rightarrow x=4\)

Vậy \(x=4\)

b) \(2^{x-1}+5.2^{x-2}=\frac{7}{32}\)

\(\Rightarrow2^x.\frac{1}{2}+5.2^x.\frac{1}{4}=\frac{7}{32}\)

\(\Rightarrow2^x.\left(\frac{1}{2}+5.\frac{1}{4}\right)=\frac{7}{32}\)

\(\Rightarrow2^x.\frac{7}{4}=\frac{7}{32}\)

\(\Rightarrow2^x=\frac{7}{32}:\frac{7}{4}\)

\(\Rightarrow2^x=\frac{1}{8}\)

\(\Rightarrow2^x=2^{-3}\)

\(\Rightarrow x=-3\)

Vậy \(x=-3\)

 

 

a. 2x-1+ 5.2x-1:2=7/32

=> 2x+1.(1+5/2)=7/32

=>2x+1.7/2=7/32

=> 2x+1=1/16=1/24

=> x+1=-4=>x=-5

8 tháng 3 2019

a. 2x-1+ 5.2x-1:2=7/32

=> 2x+1.(1+5/2)=7/32

=>2x+1.7/2=7/32

=> 2x+1=1/16=1/24

=> x+1=-4=>x=-3