1/2+1/3+1/6
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Cách 1:
Tổng số kg giấy vụn và báo cũ lớp 4A thu gom được là:
108 + 72 = 180 (kg)
Mỗi bạn thu gom được số kg là:
180 : 36 = 5 (kg)
Cách 2:
Mỗi bạn thu gom được số kg giấy vụn là:
108 : 36 = 3 (kg giấy vụn)
Mỗi bạn thu gom được số kg báo cũ là:
72 : 36 = 2 (kg báo cũ)
Vậy mỗi bạn thu gom được số kg vừa báo cũ vừa giấy vụn là:
3 + 2 = 5 (kg)
\(3-\dfrac{16}{11}=\dfrac{33}{11}-\dfrac{16}{11}\\ =\dfrac{33-16}{11}=\dfrac{17}{11}\)
.
\(\dfrac{9}{8}-1=\dfrac{9}{8}-\dfrac{8}{8}=\dfrac{9-8}{8}\\ =\dfrac{1}{8}\)
\(\dfrac{19}{5}-\dfrac{3}{15}\)
\(=\dfrac{19}{5}-\dfrac{1}{5}\)
\(=\dfrac{19-1}{5}\)
\(=\dfrac{18}{5}\)
a) \(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{3}{4}\right)\le x\le\dfrac{1}{24}-\left(\dfrac{1}{8}-\dfrac{1}{3}\right)\)
\(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{3}{4}\le x\le\dfrac{1}{24}-\dfrac{1}{8}+\dfrac{1}{3}\)
\(\dfrac{6}{12}-\dfrac{4}{12}-\dfrac{9}{12}\le x\le\dfrac{1}{24}-\dfrac{3}{24}+\dfrac{8}{24}\)
\(-\dfrac{7}{12}\le x\le\dfrac{6}{24}\)
\(-\dfrac{7}{12}\le x\le\dfrac{1}{4}\)
Giá trị x nguyên thỏa mãn là: 0
b) \(-\dfrac{1}{2}-\dfrac{3}{4}+\dfrac{1}{3}\le x\le\dfrac{5}{6}-\left(-\dfrac{3}{4}\right)+\dfrac{7}{12}\)
\(\dfrac{-6}{12}-\dfrac{9}{12}+\dfrac{4}{12}\le x\le\dfrac{5}{6}+\dfrac{3}{4}+\dfrac{7}{12}\)
\(-\dfrac{11}{12}\le x\le\dfrac{10}{12}+\dfrac{9}{12}+\dfrac{7}{12}\)
\(-\dfrac{11}{12}\le x\le\dfrac{26}{12}\)
\(-\dfrac{11}{12}\le x\le\dfrac{13}{6}\)
Các giá trị x nguyên thỏa mãn là: 0; 1; 2
a) Ta có :
\(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{3}{4}\right)\le x\le\dfrac{1}{24}-\left(\dfrac{1}{8}-\dfrac{1}{3}\right)\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{13}{12}\le x\le\dfrac{1}{24}-\dfrac{-5}{24}\)
\(\Leftrightarrow\dfrac{-7}{12}\le x\le\dfrac{1}{4}\)
mà \(x\in Z\)
\(\Rightarrow x=0\)
Vậy \(x=0\)
b) Ta có :
\(\dfrac{-1}{2}-\dfrac{3}{4}+\dfrac{1}{3}\le x\le\dfrac{5}{6}-\left(\dfrac{-3}{4}\right)+\dfrac{7}{12}\)
\(\Leftrightarrow\dfrac{-6}{12}-\dfrac{9}{12}+\dfrac{4}{12}\le x\le\dfrac{5}{6}+\dfrac{3}{4}+\dfrac{7}{12}\)
\(\Leftrightarrow\dfrac{-11}{12}\le x\le\dfrac{10}{12}+\dfrac{9}{12}+\dfrac{7}{12}\)
