Bộ 24 đề thi giữa học kì 2 môn Toán lớp 6 năm 2025

Vndoc Bộ 24 đề thi giữa học kì 2 môn Toán lớp 6 năm 2025 5,0
\frac{3}{0}\frac{{0,25}}{3}\frac{4}{7}\frac{{1,25}}{{2,7}}\frac{a}{b}\frac{c}{d}\frac{{ - 3}}{5}\frac{{ - 9}}{{15}}\frac{5}{{15}}- \frac{3}{{10}}- \frac{{10}}{5}\frac{m}{8} > \frac{{ - 7}}{8}\left( {\frac{7}{{16}} + \frac{{ - 1}}{8} + \frac{9}{{32}}} \right):\frac{5}{4}10\frac{2}{9} + 2\frac{3}{5} - 6\frac{2}{9}30\%  - 2\frac{2}{5} + 0,2.\frac{1}{2}\frac{{ - 25}}{{30}}.\frac{{37}}{{44}} + \frac{{ - 25}}{{30}}.\frac{{13}}{{44}} + \frac{{ - 25}}{{30}}.\frac{{ - 6}}{{44}}2,5x + \frac{4}{7} =  - 1,5Đề thi Giữa kì 2 Toán lớp 6 có đáp án (3 đề)\frac{5}{2} x=4\frac{8}{5}\frac{5}{8}\frac{3}{7}=\frac{-24}{y}Đề thi giữa kì 2 Toán 6 KNTT\frac{m}{n}\frac{m}{n}: aa: \frac{m}{n}\frac{m}{n} \cdot a\frac{n}{m} \cdot aĐề thi giữa kì 2 Toán 6 KNTT\frac{3}{5}+\frac{-4}{9}\frac{3}{5}+\frac{2}{5} \cdot \frac{15}{8}\frac{7}{2} \cdot \frac{8}{13}+\frac{8}{13} \cdot \frac{-5}{2}+\frac{8}{13}\frac{-5}{17} \cdot \frac{-9}{23}+\frac{9}{23} \cdot \frac{-22}{17}+11 \frac{9}{23}\frac{7}{8}+x=\frac{3}{5}\frac{1}{3}:(2 x-1)=\frac{-4}{24}\frac{17}{2}-\left|x-\frac{3}{4}\right|=\frac{-7}{4}(3 x+2)\left(\frac{-2}{5} x-7\right)=0\frac{1}{5}M=1+\frac{1}{5}+\frac{3}{35}+\ldots+\frac{3}{9603}+\frac{3}{9999}S=\frac{1}{4^{2}}+\frac{1}{6^{2}}+\frac{1}{8^{2}}+\ldots+\frac{1}{(2 n)^{2}}<\frac{1}{4} \quad(\mathrm{n} \in \mathrm{N}, \mathrm{n} \geq 2)\frac{1}{8};\frac{3}{2};\frac{2}{9}\frac{{ - 5}}{8} - \left( {\frac{3}{8} - \frac{7}{8}} \right)\frac{{ - 1}}{8}\frac{{ - 15}}{8}- \frac{3}{8}\frac{{ - 11}}{8}Đề thi giữa kì 2 Toán 6 năm học 2021 – 2022 sách Cánh Diều Đề số 1\frac{1}{3}\frac{1}{5}\frac{5}{6}\frac{2}{3}\frac{2}{3}.\frac{3}{4} + \frac{1}{2}.\frac{5}{6}\frac{1}{3}.\left( {\frac{9}{4} - \frac{6}{5}} \right) + \frac{1}{5}:\frac{1}{{10}}\frac{1}{{143}} + \frac{1}{{99}} + \frac{1}{{63}} + \frac{1}{{35}} + \frac{1}{{15}}\frac{{ - 1}}{5}.\frac{6}{7} + \frac{9}{7}.\frac{1}{{ - 5}} - \frac{1}{5}.\frac{1}{{ - 7}}\frac{x}{2} = \frac{2}{3} + \frac{{ - 3}}{{12}}2x - \frac{1}{5} = \frac{{ - 1}}{3}:\frac{5}{7}x - \frac{7}{{15}}:\frac{3}{5} = \frac{3}{{10}}\frac{2}{{\left( {x + 2} \right)\left( {x + 4} \right)}} + \frac{4}{{\left( {x + 4} \right)\left( {x + 8} \right)}} + \frac{6}{{\left( {x + 8} \right)\left( {x + 14} \right)}} = \frac{x}{{\left( {x + 2} \right)\left( {x + 14} \right)}}\frac{4}{7}\frac{0,25}{-3}\frac{5}{0}\frac{6,23}{7,4}\frac{3}{4}\frac{13}{20}\frac{25}{16}\frac{6}{8}\frac{10}{75}\frac{-16}{25}\frac{16}{25}\frac{-2}{4}<\frac{-3}{4}\frac{-4}{5}<\frac{-3}{5}\frac{1}{4}<\frac{-3}{4}\frac{-1}{6}<\frac{-5}{6}5\frac{2}{3}\frac{17}{3}\frac{3}{17}\frac{5}{3}\frac{4}{3}\frac{-31}{10}\frac{-7}{6}+\frac{18}{6}\frac{-4}{6}\frac{11}{6}\frac{-85}{72}\frac{9}{5}:\frac{-3}{5}\frac{3}{4}.\frac{4}{3}\frac{1}{4}Đề thi giữa kì 2 Toán 6Đề thi giữa kì 2 Toán lớp 6A=\frac{3}{5}.