gọi tổng này là một số A

ta có  

A=7/4x(1/1x5+1/5x9+1/9x13+1/13x17+1/17x21)

A= 7/4x(1-1/5+1/5-1/9+1/9-1/13+1/13-1/17+1/17-1/21)

A= 7/4x(1-1/21)=7/4x20/21

suy ra A=5/3

7/1×5 + 7/5x9 + 7/9x13 + 7/13x17 + 7/17x21

= 7/4x(4/1x5 + 4/5x9 + 4/9x13 + 4/13x17 + 4/17x21)

= 7/4x(1 - 1/5 + 1/5 - 1/9 + 1/9 - 1/13 + 1/13 - 1/17 + 1/17 - 1/21)

= 7/4x(1-1/20)

= 7/4x19/20

= 133/80

đặt A=7/1X5+7/5X9+7/9X13+7/13X17+7/17X21

Ta có:4A=4x(7/1x5+7/5x9+7/9x13+7/13x17+7/17x21)

4A=7x(4/1x5+4/5x9+4/9x13+4/13x17+4/17x21)

4A=7x(1-1/5+1/5-1/9+1/9-1/13+1/13-1/17+1/17-1/21)

4A=7x(1-1/21)

4A=7x20/21

A=20/3:4

A=20/3 x 1/4

A=5/3

Vậy A=5/3

Em dùng công thức toán học hoặc viết ra giấy, chụp ảnh rồi up lên chứ thế này cô không đúng đề bài để giúp em được.

Bài 2:

\(A=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)

\(3A=\frac{3}{2\times5}+\frac{3}{5\times8}+\frac{3}{8\times11}+\frac{3}{11\times14}+\frac{3}{14\times17}\)

\(3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)

\(3A=\frac{1}{2}-\frac{1}{17}=\frac{15}{34}\)

\(A=\frac{15}{34}\times\frac{1}{3}=\frac{5}{34}\)

Bài 2:

\(A=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)

\(3A=\frac{3}{2\times5}+\frac{3}{5\times8}+\frac{3}{8\times11}+\frac{3}{11\times14}+\frac{3}{14\times17}\)

\(A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)

\(3A=\frac{1}{2}-\frac{1}{17}=\frac{15}{34}\)

\(A=\frac{15}{34}\times\frac{1}{3}=\frac{5}{34}\)

Cảm ơn cái đặc cầu

\(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)

\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\)

\(\Leftrightarrow3x^2+26x+28=28\)

\(\Leftrightarrow3x^2+26x=0\)\(\Leftrightarrow x\left(3x+26\right)=0\)

Suy ra x=0 hoặc x=-26/3

cho mk hỏi ngu tí là 6x^2 ở đâu thế ạ

c1) 7x + x5  -100 =19

=> 70 + x + 10x + 5 -100 =19

=> 11x + 75 - 100 =19

=> 11x = 19+100-75

=> 11x = 44

=>x=44:11=4

c2)1x5 +   4x  - 78 = 100

=> 100 + 10x + 5 +40 +x =100+78=178

=> 145 +11x =178

=>11x=178-145=33

=> x=33:11=3

k mik nha! kết bn nha mọi người!

1. 

 7X+X5-100=19

Xx(7+5)-100=19

Xx12 -100 =19

Xx12 =19+100

Xx12=119

=> X=119:12

=> X=119/12

2.

1x5+4x-78=100

x5+4x = 100+78

Xx(5+4)=178

X9=178

=> X=178:9

=> X=178/9

1/Ta có :

7x + x5 ‐ 100 = 19

=> 12x ‐ 100 = 19

=> 12x = 100 + 19 = 119

x = 119 : 12 =119/12
2/1x5 + 4x  - 78 = 100

5 + 4x - 78 = 100
5 + 4x = 100 + 78
5 + 4x = 178
4x = 178 - 5
4x = 173
x = 173 : 4 
x = 43.25 hoặc 173/4

\(x:\dfrac{2}{3}=\dfrac{5}{4}\)

\(x=\dfrac{5}{4}\times\dfrac{2}{3}\)

\(x=\dfrac{5}{6}\)

x : 2/3 = 1/2 + 3/4 

x : 2/3 = 5/4

x = 5/4 x 2/3

x = 5/6 

\(x:\dfrac{2}{3}=\dfrac{1}{2}+\dfrac{3}{4}\)

\(x:\dfrac{2}{3}=\dfrac{2}{4}+\dfrac{3}{4}\)

\(x:\dfrac{2}{3}=\dfrac{5}{4}\)

\(x=\dfrac{5}{4}\times\dfrac{2}{3}\)

\(x=\dfrac{10}{12}\)

\(x=\dfrac{5}{6}\)

\(1,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 2,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 3,\left(x+1\right)^2+2\left(x+1\right)=0\\ \Rightarrow\left(x+1\right)\left(x+1+2\right)=0\\ \Rightarrow\left(x+1\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)

x²+4x+4=0
(x+2)²=0
x+2=0
x=+-2
câu 1 giống câu 2
(x+1)²+2(x+1)=0
(x+1+2)(x+1)=0
Th1: x+3=0           Th2: x+1=0
            x=-3                      x=-1
vậy ...

Theo đề, ta có:  \(\dfrac{1+2x}{18}=\dfrac{1+4x}{34}\)

\(\Leftrightarrow34\left(1+2x\right)=18\left(1+4x\right)\)

\(\Leftrightarrow34+68x=18+72x\)

\(\Leftrightarrow34-18=72x-68x\)

\(\Leftrightarrow16=4x\)

\(\Leftrightarrow x=4\)

Khi \(x=4\) vào ta có:   \(\dfrac{1+4.4}{34}=\dfrac{1+6.4}{2y^2}\Leftrightarrow\dfrac{1}{2}=\dfrac{25}{2y^2}\) 

\(\Leftrightarrow2y^2=50\)

\(\Leftrightarrow y^2=50\)

\(\Leftrightarrow y=\pm5\)

\(4x^2-8x+25\)

\(=\left(4x^2-8x+4\right)+21\)

\(=\left(2x-2\right)^2+21\)

Mà  \(\left(2x-2\right)^2\ge0\forall x\)

\(\Rightarrow\) giá trị nhỏ nhất của biểu thức trên là 21

Dấu " = " xảy ra khi :  \(2x-2=0\Leftrightarrow x=1\)

Vậy ...