4) \(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{4}=\dfrac{z}{5}\) và x- y - z = 28
bài 3: Tìm y
a) \(\dfrac{1}{2}\) : y x \(\dfrac{3}{5}=\dfrac{4}{3}+\dfrac{3}{4}\) b) \(\dfrac{4}{3}-\dfrac{1}{2}\) x y \(=1\) c) \(\dfrac{1}{4}+y\) : \(\dfrac{1}{3}=\dfrac{5}{6}\)
a) \(\dfrac{1}{2}:y\times\dfrac{3}{5}=\dfrac{4}{3}+\dfrac{3}{4}\)
\(\dfrac{1}{2}:y\times\dfrac{3}{5}=\dfrac{25}{12}\)
\(\dfrac{1}{2}:y=\dfrac{25}{12}:\dfrac{3}{5}\)
\(\dfrac{1}{2}:y=\dfrac{125}{36}\)
\(y=\dfrac{1}{2}:\dfrac{125}{36}\)
\(y=\dfrac{18}{125}\)
b) \(\dfrac{4}{3}-\dfrac{1}{2}\times y=1\)
\(\dfrac{1}{2}\times y=\dfrac{4}{3}-1\)
\(\dfrac{1}{2}\times y=\dfrac{1}{3}\)
\(y=\dfrac{1}{3}:\dfrac{1}{2}\)
\(y=\dfrac{2}{3}\)
c) \(\dfrac{1}{4}+y:\dfrac{1}{3}=\dfrac{5}{6}\)
\(y:\dfrac{1}{3}=\dfrac{5}{6}-\dfrac{1}{4}\)
\(y:\dfrac{1}{3}=\dfrac{7}{12}\)
\(y=\dfrac{7}{12}\cdot\dfrac{1}{3}\)
\(y=\dfrac{7}{36}\)
Bài 2: (đề 2) Tìm y
a) \(2\dfrac{2}{5}-y:2\dfrac{3}{4}=1\dfrac{1}{2}\) b) \(1\dfrac{1}{4}+2\dfrac{1}{5}\) x \(y=2\dfrac{3}{5}\)
c) \(2\dfrac{4}{5}-2\dfrac{1}{4}:y=\dfrac{3}{4}\) c) \(x:3\dfrac{1}{3}=2\dfrac{2}{5}+\dfrac{7}{10}\)
\(2\dfrac{2}{5}-y:2\dfrac{3}{4}=1\dfrac{1}{2}\\ \dfrac{12}{5}-y:\dfrac{11}{4}=\dfrac{3}{2}\\ y:\dfrac{11}{4}=\dfrac{12}{5}-\dfrac{3}{2}\\ y:\dfrac{11}{4}=\dfrac{9}{10}\\ y=\dfrac{9}{10}\times\dfrac{11}{4}=\dfrac{99}{40}\\ b,1\dfrac{1}{4}+2\dfrac{1}{5}\times y=2\dfrac{3}{5}\\ \dfrac{5}{4}+\dfrac{11}{5}\times y=\dfrac{13}{5}\\ \dfrac{11}{5}\times y=\dfrac{13}{5}-\dfrac{5}{4}\\ \dfrac{11}{5}\times y=\dfrac{27}{20}\\ y=\dfrac{27}{20}:\dfrac{11}{5}=\dfrac{27}{44}\)
\(c,2\dfrac{4}{5}-2\dfrac{1}{4}:y=\dfrac{3}{4}\\ \dfrac{14}{5}-\dfrac{9}{4}:y=\dfrac{3}{4}\\ \dfrac{9}{4}:y=\dfrac{14}{5}-\dfrac{3}{4}\\ \dfrac{9}{4}:y=\dfrac{41}{20}\\ y=\dfrac{9}{4}:\dfrac{41}{20}=\dfrac{45}{41}\\ c2,x:3\dfrac{1}{3}=2\dfrac{2}{5}+\dfrac{7}{10}\\ x:\dfrac{10}{3}=\dfrac{12}{5}+\dfrac{7}{10}\\ x:\dfrac{10}{3}=\dfrac{31}{10}\\ x=\dfrac{31}{10}\times\dfrac{10}{3}=\dfrac{31}{3}\)
a) \(...\Rightarrow\dfrac{12}{5}-y:\dfrac{11}{4}=\dfrac{3}{2}\)
\(\Rightarrow y:\dfrac{11}{4}=\dfrac{12}{5}-\dfrac{3}{2}\Rightarrow y:\dfrac{11}{4}=\dfrac{24}{10}-\dfrac{15}{10}\)
\(\Rightarrow y:\dfrac{11}{4}=\dfrac{9}{10}\Rightarrow y=\dfrac{9}{10}x\dfrac{11}{4}=\dfrac{99}{40}\)
b) \(...\Rightarrow\dfrac{5}{4}+\dfrac{11}{5}xy=\dfrac{13}{5}\Rightarrow\dfrac{11}{5}xy=\dfrac{13}{5}-\dfrac{5}{4}\)
