Ta có :

\(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+..............+\dfrac{1}{\left(2x+1\right)\left(2x+3\right)}=\dfrac{15}{93}\)

\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+..............+\dfrac{2}{\left(2x+1\right)\left(2x+3\right)}=\dfrac{30}{93}\)

\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+..............+\dfrac{1}{2x+1}-\dfrac{1}{2x+3}=\dfrac{30}{93}\)

\(\dfrac{1}{3}-\dfrac{1}{2x+3}=\dfrac{30}{93}\)

\(\Rightarrow\dfrac{1}{3}-\dfrac{30}{93}=\dfrac{1}{2x+3}\)

\(\Rightarrow\dfrac{1}{93}=\dfrac{1}{2x+3}\)

\(\Rightarrow2x+3=93\)

\(2x=90\)

\(\Rightarrow x=45\)

Vậy \(x=45\) là giá trị cần tìm

~ Chúc bn học tốt ~

(2x-1).5\(^{2021}\)=3.5\(^{2022}\)

Ta có: (2x-1)=3.(5\(^{2022}\):5\(^{2021}\))

(2x-1)=3.5

(2x-1)=15

2x-1=15

2x=15-1

2x=14

x=14:2

x=7

a)

\(280-\dfrac{\left(x-140\right)}{35}=270\\\dfrac{ \left(x-140\right)}{35}=10\\ x-140=\dfrac{10}{35}\\ x=\dfrac{10}{35}+140=\dfrac{982}{7}\)

b)

\(\dfrac{\left(190-2x\right)}{35}-32=16\\ \dfrac{\left(190-2x\right)}{35}=48\\ 190-2x=1680\\ -2x=1490\\ x=-745\)

c)

\(\dfrac{720}{41-2x+5}=2^3.5\\ \dfrac{720}{46-2x}=40\\ 720=1840-80x\\ 80x=1120\\ x=14\)

\(x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+4,5\right)=3,5\)

\(\Rightarrow\left(2x^2-x\right)\left(x+5\right)-2x^3-9x^2-x-4,5=3,5\)

\(\Rightarrow2x^3+10x^2-x^2-5x-2x^3-9x^2-x-4,5=3,5\)

\(\Rightarrow-5x-4,5=3,5\)

\(\Rightarrow-5x=8\)

\(\Rightarrow x=-\dfrac{8}{5}\)

\(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{49}{99}\\ \Leftrightarrow2\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{\left(2x-1\right)\left(2x+1\right)}\right)=2\cdot\dfrac{49}{99}\\ \Leftrightarrow\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{98}{99}\\ \Leftrightarrow\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow1-\dfrac{1}{2x+1}=1-\dfrac{1}{99}\\ \Leftrightarrow\dfrac{1}{2x+1}=\dfrac{1}{99}\\ \Rightarrow2x+1=99\\ \Leftrightarrow2x=98\\ \Leftrightarrow x=49\)

2x-5=x+1

<=> 2x-x = 1+5

<=> x = 6

\(2x-5=x+1\)

=> \(2x-x=1+5\)

=> \(1x=6\)

=> \(x=6:1\)

=> \(x=6\)

Vậy \(x=6.\)

\(5-3.5-2x=-4,5\)

=> \(5-15-2x=-4,5\)

=> \(-10-2x=-4,5\)

=> \(2x=\left(-10\right)-\left(-4,5\right)\)

=> \(2x=-\frac{11}{2}\)

=> \(x=\left(-\frac{11}{2}\right):2\)

=> \(x=-\frac{11}{4}\)

Vậy \(x=-\frac{11}{4}.\)

Chúc bạn học tốt!

Ta có \(3.5^{2x}-4=5^x.5^x-4\Rightarrow3.5^{2x}=5^{2x}\Rightarrow3.5^{2x}-5^{2x}=0\Rightarrow2.5^{2x}=0.\)

Vì \(5^{2x}\ge1\Rightarrow2.5^{2x}\ge2\Rightarrow\)

Không có giá trị x thỏa mãn.

bài toán này ko có giá trị thõa mãn

\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{49}{99}\)

\(\Leftrightarrow\) \(\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}\right)=\dfrac{49}{99}\)

\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}\right)=\dfrac{49}{99}\)\(\Leftrightarrow\dfrac{1}{2}\left(1-\dfrac{1}{2x+1}\right)=\dfrac{49}{99}\)

\(\Leftrightarrow1-\dfrac{1}{2x+1}=\dfrac{98}{99}\)

\(\Leftrightarrow\dfrac{1}{2x+1}=1-\dfrac{98}{99}\)

\(\Leftrightarrow\dfrac{1}{2x+1}=\dfrac{1}{99}\)

\(\Leftrightarrow2x+1=99\)

\(\Rightarrow x=\dfrac{99-1}{2}\)

\(\Rightarrow x=49\)

Vậy \(x=49\)

 1/3.5 + 1/5.7 +......+ 1/(2x+1)(2x+3) =100/609

 2/3.5 + 2/5.7 +......+ 1/(2x+1)(2x+3)=200/609

1/3 - 1/5 + 1/5 - 1/7 +.....+1/2x+1 - 1/2x+3=200/609

 1/3 - 1/2x+3 = 200/609

1/2x+3 = 1/3 - 200/609

Đoạn còn lại tự làm nhá!