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TD
14
27 tháng 5 2019
Bài 1:
a) b) c) sẽ có bạn giải cho em thôi vì nó dễ tính tay cũng đc
d) \(\frac{4}{2.5}+\frac{4}{5.8}+...+\frac{4}{23.26}\)
\(=\frac{4}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{23.26}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{23}-\frac{1}{26}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{26}\right)\)
\(=\frac{4}{3}.\frac{6}{13}\)
\(=\frac{8}{13}\)
Bài 2:
a) b) c)
d)\(|\frac{5}{8}x+\frac{6}{7}|-\frac{4}{7}=\frac{10}{7}\)
\(\Leftrightarrow|\frac{5}{8}x+\frac{6}{7}|=2\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x+\frac{6}{7}=2\\\frac{5}{8}x+\frac{6}{7}=-2\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x=\frac{8}{7}\\\frac{5}{8}x=\frac{-20}{7}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{64}{35}\\x=\frac{-32}{7}\end{cases}}}\)
Vậy \(x\in\left\{\frac{64}{35};\frac{-32}{7}\right\}\)
27 tháng 5 2019
Bài 1 :
a) \(\left(\frac{2}{5}-\frac{5}{8}\right):\frac{11}{30}+\frac{1}{8}\)
\(=\frac{-9}{40}:\frac{11}{30}+\frac{1}{8}\)
\(=\frac{-27}{44}+\frac{1}{8}\)
\(=\frac{-43}{88}\)
16 tháng 2 2021
a) \(x+\left(-7\right)=-20\)
\(\Rightarrow x=-20+7\)
\(\Rightarrow x=-13\)
Vậy \(x=-13\)
b) \(8-x=-12\)
\(\Rightarrow x=8-\left(-12\right)\)
\(\Rightarrow x=20\)
Vậy \(x=20\)
c) \(|x|-7=-6\)
\(\Rightarrow|x|=-6+7\)
\(\Rightarrow|x|=1\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
Vậy \(x\in\left\{1;-1\right\}\)
d) \(5^2.2^2-7.|x|=65\)
\(\Rightarrow\left(5.2\right)^2-7.|x|=65\)
\(\Rightarrow10^2-7.|x|=65\)
\(\Rightarrow100-7.|x|=65\)
\(\Rightarrow7.|x|=35\)
\(\Rightarrow|x|=5\)
\(\Rightarrow\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)
Vậy \(x\in\left\{5;-5\right\}\)
e) \(37-3.|x|=2^3-4\)
\(\Rightarrow37-3.|x|=8-4\)
\(\Rightarrow37-3.|x|=4\)
\(\Rightarrow3.|x|=33\)
\(\Rightarrow|x|=11\)
\(\Rightarrow\orbr{\begin{cases}x=11\\x=-11\end{cases}}\)
Vậy \(x\in\left\{11;-11\right\}\)
f) \(|x|+|-5|=|-37|\)
\(\Rightarrow|x|+5=37\)
\(\Rightarrow|x|=32\)
\(\Rightarrow\orbr{\begin{cases}x=32\\x=-32\end{cases}}\)
Vậy \(x\in\left\{32;-32\right\}\)
g)\(5.|x+9|=40\)
\(\Rightarrow|x+9|=8\)
\(\Rightarrow\orbr{\begin{cases}x+9=8\\x+9=-8\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-1\\x=-17\end{cases}}\)
Vậy \(x\in\left\{-1;-17\right\}\)
h) \(-\frac{5}{6}+\frac{8}{3}+\frac{-29}{6}\le x\le\frac{-1}{2}+2+\frac{5}{2}\)
\(\Rightarrow\frac{-5}{6}+\frac{16}{6}+\frac{-29}{6}\le x\le\frac{-1}{2}+\frac{4}{2}+\frac{5}{2}\)
\(\Rightarrow-3\le x\le4\)
Vậy \(-3\le x\le4\)
HQ
Hà Quang Minh
Giáo viên
7 tháng 10 2023
a) \(\frac{{ - 3}}{8} = \frac{{ - 3.3}}{{8.3}} = \frac{{ - 9}}{{24}}\)
Vì -9 < -5 nên \(\frac{{ - 9}}{{24}} < \frac{{ - 5}}{{24}}\)
Vậy \(\frac{{ - 3}}{8} < \frac{{ - 5}}{{24}}\).
b) Cách 1: \(\frac{{ - 2}}{{ - 5}} = \frac{2}{5}; \frac{3}{{ - 5}} = \frac{-3}{{5}}\)
Vì 2 > -3 nên \(\frac{2}{5} > \frac{-3}{{5}}\)
Vậy \(\frac{{ - 2}}{{ - 5}} > \frac{3}{{ - 5}}\).
