\(S=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{101}}\)

\(3S=3+1+\frac{1}{3}+...+\frac{1}{3^{100}}\)

\(2S=3-\frac{1}{3^{100}}\)

\(S=\frac{3-\frac{1}{3^{101}}}{2}\)

Xong ^.^

3^x-1+5.3^x-1=162

3^x-1 .(5 + 1)=162

3^x-1 . 6 =162

3^x-1  =162:6

3^x-1 =27

3^x-1  =3\(^3\)

x - 1 = 3

x       =  3 + 1

x      = 4

3^x-1+5.3^x-1=162

3^x-1.1+5.3^x-1=162

3^x-1.(1+5)=162.Suy ra:3^x-1=162:6=27

3^x:3=27.Suy ra:3^x=3.27=81=3⁴(81=3⁴)

Vậy x=4

CHÚC BẠN HỌC TỐT

5
x
+5
x+2
=650

5^x\left(1+5^2\right)=6505
x
(1+5
2
)=650

5^x\cdot26=6505
x
⋅26=650

5^x=650\text{ : }265
x
=650 : 26

5^x=255
x
=25

5^x=5^25
x
=5
2


\Rightarrow\text{ }x=2⇒ x=2

\(5^x+5^{x+2}=650\)

\(5^x\left(1+5^2\right)=650\)

\(5^x\cdot26=650\)

\(5^x=650\text{ : }26\)

\(5^x=25\)

\(5^x=5^2\)

\(\Rightarrow\text{ }x=2\)

=\(\frac{1}{64}\)

Vì ko có đề nen mk làm theo hai cách

C1 : Phân tích\(\frac{\left(\frac{1}{9}\right)^{25}}{\left(\frac{1}{3}\right)^{30}}=\frac{\frac{1^{25}}{9^{25}}}{\frac{1^{30}}{3^{30}}}=\frac{\frac{1}{9^{25}}}{\frac{1}{3^{30}}}\)

C2 : So sánh \(\left(\frac{1}{9}\right)^{25}=\frac{1^{25}}{9^{25}}=\frac{1}{9^{25}}\)

                       \(\left(\frac{1}{3}\right)^{30}=\frac{1^{30}}{3^{30}}=\frac{1}{3^{30}}=\frac{1}{\left(3^2\right)^{15}}=\frac{1}{9^{15}}\)

Vì tử bằng tử nên mẫu bé hơn thì lớn hơn

Vì \(9^{25}>9^{15}\)

\(=>\frac{1}{9^{25}}< \frac{1}{9^{15}}=>\left(\frac{1}{9}\right)^{25}< \left(\frac{1}{3}\right)^{30}\)

Hết !!!

\(\frac{\left[\frac{1}{9}\right]^{25}}{\left[\frac{1}{3}\right]^{30}}=\frac{\left[\frac{1}{3}\cdot\frac{1}{3}\right]^{25}}{\left[\frac{1}{3}\right]^{30}}=\frac{\left[\frac{1}{3}\right]^{25}\cdot\left[\frac{1}{3}\right]^{25}}{\left[\frac{1}{3}\right]^{30}}=\frac{\left[\frac{1}{3}\right]^{50}}{\left[\frac{1}{3}\right]^{30}}=\left[\frac{1}{3}\right]^{20}\)

mình ko hiểu

Tìm a,b,c biết rằng : \(\frac{2a}{3}=\frac{3b}{4}=\frac{4c}{5}\) và a + b + c = 49

                                     Giải:

Ta có : \(\frac{2a}{3}=\frac{a}{\frac{3}{2}}\), \(\frac{3b}{4}=\frac{b}{\frac{4}{3}}\), \(\frac{4c}{5}=\frac{c}{\frac{5}{4}}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\frac{a}{\frac{3}{2}}=\frac{b}{\frac{4}{3}}=\frac{c}{\frac{5}{4}}=\frac{a+b+c}{\frac{3}{2}+\frac{4}{3}+\frac{5}{4}}=\frac{49}{\frac{49}{12}}=12\)

=> \(\hept{\begin{cases}\frac{a}{\frac{3}{2}}=12\\\frac{b}{\frac{4}{3}}=12\\\frac{c}{\frac{5}{4}}=12\end{cases}}\)=> \(\hept{\begin{cases}a=18\\b=16\\c=15\end{cases}}\)