20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)

Vậy...

1) 16 - 8x = 0 ⇔ 8(2 - x) = 0⇔ 2 - x = 0 ⇔ x = 2

Vậy phương trình có nghiệm là x = 2

1) \(2^{x+1}\cdot2^{2014}=2^{2015}\)\(\Leftrightarrow2^{2014x+2014}=2^{2015}\)\(\Leftrightarrow2014x+2014=2015\)\(\Leftrightarrow x=\frac{1}{2014}\)

2) \(7x-2x=\frac{6^{17}}{6^{15}}+\frac{44}{11}\)\(\Leftrightarrow5x=6^2+4=36+4=40\)\(\Leftrightarrow x=\frac{40}{5}=8\)

3) \(3^x=9\)\(\Leftrightarrow3^x=3^2\)\(\Leftrightarrow x=2\)

4) \(7x-x=\frac{5^{21}}{5^{19}}+3\cdot2^2-7^0\)\(\Leftrightarrow6x=5^2+3\cdot4-1=25+12-1=36\)\(\Leftrightarrow x=6\)

5) \(4^x=64\)\(\Leftrightarrow4^x=4^3\)\(\Leftrightarrow x=3\)

6) \(9^{x-1}=9\)\(\Leftrightarrow x-1=1\)\(\Leftrightarrow x=0\)

7) \(\frac{2^x}{2^5}=1\)\(\Leftrightarrow2^{x-5}=2^0\)\(\Leftrightarrow x-5=0\)\(\Leftrightarrow x=5\)

8) \(\left(5x-9\right)^3=216\)\(\Leftrightarrow\left(5x-9\right)^3=6^3\)\(\Leftrightarrow5x-9=6\)\(\Leftrightarrow5x=15\)\(\Leftrightarrow x=3\)

9) \(5\cdot3^{7x-11}=135\)\(\Leftrightarrow5.3^{7x-11}=5.3^3\)\(\Leftrightarrow3^{7x-11}=3^3\)\(\Leftrightarrow7x-11=3\)\(\Leftrightarrow7x=14\Leftrightarrow x=2\)

10) \(2.3^x=19\cdot3^8-81^2\)\(\Leftrightarrow2.3^x=19\cdot3^8-3^8=18.3^8=2.3^{11}\)\(\Leftrightarrow3^x=3^{11}\Leftrightarrow x=11\)

Đây là cách làm của mình. Bạn có thể chỉnh sửa tuỳ ý theo cách làm của bạn nhé ^^

Học tốt ^3^

c , d giải nốt:

c) 2x + 3/6 = 7x - 13/15

=> (2x + 3) . 15 = (7x - 13) x 6

=> 30x +45 = 42x - 78

=> 30x - 42x = -78 - 45

=> -12x = -123

=> x = 41/4

d) 2-x/4 = 3x - 1/ - 3

=> (2-x) . -3 = 4. (3x - 1)

=> -6 - (-3x) = 12x - 4

(-6) + 4 = 12x - - (-3x)

=> -2 = 9x

=> x = -2/9    

b: \(x+\dfrac{5}{6}=\dfrac{3}{8}\)

=>\(x=\dfrac{3}{8}-\dfrac{5}{6}\)

=>\(x=\dfrac{9}{24}-\dfrac{20}{24}=-\dfrac{11}{24}\)

c: \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{5}{6}\)

=>\(\dfrac{2}{3}x=\dfrac{5}{6}+\dfrac{1}{2}=\dfrac{8}{6}=\dfrac{4}{3}\)

=>2x=4

=>x=4/2=2

d: \(\dfrac{x}{7}=\dfrac{6}{-21}\)

=>\(\dfrac{x}{7}=\dfrac{-2}{7}\)

=>x=-2

e: \(\left(\dfrac{7}{3}x-0,6\right):\dfrac{3^2}{5}=1\)

=>\(\dfrac{7}{3}x-0,6=\dfrac{3^2}{5}=1,8\)