\frac{5}{4}-\frac{3}{5}.\frac{1}{4}x-\frac{3}{10}=\frac{7}{15}.\frac{3}{5}P=\frac{1}{1.2}+\frac{1}{2.3}+\ \frac{1}{3.4}+...+\frac{1}{99.100}\frac{{ - 1515}}{{2828}} = \frac{{ - 151515}}{{282828}}\frac{{ - 21}}{{28}} = \frac{{ - 39}}{{52}}\frac{{52}}{{91}} = \frac{{28}}{{49}}\frac{{165}}{{143}} = \frac{{26}}{{30}}\frac{{ - 32}}{{60}}\frac{{ - 16}}{{30}}\frac{{16}}{{30}}\frac{8}{{15}}\frac{{ - 8}}{{15}}\frac{7}{{ - 12}};\frac{7}{{ - 10}};\frac{{ - 7}}{8};\frac{7}{{12}}\frac{7}{{ - 12}}\frac{7}{{ - 10}}\frac{{ - 7}}{8}\frac{7}{{12}}\frac{6}{7}{m^2}\frac{3}{{14}}m\frac{{15}}{7}\frac{{15}}{{14}}\frac{9}{{49}}\frac{{35}}{{15}} = \frac{x}{3}Đề thi giữa kì 2 Toán 6 năm học 2021 - 2022 sách Chân trời sáng tạo Đề số 12\frac{{17}}{{20}} - 1\frac{{11}}{{15}} + 6\frac{1}{{20}}:3\left( {31\frac{6}{{13}} + 5\frac{9}{{41}}} \right) - 36\frac{6}{{13}}\frac{{ - 5}}{{46}} + \frac{{ - 7}}{{25}} + \frac{{35}}{{19}} + \frac{5}{{46}} + \frac{{ - 16}}{{19}} + \frac{7}{{25}}\frac{1}{{5.6}} + \frac{1}{{6.7}} + ... + \frac{1}{{24.25}}2\frac{3}{4}x = 1\frac{{x + 2}}{8} = \frac{{ - 15}}{4}75\%  - \frac{1}{5}.x = 3,75Đề thi giữa kì 2 Toán 6 năm học 2021 - 2022 sách Chân trời sáng tạo Đề số 1\frac{1}{5} + \frac{1}{{13}} + \frac{1}{{25}} + .... + \frac{1}{{{{10}^2} + {{11}^2}}} < \frac{9}{{20}}\frac{3}{{10}}\frac{7}{2};\frac{5}{6};\frac{4}{9}\frac{{14}}{{18}};\frac{{10}}{{18}};\frac{8}{{18}}\frac{{21}}{{18}};\frac{{15}}{{18}};\frac{{12}}{{18}}\frac{{63}}{{18}};\frac{{15}}{{18}};\frac{8}{{18}}\frac{{63}}{{18}};\frac{{45}}{{18}};\frac{{63}}{{18}}\frac{0}{9};\frac{{12}}{{15}};\frac{{11}}{5};\frac{{ - 4}}{{ - 5}}\frac{0}{9}\frac{{12}}{{15}}\frac{{11}}{5}\frac{{ - 4}}{{ - 5}}\frac{1}{27}-\frac{1}{9}\frac{1}{27}-\frac{1}{9}=\frac{0}{18}\frac{1}{27}-\frac{3}{27}=\frac{-2}{0}\frac{1}{27}-\frac{3}{27}=\frac{2}{27}\frac{1}{27}-\frac{3}{27}=\frac{1-3}{27}=\frac{-2}{27}\frac{-1}{4} \cdot \frac{1}{2}\frac{-1}{4} \cdot \frac{1}{2}=\frac{-1.2}{4.4}=\frac{-2}{4}\frac{-1}{4} \cdot \frac{1}{2}=\frac{-1}{4} \cdot \frac{2}{4}=\frac{-2}{16}\frac{-1}{4} \cdot \frac{1}{2}=\frac{-0}{8}\frac{-1}{4} \cdot \frac{1}{2}=\frac{-1}{8}\frac{4}{16}\frac{2}{8}\frac{4}{8}\frac{1}{4}\frac{1}{8}\frac{-7}{8}\frac{8}{7}\frac{7}{8}\frac{7}{-8}\frac{-8}{7}-\frac{5}{8} \cdot \frac{(-4)^{2}}{10}\frac{-2020}{2021} \cdot \frac{9}{11}+\frac{-2020}{2021} \cdot \frac{2}{11}\frac{-5}{7} \cdot \frac{2}{11}+\frac{-5}{7} \cdot \frac{9}{11}+\frac{5}{7}\frac{-3}{8} \cdot \frac{1}{2}+\frac{1}{6} \cdot \frac{-3}{8}+\frac{1}{3}: \frac{-8}{3}a) x-\frac{-1}{5}=3+\frac{-3}{2}b) \frac{1}{2}-\left(x-\frac{5}{11}\right)=\frac{-3}{4}c) \frac{3}{4}+\left(\frac{2}{5}-x\right)=\frac{1}{4}\frac{1}{3}A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+\ldots+\frac{9}{98.99}+\frac{9}{99.100}Đề thi giữa kì 2 Toán 6 CTSTĐề thi giữa kì 2 Toán 6 CTST
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