\(\dfrac{5}{12}x\) - \(\dfrac{1}{4}x\) = \(\dfrac{-13}{18}\)
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17 tháng 4 2021
a) Ta có: \(\dfrac{x-2}{15}+\dfrac{x-3}{14}+\dfrac{x-4}{13}+\dfrac{x-5}{12}=4\)
\(\Leftrightarrow\dfrac{x-2}{15}-1+\dfrac{x-3}{14}-1+\dfrac{x-4}{13}-1+\dfrac{x-5}{12}-1=0\)
\(\Leftrightarrow\dfrac{x-17}{15}+\dfrac{x-17}{14}+\dfrac{x-17}{13}+\dfrac{x-17}{12}=0\)
\(\Leftrightarrow\left(x-17\right)\left(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}\right)=0\)
mà \(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}>0\)
nên x-17=0
hay x=17
Vậy: x=17
17 tháng 4 2021
b) Ta có: \(\dfrac{x+1}{19}+\dfrac{x+2}{18}+\dfrac{x+3}{17}+...+\dfrac{x+18}{2}+18=0\)
\(\Leftrightarrow\dfrac{x+1}{19}+1+\dfrac{x+2}{18}+1+\dfrac{x+3}{17}+1+...+\dfrac{x+18}{2}+1=0\)
\(\Leftrightarrow\dfrac{x+20}{19}+\dfrac{x+20}{18}+\dfrac{x+20}{17}+...+\dfrac{x+20}{2}=0\)
\(\Leftrightarrow\left(x+20\right)\left(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}\right)=0\)
mà \(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}>0\)
nên x+20=0
hay x=-20
Vậy: x=-20
VQ
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 :
\(\dfrac{2}{3}x\dfrac{2}{10}+\dfrac{2}{3}x\dfrac{5}{10}x\dfrac{3}{3}\)
\(\dfrac{2}{3}x\left(\dfrac{2}{10}+\dfrac{5}{10}\right)x\dfrac{2}{3}\)
=\(\dfrac{2}{3}x1x\dfrac{2}{3}\)
\(=\dfrac{2}{3}x\dfrac{2}{3}\)
=\(\dfrac{4}{9}\)
LM
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
\(21x9=189\left(cm2\right)\)
O
2 tháng 4 2022
a) \(\dfrac{-5}{9}-\dfrac{-5}{12}=\dfrac{-5}{9}+\dfrac{5}{12}=\dfrac{-20}{36}+\dfrac{15}{36}=-\dfrac{5}{36}\)
b) \(\dfrac{-5}{12}:\dfrac{15}{4}=\dfrac{-5}{12}\times\dfrac{4}{15}=\dfrac{-1}{9}\)
c) \(\dfrac{1}{13}\cdot\dfrac{8}{13}+\dfrac{5}{13}\cdot\dfrac{1}{13}-\dfrac{14}{13}=\dfrac{1}{13}\cdot\left(\dfrac{8}{13}+\dfrac{5}{13}\right)-\dfrac{14}{13}=\dfrac{1}{13}\cdot1-\dfrac{14}{13}=\dfrac{1}{13}-\dfrac{14}{13}=-1\)
11 tháng 8 2021
1: Ta có: \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)
\(\Leftrightarrow5x+20+12x-28=7x+2\)
\(\Leftrightarrow17x-7x=2+8=10\)
hay x=1
2: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)
\(\Leftrightarrow\dfrac{6x}{36}+\dfrac{4\left(1-3x\right)}{36}=\dfrac{3\left(-x+1\right)}{36}\)
\(\Leftrightarrow6x+4-12x=-3x+3\)
\(\Leftrightarrow-6x+3x=3-4\)
hay \(x=\dfrac{1}{3}\)
3: Ta có: \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)
\(\Leftrightarrow4x-12-x-2=6x-3\)
\(\Leftrightarrow3x-14-6x+3=0\)
\(\Leftrightarrow-3x=11\)
hay \(x=-\dfrac{11}{3}\)
11 tháng 8 2021
4: Ta có: \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)
\(\Leftrightarrow3x-6-8x-12=x+6\)
\(\Leftrightarrow-5x-x=6+18\)
hay x=-4
5: Ta có: \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)
\(\Leftrightarrow6x-3+2x-6=-1\)
\(\Leftrightarrow8x=8\)
hay x=1
NT
Nguyễn thành Đạt
CTVHS
25 tháng 2 2023
\(1.\dfrac{-7}{18}+\dfrac{-5}{12}-\dfrac{-13}{18}\text{=}\left(\dfrac{-7}{18}-\dfrac{-13}{18}\right)+\dfrac{-5}{12}\text{=}\dfrac{1}{3}+\dfrac{-5}{12}\text{=}\dfrac{-1}{12}\)
