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15 tháng 4 2020

63:(39-2.(2x+1)^2)+4^3=67

\(63:\left(39-2\left(2x+1\right)^2\right)+64=67\)

\(63:\left(39-2\left(2x+1\right)^2\right)=3\)

\(39-2\left(2x+1\right)^2=21\)

\(2\left(2x+1\right)^2=18\)

\(\left(2x+1\right)^2=9\)

\(2x+1=3\)

\(2x=2\)

\(x=1\)

19 tháng 12 2018

63 : [ 39 - 2(2x+1)2] +43 = 67

63 : [ 39 - 2(2x+1)2] +64 = 67

63: [ 39 - 2(2x+1)2] = 67 - 64

63: [ 39 - 2(2x+1)2] = 3

39 - 2(2x+1)2 = 63 : 3

39 - 2(2x+1)2 = 21

2(2x+1)2 = 39 - 21

2(2x+1)2 = 18

(2x+1)2 = 18 : 2

(2x+1)2 = 9

(2x+1)2 = 32

2x+1 = 3

2x = 3 - 1

2x = 2

x = 2 : 2

x = 1

19 tháng 12 2018

cảm ơn bạnyeu

6 tháng 11 2016

2/15-2/65-4/39=4/39-4/39=0

32/13-(7/23+8/23)=32/13-15/23=541/299

(1/3+12/67+13/41) - (79/67 - 28/41)=6836/8241-1363/2747=1/3

38/145 - (8/45-17/51-3/11)=38/145-(-212/495)=1982/2871

8 tháng 8 2017

\(\dfrac{2x}{15}+\dfrac{2x}{35}+\dfrac{2x}{63}+...+\dfrac{2x}{195}=\dfrac{4}{5}\\ x\cdot\left(\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+...+\dfrac{2}{195}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{13\cdot15}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{1}{3}-\dfrac{1}{15}\right)=\dfrac{4}{5}\\ x\cdot\dfrac{4}{15}=\dfrac{4}{5}\\ x=\dfrac{4}{5}:\dfrac{4}{15}\\ x=3\)

Gọi \(D=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\)

\(2D=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\\ 2D+D=\left(1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\right)\\ 3D=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\\ 3D=1-\dfrac{1}{64}< 1\\ \Rightarrow D=\dfrac{1-\dfrac{1}{64}}{3}< \dfrac{1}{3}\)

Vậy \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}< \dfrac{1}{3}\)

22 tháng 5 2020

Tìm x:
b) 1/3.x+2/5.(x-1)=0

\(<=> \dfrac{1}{3} .x +\dfrac{2}{5}x - \dfrac{2}{5} =0\)

\(<=> \dfrac{11}{15}x = \dfrac{2}{5}\)

\(<=> x= \dfrac{6}{11}\)

Vậy \( x= \dfrac{6}{11}\)
c) (2x-3).(6-2x)=0
\(<=> \begin{cases} 2x-3=0 \\ 6-2x=0 \end{cases}\) \(<=> \begin{cases} 2x=3 \\ -2x=-6 \end{cases}\) \(<=>\begin{cases} x=\dfrac{3}{2} \\ x=3 \end{cases}\)

Vậy \(x=( \dfrac{3}{2} ; 3)\)
d) -2/3-1/3.(2x-5)= 3/2

\(<=> 2x-5= \dfrac{5}{2}\)

\(<=> 2x= \dfrac{15}{2}\)

\(<=> x= \dfrac{15}{4}\)

Vậy \(x= \dfrac{15}{4}\)
f) 1/3.x-1/2=4 và 1/2 (Hỗn số ý '^')

\(<=> \dfrac{1}{3} x -\dfrac{1}{2} = \dfrac{9}{2}\)

\(<=> \dfrac{1}{3}x =5\)

\(<=> x= 15\)

Vậy \(x= 15\)

23 tháng 5 2020

Cảm ơn cậu nhiều nhé!