(\(\frac{7}{3}\) x \(x\) - 0,6) : \(3\frac{2}{5}\) = 1
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a, \(\left(\frac{5}{8}.0,6-5:3\frac{1}{3}\right).\left(\frac{1}{5}-1,4\right).\left(-5\right)^2.x=120\)
\(\left(\frac{5}{8}.\frac{3}{5}-5:\frac{10}{3}\right)\left(\frac{1}{5}-\frac{7}{5}\right).25x=120\)
\(\left(\frac{3}{8}-\frac{3}{2}\right)\left(-\frac{6}{5}\right).25x=120\)
\(\left(-\frac{9}{8}\right).\left(-\frac{6}{5}\right).25x=120\)
\(\frac{27}{20}.25x=120\)\(25x=120:\frac{27}{20}=\frac{800}{9}\)
\(x=\frac{800}{9}:25=\frac{32}{9}\)
đợi tí đi nấu cơm đã
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a) -0,6 . x - 7/3 = 5,4 (sửa dấu phẩy thành dấu nhân)
=> \(-\frac{3}{5}x-\frac{7}{3}=\frac{27}{5}\)
=> \(-\frac{3}{5}x=\frac{27}{5}+\frac{7}{3}=\frac{116}{15}\)
=> \(x=\frac{116}{15}:\left(-\frac{3}{5}\right)=\frac{116}{15}\cdot\left(-\frac{5}{3}\right)=-\frac{116}{9}\)
b) 2,8 : \(\left(\frac{1}{5}-3x\right)=1\frac{2}{5}\)
=> \(\frac{1}{5}-3x=2,8:1\frac{2}{5}\)
=> \(\frac{1}{5}-3x=\frac{14}{5}:\frac{7}{5}=\frac{14}{5}\cdot\frac{5}{7}=2\)
=> 3x = \(\frac{1}{5}-2=-\frac{9}{5}\)
=> \(x=\left(-\frac{9}{5}\right):3=\left(-\frac{9}{5}\right)\cdot\frac{1}{3}=-\frac{3}{5}\)
Nên sửa lại đề câu a đi nhá :))
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\(VP=1+\frac{2014}{2}+\frac{2015}{3}+...+\frac{4023}{2011}+\frac{4024}{2012}\)
\(=1-1+\left(\frac{2014}{2}-1\right)+\left(\frac{2015}{3}-1\right)+...+\left(\frac{4023}{2011}-1\right)+\left(\frac{40024}{2012}-1\right)+2012\)
\(=\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}+\frac{2012}{1}\)
\(=2012.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)\)
\(\Rightarrow2012=503.x\Rightarrow x=\frac{2012}{503}=4\)
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\(a,\left[\frac{4}{5}+\frac{2}{3}\right]:\frac{1}{5}-1,4\cdot\left[\frac{-5}{7}\right]^2\)
\(=\left[\frac{4\cdot3}{15}+\frac{2\cdot5}{15}\right]:\frac{1}{5}-1,4\cdot\frac{-5}{7}\cdot\frac{-5}{7}\)
\(=\left[\frac{12}{15}+\frac{10}{15}\right]:\frac{1}{5}-\frac{14}{10}\cdot\frac{25}{49}\)
\(=\frac{22}{15}:\frac{1}{5}-\frac{7}{5}\cdot\frac{25}{49}\)
\(=\frac{22}{15}\cdot\frac{5}{1}-\frac{7}{5}\cdot\frac{25}{49}\)
\(=\frac{22\cdot5}{15\cdot1}-\frac{7\cdot25}{5\cdot49}=\frac{22\cdot1}{3\cdot1}-\frac{1\cdot5}{1\cdot7}=\frac{22}{3}-\frac{5}{7}\)
= ...
