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1.
a.= (-5/24+3/4+7/12): -9/4
=(-5/24+18/24+14/24) . -4/9
= 9/8 .-4/9=-1/2
b. =(3/5+83/200-3/200).8/3 .1/4
=(120/200+83/200-3/200) .2/3
=1.2/3=2/3
2.
a. 1/3(2x-5)=-2/3-3/2
1/3(2x-5)=-13/6
2x-5=-13/2
2x=-3/2
x=-3/2:2=-3/4
b. 1/3x-1/2x=3/4
x(1/3-1/2)=3/4
x.1/6=3/4
x=9/2
a) \(22\frac{1}{2}\cdot\frac{7}{9}+50\%-1,25\)
\(=\frac{45}{2}\cdot\frac{7}{9}+\frac{50}{100}-\frac{125}{100}\)
\(=\frac{5}{2}\cdot\frac{7}{1}+\frac{1}{2}-\frac{5}{4}\)
\(=\frac{35}{2}+\frac{1}{2}-\frac{5}{4}=18-\frac{5}{4}=\frac{67}{4}\)
b) \(1,4\cdot\frac{15}{49}-\left(\frac{4}{5}+\frac{2}{3}\right):2\frac{1}{5}\)
\(=\frac{7}{5}\cdot\frac{15}{49}-\frac{22}{15}:\frac{11}{15}\)
\(=\frac{1}{1}\cdot\frac{3}{7}-\frac{22}{15}\cdot\frac{15}{11}\)
\(=\frac{3}{7}-2=\frac{3-14}{7}=\frac{-11}{7}\)
c) \(\left(-\frac{1}{2}\right)^2-\frac{7}{16}:\frac{7}{4}+75\%\)
\(=\frac{1}{4}-\frac{7}{16}\cdot\frac{4}{7}+\frac{75}{100}\)
\(=\frac{1}{4}-\frac{1}{4}+\frac{3}{4}=\frac{3}{4}\)
Bài 2 Bạn tự làm nhé
1.a,\(22\frac{1}{2}.\frac{7}{9}+50\%-1,25\)
\(=\frac{45}{2}.\frac{7}{9}+\frac{1}{2}-\frac{5}{4}\)
\(=\frac{35}{2}+\frac{1}{2}-\frac{5}{4}\)
\(=\frac{67}{4}\)
b,Các phép tính khác làm tương tự
Đổi các số ra hết thành phân số,có ngoặc thì lm ngoặc trc,Xoq đến nhân chia trước dồi mới cộng trừ
c,tương tự
2.
a,\(1\frac{3}{5}+\frac{7}{12}\div x=\frac{-9}{4}\)
\(\frac{8}{5}+\frac{7}{12}\div x=\frac{-9}{4}\)
\(\frac{7}{12}\div x=\frac{-77}{20}\)
Đến đây dễ bạn tự làm
b,\(\left(2\frac{4}{5}.x+50\right)\div\frac{2}{3}=-51\)
\(\left(\frac{14}{5}x+50\right)\div\frac{2}{3}=-51\)
\(\frac{14}{5}x+50=-34\)
\(\frac{14}{5}x=-84\)
Tự làm tiếp
c,\(\left|\frac{3}{4}.x-\frac{1}{2}\right|=\frac{1}{4}\)\(\Rightarrow\left|\frac{3}{4}x-\frac{1}{2}\right|=\varnothing\)
\(a,\)\(-\frac{3}{5}\cdot x=\frac{1}{4}+0,75\)
\(-\frac{3}{5}\cdot x=\frac{1}{4}+\frac{3}{4}=\frac{4}{4}=1\)
\(x=1\div\left(-\frac{3}{5}\right)\)
\(x=-\frac{5}{3}\)
\(b,\)\(\left(\frac{1}{7}-\frac{1}{3}\right)\cdot x=\frac{28}{5}\times\left(\frac{1}{4}-\frac{1}{7}\right)\)
\(\left(\frac{3}{21}-\frac{7}{21}\right)\cdot x=\frac{28}{5}\cdot\left(\frac{7}{28}-\frac{4}{28}\right)\)
\(-\frac{4}{21}\cdot x=\frac{28}{5}\cdot\frac{3}{28}\)
\(-\frac{4}{21}\cdot x=\frac{3}{5}\)
\(x=\frac{3}{5}\div\left(-\frac{4}{21}\right)\)
\(x=-\frac{63}{20}\)
\(c,\)\(\frac{5}{7}\cdot x=\frac{9}{8}-0,125\)
\(\frac{5}{7}\cdot x=\frac{9}{8}-\frac{1}{8}\)
\(\frac{5}{7}\cdot x=1\)
\(x=1\div\frac{5}{7}\)
\(x=\frac{7}{5}\)
\(d,\)\(\left(\frac{2}{11}+\frac{1}{3}\right)\cdot x=\left(\frac{1}{7}-\frac{1}{8}\right)\cdot36\)
\(\left(\frac{6}{33}+\frac{11}{33}\right)\cdot x=\left(\frac{8}{56}-\frac{7}{56}\right)\cdot36\)
