4/11*y-6/11=2/5
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9/11 : y = 2/5 + 2/7
9/11 :y = 24/35
y = 9/11 : 24/35
y = 105/88
y * 3/11 = 3/4 - 6/11
y * 3/11 = 9/44
y = 9/44 : 3/11
y = 3/4
Bạn thamkhaor nhé
a: \(=\dfrac{-3-5-7}{4}=\dfrac{-15}{4}\)
b: \(=\dfrac{-6-5-2+6}{11}=\dfrac{-7}{11}\)
\(\left(-\dfrac{1}{2}\right)^2\div\dfrac{1}{4}-2\times\left(-\dfrac{1}{2}\right)^2\\= \dfrac{1}{4}\div\dfrac{1}{4}-2\times\dfrac{1}{4}\\ =1-\dfrac{1}{2}\\ =\dfrac{1}{2}\)
\(\left(-2\right)^3\times-\dfrac{1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right)\div\dfrac{5}{12}\)
= \(-6\times-\dfrac{1}{24}+\left(\dfrac{4}{3}-\dfrac{11}{6}\right)\div\dfrac{5}{12}\)
= \(\dfrac{1}{4}+-\dfrac{1}{2}\div\dfrac{5}{12}\)
= \(\dfrac{1}{4}+-\dfrac{6}{5}\)
= \(\dfrac{1}{4}-\dfrac{6}{5}\)
= \(-\dfrac{19}{20}\)
\(\left(6\dfrac{4}{9}+\dfrac{7}{11}\right)-\left(4\dfrac{4}{9}-2\dfrac{4}{11}\right)\\ =\dfrac{58}{9}+\dfrac{7}{11}-\dfrac{40}{9}+\dfrac{26}{11}\\ =\dfrac{58}{9}-\dfrac{40}{9}+\dfrac{7}{11}+\dfrac{26}{11}\\ =12+3\\ =15\)
\(a,\left(\dfrac{-1}{2}\right)^2:\dfrac{1}{4}-2\left(-\dfrac{1}{2}\right)^2\)
\(=\left(-\dfrac{1}{2}\right)^2\left(4-2\right)\)
\(=\dfrac{1}{4}.2=\dfrac{1}{2}\)
\(b,\left(-2\right)^3.\dfrac{-1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right):\dfrac{5}{12}\)
\(=\left(-8\right).\dfrac{-1}{24}+\left(-\dfrac{1}{2}\right).\dfrac{12}{5}\)
\(=\dfrac{1}{3}+\left(-\dfrac{1}{5}\right)=\dfrac{2}{15}\)
\(c,\left(6\dfrac{4}{9}+\dfrac{7}{11}\right)-\left(4\dfrac{4}{9}-2\dfrac{4}{11}\right)\)
\(=\dfrac{701}{99}-\dfrac{206}{99}=\dfrac{495}{99}=5\)
\(d,10\dfrac{1}{5}-5\dfrac{1}{2}.\dfrac{60}{11}+\dfrac{3}{15\%}\)
\(=\dfrac{51}{5}-30+20=\dfrac{1}{5}\)
\(e,\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\)
\(=\dfrac{5}{7}\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)=\dfrac{5}{7}.\left(-\dfrac{7}{11}\right)\)
\(=-\dfrac{5}{11}\)
\(f,\dfrac{-5}{7}.\dfrac{2}{11}+\left(-\dfrac{5}{7}\right).\dfrac{9}{11}+1\dfrac{5}{7}\)
\(=\left(-\dfrac{5}{7}\right)\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\)
\(=\left(-\dfrac{5}{7}\right)+\dfrac{12}{7}=1\)
\(4.3^{x-1}+2.3^{x+2}=4.3^6+2.3^9\)
\(3^{x-1}.\left(4+2.3^3\right)=3^6.\left(4+2.3^3\right)\)
\(\Leftrightarrow3^{x-1}=3^6\)
\(\Leftrightarrow x-1=6\)
\(\Leftrightarrow x=7\)
Vậy \(x=7\)
\(\frac{4}{11}.y-\frac{6}{11}=\frac{2}{5}\Leftrightarrow\frac{4}{11}y=\frac{2}{5}+\frac{6}{11}\)
\(\Leftrightarrow\frac{4}{11}y=\frac{52}{55}\Rightarrow y=\frac{52}{55}:\frac{4}{11}\)
\(\Rightarrow y=\frac{13}{5}\)