=>\(\dfrac{7}{3}x=2,4\)

=>\(x=2,4:\dfrac{7}{3}=2.4\cdot\dfrac{3}{7}=\dfrac{7.2}{7}=\dfrac{36}{35}\)

f: \(\dfrac{x}{45}=\dfrac{5}{6}+\dfrac{-29}{30}\)

=>\(\dfrac{x}{45}=\dfrac{25}{30}-\dfrac{29}{30}=-\dfrac{4}{30}=-\dfrac{2}{15}\)

=>\(x=-\dfrac{2}{15}\cdot45=-6\)

g: \(\left(4,5-2x\right)\cdot\left(-\dfrac{1^4}{7}\right)=\dfrac{11}{14}\)

=>\(4,5-2x=\dfrac{11}{14}:\dfrac{-1}{7}=\dfrac{-11}{2}\)

=>\(2x=4,5+\dfrac{11}{2}=\dfrac{20}{2}=10\)

=>x=10/2=5

h: \(-\dfrac{2}{7}+\dfrac{4}{7}x=\dfrac{5}{7}\)

=>\(\dfrac{4}{7}x=\dfrac{5}{7}+\dfrac{2}{7}=\dfrac{7}{7}\)

=>4x=7

=>\(x=\dfrac{7}{4}\)

1.x=4

2.x=6

3.x=8

a: \(2^{x^2-2x+1}=1\)

=>\(2^{\left(x-1\right)^2}=2^0\)

=>\(\left(x-1\right)^2=0\)

=>x-1=0

=>x=1

b: \(7^{x^2+7x}=5764801\)

=>\(7^{x^2+7x}=7^8\)

=>\(x^2+7x=8\)

=>\(x^2+7x-8=0\)

=>(x+8)(x-1)=0

=>\(\left[{}\begin{matrix}x+8=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=1\end{matrix}\right.\)

c: \(6^{x^2+12x}=6^{7x}\)

=>\(x^2+12x=7x\)

=>\(x^2+5x=0\)

=>x(x+5)=0

=>\(\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

d: \(\left(\dfrac{1}{3}\right)^{x-1}=3^{2x-5}\)

=>\(3^{-x+1}=3^{2x-5}\)

=>-x+1=2x-5

=>-x-2x=-5-1

=>-3x=-6

=>x=2

e: \(\left(\dfrac{1}{5}\right)^{3x+5}=5^{2x+1}\)

=>\(5^{-3x-5}=5^{2x+1}\)

=>-3x-5=2x+1

=>-5x=6

=>\(x=-\dfrac{6}{5}\)

a) \(1678\cdot24+77\cdot1678-1678=1678\left(24+77-1\right)=167800\)

b) Ta có: \(\dfrac{2}{5}+\dfrac{4}{11}+\dfrac{3}{5}+\dfrac{18}{11}+\dfrac{1}{4}+\dfrac{1}{6}\)

\(=\left(\dfrac{2}{5}+\dfrac{3}{5}\right)+\left(\dfrac{4}{11}+\dfrac{18}{11}\right)+\dfrac{1}{4}+\dfrac{1}{6}\)

\(=3+\dfrac{1}{4}+\dfrac{1}{6}\)

\(=\dfrac{36}{12}+\dfrac{3}{12}+\dfrac{2}{12}=\dfrac{41}{12}\)

c) \(\dfrac{7}{9\cdot7}+\dfrac{7}{9\cdot4}-\dfrac{7}{9\cdot10}\)

\(=\dfrac{7}{9}\left(\dfrac{1}{7}+\dfrac{1}{4}-\dfrac{1}{10}\right)\)

\(=\dfrac{7}{9}\cdot\left(\dfrac{20}{140}+\dfrac{35}{140}-\dfrac{14}{140}\right)\)

\(=\dfrac{7}{9}\cdot\dfrac{41}{140}=\dfrac{41}{180}\)