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2 tháng 4 2017

1. Tìm \(x\):

a) \(\dfrac{x}{5}=\dfrac{5}{6}+\dfrac{-19}{30}\)

\(\dfrac{x}{5}=\dfrac{1}{5}\)

\(\Rightarrow x=1\)

b) \(\dfrac{-5}{6}-x=\dfrac{7}{12}-\dfrac{1}{3}.x\)

\(\dfrac{-5}{6}-\dfrac{7}{12}=x-\dfrac{1}{3}.x\)

\(x-\dfrac{1}{3}.x=\dfrac{-17}{12}\)

\(\dfrac{2}{3}.x=\dfrac{-17}{12}\)

\(x=\dfrac{-17}{12}:\dfrac{2}{3}\)

\(x=\dfrac{-17}{8}\)

c) \(2016^3.2016^x=2016^8\)

\(2016^x=2016^8:2016^3\)

\(2016^x=2016^{8-3}\)

\(2016^x=2016^5\)

\(\Rightarrow x=5\)

d) \(\left(x+\dfrac{3}{4}\right):\dfrac{5}{2}=3\dfrac{1}{2}\)

\(\left(x+\dfrac{3}{4}\right):\dfrac{5}{2}=\dfrac{7}{2}\)

\(\left(x+\dfrac{3}{4}\right)=\dfrac{7}{2}.\dfrac{5}{2}\)

\(x+\dfrac{3}{4}=\dfrac{35}{4}\)

\(x=\dfrac{35}{4}-\dfrac{3}{4}\)

\(x=\dfrac{32}{4}=8\)

e) \(\left(2,8.x-2^5\right):\dfrac{2}{3}=3^2\)

\(\left(2,8.x-2^5\right)=9.\dfrac{2}{3}\)

\(2,8.x-2^5=6\)

\(2,8.x=6+32\)

\(2,8.x=38\)

\(x=38:2,8\)

\(x=\dfrac{95}{7}\)

f) \(\dfrac{4}{7}.x-\dfrac{2}{3}=\dfrac{2}{5}\)

\(\dfrac{4}{7}.x=\dfrac{2}{5}+\dfrac{2}{3}\)

\(\dfrac{4}{7}.x=\dfrac{16}{15}\)

\(x=\dfrac{16}{15}:\dfrac{4}{7}\)

\(x=\dfrac{28}{15}\)

g) \(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{28}\)

\(\left(\dfrac{3x}{7}+1\right)=\dfrac{-1}{28}.\left(-4\right)\)

\(\dfrac{3x}{7}+1=\dfrac{1}{7}\)

\(\dfrac{3x}{7}=\dfrac{1}{7}-1\)

\(\dfrac{3x}{7}=\dfrac{-6}{7}\)

\(\Rightarrow3x=-6\)

\(x=\left(-6\right):3\)

\(x=-2\)

2 tháng 4 2017

2. Thực hiện phép tính:

a) \(\dfrac{1}{2}+\dfrac{1}{2}.\dfrac{2}{3}-\dfrac{1}{3}:\dfrac{3}{4}+1\dfrac{4}{5}\)

\(=\dfrac{1}{2}.\left(\dfrac{2}{3}+1\right)-\dfrac{1}{3}:\dfrac{3}{4}+\dfrac{9}{5}\)

\(=\dfrac{1}{2}.\dfrac{5}{3}-\dfrac{1}{3}:\dfrac{3}{4}+\dfrac{9}{5}\)

\(=\dfrac{5}{6}-\dfrac{4}{9}+\dfrac{9}{5}\)

\(=\dfrac{7}{18}+\dfrac{9}{5}\)

\(=\dfrac{197}{90}\)

b) \(\dfrac{7.5^2-7^2}{7.24+21}\)

\(=\dfrac{7.25-7.7}{7.24+7.3}\)

\(=\dfrac{7.\left(25-7\right)}{7.\left(24+3\right)}\)

\(=\dfrac{7.18}{7.27}\)