\(\Rightarrow\dfrac{11}{5}xy=\dfrac{52}{20}-\dfrac{25}{20}\Rightarrow\dfrac{11}{5}xy=\dfrac{27}{20}\)
\(\Rightarrow y=\dfrac{27}{20}:\dfrac{11}{5}=\dfrac{27}{20}x\dfrac{5}{11}=\dfrac{27}{44}\)
c) \(...\Rightarrow\dfrac{14}{5}-\dfrac{9}{4}:y=\dfrac{3}{4}\Rightarrow\dfrac{9}{4}:y=\dfrac{14}{5}-\dfrac{3}{4}\)
\(\Rightarrow\dfrac{9}{4}:y=\dfrac{56}{20}-\dfrac{15}{20}\Rightarrow\dfrac{9}{4}:y=\dfrac{39}{20}\)
\(\Rightarrow y=\dfrac{9}{4}:\dfrac{39}{20}\Rightarrow y=\dfrac{9}{4}x\dfrac{20}{39}=\dfrac{15}{13}\)
d) \(...\Rightarrow x:\dfrac{10}{3}=\dfrac{12}{5}+\dfrac{7}{10}\Rightarrow x:\dfrac{10}{3}=\dfrac{24}{10}+\dfrac{7}{10}\)
\(\Rightarrow x:\dfrac{10}{3}=\dfrac{31}{10}\Rightarrow x=\dfrac{31}{10}x\dfrac{10}{3}=\dfrac{31}{3}\)
Tìm x,y,z biết:
a, x : y : z = 10 : 3 : 4 và x + 2y - 3z = -20
b, \(\dfrac{x}{2}\) = \(\dfrac{y}{3}\) và \(\dfrac{y}{5}\) = \(\dfrac{z}{4}\) và x - y + z = -49
c, \(\dfrac{x}{2}\)= \(\dfrac{y}{3}\) =\(\dfrac{z}{4}\) và xy + \(z^2\)= 88
d, \(\dfrac{x}{5}\)= \(\dfrac{y}{7}\) = \(\dfrac{z}{3}\) và \(x^2\) + \(y^2\) + \(z^2\) = 415
Giải hộ mk nha
Tìm y
a) 3 \(\dfrac{1}{5}\) : 2\(\dfrac{1}{3}\): y = \(\dfrac{12}{7}\)
b) 3 : y x 3 \(\dfrac{1}{2}\)= \(\dfrac{2}{3}x\dfrac{3}{4}\)
c) \(3\dfrac{2}{3}-y+1\dfrac{3}{4}=2\)
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 = \(\dfrac{4}{5}\)
b) 3 : y x \(\dfrac{7}{2}\)= \(\dfrac{1}{2}\)
3 : y = \(\dfrac{1}{2}:\dfrac{7}{2}\)
3 : y = \(\dfrac{1}{7}\)
y = 3 : \(\dfrac{1}{7}\)
y = 21
bài 1:
2 : y x \(\dfrac{3}{5}\) = \(\dfrac{9}{10}\) \(\dfrac{5}{4}-\dfrac{2}{5}:\) y = 1 \(\dfrac{3}{4}x\) (\(\dfrac{7}{2}\) - y) =\(\dfrac{3}{2}\)
2: y \(\times\) \(\dfrac{3}{5}\) = \(\dfrac{9}{10}\)
2:y = \(\dfrac{9}{10}\) : \(\dfrac{3}{5}\)
2: y = \(\dfrac{3}{2}\)
y = 2 : \(\dfrac{3}{2}\)
y = \(\dfrac{4}{3}\)
\(\dfrac{5}{4}\) - \(\dfrac{2}{5}\) : y = 1
\(\dfrac{2}{5}\) : y = \(\dfrac{5}{4}\) - 1
\(\dfrac{2}{5}\): y = \(\dfrac{1}{4}\)
y = \(\dfrac{2}{5}\) : \(\dfrac{1}{4}\)
y = \(\dfrac{8}{5}\)
\(\dfrac{3}{4}\) \(\times\) ( \(\dfrac{7}{2}\) - y) = \(\dfrac{3}{2}\)
\(\dfrac{7}{2}\) - y = \(\dfrac{3}{2}\) : \(\dfrac{3}{4}\)
\(\dfrac{7}{2}\) - y = 2
y = \(\dfrac{7}{2}\) - 2
y = \(\dfrac{3}{2}\)
Bài 3: (Đề 2) Tìm y
a) \(2\dfrac{2}{5}:\) y x \(1\dfrac{3}{4}=\dfrac{7}{8}\) b)\(3\dfrac{2}{5}:y:1\dfrac{1}{4}=2\dfrac{3}{5}\) c) \(\dfrac{12}{5}-2\dfrac{2}{5}x\) y \(=1\dfrac{1}{4}\)