\(\frac{17}{33}\cdot x=\frac{1}{56}\cdot36\)
\(\frac{17}{33}\cdot x=\frac{9}{14}\)
\(x=\frac{9}{14}\div\frac{17}{33}\)
\(x=\frac{9}{14}\cdot\frac{33}{17}=\frac{297}{238}\)
Bài 1:
\(a,22\frac{1}{2}.\frac{7}{9}+50\%-1,25\)
=\(\frac{45}{2}.\frac{7}{9}+\frac{1}{2}-\frac{5}{4}\)
=\(\frac{35}{2}+\frac{1}{2}-\frac{5}{4}\)
=\(\frac{70}{4}+\frac{2}{4}-\frac{5}{4}\)
=\(\frac{67}{4}\)
\(b,1,4.\frac{15}{49}-\left(\frac{4}{5}+\frac{2}{3}\right):2\frac{1}{5}\)
=\(\frac{7}{5}.\frac{15}{49}-\left(\frac{12}{15}+\frac{10}{15}\right):\frac{11}{5}\)
=\(\frac{3}{7}-\frac{22}{15}.\frac{5}{11}\)
=\(\frac{3}{7}-\frac{2}{3}\)
=\(-\frac{5}{21}\)
\(c,125\%.\left(-\frac{1}{2}\right)^2:\left(1\frac{5}{6}-1,6\right)+2016^0\)
=\(\frac{5}{4}.\frac{1}{4}:\left(\frac{11}{6}-\frac{8}{5}\right)+1\)
=\(\frac{5}{16}:\frac{7}{30}+1\)
=\(\frac{131}{56}\)
\(d,1,4.\frac{15}{49}-\left(20\%+\frac{2}{3}\right):2\frac{1}{5}\)
=\(\frac{7}{5}.\frac{15}{49}-\left(\frac{1}{5}+\frac{2}{3}\right):\frac{11}{5}\)
=\(\frac{3}{7}-\frac{13}{15}:\frac{11}{5}\)
=\(\frac{3}{7}-\frac{13}{33}\)
=\(\frac{8}{231}\)
Bài đ làm giống hệt như bài c
Bài 2 :
\(a,\left|\frac{3}{4}.x-\frac{1}{2}\right|=\frac{1}{4}\)
=>\(\left[{}\begin{matrix}\frac{3}{4}.x-\frac{1}{2}=\frac{1}{4}\\\frac{3}{4}.x-\frac{1}{2}=-\frac{1}{4}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}\frac{3}{4}.x=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}\\\frac{3}{4}.x=-\frac{1}{4}+\frac{1}{2}=\frac{1}{4}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\frac{3}{4}:\frac{3}{4}=1\\x=\frac{1}{4}:\frac{3}{4}=\frac{1}{3}\end{matrix}\right.\)
Vậy x ∈{1;\(\frac{1}{3}\)}
\(b,\frac{5}{3}.x-\frac{2}{5}.x=\frac{19}{10}\)
=>\(\frac{19}{15}.x=\frac{19}{10}\)
=>\(x=\frac{19}{10}:\frac{19}{15}=\frac{3}{2}\)
Vậy x ∈ {\(\frac{3}{2}\)}
c,\(\left|2.x-\frac{1}{3}\right|=\frac{2}{9}\)
=>\(\left[{}\begin{matrix}2.x-\frac{1}{3}=\frac{2}{9}\\2.x-\frac{1}{3}=-\frac{2}{9}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2.x=\frac{2}{9}+\frac{1}{3}=\frac{5}{9}\\2.x=-\frac{2}{9}+\frac{1}{3}=\frac{1}{9}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\frac{5}{9}:2=\frac{5}{18}\\x=\frac{1}{9}:2=\frac{1}{18}\end{matrix}\right.\)
Vậy x∈{\(\frac{5}{18};\frac{1}{18}\)}
\(d,x-30\%.x=-1\frac{1}{5}\)
=\(70\%x=-\frac{6}{5}\)
=\(\frac{7}{10}.x=-\frac{6}{5}\)
=>\(x=-\frac{6}{5}:\frac{7}{10}=-\frac{12}{7}\)
Vậy x∈{\(-\frac{12}{7}\)}
Bài 2
a/
\(\Rightarrow\left[{}\begin{matrix}\frac{3}{4}.x-\frac{1}{2}=\frac{1}{4}\\\frac{3}{4}.x-\frac{1}{2}=-\frac{1}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{3}{4}.x=\frac{1}{4}+\frac{1}{2}\\\frac{3}{4}.x=-\frac{1}{4}+\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{3}{4}.x=\frac{3}{4}\\\frac{3}{4}.x=\frac{1}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{3}{4}:\frac{3}{4}\\x=\frac{1}{4}:\frac{3}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy \(x=1\) hoặc \(x=\frac{1}{3}\)
