2^100.x=2^103

chia cả 2 vế cho 2^100 ta có

x=2^3

x=8

vậy x=8

\(2^{100}.x=2^{103}\)

\(x=2^{103}:2^{100}\)

\(x=2^3\)

\(\frac{2x^2+1}{x+2}=\frac{2x^2+4x-4x-8+9}{x+2}=\frac{2x\left(x+2\right)-4\left(x+2\right)+9}{x+2}=2x-4+\frac{9}{x+2}\)

\(\Rightarrow x+2\inƯ\left(9\right)\Rightarrow x+2\in\left\{-9;-3;-1;1;3;9\right\}\Rightarrow x\in\left\{-11;-5;-3;-1;1;7\right\}\)

Cách 2:

\(\frac{2x^2+1}{x+2}=\frac{2\left(x^2-2^2\right)+9}{x+2}=\frac{2\left(x-2\right)\left(x+2\right)+9}{x+2}=2\left(x-1\right)+\frac{9}{x+2}\)

\(\Rightarrow x+2\inƯ\left(9\right)\Rightarrow x+2\in\left\{-9;-3;-1;1;3;9\right\}\Rightarrow x\in\left\{-11;-5;-3;-1;1;7\right\}\)

\(B=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}...\frac{779}{780}\)

\(B=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}...\frac{1558}{1560}\)

\(B=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{38.41}{39.40}\)

\(B=\frac{1.2.3...38}{3.4.5...40}.\frac{4.5.6...41}{2.3.4...39}\)

\(B=\frac{2}{39.40}.\frac{40.41}{2.3}\)

\(B=\frac{41}{39.3}=\frac{41}{297}\)

(2\(\hept{\begin{cases}2\\3\end{cases}}^5\))

\(cam\in A\)

\(táo\in A\); \(táo\notin B\)

\(ổi\in B\); \(ổi\notin A\)

\(chanh\in B\); \(chanh\notin A\)

\(cam\in B\)

\(CAM\in A\)

\(\text{TÁO}\in A\)

\(\text{ỔI}\in B\)

\(CHANH\in B\)

\(CAM\in B\)

a) 5^x + 5^​x+2 ​=650

5^x ( 1 + 5²)=650

5^x . 26 = 650

5^x = 650 : 26 = 25= 5²

=> x = 2

b) (7x + 2)^-1 = 3^-2

\(\frac{1}{7x+2}=\frac{1}{3^2}\)

\(\Rightarrow7x+2=9\)

\(7x=9-2=7\)

\(x=\frac{7}{7}=1\)

2.(x-3)+3(x-1)=0

=>2x-6+3x-3=0

=>(2x+3x)-6-3=0

=>5x-9=0

=>5x=9

=>x=9/5

Ta có: 2.(x-3)+3.(x-1) = 0

           2x-6 + 3x -3  = 0

            (2x+3x)-6-3=0

                5x    - 9   = 0

                5x           = 9

                 x            = 9 : 5

                x               = 9/5