\(X=1\)

\(X=1\)

\(X=7\)

MK LÀM ĐÚNG ĐÓ

a. \(x^{10}=1^x\Leftrightarrow x^{10}=1\)

Vì số mũ \(\ne0\)nên \(x=1\)

b. \(x^{10}=x\)

\(\Rightarrow x^{10}-x=0\)

\(\Rightarrow x\left(x^9-1\right)=0\)

\(\Rightarrow\)\(\orbr{\begin{cases}x=0\\x^9-1=0\Rightarrow x=1\end{cases}}\)

Vậy x=0 hoặc x=1

KO HIỂU J CỨ HỎI MIK NHA

987=9.10² +8.10¹ +7.10⁰

2564=2.10³ +5.10² +6.10¹ 

abcde=a.10⁴ +b.10³ +c.10² +d.10¹ +e.10⁰

ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{n^2}< \frac{1}{\left(n-1\right).n}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\)

                                                                        \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)

                                                                           \(=1-\frac{1}{n}< 1\)

=> đ p c m

Ta có: \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+....+\frac{1}{n.n}\)

\(< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n.\left(n-1\right)}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n-1}\)

                                                                \(1-\frac{1}{n-1}< 1\)

Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< 1\)

Cho A = 2^2 + 2^3  + 2^4 + ...+ 2^98 + 2^99 và B = 2^100 - 5. So sánh A và B

Bài giải

ta có: A = 2^2+2^3+2^4+...+2^98+2^99

=> 2A = 2^3+2^4+2^5+...+2^99+2^100

=> 2A-A = 2^100 - 2^2

A = 2^100 - 4 > B = 2^100 - 5

=> A >B

Ta có : A = 2² + 2³ + ...... + 2⁹⁹

=> 2A = 2³ + 2⁴ + ...... + 2¹⁰⁰

=> 2A  - A = 2¹⁰⁰ - 2² 

=> A = 2¹⁰⁰ - 4 > 2¹⁰⁰ - 5

Vậy A > B 

x.2,8 + 5,2 = 48

x.2,8 = 42,8

x = 107/7

b) x.12,25 - x +x.2,75 = 1050

x.(12,25-1+2,75) = 1050

x.14 = 1050

x = 75

2 câu cn lại bn dựa vào mak lm nha!
 

x:15,28571429

x:75

x:1,928571429

x:49,52

\(91924\)\(:1372=67\)

\(163840:98304=\frac{5}{3}\)

k mk nha

cần gấp mn

Đặt \(2^x=t\left(t>0\right)\)

\(\Rightarrow4t+\frac{1}{2}t+\frac{1}{4}t=152\)

\(\Rightarrow t=32\Rightarrow2^x=32\Rightarrow x=5\)

2^x+2 + 2^x-1 + 2^x-2 = 152

2^x.4 + 2^x.1/2 + 2^x.1/4 = 152

2^x.(4+1/2+1/4) = 152

2^x.19/4 = 152

2^x = 32 = 2⁵

=> x = 5

Ta có : \(\left(5x-3\right)^2-\frac{1^2}{64}=0\)

\(\Leftrightarrow\left(5x-3\right)^2=\frac{1}{64}\)

\(\Leftrightarrow\left(5x-3\right)^2=\left(\frac{1}{8}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}5x-3=\frac{1}{8}\\5x-3=-\frac{1}{8}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}5x=\frac{1}{8}+3\\5x=-\frac{1}{8}+3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}5x=\frac{25}{8}\\5x=\frac{23}{8}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{25}{8}.\frac{1}{5}\\x=\frac{23}{8}.\frac{1}{5}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{8}\\x=\frac{23}{40}\end{cases}}\)

b) 3x - 7.(5x-1) = 6 - 2.(4-3x)

=> 3x - 35x + 7 = 6 - 8 + 6x

=> 3x - 35x - 6x = 6-8 -7

-38x = -9

x = 9/38

\(A=2^0+2^3+2^5+...+2^{99\text{​​}}\)

\(\Rightarrow4A=2^3+2^5+2^7+...+2^{101}​​\)

\(\Rightarrow3A=2^{101}-1\)

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