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\(\left(\frac{13}{4}-y\cdot2\right)\cdot3=4\)
\(\frac{13}{4}-y\cdot2=\frac{4}{3}\)
\(y\cdot2=\frac{13}{4}-\frac{4}{3}\)
\(y\cdot2=\frac{39}{12}-\frac{16}{12}\)
\(y\cdot2=\frac{23}{12}\)
\(y=\frac{23}{12}:2\)
\(y=\frac{23}{12}\cdot\frac{1}{2}\)
\(y=\frac{23}{24}\)
y + \(\dfrac{3}{2}y\) +\(y:\dfrac{2}{7}\)= 720
\(y+\dfrac{3}{2}y+\dfrac{7}{2}y=720\)
\(6y=720\)
y = 120
(1981 x 1982 - 990) : (1980 x 1982 + 992)
=(1980 x 1982+1982 -990) : (1980 x 1982 +992)
=(1980 x 1982 + 992) : ( 1980 x 1982 + 992)
=1
\(y\times3+y\times2+y\times5=2022\)
\(\Rightarrow y\times\left(3+2+5\right)=2022\)
\(\Rightarrow y\times10=2022\)
\(\Rightarrow y=2022:10\)
\(\Rightarrow y=202,2\)
y x3 +y x2 +y x5 =2022
y x ( 3 + 2 + 5 ) = 2022
y x 10 = 2022
y = 2022 : 10
y =202,2
Mình nghĩ là đề bài sai đó
Ở chỗ x + y = 2
Nếu thế thì x2 và y2 cộng vào không bằng 10 được
mình ko chép đề bài nha
a) \(\dfrac{16}{5}\): \(\dfrac{7}{3}\) : y =\(\dfrac{12}{7}\)
\(\dfrac{48}{35}\): y = \(\dfrac{12}{7}\)
y = \(\dfrac{48}{35}\): \(\dfrac{12}{7}\)
y = \(\dfrac{4}{5}\)
\(2\dfrac{2}{5}-y:2\dfrac{3}{4}=1\dfrac{1}{2}\\ \dfrac{12}{5}-y:\dfrac{11}{4}=\dfrac{3}{2}\\ y:\dfrac{11}{4}=\dfrac{12}{5}-\dfrac{3}{2}\\ y:\dfrac{11}{4}=\dfrac{9}{10}\\ y=\dfrac{9}{10}\times\dfrac{11}{4}=\dfrac{99}{40}\\ b,1\dfrac{1}{4}+2\dfrac{1}{5}\times y=2\dfrac{3}{5}\\ \dfrac{5}{4}+\dfrac{11}{5}\times y=\dfrac{13}{5}\\ \dfrac{11}{5}\times y=\dfrac{13}{5}-\dfrac{5}{4}\\ \dfrac{11}{5}\times y=\dfrac{27}{20}\\ y=\dfrac{27}{20}:\dfrac{11}{5}=\dfrac{27}{44}\)
\(c,2\dfrac{4}{5}-2\dfrac{1}{4}:y=\dfrac{3}{4}\\ \dfrac{14}{5}-\dfrac{9}{4}:y=\dfrac{3}{4}\\ \dfrac{9}{4}:y=\dfrac{14}{5}-\dfrac{3}{4}\\ \dfrac{9}{4}:y=\dfrac{41}{20}\\ y=\dfrac{9}{4}:\dfrac{41}{20}=\dfrac{45}{41}\\ c2,x:3\dfrac{1}{3}=2\dfrac{2}{5}+\dfrac{7}{10}\\ x:\dfrac{10}{3}=\dfrac{12}{5}+\dfrac{7}{10}\\ x:\dfrac{10}{3}=\dfrac{31}{10}\\ x=\dfrac{31}{10}\times\dfrac{10}{3}=\dfrac{31}{3}\)
Câu đầu em xem lại đề bài sao có hai dấu bằng.
Câu 2:
\(\dfrac{3}{2}\) \(\times\)y - \(\dfrac{3}{4}\) \(\times\)y + y = \(\dfrac{4}{5}\)
y \(\times\) ( \(\dfrac{3}{2}\) - \(\dfrac{3}{4}\) + 1) = \(\dfrac{4}{5}\)
y \(\times\) (\(\dfrac{6}{4}\) - \(\dfrac{3}{4}\) + \(\dfrac{4}{4}\)) = \(\dfrac{4}{5}\)
y \(\times\) \(\dfrac{7}{4}\) = \(\dfrac{4}{5}\)
y = \(\dfrac{4}{5}\): \(\dfrac{7}{4}\)
y = \(\dfrac{16}{35}\)
bài này violympic quốc gia lần trước mình cũng gặp :
\(\frac{y\cdot3}{1}+\frac{y}{2}+\frac{y}{4}=\frac{3}{2}=1,5\)
\(y\cdot\left(3+\frac{1}{2}+\frac{1}{4}\right)=1,5\)
\(y\cdot\frac{15}{4}=\frac{3}{2}\)
\(y=\frac{3}{2}:\frac{15}{4}=\frac{3}{2}\cdot\frac{4}{15}=\frac{12}{30}=\frac{2}{5}\)