\(\frac{4}{11}\times y-\frac{6}{11}=\frac{2}{5}\)
\(\Rightarrow\frac{4}{11}\times y=\frac{2}{6}+\frac{6}{11}\)
\(\Rightarrow\frac{4}{11}\times y=\frac{29}{33}\)
\(\Rightarrow y=\frac{29}{33}:\frac{4}{11}\)
\(\Rightarrow y=\frac{29}{33}\times\frac{11}{4}\)
\(\Rightarrow y=\frac{29}{12}\)
1/5+8/15:y=1/2
8/15:y=1/5-1/2
8/15:y=-3/10
y=8/15:1/2
y=16/15
vậy y=16/15
4/11.y-6/11=2/5
4/11.y=2/5+6/11
4/11.y=52/55
y=52/55:4/11
y=13/5
vậy y=13/5
\(Đặt:\left\{{}\begin{matrix}a=\dfrac{1}{\left|x+3\right|}\left(ĐK:x\ne-3\right)\\b=\dfrac{1}{\left|y\right|-2}\left(ĐK:y\ne\pm2\right)\end{matrix}\right.\\ Có:\left\{{}\begin{matrix}\dfrac{1}{\left|x+3\right|}+\dfrac{4}{\left|y\right|-2}=\dfrac{11}{6}\\\dfrac{5}{\left|x+3\right|}+\dfrac{2}{\left|y\right|-2}=\dfrac{11}{6}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+4b=\dfrac{11}{6}\\5a+2b=\dfrac{11}{6}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a+4b=\dfrac{11}{6}\\10a+4b=\dfrac{22}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-9a=-\dfrac{11}{6}\\a+4b=\dfrac{11}{6}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{11}{54}\\b=\dfrac{\dfrac{11}{6}-\dfrac{11}{54}}{4}=\dfrac{11}{27}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{\left|x+3\right|}=a=\dfrac{11}{54}\\\dfrac{1}{\left|y\right|-2}=b=\dfrac{11}{27}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}11\left|x+3\right|=54\\11\left(\left|y\right|-2\right)=27\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left|x+3\right|=\dfrac{54}{11}\\\left|y\right|=\dfrac{27}{11}+2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x+3=\dfrac{54}{11}\\x+3=\dfrac{-54}{11}\end{matrix}\right.\\\left[{}\begin{matrix}y=\dfrac{27}{11}+2\\y=-\left(\dfrac{27}{11}+2\right)\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=\dfrac{21}{11}\left(TM\right)\\x=\dfrac{-87}{11}\left(TM\right)\end{matrix}\right.\\\left[{}\begin{matrix}y=\dfrac{49}{11}\left(TM\right)\\y=-\dfrac{49}{11}\left(TM\right)\end{matrix}\right.\end{matrix}\right.\\ Vậy:\left(x;y\right)=\left\{\left(\dfrac{21}{11};\dfrac{49}{11}\right);\left(\dfrac{-87}{11};\dfrac{49}{11}\right);\left(\dfrac{21}{11};\dfrac{-49}{11}\right);\left(\dfrac{-87}{11};\dfrac{-49}{11}\right)\right\}\)
\(\frac{4}{11}.y-\frac{6}{11}=\frac{2}{5}\Rightarrow\frac{4}{11}.y=\frac{52}{55}\Rightarrow y=\frac{13}{5}\)
Vậy :\(y=\frac{13}{5}\)
\(\frac{4}{11}\cdot y-\frac{6}{11}=\frac{2}{5}\)
\(\Rightarrow\frac{4}{11}\cdot y=\frac{2}{5}+\frac{6}{11}\)
\(\Rightarrow y=\frac{52}{55}:\frac{4}{11}\)
\(\Rightarrow y=\frac{13}{5}=2,6\)
Good luck!