tìm x
a )\(x+7\frac{4}{5}=9\frac{7}{15}\) b)\(x-4\frac{9}{11}=\frac{2121}{2222}\)
giúp mik nha
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a ) \(x+7\frac{4}{5}=9\frac{7}{15}\)
\(x+\frac{39}{5}=\frac{142}{15}\)
\(x=\frac{142}{15}-\frac{39}{5}\)
\(x=\frac{5}{3}\)
b ) \(x-4\frac{9}{11}=\frac{2121}{2222}\)
\(x-\frac{53}{11}=\frac{21}{22}\)
\(x=\frac{21}{22}+\frac{53}{11}\)
\(x=\frac{127}{22}\)
\(x+7\frac{4}{5}=9\frac{7}{15}\)
\(x=9\frac{7}{15}-7\frac{4}{5}\)
\(x=9\frac{7}{15}-7\frac{12}{15}\)
\(x=-2\frac{1}{3}\)
\(x-4\frac{9}{11}=\frac{2121}{2222}\)
\(x-4\frac{9}{11}=\frac{101.21}{101.22}\)
\(x-\frac{53}{11}=\frac{21}{22}\)
\(x=\frac{21}{22}+\frac{53}{11}\)
\(x=\frac{127}{22}=5\frac{17}{22}\)
\(x+7\dfrac{4}{5}=9\dfrac{7}{15}\)
\(=>x+\dfrac{39}{5}=\dfrac{142}{15}\)
\(=>x=\dfrac{142}{15}-\dfrac{39}{5}=\dfrac{142}{15}-\dfrac{117}{15}\)
\(=>x=\dfrac{25}{15}=\dfrac{5}{3}\)
___________
\(x-4\dfrac{9}{11}=\dfrac{2121}{2222}\)
\(=>x-\dfrac{53}{11}=\dfrac{21}{22}\)
\(=>x=\dfrac{21}{22}+\dfrac{53}{11}=\dfrac{21}{22}+\dfrac{106}{22}\)
\(=>x=\dfrac{127}{22}\)
\(x+7\dfrac{4}{5}=9\dfrac{7}{15}\)
\(x=9\dfrac{7}{15}-7\dfrac{4}{5}\)
\(x=\left(9-7\right)+\left(\dfrac{7}{15}-\dfrac{4}{5}\right)\)
\(x=2\dfrac{-1}{3}=\dfrac{5}{3}\)
\(x=\dfrac{2121:101}{2222:101}+4\dfrac{9}{11}\)
\(x=\dfrac{21}{22}+4\dfrac{18}{22}\)
\(x=\dfrac{21+4\cdot22+18}{22}=\dfrac{127}{22}\)
\(a.\left(\frac{6}{11}+\frac{5}{11}\right).\frac{3}{7}=1\cdot\frac{3}{7}=\frac{3}{7}b.\frac{3}{5}\cdot\frac{7}{9}+\frac{3}{5}\cdot\frac{2}{9}=\frac{3}{5}\cdot\left(\frac{7}{9}+\frac{2}{9}\right)=\frac{3}{5}\cdot1=\frac{3}{5}\)
\(a)x+30\%x=-1,31\)
\(\Leftrightarrow x+\frac{3x}{10}=-1,31\)
\(\Leftrightarrow10x+3x=-13,1\)
\(\Leftrightarrow13x=-13,1\Leftrightarrow x=-\frac{131}{130}\)
\(b)\left(x-\frac{1}{2}\right):\frac{1}{3}+\frac{5}{7}=9\frac{5}{7}\)
\(\Leftrightarrow\frac{2x-1}{2}.3+\frac{5}{7}=\frac{68}{7}\)
\(\Leftrightarrow\frac{6x-3}{2}=\frac{63}{7}\)
\(\Leftrightarrow\frac{6x-3}{2}=9\)
\(\Leftrightarrow6x-3=18\)
\(\Leftrightarrow x=\frac{7}{2}\)
\(a)\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{3}{5}+\frac{5}{7}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{-2}{5}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{-1}{4}+\frac{2}{7}+\frac{5}{7}+\frac{3}{5}\)
\(\Rightarrow\frac{2}{6}+\frac{1}{6}+\frac{-3}{5}\le x< -1+1+\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}+\frac{-3}{5}\le x< \frac{3}{5}\)
\(\Rightarrow\frac{-1}{10}\le x< \frac{6}{10}\)
\(\Rightarrow-1\le x< 6\)
\(\Rightarrow x\in\left\{-1;0;1;2;3;4;5\right\}\)
Bài b tương tự
1)
a)
\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)
\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)
\(\frac{-20}{5}< x< \frac{-3}{10}\)
\(\frac{-40}{10}< x< \frac{-3}{10}\)
\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)
\(A=\left(-\frac{5}{11}\right).\frac{7}{15}+\frac{11}{-5}.\frac{30}{33}\)
\(A=-\frac{7}{33}+-2\)
\(A=-\frac{73}{33}\)
[ A] = -2
a) x + \(7\frac{4}{5}\)= \(9\frac{7}{15}\)
x + 39/5 = 95/15
x = 95/15 - 39/5
x = 95/15 - 127/15
x = -32/15
b) Tương tự
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