Cách 2: \(\frac{{ - 2}}{{ - 5}} = \frac{2}{5} > 0\) mà \(\frac{3}{{ - 5}} < 0\)
\(\Rightarrow\) \(\frac{{ - 2}}{{ - 5}} > \frac{3}{{ - 5}}\).
c) \(\frac{{ - 3}}{{ - 10}} = \frac{3}{{10}} = \frac{{3.2}}{{10.2}} = \frac{6}{{20}}\)
\(\frac{{ - 7}}{{ - 20}} = \frac{7}{{20}}\)
Vì 6 < 7 nên \(\frac{6}{{20}} < \frac{7}{{20}}\) nên \(\frac{{ - 3}}{{ - 10}} < \frac{{ - 7}}{{ - 20}}\).
d) \(\frac{{ - 5}}{4} = \frac{{ - 5.5}}{{4.5}} = \frac{{ - 25}}{{20}}; \frac{{ 23}}{{-20}}=\frac{{-23}}{{20}} \)
Vì -25 < -23 nên \( \frac{{ - 25}}{{20}} < \frac{{-23}}{{20}} \)
Vậy \(\frac{{ - 5}}{4} < \frac{{23}}{{ - 20}}\).
15 tháng 6 2020
a, Câu hỏi của Nguyễn Ánh Ngân - Toán lớp 6 - Học toán với OnlineMath
b, Câu hỏi của Vũ Xuân Hiếu - Toán lớp 6 | Học trực tuyến
c)
8 tháng 2 2019
\(a)\frac{8}{9}x-\frac{2}{3}=\frac{1}{3}x+1\frac{1}{3}\)
\(\Rightarrow\frac{8}{9}x-\frac{1}{3}x=\frac{2}{3}+1\frac{1}{3}\)
\(\Rightarrow\frac{5}{9}x=\frac{2}{3}+\frac{4}{3}\)
\(\Rightarrow\frac{5}{9}x=2\Rightarrow x=2\div\frac{5}{9}=\frac{18}{5}\)
8 tháng 2 2019
\(b)(\frac{-2}{5}+\frac{3}{7})-(\frac{4}{9}+\frac{12}{20}-\frac{13}{25})+\frac{7}{35}\)
\(=\frac{1}{35}-(\frac{4}{9}+\frac{3}{5}-\frac{13}{25})+\frac{1}{5}\)
\(=\frac{1}{35}-(\frac{4}{9}+\frac{15}{25}-\frac{13}{25})+\frac{1}{5}\)
\(=\frac{1}{35}-(\frac{4}{9}+\frac{2}{25})+\frac{1}{5}\)
\(=\frac{1}{35}-\frac{118}{25}+\frac{1}{5}\)
Làm nốt
9 tháng 8 2019
\(bai1:a,\frac{3}{7}\cdot\frac{-5}{9}+\frac{4}{9}\cdot\frac{3}{7}-\frac{3}{7}\cdot\frac{8}{9}\)
\(< =>\frac{-15}{63}+\frac{12}{63}-\frac{24}{63}\)
\(< =>\frac{-15+12-24}{63}\)
\(< =>\frac{-3}{7}\)
\(b,1\frac{13}{15}\cdot0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\)
\(< =>\frac{28}{15}\cdot\frac{3}{4}-\left(\frac{11}{20}+\frac{1}{4}\right):\frac{7}{5}\)
\(< =>\frac{7}{5}-\frac{4}{5}:\frac{7}{5}\)
\(< =>\frac{7}{5}-\frac{4}{7}\)
\(< =>\frac{29}{35}\)
\(bai2:\)
\(a,\frac{-3}{4}\cdot x-\frac{4}{10}=\frac{1}{5}\)
\(< =>\frac{-3}{4}\cdot x=\frac{1}{5}+\frac{4}{10}\)
\(< =>\frac{-3}{4}\cdot x=\frac{3}{5}\)
\(< =>x=\frac{3}{5}:\frac{-3}{4}\)
\(< =>x=\frac{-4}{5}\)
\(b,3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)
\(< =>3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{12}\)
\(< =>\left[3\left(x-\frac{1}{3}\right)\right]=\frac{1}{12}< =>x-\frac{1}{3}=\frac{1}{12}:3=\frac{1}{36}=>x=\frac{1}{36}+\frac{1}{3}=>x=\frac{13}{36}\)
\(< =>\left[\frac{1}{3}\cdot x\right]=\frac{1}{12}< =>x=\frac{1}{12}:\frac{1}{3}=>x=\frac{1}{4}\)
MM
9 tháng 8 2019
Bài 1:
a)\(\frac{3}{7}.\frac{-5}{9}+\frac{4}{9}.\frac{3}{7}-\frac{3}{7}.\frac{8}{9}\) b,\(1\frac{13}{15}.0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\)
\(=\frac{3}{7}.(\frac{-5}{9}+\frac{4}{9}-\frac{8}{9})\) \(=\frac{28}{15}.\frac{3}{4}-\left(\frac{11}{20}+\frac{5}{20}\right):\frac{7}{5}\)
\(=\frac{3}{7}.\frac{-9}{9}\) \(=\frac{7}{5}-\frac{4}{5}:\frac{7}{5}\)
\(=\frac{-3}{7}\) \(=\frac{7}{5}-\frac{4}{7}\)
\(=\frac{29}{35}\)
Bài 2:
a)\(\frac{-3}{4}x-\frac{4}{10}=\frac{1}{5}\) b,\(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)
