1) (x-3)(x-5) = 0

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=5\end{cases}}\)

(x+7).35 = 2.35

\(\Rightarrow x+7=2\)

\(\Rightarrow x=2-7=-5\)

Vậy x = -5

2) 1.2 + 2.3 + 3.4 + .... + 99.100

Đặt A = 1.2 + 2.3 + .... + 99.100

3A = 1.2.3 + 2.3.4 + 3.4.3 + .... + 99.100.3

3A = 1.2.(3-0) + 2.3.(4-1) + 3.4.(5-2)+ .... + 99.100.(101-98)

3A = ( 1.2.3 + 2.3.4 + 3.4.5 +.... + 99.100.101 ) - ( 0.1.2 + 1.2.3 + 2.3.4 + ... + 98.99.100 )

3A = 99.100.101 - 0.1.2

3A = 999900 - 0

3A = 999900

A = 999900 : 3

A = 333300

Vậy A = 333300

1

\(\left(x-3\right)\left(x-5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\x=-5\end{cases}}}\)

\(\left(x+7\right).35=2.35\)

\(x+7=2=>x=-5\)

A=1.2+2.3+3.4...+99.100
=> 3A= 1.2.3+2.3.3+3.4.3...+99.100.3
=> 3A=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+...+99.100(101=98)
=> 3A= (1.2.3+2.3.4+3.4.5+...+99.100.101)-(0.1.2+1.2.3+..+98.99.200)
=> 3A= 99.100.101-0.1.2
=> 3A= 999900-0=999900
=> A= 999900:3=333300

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

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

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

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

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

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

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

\(\text{Vậy }x=64\)

Số x cần tìm là : 99

Ta có : \(\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+.....+\frac{5}{x\left(x+1\right)}=\frac{99}{20}\)

\(\Rightarrow5\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{x\left(x+1\right)}\right)=\frac{99}{20}\)

\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{x}-\frac{1}{x+1}=\frac{99}{20}.\frac{1}{5}\)

\(\Rightarrow1-\frac{1}{x+1}=\frac{99}{100}\)

\(\Rightarrow\frac{1}{x+1}=1-\frac{99}{100}=\frac{1}{100}\)

=> x + 1 = 100

=> x = 99

Ta có  \(\frac{5}{1.2}+\frac{5}{2.3}+...+\frac{5}{x\left(x+1\right)}=\frac{99}{20}\)

   =>  \(5\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{99}{20}\)

   =>    \(5\left(1-\frac{1}{x+1}\right)=\frac{99}{20}\)

   =>   \(1-\frac{1}{x+1}=\frac{99}{100}\)

  =>   \(\frac{1}{x+1}=\frac{1}{100}\)

 =>   x+1 = 100

=>  x = 99

Ta có: \(\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\Leftrightarrow100\left(\dfrac{1}{1}-\dfrac{1}{10}\right)-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]\cdot2=89\)

\(\Leftrightarrow\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)=\dfrac{1}{2}\)

\(\Leftrightarrow x+\dfrac{103}{50}=5\)

hay \(x=\dfrac{147}{50}\)

\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{8.9}+\frac{1}{9.10}\right)\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-........-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}\)

\(=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}\)

\(\Leftrightarrow\frac{9}{10}.100-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right):\frac{1}{2}\right]=89\)

\(\Leftrightarrow90-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right):\frac{1}{2}\right]=89\)

\(\Leftrightarrow\frac{5}{2}:\left(x+\frac{206}{100}\right):\frac{1}{2}=90-89=1\)

\(\Leftrightarrow\frac{5}{2}:\left(x+\frac{206}{100}\right)=1.\frac{1}{2}=\frac{1}{2}\)

\(\Leftrightarrow x+\frac{206}{100}=\frac{5}{2}:\frac{1}{2}\)

\(\Leftrightarrow x+\frac{103}{50}=\frac{5}{2}.2\)

\(\Leftrightarrow x+\frac{103}{50}=5\)

\(\Leftrightarrow x=5-\frac{103}{50}\)

\(\Leftrightarrow x=\frac{250}{50}-\frac{103}{50}\)

\(\Leftrightarrow x=\frac{147}{50}\)

Cai phan 1+3+5+...+99 chac em biet lam roi phai ko? Con 3/1.2+3/2.3+3/3.4+...+3/99.100 thi em cu tach lam sao cho tro thanh dang ban dau thi lam . Anh phai nghi roi !~ Neu chieu anh ranh ranh thi len giai tiep . BYE BYE

\(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{99.100}+4x=1+3+5+...+99\)

\(3\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)+4x=\left(1+99\right)+\left(3+97\right)+\left(5+95\right)+...+\left(49+51\right)\)\(3\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)+4x=100+100+100+...+100\)\(3\left(1-\frac{1}{100}\right)+4x=100.25\)

\(3.\frac{99}{100}+4x=2500\)

\(\frac{297}{100}+4x=2500\)

\(4x=2500-\frac{297}{100}\)

\(4x=2500-2,97\)

\(4x=2497,03\)

\(x=624,2575\)

\(x=2497,03:4\)

HD;Nhìn chung thấy có (x.99.100) cuối cùng => có 99 x Và tách ra ta đc 1.2.3......99.100+99x nha! 2