\(=\dfrac{2}{3}\)

c) \(\dfrac{2}{3}+\dfrac{1}{3}.\left(\dfrac{-4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}\)

\(=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}\)

\(=\dfrac{2}{3}+\dfrac{7}{54}:\dfrac{7}{12}\)

\(=\dfrac{2}{3}+\dfrac{2}{9}\)

\(=\dfrac{8}{9}\)

7 tháng 5 2022

bài 1 :

\(a,\dfrac{2}{7}+\dfrac{1}{3}=\dfrac{6}{21}+\dfrac{7}{21}=\dfrac{13}{21}\)

\(b,\dfrac{3}{5}-\dfrac{1}{3}=\dfrac{9}{15}-\dfrac{5}{15}=\dfrac{4}{15}\)

\(c,\dfrac{13}{4}:5=\dfrac{13}{4}:\dfrac{5}{1}=\dfrac{13}{4}x\dfrac{1}{5}=\dfrac{13}{20}\)

\(d,\dfrac{6}{23}x\dfrac{1}{18}=\dfrac{1}{69}\)

bài 2 :

\(a,x+\dfrac{1}{3}=\dfrac{5}{12}\)

   \(x=\dfrac{5}{12}-\dfrac{1}{3}\)

  \(x=\dfrac{1}{12}\)

 

\(b,x:\dfrac{7}{4}=\dfrac{2}{5}\)

   \(x=\dfrac{2}{5}x\dfrac{7}{4}\)

   \(x=\dfrac{7}{10}\)

bài 3 :

đổi : 2 dm 1cm = 21cm

chiều cao hình bình hành là;

       21 x\(\dfrac{3}{7}=\)9(cm)

diện tích hình bình hành là;

       21 x 9 =189 (cm2)

                 đáp số : 189 cm2

bài 4 :

\(\dfrac{2}{3}x\dfrac{2}{10}+\dfrac{2}{3}x\dfrac{5}{10}x\dfrac{3}{3}\)

\(\dfrac{2}{3}x\left(\dfrac{2}{10}+\dfrac{5}{10}\right)x\dfrac{2}{3}\)

=\(\dfrac{2}{3}x1x\dfrac{2}{3}\)

\(=\dfrac{2}{3}x\dfrac{2}{3}\)

=\(\dfrac{4}{9}\)

 

7 tháng 5 2022

Bài 1)

a) \(\dfrac{6}{21}+\dfrac{7}{21}=\dfrac{13}{21}\)

b) \(\dfrac{9}{15}-\dfrac{5}{15}=\dfrac{4}{15}\)

c) \(\dfrac{13}{4}x\dfrac{1}{5}=\dfrac{13}{20}\)

d) \(\dfrac{6}{414}=\dfrac{1}{69}\)

Bài 2)

a) \(x=\dfrac{5}{12}-\dfrac{1}{3}\)

\(x=\dfrac{1}{12}\)

b) \(x=\dfrac{2}{5}x\dfrac{7}{4}\)

\(x=\dfrac{7}{10}\)

Bài 3)

2dm 1cm = 21 cm

Chiều cao tấm bìa la

\(21x\dfrac{3}{7}=9\left(cm\right)\)

Diện tích tấm bìa là

\(21x9=189\left(cm2\right)\)

2:

a: =>2/3:x=1,4-2,4=-1

=>x=-2/3

b: =>x/5=25/30-19/30=6/30=1/5

=>x=1

3:

Số học sinh giỏi là 40*1/4=10 bạn

Số học sinh khá là 30*3/5=18 bạn

Số học sinh TB là 30-18=12 bạn

 

a: \(\Leftrightarrow\dfrac{32}{x}=\dfrac{2}{15}+\dfrac{2}{35}+...+\dfrac{2}{99}\)