\(a,2\dfrac{2}{5}:y\times1\dfrac{3}{4}=\dfrac{7}{8}\\ \dfrac{12}{5}:y\times\dfrac{7}{4}=\dfrac{7}{8}\\ \dfrac{12}{5}:y=\dfrac{7}{8}:\dfrac{7}{4}\\ \dfrac{12}{5}:y=\dfrac{1}{2}\\ y=\dfrac{12}{5}:\dfrac{1}{2}=\dfrac{24}{5}\\ b,3\dfrac{2}{5}:y:1\dfrac{1}{4}=2\dfrac{3}{5}\\ \dfrac{17}{5}:y:\dfrac{5}{4}=\dfrac{13}{5}\\ y:\dfrac{5}{4}=\dfrac{17}{5}:\dfrac{13}{5}\\ y:\dfrac{5}{4}=\dfrac{17}{13}\\ y=\dfrac{17}{13}\times\dfrac{5}{4}=\dfrac{85}{52}\)
\(c,\dfrac{12}{5}-2\dfrac{2}{5}\times y=1\dfrac{1}{4}\\ \dfrac{12}{5}-\dfrac{12}{5}\times y=\dfrac{5}{4}\\ \dfrac{12}{5}\times y=\dfrac{12}{5}-\dfrac{5}{4}\\ \dfrac{12}{5}\times y=\dfrac{23}{20}\\ y=\dfrac{23}{20}:\dfrac{12}{5}\\ y=\dfrac{23}{48}\)
a, 2\(\dfrac{2}{5}\): y \(\times\)1\(\dfrac{3}{4}\) = \(\dfrac{7}{8}\)
\(\dfrac{12}{5}\) : y \(\times\dfrac{7}{4}\) = \(\dfrac{7}{8}\)
\(\dfrac{12}{5}\) : y = \(\dfrac{7}{8}\) : \(\dfrac{7}{4}\)
\(\dfrac{12}{5}\) : y = \(\dfrac{1}{2}\)
y = \(\dfrac{12}{5}\) : \(\dfrac{1}{2}\)
y = \(\dfrac{24}{5}\)
b, 3\(\dfrac{2}{5}\): y : 1\(\dfrac{1}{4}\) = 2\(\dfrac{3}{5}\)
\(\dfrac{17}{5}\): y: \(\dfrac{5}{4}\) = \(\dfrac{13}{5}\)
\(\dfrac{17}{5}\):y = \(\dfrac{13}{5}\times\dfrac{5}{4}\)
\(\dfrac{17}{5}\) : y = \(\dfrac{13}{4}\)
y = \(\dfrac{17}{5}\) : \(\dfrac{13}{4}\)
y = \(\dfrac{68}{65}\)
c, \(\dfrac{12}{5}\) - 2\(\dfrac{2}{5}\)\(\times y\) = 1\(\dfrac{1}{4}\)
\(\dfrac{12}{5}\) - \(\dfrac{12}{5}\)\(\times\)y = \(\dfrac{5}{4}\)
\(\dfrac{12}{5}\times y\) = \(\dfrac{12}{5}\) - \(\dfrac{5}{4}\)
\(\dfrac{12}{5}\) \(\times\) y = \(\dfrac{23}{20}\)
\(y\) = \(\dfrac{23}{20}\): \(\dfrac{12}{5}\)
y = \(\dfrac{23}{48}\)
Rút gọn biểu thức:
A = \([(32)^{\dfrac{2}{3}}]^{\dfrac{-2}{5}}\)
B= \(\dfrac{x^{-2}+y^{-2}}{x^{-1}+y^{-1}}\)
C = \((a^{\dfrac{1}{3}}-b^{\dfrac{2}{3}})(a^{\dfrac{2}{3}}+a^{\dfrac{1}{3}}×b^{\dfrac{4}{3}}+b^{\dfrac{4}{9}})\)
D = \((x+y^\dfrac{3}{2}÷\sqrt{x})^\dfrac{2}{3}÷[\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}}+\dfrac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}]^\dfrac{2}{3}\)
E = \([\dfrac{1}{x^{\dfrac{1}{2}}-4x^{\dfrac{-1}{2}}}-\dfrac{2\sqrt[3]{x}}{x\sqrt[3]{x}-4\sqrt[3]{x}}]^{-2}-\sqrt{x^2+8x+16}\)
\(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{4}\)và 2x+y-z=81
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{2}\)và 5x-y+3z=124
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\)và x.y.z=810
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{6}\)và\(x^2.y^2.z^2=288^2\)
a.