\(\Leftrightarrow\dfrac{-11}{12}\le x\le\dfrac{26}{12}\)
\(\Leftrightarrow\dfrac{-11}{12}\le x\le\dfrac{13}{6}\)
mà \(x\in Z\)
\(\Rightarrow x\in\left\{0;1;2\right\}\)
Vậy \(x\in\left\{0;1;2\right\}\)
Số số hạng là: ( 2025 - 5 ) : 5 + 1 = 405
Tổng của dãy số trên là: ( 2025 + 5 ) x 405 : 2 = 411075
Đ/s : 411075
\(\dfrac{4}{18}-\dfrac{1}{9}\)
\(=\dfrac{4}{18}-\dfrac{2}{18}\)
\(=\dfrac{4-2}{18}\)
\(=\dfrac{2}{18}\)
\(=\dfrac{1}{9}\)
\(\dfrac{4}{18}-\dfrac{1}{9}\)
\(=\dfrac{2}{9}-\dfrac{1}{9}\)
\(=\dfrac{2-1}{9}\)
\(=\dfrac{1}{9}\)
\(\dfrac{9}{5}+\dfrac{5}{8}=\dfrac{72}{40}+\dfrac{25}{40}=\dfrac{97}{40}\)
\(\dfrac{3}{5}+\dfrac{13}{9}=\dfrac{27}{45}+\dfrac{65}{45}=\dfrac{82}{45}\)
\(\dfrac{2}{3}+\dfrac{11}{2}=\dfrac{4}{6}+\dfrac{33}{6}=\dfrac{37}{6}\)
\(\dfrac{9}{4}+\dfrac{17}{20}=\dfrac{45}{20}+\dfrac{17}{20}=\dfrac{62}{20}=\dfrac{31}{10}\)
\(\dfrac{43}{6}+\dfrac{45}{8}=\dfrac{172}{24}+\dfrac{135}{24}=\dfrac{307}{24}\)
\(\dfrac{5}{9}+\dfrac{7}{15}=\dfrac{25}{45}+\dfrac{21}{45}=\dfrac{46}{45}\)
\(\dfrac{9}{5}+\dfrac{5}{8}\\ =\dfrac{72}{40}+\dfrac{25}{40}\\ =\dfrac{97}{40}\\ ----------\\ \dfrac{3}{5}+\dfrac{13}{9}\\ =\dfrac{27}{45}+\dfrac{65}{45}\\ =\dfrac{92}{45}\\ --------\\ \dfrac{2}{3}+\dfrac{11}{2}\\ =\dfrac{4}{6}+\dfrac{33}{6}\\ =\dfrac{37}{6}\\ --------\\ \dfrac{9}{4}+\dfrac{17}{20}\\ =\dfrac{45}{20}+\dfrac{17}{20}\\ =\dfrac{62}{20}=\dfrac{31}{10}\\ ----------\\ \dfrac{43}{6}+\dfrac{45}{8}\\ =\dfrac{172}{24}+\dfrac{135}{24}\\ =\dfrac{307}{24}\)
\(-----------\\ \dfrac{5}{9}+\dfrac{7}{15}\\ =\dfrac{25}{45}+\dfrac{21}{45}\\ =\dfrac{46}{45}\)
\(\dfrac{8}{3}+\dfrac{10}{7}=\dfrac{56}{21}+\dfrac{30}{21}\\ =\dfrac{56+30}{21}=\dfrac{86}{21}\)
\(\dfrac{8}{3}+\dfrac{10}{7}\)
\(=\dfrac{56}{21}+\dfrac{30}{21}\)
\(=\dfrac{56+30}{21}\)
\(=\dfrac{86}{21}\)
\(\dfrac{4}{5}+\dfrac{3}{4}=\dfrac{16}{20}+\dfrac{15}{20}\\ =\dfrac{16+15}{20}=\dfrac{31}{20}\)
\(\dfrac{4}{5}+\dfrac{3}{4}\)
\(=\dfrac{16}{20}+\dfrac{15}{20}\)
\(=\dfrac{16+15}{20}\)
\(=\dfrac{31}{20}\)
\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}\)
\(=\dfrac{3}{6}+\dfrac{2}{6}+\dfrac{1}{6}\)
\(=\dfrac{3+2+1}{6}\)
\(=\dfrac{6}{6}\)
\(=1\)
\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}=\dfrac{5}{6}+\dfrac{1}{6}=1\)