2 ) 1.2 + 2.3 + 3.4 + ... + x.( x + 1 ) = 5850

Đặt A = 1.2 + 2.3 + 3.4 + ... + x.( x + 1 )

=> 3A = 1.2.3 + 2.3.( 4 - 1 ) + 3.4.(5 - 2 ) + ... + x.( x + 1 ). [ ( x + 2 ) - ( x - 1 ) ] 

= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + x . ( x + 1 ) . ( x + 2 ) - ( x - 1 ). x . ( x + 1 )

= x . ( x + 1 ) . ( x + 2 )

=> A = [ x . ( x + 1 ) . ( x + 2 ) ] : 3

Mà tổng A = 5850

=> [ x . ( x + 1 ) . ( x + 2 ) ] : 3 = 5850

=> x . ( x + 1 ) . ( x + 2 ) = 17550

x . ( x + 1 ) . ( x + 2 ) = 25 . 26 . 27 

=> x = 25

Vậy x = 25

help 

lớp mấy đây ạ

Bài 3:

a: \(S=1+5^2+5^4+\cdots+5^{200}\)

=>25S=\(5^2+5^4+5^6+\cdots+5^{202}\)

=>25S-S=\(5^2+5^4+\cdots+5^{202}-1-5^2-\cdots-5^{200}\)

=>24S=\(5^{202}-1\)

=>\(S=\frac{5^{202}-1}{24}\)

b: \(4^{30}=\left(2^2\right)^{30}=2^{60}=2^{30}\cdot2^{30}=8^{10}\cdot4^{15}\)

\(3\cdot24^{10}=3\cdot3^{10}\cdot8^{10}=8^{10}\cdot3^{11}\)

mà \(4^{15}>3^{11}\)

nên \(4^{30}>3\cdot24^{10}\)

=>\(2^{30}+3^{30}+4^{30}>3\cdot24^{10}\)

Bài 2:

a: |2x-3|>5

=>\(\left[\begin{array}{l}2x-3>5\\ 2x-3<-5\end{array}\right.\Rightarrow\left[\begin{array}{l}2x>8\\ 2x<-2\end{array}\right.\Rightarrow\left[\begin{array}{l}x>4\\ x<-1\end{array}\right.\)

c: |3x-1|<=7

=>-7<=3x-1<=7

=>-6<=3x<=8

=>\(-2\le x\le\frac83\)

d: \(\left|3x-5\right|+\left|2x+3\right|=7\) (1)

TH1: \(x<-\frac32\)

=>2x+3<0; 3x-5<0

(1) sẽ trở thành: -2x-3-3x+5=7

=>-5x+2=7

=>-5x=5

=>x=-1(loại)

TH2: -3/2<=x<5/3

=>2x+3>=0; 3x-5<0

(1) sẽ trở thành: 2x+3-3x+5=7

=>-x+8=7

=>-x=-1

=>x=-1(nhận)

TH3: x>=5/3

=>2x+3>0; 3x-5>=0

(1) sẽ trở thành: 2x+3+3x-5=7

=>5x-2=7

=>5x=9

=>x=9/5(nhận)

\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

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

\(\Rightarrow\left(100-10\right)-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=1\)

\(\Rightarrow\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)=1.2=2\)

\(\Rightarrow\left(x+\dfrac{206}{100}\right)=\dfrac{5}{2}:2=\dfrac{5}{2}.\dfrac{1}{2}=\dfrac{5}{4}\)

\(\Rightarrow x=\dfrac{5}{4}-\dfrac{206}{100}=\dfrac{125}{100}-\dfrac{206}{100}\)

\(\Rightarrow x=-\dfrac{81}{100}\)

\(\left(d\right)30-x=200-175\)

\(30-x=25\)

\(x=5\)

\(\left(e\right)5\left(x+12\right)=70\)

\(x+12=14\)

\(x=2\)

tương tự 2 câu trên bạn tự làm câu cuối nhé, mình không ghi lại đề đâu

#quankun^^

 d,175+(30-x)=200

    (30-x)       = 200 - 175

     30 - x      = 25

           x       = 30 - 25 

           x       = 5. 

e,5(x+12)+22=92

   5(x+12)      = 92 -  22

   5(x+12)      = 70

       x+12       = 70 : 5

         x + 12   = 14

         x           = 14 - 12

          x          = 2.

f,95-5(x+2)=45

       5(x+2) = 95 - 45

        5(x+2)= 50

         x + 2 = 50 : 5

         x + 2 = 10

          x      =10 -2

          x      = 8

Hok tốt !

b giúp mk câu hỏi của mkn nha

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

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

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

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

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

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

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

\(\text{Vậy }x=64\)