=>32/x=1/3-1/5+1/5-1/7+...+1/9-1/11

=>32/x=1/3-1/11=8/33

=>x=32:8/33=132

b: \(\Leftrightarrow1-\dfrac{1}{6}+1-\dfrac{1}{12}+...+1-\dfrac{1}{56}=\dfrac{x}{16}\)
\(\Leftrightarrow6-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\right)=\dfrac{x}{16}\)

=>x/16=6-1/2+1/8=11/2+1/8=45/8=90/16

=>x=90

c: \(\Leftrightarrow\dfrac{22}{x}=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{10}\right)\left(1+\dfrac{1}{10}\right)\)

=>22/x=1/2*2/3*...*9/10*3/2*4/3*...*11/10

=>22/x=1/10*11/2=11/20=22/40

=>x=40

16 tháng 7 2023

a) \(\dfrac{x}{3}=\dfrac{2}{3}+\dfrac{-1}{7}\)

\(\Rightarrow\dfrac{x}{3}=\dfrac{11}{21}\)

\(\Rightarrow x=\dfrac{3\cdot11}{21}\)

\(\Rightarrow x=\dfrac{33}{21}\)

\(\Rightarrow x=\dfrac{11}{7}\)

b) \(\dfrac{x}{5}=\dfrac{5}{6}+\dfrac{-19}{30}\)

\(\Rightarrow\dfrac{x}{5}=\dfrac{1}{5}\)

\(\Rightarrow x=\dfrac{5\cdot1}{5}\)

\(\Rightarrow x=1\)

1 tháng 8 2021

1/ \(\left(\dfrac{2021}{2020}+\dfrac{2020}{2021}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)

=\(\left(\dfrac{2021}{2020}+\dfrac{2020}{2021}\right).0\)

=\(0\)

 

26 tháng 10 2021

mink chịu bài này nó rất khó

8 tháng 5 2017

1,

x =( -12 . ( -3) ) : 2

x = 18

2,

a, -7/9 . 6/11 + (-2/9) = -14/33 + (-2/9) = -64/99

b, -4/7 : 2 = -4/7 . 1/2 = -2/7

c, 115 - (24 - 5. 3) = 115 - ( 24 - 15) = 115 - 9 = 106

d,= -3/7. (5/9 + 4/9) + 17/7 = -3/7 . 1 +17/7 = -3/7 . 17/7 = -51/49

e, ??? mình cx k biếtleuleu

8 tháng 5 2017

Lời giải:

\(\dfrac{x}{-12}=\dfrac{-3}{2}\)

\(\Rightarrow2x=-12.\left(-3\right)\)

\(\Rightarrow2x=36\)

\(\Rightarrow x=18\)

6 tháng 5 2017

a) Vì \(\dfrac{x+5}{3}\)= \(\dfrac{x-6}{7}\) nên 7(x+5) = 3(x-6)

=> 7x+ 35 = 3x - 18

7x - 3x = -18 -35

4x = -53

x = -53:4

x = \(\dfrac{-53}{4}\)

26 tháng 7 2017

a, \(4\times\left(-\dfrac{1}{2}\right)^3-2\times\left(-\dfrac{1}{2}\right)^2+3\times\left(-\dfrac{1}{2}\right)+1\)

\(=\left(-\dfrac{1}{2}\right)\left[\left(4\times-\dfrac{1}{2}\right)-\left(2\times-\dfrac{1}{2}\right)+3\right]+1\)

\(=\left(-\dfrac{1}{2}\right)\left(-2+1+3\right)+1\)

\(=\left(-\dfrac{1}{2}\right)2+1\)

\(=-1+1\)

\(=0\)

@Trịnh Thị Thảo Nhi

29 tháng 4 2018

a, 4×(−12)3−2×(−12)2+3×(−12)+14×(−12)3−2×(−12)2+3×(−12)+1

=(−12)[(4×−12)−(2×−12)+3]+1=(−12)[(4×−12)−(2×−12)+3]+1

=(−12)(−2+1+3)+1=(−12)(−2+1+3)+1

=(−12)2+1=(−12)2+1

=−1+1=−1+1

=0=0