Question

Cần gấp, giúp nhanh ạ ! (làm bài 2 câu k,l,m và bài 3)

Answer 1

Bài 3:

1: \(2^x=8^3\)

=>\(2^x=\left(2^3\right)^3=2^9\)

=>x=9

2: \(2^x=8^9\)

=>\(2^x=\left(2^3\right)^9=2^{27}\)

=>x=27

3: \(\left(\dfrac{1}{5}\right)^x=\left(\dfrac{1}{125}\right)^5\)

=>\(\left(\dfrac{1}{5}\right)^x=\left(\dfrac{1}{5}\right)^{5\cdot3}=\left(\dfrac{1}{5}\right)^{15}\)

=>x=15

4: \(2^{7x+4}=32^{12}\)

=>\(2^{7x+4}=\left(2^5\right)^{12}=2^{60}\)

=>7x+4=60

=>7x=56

=>x=8

5: \(3^{3-2x}=27^{13}\)

=>\(3^{-2x+3}=3^{3\cdot13}=3^{39}\)

=>-2x+3=39

=>-2x=36

=>x=-18

6: \(3^{6-3x}=81^3\)

=>\(3^{-3x+6}=3^{12}\)

=>-3x+6=12

=>-3x=6

=>x=-2

7: \(4^{-5x-1}=16^{12}\)

=>\(4^{-5x-1}=4^{24}\)

=>-5x-1=24

=>-5x=25

=>x=-5

8: \(5^{2x+1}=125^{x+25}\)

=>\(5^{2x+1}=5^{3\cdot\left(x+25\right)}\)

=>3(x+25)=2x+1

=>3x+75=2x+1

=>x=-74

Answer 2

l, \(2\left(\dfrac{3}{2}-x\right)-\dfrac{1}{3}=7x-\dfrac{1}{4}\)
\(2.\dfrac{3}{2}-2x-\dfrac{1}{3}=7x-\dfrac{1}{4}\)
\(3-2x-\dfrac{1}{3}=7x-\dfrac{1}{4}\)
\(-2x-7x=-\dfrac{1}{4}-3+\dfrac{1}{3}\)
\(x\left[\left(-2\right)-7\right]=\dfrac{-3}{12}-\dfrac{36}{12}+\dfrac{4}{12}\)
\(-9x=\dfrac{-35}{12}\)
\(x=\dfrac{-35}{12}:\left(-9\right)\)
\(x=\dfrac{-35}{12}.\dfrac{-1}{9}\)
\(x=\dfrac{35}{108}\)
Vậy \(x=\dfrac{35}{108}\)

Answer 3

\(4x-\left(2x+1\right)=3-\dfrac{1}{3}+x\)
\(4x-2x-1=\dfrac{9}{3}-\dfrac{1}{3}+x\)
\(x\left(4-2\right)-1=\dfrac{8}{3}+x\)
\(x.2-x=\dfrac{8}{3}+1\)
\(x\left(2-1\right)=\dfrac{8}{3}+\dfrac{3}{3}\)
\(x=\dfrac{11}{3}\)
Vậy \(x=\dfrac{11}{3}\)