b/ Đặt x làm thừa số chung rồi tính như bình thường
c/ Tương tự câu a
d/ Tương tự câu b
B1
a) \(1-\left(5\frac{3}{8}+x-7\frac{5}{24}\right):16\frac{2}{3}=0\)
\(1-\left(\frac{43}{8}+x-\frac{173}{24}\right):\frac{50}{3}=0\)
\(1-\left(x-\frac{11}{6}\right).\frac{3}{50}=0\)
\(\left(x-\frac{11}{6}\right).\frac{3}{50}=1-0\)
\(\left(x-\frac{11}{6}\right).\frac{3}{50}=1\)
\(x-\frac{11}{6}=1:\frac{3}{50}\)
\(x-\frac{11}{6}=\frac{50}{3}\)
\(x=\frac{50}{3}+\frac{11}{6}\)
\(x=\frac{37}{2}\)
b) \(\frac{3}{5}+\frac{5}{7}:x=\frac{1}{3}\)
\(\frac{5}{7}:x=\frac{1}{3}-\frac{3}{5}\)
\(\frac{5}{7}:x=-\frac{4}{15}\)
\(x=\frac{5}{7}:\left(-\frac{4}{15}\right)\)
\(x=-\frac{75}{28}\)
c) \(\left(4\frac{1}{2}-\frac{2}{5}.x\right):\frac{7}{4}=\frac{11}{9}\)
\(\left(\frac{9}{2}-\frac{2}{5}.x\right):\frac{7}{4}=\frac{11}{9}\)
\(\frac{9}{2}-\frac{2}{5}.x=\frac{11}{9}.\frac{7}{4}\)
\(\frac{9}{2}-\frac{2}{5}.x=\frac{11}{2}\)
\(\frac{2}{5}.x=\frac{9}{2}-\frac{11}{2}\)
\(\frac{2}{5}.x=-1\)
\(x=-1:\frac{2}{5}\)
\(x=-\frac{5}{2}\)
B2
a) \(\left(\frac{1}{2}+\frac{1}{3}+\frac{2}{6}\right).24:5-\frac{9}{22}:\frac{15}{121}\)
\(=\left(\frac{3}{6}+\frac{2}{6}+\frac{2}{6}\right).24:5-\frac{9}{22}.\frac{121}{15}\)
\(=\frac{7}{6}.24:5-\frac{33}{10}\)
\(=28:5-\frac{33}{10}\)
\(=\frac{28}{5}-\frac{33}{10}\)
\(=\frac{56}{10}-\frac{33}{10}\)
\(=\frac{23}{10}\)
b) \(\frac{5}{14}+\frac{18}{35}+\left(1\frac{1}{4}-\frac{5}{4}\right):\left(\frac{5}{12}\right)^2\)
\(=\frac{25}{70}+\frac{36}{70}+\left(\frac{5}{4}-\frac{5}{4}\right):\frac{25}{144}\)
\(=\frac{61}{70}+0:\frac{25}{144}\)
\(=\frac{61}{70}+0\)
\(=\frac{61}{70}\)
\(a,\left[\frac{4}{5}+\frac{2}{3}\right]:\frac{1}{5}-1,4\cdot\left[\frac{-5}{7}\right]^2\)
\(=\left[\frac{4\cdot3}{15}+\frac{2\cdot5}{15}\right]:\frac{1}{5}-1,4\cdot\frac{-5}{7}\cdot\frac{-5}{7}\)
\(=\left[\frac{12}{15}+\frac{10}{15}\right]:\frac{1}{5}-\frac{14}{10}\cdot\frac{25}{49}\)
\(=\frac{22}{15}:\frac{1}{5}-\frac{7}{5}\cdot\frac{25}{49}\)
\(=\frac{22}{15}\cdot\frac{5}{1}-\frac{7}{5}\cdot\frac{25}{49}\)
\(=\frac{22\cdot5}{15\cdot1}-\frac{7\cdot25}{5\cdot49}=\frac{22\cdot1}{3\cdot1}-\frac{1\cdot5}{1\cdot7}=\frac{22}{3}-\frac{5}{7}\)
= ...
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, \(0,6-160\%< x\le3\frac{2}{3}:\frac{22}{18}\)
\(\Rightarrow0,6-\frac{160}{100}< x\le\frac{11}{3}:\frac{22}{18}\)
\(\Rightarrow0,6-\frac{8}{5}< x\le\frac{11}{3}\cdot\frac{18}{22}\)
\(\Rightarrow0,6-1,6< x\le3\)
\(\Rightarrow-1< x\le3\)
\(\Rightarrow x\in\left\{0;1;2;3\right\}\)