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9 tháng 6 2017

\(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{4}\right)< ...........< \dfrac{1}{48}-\left(\dfrac{1}{16}-\dfrac{1}{6}\right)\)

\(\dfrac{-1}{12}< .........< \dfrac{1}{8}\)và số ghi trong ô trống là số nguyên nên số đó phải là 0 .

20 tháng 9 2018

Giải sách bà i tập Toán 7 | Giải sbt Toán 7Giải sách bà i tập Toán 7 | Giải sbt Toán 7

3 tháng 10 2017

Số thích hợp:

... < 1 < ....

... > 2 > ...

27 tháng 10 2017

a. \(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{4}\right)< x< \dfrac{1}{48}-\left(\dfrac{1}{16}-\dfrac{1}{6}\right)\)

\(\dfrac{1}{2}-\dfrac{7}{12}< x< \dfrac{1}{48}-\dfrac{-5}{48}\)

\(\dfrac{-1}{12}< x< \dfrac{1}{8}\) hay \(-0,08333...< x< 0,125\)

\(x\in Z\Rightarrow x\in\left\{0\right\}\)

a: Gọi số nguyên cần tìm là x

Theo đề, ta có: \(\dfrac{1}{3}+\left(\dfrac{2}{4}-1\dfrac{2}{5}\right)< x< 2\dfrac{1}{7}+\left(\dfrac{-2}{5}-\dfrac{1}{4}\right)\)

\(\Leftrightarrow\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{7}{5}< x< \dfrac{15}{7}-\dfrac{2}{5}-\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{20}{60}+\dfrac{30}{60}-\dfrac{84}{60}< x< \dfrac{15\cdot20-2\cdot28-35}{140}\)

\(\Leftrightarrow-\dfrac{34}{60}< x< \dfrac{209}{140}\)

mà x là số nguyên

nên \(x\in\left\{0;1\right\}\)

b: Gọi số nguyên cần tìm là x

Theo đề, ta có: \(\dfrac{7}{3}+\dfrac{3}{4}-\dfrac{1}{5}>x>\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{2}{7}\)

\(\Leftrightarrow\dfrac{7\cdot20+3\cdot15-12}{60}>x>\dfrac{56-21+2\cdot12}{84}\)

\(\Leftrightarrow\dfrac{173}{60}>x>\dfrac{59}{84}\)

mà x là số nguên

nên \(x\in\left\{2;1\right\}\)

9 tháng 6 2017

\(\left\{\dfrac{-5< 0< -0,4}{x\in Z}\right\}\Rightarrow x\in\left\{-4;-3;-2;-1\right\}\)

31 tháng 8 2017

chiu thôi khocroikhocroikhocroileo nheo qua

a: \(\Leftrightarrow-\dfrac{23}{5}\cdot\dfrac{50}{23}< x< \dfrac{-13}{5}:\dfrac{21}{15}=\dfrac{-13}{5}\cdot\dfrac{5}{7}=\dfrac{-13}{7}\)

=>-10<x<-13/7

hay \(x\in\left\{-9;-8;-7;-6;-5;-4;-3;-2\right\}\)

b: \(\Leftrightarrow-\dfrac{13}{3}\cdot\dfrac{1}{3}< x< \dfrac{-2}{3}\cdot\dfrac{4-3-9}{12}\)

\(\Leftrightarrow-\dfrac{13}{9}< x< \dfrac{4}{9}\)

mà x là số nguyên

nên \(x\in\left\{-1;0\right\}\)

20 tháng 8 2017

1.Tính

a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)

b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)

c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)

d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)

e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)

Bài 2

a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)

\(x=\dfrac{13}{49}\)

b.\(\left|x-1,5\right|=2\)

Xảy ra 2 trường hợp

TH1

\(x-1,5=2\)

\(x=3,5\)

TH2

\(x-1,5=-2\)

\(x=-0,5\)

Vậy \(x=3,5\) hoặc \(x=-0,5\) .

Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.

