\(\dfrac{-2}{17}+\dfrac{15}{23}+\dfrac{-15}{17}+\dfrac{4}{19}+\dfrac{8}{23}\)

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6 tháng 2 2023

`f)(-2)/17 + 15/23 + (-15)/17 + 4/19 + 8/23`

`= (-2/17+ -15/17)+(15/23+8/23)+4/19`

`= -1+1+4/19`

`= 0 +4/19`

`= 0`

`g)(-1)/2 + 3/21 + (-2)/6 + (-5)/30`

`= (-1)/2 + 1/7 + (-1)/3 + (-1)/6`

`= (-21)/42 + 6/42 + (-14)/42 + (-7)/42`

`=(-36)/42`

`=(-6)/7`

6 tháng 2 2023

f)\(-\dfrac{2}{17}+\dfrac{15}{23}+-\dfrac{15}{17}+\dfrac{4}{19}+\dfrac{8}{23}\)
\(=\left(-\dfrac{2}{17}+-\dfrac{15}{17}\right)+\left(\dfrac{15}{23}+\dfrac{8}{23}\right)+\dfrac{4}{19}\)
\(=-1+1+\dfrac{4}{19}\)
\(=0+\dfrac{4}{19}=\dfrac{4}{19}\)
g)\(-\dfrac{1}{2}+\dfrac{3}{21}+-\dfrac{2}{6}+-\dfrac{5}{30}\)
\(=-\dfrac{1}{2}+\dfrac{1}{7}+-\dfrac{1}{3}+-\dfrac{1}{6}\)
\(=\left(-\dfrac{1}{2}+-\dfrac{1}{3}+-\dfrac{1}{6}\right)+\dfrac{1}{7}\)
\(=-\dfrac{3+2+1}{6}+\dfrac{1}{7}\)
\(=\dfrac{1}{7}-1\)
\(=\dfrac{1}{7}-\dfrac{7}{7}=-\dfrac{6}{7}\)

14 tháng 3 2017

đây là tính nhanh à nếu tính bình thường thì tính may tính là ra

14 tháng 3 2017

a) 17/23 . 8/16 . 23/17. (-80) . 3/4

= (17/23 . 23/17) . (8/16 . 3/4) . (-80)

= 1 . 3/8 . (-80)

= 3/8 . (-80)

= -30

b) 5/11 . 18/29 - 5/11 . 8/29 + 5/11 . 19/29

= 5/11 . (18/29 - 8/29 + 19/29)

= 5/11 . 1

= 5/11

c)(13/23 + 1313/2323 - 131313/232323).(1/3+1/4 -7/12)

= (13/23 + 1313/2323 - 131313/232323).0

= 0

d) 12/2x2 . 22/2x3 . 32/3x4 . 42/4x5 . 52/5x6 . 62/6x7 . 72/7x8 . 82/8x9 . 92/9x10

= 1/2 . 2/3 . 3/4 . 4/5 . 5/6 . 6/7 . 7/8 . 8/9 .9/10

= 1/10

Khó nhìn quá. Bạn thông cảm nhé! vui

2 tháng 5 2017

Bài 1:

a)

\(\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Leftrightarrow\dfrac{x-1}{9}=\dfrac{24}{9}\\ \Leftrightarrow x-1=24\\ x=24+1\\ x=25\)

b)

\(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{8}\\ \dfrac{3x}{7}+1=\dfrac{-1}{8}\cdot\left(-4\right)\\ \dfrac{3x}{7}+1=\dfrac{1}{2}\\ \dfrac{3x}{7}=\dfrac{1}{2}-1\\ \dfrac{3x}{7}=\dfrac{-1}{2}\\ 3x=\dfrac{-1}{2}\cdot7\\ 3x=\dfrac{-7}{2}\\ x=\dfrac{-7}{2}:3\\ x=\dfrac{-7}{6}\)

c)

\(x+\dfrac{7}{12}=\dfrac{17}{18}-\dfrac{1}{9}\\ x+\dfrac{7}{12}=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{12}\\ x=\dfrac{1}{4}\)

d)

\(0,5x-\dfrac{2}{3}x=\dfrac{7}{12}\\ \dfrac{1}{2}x-\dfrac{2}{3}x=\dfrac{7}{12}\\ x\cdot\left(\dfrac{1}{2}-\dfrac{2}{3}\right)=\dfrac{7}{12}\\ \dfrac{-1}{6}x=\dfrac{7}{12}\\ x=\dfrac{7}{12}:\dfrac{-1}{6}\\ x=\dfrac{-7}{2}\)

e)