Tự tính
Bài 2 : \(a,3-\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{1}{2}\)
\(\Rightarrow\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{1}{2}+3\)
\(\Rightarrow\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{2}\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{2}x-\frac{1}{3}=\frac{7}{2}\\\frac{1}{2}x-\frac{1}{3}=-\frac{7}{2}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{23}{3}\\x=\frac{-19}{3}\end{cases}}\)
Vậy \(x\in\left\{\frac{23}{3};\frac{-19}{3}\right\}\)
b, \(0,6-160\%< x\le3\frac{2}{3}:\frac{22}{18}\)
\(\Rightarrow0,6-\frac{160}{100}< x\le\frac{11}{3}:\frac{22}{18}\)
\(\Rightarrow0,6-\frac{8}{5}< x\le\frac{11}{3}\cdot\frac{18}{22}\)
\(\Rightarrow0,6-1,6< x\le3\)
\(\Rightarrow-1< x\le3\)
\(\Rightarrow x\in\left\{0;1;2;3\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1+0,6-\frac{3}{7}}{\frac{8}{3}+\frac{8}{5}-\frac{8}{7}}=\frac{\frac{3}{3}+\frac{3}{5}-\frac{3}{7}}{\frac{8}{3}+\frac{8}{5}-\frac{8}{7}}=\frac{3.\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{7}\right)}{8.\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{7}\right)}=\frac{3.1}{8.1}=\frac{3}{8}\)
\(\frac{\frac{1}{3}+0,25-\frac{1}{5}+0,125}{\frac{7}{6}+\frac{7}{8}-0,7+\frac{7}{16}}=\frac{\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{8}}{\frac{7}{6}+\frac{7}{8}-\frac{7}{10}+\frac{7}{16}}=\frac{1.\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{8}\right)}{7.\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{8}\right)}=\frac{1.1}{7.1}=\frac{1}{7}\)
=>\(\frac{3}{8}-\frac{1}{7}=\frac{13}{56}\)
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Sửa đề \(\frac{11}{13}\)chứ không phải \(\frac{11}{3}\)
\(\frac{2,75-2,2+\frac{11}{7}+\frac{11}{13}}{0,75-0,6+\frac{3}{7}+\frac{3}{13}}-x-\frac{1}{9}=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\)
+) Đặt \(A=\frac{2,75-2,2+\frac{11}{7}+\frac{11}{13}}{0,75-0,6+\frac{3}{7}+\frac{3}{13}}\)
\(A=\frac{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}\)
\(A=\frac{11\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)
\(A=\frac{11}{3}\)(1)
+) Đặt \(B=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\)
\(B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}\)
\(B=\frac{2}{2}\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}\right)\)
\(B=\frac{2}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{9}\right)\)
\(B=\frac{2}{2}\left(1-\frac{1}{9}\right)=1\cdot\frac{8}{9}=\frac{8}{9}\)(2)
Từ (1) và (2) => \(A-x-\frac{1}{9}=B\)
=> \(\frac{11}{3}-x-\frac{1}{9}=\frac{8}{9}\)
=> \(\frac{11}{3}-x=1\)
=> \(x=\frac{11}{3}-1=\frac{8}{3}\)
Vậy x = 8/3
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b)5.(1-1/3+1/9)/1(1+1/9-1/3):3/4-3/7+3/5/11/4-11/7+11/5
=5/8:3.(1/4-1/7+1/5)/11.(1/4-1/7+1/5)
=5/8:3/11=55/24
^_^
a) Bạn tách ra rồi làm như thế này nhé:
~ \(4\frac{5}{9}:2\frac{5}{18}-7\)\(\Rightarrow\)\(4\frac{5}{9}:2\frac{5}{18}-7=2-7=-5\)
~\(\left(3\frac{1}{5}:3,2+4,5-1\frac{31}{45}\right):\left(-21\frac{1}{2}\right)\)........bạn tính phần này ra nhé. ^_^(mk hơi ngại làm hih)
tiếp theo bạ làm như này:
\(-5< x< ...\)(chỗ ... là két quả của phần 2 nhé.)
rồi bạn tìm tập hợp các số x nằm giữa -5 và....
TK MK NHA. cHÚC BẠN HỌC TỐT. ^_^