\(\frac{-3}{4}x\) \(=\frac{1}{5}+\frac{4}{10}\) \(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{12}\)
\(\frac{-3}{4}x\) \(=\frac{3}{5}\) \(\left(x.3-\frac{1}{3}.3\right)+\frac{1}{3}x=\frac{1}{12}\)
\(x\) \(=\frac{3}{5}:\frac{-3}{4}\) \(\left(x.3-1\right)+\frac{1}{3}x=\frac{1}{12}\)
\(x\) \(=\frac{4}{-5}\) \(x.\left(3+\frac{1}{3}\right)-1=\frac{1}{12}\)
\(x.\left(3+\frac{1}{3}\right)=\frac{1}{12}+1\)
\(x.\frac{10}{3}=\frac{13}{12}\)
\(x=\frac{13}{12}:\frac{10}{3}\)
\(x=\frac{13}{40}\)
16 tháng 4 2019
a) \(\frac{2}{5}x-x=\frac{\left(-2018\right)^0}{5^2}\\ x\left(\frac{2}{5}-1\right)=\frac{1}{25}\\ x\left(\frac{2}{5}-\frac{5}{5}\right)=\frac{1}{25}\\ x\cdot\frac{-3}{5}=\frac{1}{25}\\ x=\frac{1}{25}:\frac{-3}{5}\\ x=\frac{1}{25}\cdot\frac{-5}{3}\\ x=\frac{-1}{15}\)Vậy \(x=\frac{-1}{15}\)
b) \(\left|-1\frac{1}{2}x+2x\right|-\frac{7}{4}=0,5\\ \left|x\left(-1\frac{1}{2}+2\right)\right|-\frac{7}{4}=\frac{1}{2}\\ \left|x\cdot\frac{1}{2}\right|=\frac{1}{2}+\frac{7}{4}\\ \left|x\cdot\frac{1}{2}\right|=\frac{2}{4}+\frac{7}{4}\\ \left|x\cdot\frac{1}{2}\right|=\frac{9}{4}\\ \Rightarrow\left[{}\begin{matrix}x\cdot\frac{1}{2}=\frac{9}{4}\\x\cdot\frac{1}{2}=\frac{-9}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{9}{4}:\frac{1}{2}\\x=\frac{-9}{4}:\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{9}{4}\cdot2\\x=\frac{-9}{4}\cdot2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{9}{2}\\x=\frac{-9}{2}\end{matrix}\right.\)Vậy \(x\in\left\{\frac{9}{2};\frac{-9}{2}\right\}\)
c) \(x+\left(x+\frac{2}{7}\right)+\frac{-5}{11}=\frac{4}{11}\\ x+x+\frac{2}{7}=\frac{4}{11}-\frac{-5}{11}\\ 2x+\frac{2}{7}=\frac{4}{11}+\frac{5}{11}\\ 2x+\frac{2}{7}=\frac{9}{11}\\ 2x=\frac{9}{11}-\frac{2}{7}\\ 2x=\frac{63}{77}-\frac{22}{77}\\ 2x=\frac{41}{77}\\ x=\frac{41}{77}:2\\ x=\frac{41}{77\cdot2}\\ x=\frac{41}{154}\)Vậy \(x=\frac{41}{154}\)
d) \(\left|0,25x-20\%\right|+\frac{3}{8}=1\frac{3}{8}\\ \left|\frac{1}{4}x-\frac{1}{5}\right|=1\frac{3}{8}-\frac{3}{8}\\ \left|\frac{1}{4}x-\frac{1}{5}\right|=1\\ \Rightarrow\left[{}\begin{matrix}\frac{1}{4}x-\frac{1}{5}=1\\\frac{1}{4}x-\frac{1}{5}=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{1}{4}x=1+\frac{1}{5}\\\frac{1}{4}x=\left(-1\right)+\frac{1}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{1}{4}x=\frac{5}{5}+\frac{1}{5}\\\frac{1}{4}x=\frac{-5}{5}+\frac{1}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{1}{4}x=\frac{6}{5}\\\frac{1}{4}x=\frac{-4}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{6}{5}:\frac{1}{4}\\x=\frac{-4}{5}:\frac{1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{6}{5}\cdot4\\x=\frac{-4}{5}\cdot4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{24}{5}\\x=\frac{-16}{5}\end{matrix}\right.\)Vậy \(x\in\left\{\frac{24}{5};\frac{-16}{5}\right\}\)
\(x-\frac{8}{5}=\frac{-7}{20}.\frac{4}{5}\)
\(\Rightarrow x-\frac{8}{5}=\frac{-7}{25}\)
\(\Rightarrow x=\frac{-7}{25}+\frac{8}{5}=\frac{33}{25}\)
\(x-\frac{8}{5}=\frac{-7}{20}.\frac{4}{5}\)
\(x-\frac{8}{5}=-\frac{7}{25}\)
\(x=-\frac{7}{25}+\frac{8}{5}\)
\(x=-\frac{7}{25}+\frac{40}{25}\)
\(x=\frac{33}{25}\)