5^2x-1=5^3.5^4
5^2x-1=5^7
=>2x-1=7
còn lại bn lm tiếp
Câu b hình như bn vik sai đề bài , đề bài đúng phải thế này :
7^x-3.4=296
7^x-12=296
7^x    =296+12
7^x    = 308
7^x     = 7^3
=>x=3
c) 2^x.2^x+3=64^2:2^5
   (2^x)^2 +3 = (2^6)^2 : 2^5 
    (2^x)^2 +3 = 2^12 : 2^5
    (2^x.2) +3= 2^7

đặt 5 ra ngoài rồi làm như bình thường

5(1/2.3+1/3.4+...+1/x(x+1)=64/13

5.(1/2-1/3+1/3-1/4+...+1/x-1/x+1)=64/13

5(1/2-1/x+10)=64/13

bạn tự làm tiếp nha mình bận rồi

\(\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+...+\frac{5}{x\left(x+1\right)}=\frac{64}{13}\)

\(\Leftrightarrow5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{64}{13}\)

\(\Leftrightarrow1-\frac{1}{x+1}=\frac{64}{13}\div5\)

\(\Leftrightarrow1-\frac{1}{x+1}=\frac{64}{65}\)

\(\Leftrightarrow\frac{1}{x+1}=1-\frac{64}{65}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{65}\)

\(\Rightarrow x+1=65\Rightarrow x=65-1=64\)

\(\text{Vậy }x=64\)

Đặt \(S=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2.}{4.5}+...+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)

\(\Rightarrow\frac{S}{2}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{1999}{4002}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1999}{4002}\)

\(=\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{4002}\)

\(\frac{1}{x+1}=\frac{1}{2001}\)

\(\Rightarrow\)x+1=2001

x=2000

Vậy x=2000.

khó lắm:))

`@` `\text {Ans}`

`\downarrow`

\(\left(2\cdot x+2\right)^2=64\)

`\Rightarrow`\(\left(2x+2\right)^2=\left(\pm8\right)^2\)

`\Rightarrow`\(\left[{}\begin{matrix}2x+2=8\\2x+2=-8\end{matrix}\right.\)

`\Rightarrow`\(\left[{}\begin{matrix}2x=8+2\\2x=-8+2\end{matrix}\right.\)

`\Rightarrow`\(\left[{}\begin{matrix}2x=10\\2x=-6\end{matrix}\right.\)

`\Rightarrow`\(\left[{}\begin{matrix}x=10\div2\\x=-6\div2\end{matrix}\right.\)

`\Rightarrow`\(\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

Vậy, `x \in {5; -3}`

`@` `\text {Kaizuu lv uuu}`