\(\dfrac{29}{30}-\left(\dfrac{13}{23}+x\right)=\dfrac{7}{46}\\ \dfrac{29}{30}-\dfrac{13}{23}-x=\dfrac{7}{46}\\ \dfrac{277}{690}-x=\dfrac{7}{46}\\ x=\dfrac{277}{690}-\dfrac{7}{46}\\ x=\dfrac{86}{345}\)

f)

\(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(2+\dfrac{1}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\\ \left(x-\dfrac{1}{12}\right):\dfrac{23}{12}=\dfrac{7}{46}\\ x-\dfrac{1}{12}=\dfrac{7}{46}\cdot\dfrac{23}{12}\\ x-\dfrac{1}{12}=\dfrac{7}{24}\\ x=\dfrac{7}{24}+\dfrac{1}{12}\\ x=\dfrac{3}{8}\)

g)

\(\dfrac{13}{15}-\left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{13}{15}-\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{1}{6}\\ \dfrac{13}{21}+x=\dfrac{1}{6}:\dfrac{7}{12}\\ \dfrac{13}{21}+x=\dfrac{2}{7}\\ x=\dfrac{2}{7}-\dfrac{13}{21}\\ x=\dfrac{-1}{3}\)

h)

\(2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{4}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}+\dfrac{3}{2}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}:2\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{8}\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\end{matrix}\right.\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\ \dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{29}{24}\\ x=\dfrac{29}{24}:\dfrac{1}{2}\\ x=\dfrac{29}{12}\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\\ \dfrac{1}{2}x=\dfrac{-7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{-13}{24}\\ x=\dfrac{-13}{24}:\dfrac{1}{2}\\ x=\dfrac{-13}{12}\)

i)

\(3\cdot\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=0-\dfrac{1}{9}\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}:3\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{27}\\ \left(3x-\dfrac{1}{2}\right)^3=\left(\dfrac{-1}{3}\right)^3\\ \Leftrightarrow3x-\dfrac{1}{2}=\dfrac{-1}{3}\\ 3x=\dfrac{-1}{3}+\dfrac{1}{2}\\ 3x=\dfrac{1}{6}\\ x=\dfrac{1}{6}:3\\ x=\dfrac{1}{18}\)

Ta có: \(3\frac{1}{3}+7\frac{3}{17}\cdot\frac{2}{15}-2\frac{3}{17}\cdot\frac{2}{15}\)

\(=\frac{10}{3}+\frac{122}{17}\cdot\frac{2}{15}-\frac{37}{17}\cdot\frac{2}{15}\)

\(=\frac{10}{3}+\frac{244}{255}-\frac{74}{255}\)

\(=\frac{10}{3}+\frac{2}{3}=\frac{12}{3}=4\)

30 tháng 6 2017

a ) \(5\left(x^2\right)+7x+2\)

\(\Leftrightarrow5x^2+7x+2=0\)

\(\Leftrightarrow5x^2+5x+2x+2=0\)

\(\Leftrightarrow\left(5x+2\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{5}\\x=-1\end{matrix}\right.\)

Vậy .............

b ) \(\dfrac{x+1}{17}+\dfrac{x+2}{16}=\dfrac{x+3}{15}+\dfrac{x+4}{14}\)

\(\Leftrightarrow\dfrac{x+1}{17}+1+\dfrac{x+2}{16}+1=\dfrac{x+3}{15}+1+\dfrac{x+4}{14}+1\)

\(\Leftrightarrow\dfrac{x+18}{17}+\dfrac{x+18}{16}=\dfrac{x+18}{15}+\dfrac{x+18}{14}\)

\(\Leftrightarrow\dfrac{x+18}{17}+\dfrac{x+18}{16}-\dfrac{x+18}{15}-\dfrac{x+18}{14}=0\)

\(\Leftrightarrow\left(x+18\right)\left(\dfrac{1}{17}+\dfrac{1}{16}-\dfrac{1}{15}-\dfrac{1}{14}\right)=0\)

\(\left(\dfrac{1}{17}+\dfrac{1}{16}-\dfrac{1}{15}-\dfrac{1}{14}\right)\ne0\)

Ta có : \(x+18=0\Leftrightarrow x=-18\)

Vậy ......

c ) \(\dfrac{x-1}{x-3}=\dfrac{x-4}{x-7}\)

\(\Leftrightarrow\left(x-1\right)\left(x-7\right)=\left(x-3\right)\left(x-4\right)\)

\(\Leftrightarrow x^2-7x-x+7=x^2-4x-3x+12\)

\(\Leftrightarrow-x=5\)

\(\Leftrightarrow x=-5\)

Vậy ..