Đặt \(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{4}=k\Rightarrow\left\{{}\begin{matrix}x=5k\\y=3k\\z=4k\end{matrix}\right.\)
Thế vào \(2x+y-z=81\)
\(\Rightarrow2.5k+3k-4k=81\)
\(\Rightarrow9k=81\)
\(\Rightarrow k=9\)
\(\Rightarrow\left\{{}\begin{matrix}x=5k=45\\y=3k=27\\z=4k=36\end{matrix}\right.\)
b.
Đặt \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{2}=k\Rightarrow\left\{{}\begin{matrix}x=3k\\y=5k\\z=2k\end{matrix}\right.\)
Thế vào \(5x-y+3z=124\)
\(\Rightarrow5.3k-5k+3.2k=124\)
\(\Rightarrow16k=124\)
\(\Rightarrow k=\dfrac{31}{4}\) \(\Rightarrow\left\{{}\begin{matrix}x=3k=\dfrac{93}{4}\\y=5k=\dfrac{155}{4}\\z=2k=\dfrac{31}{2}\end{matrix}\right.\)
c.
Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)
Thế vào \(xyz=810\)
\(\Rightarrow2k.3k.5k=810\)
\(\Rightarrow k^3=27\)
\(\Rightarrow k=3\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k=6\\y=3k=9\\z=5k=15\end{matrix}\right.\)
d.
Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{6}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=6k\end{matrix}\right.\)
Thế vào \(x^2y^2z^2=288^2\)
\(\Rightarrow\left(2k\right)^2.\left(3k\right)^2.\left(6k\right)^2=288^2\)
\(\Rightarrow\left(k^2\right)^3=64\)
\(\Rightarrow k^2=4\)
\(\Rightarrow k=\pm2\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k=4\\y=3k=6\\z=6k=12\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=2k=-4\\y=3k=-6\\z=6k=-12\end{matrix}\right.\)
Sos mn ơi
Tìm Y
\(\dfrac{11}{4}-y-\dfrac{2}{5}=\dfrac{1}{3}\). \(\dfrac{5}{3}x\left(\dfrac{3}{4}-y:\dfrac{2}{5}\right)=\dfrac{5}{8}\)
\(\dfrac{3}{2}xy-\dfrac{3}{4}xy+y=\dfrac{4}{5}\)
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 = \(\dfrac{16}{35}\)
Tìm y
\(\dfrac{2}{5}\) X y : \(\dfrac{7}{4}=\dfrac{7}{8}\)
2\(\dfrac{2}{5}\) : y x 1\(\dfrac{1}{4}\) = 2\(\dfrac{3}{5}\)
\(\dfrac{12}{5}-1\dfrac{2}{5}x\) y = 1\(\dfrac{1}{4}\)
\(\dfrac{2}{5}\) x y : \(\dfrac{7}{4}\) = \(\dfrac{7}{8}\)
\(\dfrac{2}{5}\) x y = \(\dfrac{7}{8}\) x \(\dfrac{7}{4}\)
\(\dfrac{2}{5}\) x y = \(\dfrac{49}{32}\)
y = \(\dfrac{49}{32}\) : \(\dfrac{2}{5}\)
y = \(\dfrac{245}{64}\)
2\(\dfrac{2}{5}\): y x 1\(\dfrac{1}{4}\) = 2\(\dfrac{3}{5}\)
\(\dfrac{12}{5}\): y x \(\dfrac{5}{4}\) = \(\dfrac{13}{5}\)
\(\dfrac{12}{5}\): y = \(\dfrac{13}{5}\): \(\dfrac{5}{4}\)
\(\dfrac{12}{5}\): y = \(\dfrac{52}{25}\)
y = \(\dfrac{12}{5}\): \(\dfrac{52}{25}\)
y = \(\dfrac{15}{13}\)
\(\dfrac{12}{5}\) - 1\(\dfrac{2}{5}\) \(\times\) y = 1\(\dfrac{1}{4}\)
\(\dfrac{12}{5}\) - \(\dfrac{7}{5}\) \(\times\) y = \(\dfrac{5}{4}\)
\(\dfrac{7}{5}\) \(\times\) y = \(\dfrac{12}{5}\) - \(\dfrac{5}{4}\)
\(\dfrac{7}{5}\) \(\times\) y = \(\dfrac{23}{20}\)
y = \(\dfrac{23}{20}\) : \(\dfrac{7}{5}\)
y = \(\dfrac{23}{28}\)