\(2.\dfrac{-13}{17}+\dfrac{-13}{21}+\dfrac{-4}{17}\text{=}\left(\dfrac{-13}{17}+\dfrac{-4}{17}\right)+\dfrac{-13}{21}\text{=}-1+\dfrac{-13}{21}\text{=}\dfrac{-34}{21}\)
\(3.\dfrac{-13}{10}-\dfrac{-4}{13}+\dfrac{-11}{10}\text{=}\dfrac{-12}{5}-\dfrac{-4}{13}\text{=}\dfrac{-136}{65}\)
\(4.\dfrac{13}{17}\times\left(\dfrac{-4}{5}+\dfrac{-3}{4}\right)\text{=}\dfrac{13}{17}\times\dfrac{-31}{20}\text{=}\dfrac{-403}{340}\)
\(5.\left(\dfrac{-5}{12}\times\dfrac{-9}{20}\right)\times\dfrac{-7}{17}\text{=}\dfrac{3}{16}\times\dfrac{-7}{17}\text{=}\dfrac{-21}{272}\)
\(6.\dfrac{11}{23}\times\left(\dfrac{5}{9}+\dfrac{17}{9}-\dfrac{13}{9}\right)\text{=}\dfrac{11}{23}\times1\text{=}\dfrac{11}{23}\)
8 tháng 3 2022
\(a,\dfrac{5}{13}\times\dfrac{4}{15}\times13=\dfrac{5\times4\times13}{13\times5\times3}=\dfrac{4}{3}\\ b,\left(\dfrac{3}{7}+\dfrac{5}{2}\right)\times\dfrac{7}{5}=\dfrac{3}{7}\times\dfrac{7}{5}+\dfrac{5}{2}\times\dfrac{7}{5}=\dfrac{3}{5}+\dfrac{7}{2}=\dfrac{6}{10}+\dfrac{35}{10}=\dfrac{41}{10}\\ c,\dfrac{1}{5}\times\dfrac{11}{18}+\dfrac{11}{18}\times\dfrac{3}{5}=\dfrac{11}{18}\times\left(\dfrac{1}{5}+\dfrac{3}{5}\right)=\dfrac{11}{18}\times\dfrac{4}{5}=\dfrac{22}{45}\)
25 tháng 3 2022
\(\dfrac{1}{5}+\dfrac{2}{11}< \dfrac{x}{55}< \dfrac{2}{5}+\dfrac{1}{5}\)
\(\dfrac{11+10}{55}< \dfrac{x}{55}< \dfrac{3}{5}\)
\(\dfrac{21}{55}< \dfrac{x}{55}< \dfrac{33}{55}\)
Vậy \(x\in\left\{22;23;24;...\right\}\)
VN
20 tháng 12 2022
\(\dfrac{15}{12}+\dfrac{5}{13}-\dfrac{3}{12}-\dfrac{18}{13}-\dfrac{1}{3}\)
\(=\left(\dfrac{15}{12}-\dfrac{3}{12}\right)+\left(\dfrac{5}{13}-\dfrac{18}{13}\right)-\dfrac{1}{3}\)
\(=-1+1-\dfrac{1}{3}\)
\(=0-\dfrac{1}{3}\)
\(=\dfrac{-1}{3}\)
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\(14.\dfrac{3}{2}+\dfrac{6}{5}:\left(-\dfrac{2}{5}\right)\)
\(=14.\dfrac{3}{2}+\dfrac{6}{5}.\dfrac{-5}{2}\)
\(=21+\dfrac{6}{5}.\dfrac{-5}{2}\)
\(=21+\left(-3\right)\)
\(=18\)
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\(\sqrt{\dfrac{1}{4}+\dfrac{2}{3}-\left(\dfrac{1}{3}\right)^2}\)
\(=\sqrt{\dfrac{1}{4}+\dfrac{2}{3}-\dfrac{1}{9}}\)
\(=\sqrt{\dfrac{3}{12}+\dfrac{8}{12}-\dfrac{1}{9}}\)
\(=\sqrt{\dfrac{11}{12}-\dfrac{1}{9}}\)
\(=\sqrt{\dfrac{99}{108}-\dfrac{12}{108}}\)
\(=\sqrt{\dfrac{29}{36}}\)
\(=\dfrac{\sqrt{29}}{6}\)
S
20 tháng 12 2022
\(\dfrac{15}{12}+\dfrac{5}{13}-\dfrac{3}{12}-\dfrac{18}{13}-\dfrac{1}{3}\)
\(=\dfrac{5}{4}+\dfrac{5}{13}-\dfrac{1}{4}-\dfrac{18}{13}-\dfrac{1}{3}\)
\(=\left(\dfrac{5}{4}-\dfrac{1}{4}\right)+\left(\dfrac{5}{13}-\dfrac{18}{13}\right)-\dfrac{1}{3}\)
\(=1+\left(-1\right)-\dfrac{1}{3}=0-\dfrac{1}{3}=-\dfrac{1}{3}\)
\(\dfrac{5}{12}\)\(x\) - \(\dfrac{1}{4}\)\(x\) = - \(\dfrac{13}{18}\)
(\(\dfrac{5}{12}\) - \(\dfrac{1}{4}\))\(x\) = - \(\dfrac{13}{18}\)
\(\dfrac{1}{6}\)\(x\) = - \(\dfrac{13}{18}\)
\(x\) = - \(\dfrac{13}{18}\)\(\times\) 6
\(x\)= -\(\dfrac{13}{3}\)