30 tháng 6 2017

cảm ơn nhiều nha

8 tháng 9 2017

a. \(\dfrac{5}{7}+\dfrac{9}{23}+\dfrac{-12}{7}+\dfrac{14}{23}\)

= \(\left(\dfrac{5}{7}+\dfrac{-12}{7}\right)+\left(\dfrac{9}{23}+\dfrac{14}{23}\right)\)

= \(\left(-1\right)+1=0\)

b. \(\left(7-2\dfrac{3}{5}\right).2\dfrac{4}{33}-2\dfrac{3}{5}:\dfrac{1}{2}+\dfrac{7}{5}\)

= \(\dfrac{22}{5}.2\dfrac{4}{33}-\dfrac{26}{5}+\dfrac{7}{5}\)

= \(\dfrac{28}{3}-\dfrac{33}{5}=\dfrac{41}{15}\)

c\(\dfrac{-7}{9}:\dfrac{8}{15}+\dfrac{-7}{9}.\dfrac{7}{15}+5\dfrac{7}{9}\)

= \(\dfrac{-56}{135}+\dfrac{-49}{135}+5\dfrac{7}{9}\)

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

d. \(\left(16\dfrac{3}{8}-19\dfrac{3}{4}\right)-\left(12\dfrac{3}{8}-17\dfrac{3}{4}\right)\)

= \(16\dfrac{3}{8}-19\dfrac{3}{4}-12\dfrac{3}{8}+17\dfrac{3}{4}\)

= \(\left(16\dfrac{3}{8}-12\dfrac{3}{8}\right)-\left(19\dfrac{3}{4}-17\dfrac{3}{4}\right)\)

=\(4+2=6\)

a)\(\dfrac{5}{7}+\dfrac{9}{23}+\dfrac{-12}{7}+\dfrac{14}{23}\)

=\(\left(\dfrac{5}{7}-\dfrac{12}{7}\right)+\left(\dfrac{9}{23}+\dfrac{14}{23}\right)=-1+1=0\)

b)\(\left(7-2\dfrac{3}{5}\right).2\dfrac{4}{33}-2\dfrac{3}{5}:\dfrac{1}{2}+\dfrac{7}{5}\)

=\(\dfrac{22}{5}.\dfrac{70}{33}-\dfrac{13}{5}.2+\dfrac{7}{5}\)

=\(\dfrac{28}{3}-\dfrac{26}{5}+\dfrac{7}{5}=\dfrac{83}{15}\)

Các câu sau tương tự

12 tháng 3 2017

Hỏi đáp Toán

9 tháng 5 2017

\(\dfrac{-11}{23}\).\(\dfrac{62}{15}\)+\(\dfrac{8}{5}\).\(\dfrac{-11}{23}\)-\(\dfrac{1}{23}\)

=\(\dfrac{-11}{23}\).\(\left(\dfrac{62}{15}+\dfrac{8}{15}\right)\)-\(\dfrac{1}{23}\)

=\(\dfrac{-11}{23}\).\(\dfrac{14}{3}\)-\(\dfrac{1}{23}\)

=\(\dfrac{-154}{69}\)-\(\dfrac{1}{23}\)

=\(\dfrac{-154}{69}\)-\(\)\(\dfrac{3}{69}\)

=\(\dfrac{-151}{69}\)

Chúc bn hc tốt!

9 tháng 5 2017

Đề bài là \(\dfrac{8}{5}\) mà bạn có phải \(\dfrac{8}{15}\) đâu

4 tháng 8 2018

bài 2:tính hợp lý

1.a) Dễ nhận thấy đề toán chỉ giải được khi đề là tìm x,y. Còn nếu là tìm x ta nhận thấy ngay vô nghiệm. Do đó: Sửa đề: \(\left|x-3\right|+\left|2-y\right|=0\)

\(\Leftrightarrow\left|x-3\right|=\left|2-y\right|=0\)

\(\left|x-3\right|=0\Rightarrow\left\{{}\begin{matrix}x-3=0\\-\left(x-3\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\) (1)

\(\left|2-y\right|=0\Rightarrow\left\{{}\begin{matrix}2-y=0\\-\left(2-y\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\) (2)

Từ (1) và (2) có: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x_1=3\\x_2=-3\end{matrix}\right.\\\left\{{}\begin{matrix}y_1=2\\y_2=-2\end{matrix}\right.\end{matrix}\right.\)