20 tháng 8 2017

Ths bn nhé

tính a) \(\left[\dfrac{0.8\div\left(\dfrac{4}{5}\cdot1025\right)}{0.64-1}+\dfrac{\left(1.08-\dfrac{2}{25}\right)\div\dfrac{4}{7}}{\left(6\dfrac{5}{7}-3\dfrac{1}{4}\right)\cdot2\dfrac{2}{17}}+\left(1.2\cdot0.5\right)\div\dfrac{4}{5}\right]\) b) \(\left(0.2\right)^{-3}\left[\left(-\dfrac{1}{5}\right)^{-2}\right]^{-1}+\left[\left(\dfrac{1}{2}\right)^{-3}\right]^{-2}\div\left(2^{-3}\right)^{-1}-\left(0.175\right)^{-2}\) c) \(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\) d)...
Đọc tiếp

tính

a) \(\left[\dfrac{0.8\div\left(\dfrac{4}{5}\cdot1025\right)}{0.64-1}+\dfrac{\left(1.08-\dfrac{2}{25}\right)\div\dfrac{4}{7}}{\left(6\dfrac{5}{7}-3\dfrac{1}{4}\right)\cdot2\dfrac{2}{17}}+\left(1.2\cdot0.5\right)\div\dfrac{4}{5}\right]\)

b) \(\left(0.2\right)^{-3}\left[\left(-\dfrac{1}{5}\right)^{-2}\right]^{-1}+\left[\left(\dfrac{1}{2}\right)^{-3}\right]^{-2}\div\left(2^{-3}\right)^{-1}-\left(0.175\right)^{-2}\)

c) \(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)

d) \(\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{3}\)

e) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2\div2\)

f) \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

g) \(\dfrac{1}{-\left(2017\right)\left(-2015\right)}+\dfrac{1}{\left(-2015\right)\left(-2013\right)}+...+\dfrac{1}{\left(-3\right)\cdot\left(-1\right)}\)

h) \(\left(1-\dfrac{1}{1\cdot2}\right)+\left(1-\dfrac{1}{2\cdot3}+...+\left(1-\dfrac{1}{2017\cdot2018}\right)\right)\)

3
7 tháng 10 2017

c)

Ta có :\(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)

\(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{3}{2}}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{2}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{\dfrac{8}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{3}{8}}\) \(=2+\dfrac{1}{\dfrac{11}{8}}\) \(=2+\dfrac{8}{11}\) \(=\dfrac{30}{11}\)

7 tháng 10 2017

d) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)

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

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

\(=3-1+\dfrac{1}{8}\)

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

10 tháng 12 2017

a)\(3\dfrac{1}{3}:2\dfrac{1}{2}-1< x< 7\dfrac{2}{3}.\dfrac{3}{7}+\dfrac{5}{2}\)

\(\dfrac{4}{3}-1< x< \dfrac{23}{7}+\dfrac{5}{2}\)

\(\dfrac{1}{3}< x< \dfrac{81}{14}\)

\(\dfrac{1}{3}=0,333333333333333333333333...\)

\(\dfrac{81}{14}=5,785714286\)

=>\(x=\left\{1;2;3;4;5\right\}\)

b)\(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{4}\right)< x< \dfrac{1}{48}-\left(\dfrac{1}{16}-\dfrac{1}{6}\right)\)

\(\dfrac{1}{2}-\dfrac{7}{12}< x< \dfrac{1}{48}+\dfrac{5}{48}\)

\(-\dfrac{1}{12}< x< \dfrac{1}{8}\)

\(-\dfrac{1}{12}=-0.08333333333333333\)

\(\dfrac{1}{8}=0.125\)

=> \(x=\left\{0\right\}\)

10 tháng 12 2017

a.\(3\dfrac{1}{3}:2\dfrac{1}{2}-1< x< 7\dfrac{2}{3}.\dfrac{3}{7}+\dfrac{5}{2}\)

\(\dfrac{4}{3}-1< x< \dfrac{23}{7}+\dfrac{5}{2}\)

\(\dfrac{1}{3}< x< \dfrac{81}{14}\)

\(0,3333...< x< 5,7857...\)

\(x\in Z\Rightarrow x\in\left\{1;2;3;4;5\right\}\)

Vậy........

b. \(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{4}\right)< x< \dfrac{1}{48}-\left(\dfrac{1}{16}-\dfrac{1}{6}\right)\)

\(\dfrac{-1}{12}< x< \dfrac{1}{8}\)

\(-0,0833...< x< 0,125\)

\(x\in Z\Rightarrow x\in\left\{0\right\}\)

Vậy............

a: \(=\left|\dfrac{3}{2}-\dfrac{7}{3}\right|^2+\dfrac{1}{4}=\dfrac{17}{18}\)

b: \(=\left|1-2-\dfrac{1}{3}\right|+\dfrac{5}{6}=1+\dfrac{1}{3}+\dfrac{5}{6}=\dfrac{13}{6}\)

c: \(=\left|\dfrac{3}{2}-\dfrac{7}{4}\right|-\dfrac{7}{4}=-\dfrac{3}{2}\)

d: =x-5+8-x=3