Câu 3 Tìm y
a.[y-2\5]/4\3=3\8
5\4-y*5\6=1\4+1\3
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câu 2:
(2x+1)3=125
(2x+1)3=53
=>2x+1=5
=>2x=5-1
=>2x=4
=>x=2
hc tốt!
bài 1 : a,ta có 3/x-1 =4/y-2=5/z-3 => x-1/3=y-2/4=z-3/5
áp dụng .... => x-1+y-2+z-3 / 3+4+5 = x+y+z-1-2-3/3+4+5 = 12/12=1
do x-1/3 = 1 => x-1 = 3 => x= 4 ( tìm y,z tương tự
Bài 1:
a) Ta có: 3/x - 1 = 4/y - 2 = 5/z - 3 => x - 1/3 = y - 2/4 = z - 3/5 áp dụng ... =>x - 1 + y - 2 + z - 3/3 + 4 + 5 = x + y + z - 1 - 2 - 3/3 + 4 + 5 = 12/12 = 1 do x - 1/3 = 1 => x - 1 = 3 => x = 4 ( tìm y, z tương tự )
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 1:
a.
$\frac{2}{3}\times \frac{x}{y}=\frac{8}{15}$
$\frac{x}{y}=\frac{8}{15}: \frac{2}{3}=\frac{4}{5}$
b.
$\frac{x}{y}: \frac{3}{4}=\frac{2}{5}$
$\frac{x}{y}=\frac{3}{4}\times \frac{2}{5}=\frac{3}{10}$
c.
$\frac{3}{5}: \frac{x}{y}=\frac{4}{7}$
$\frac{x}{y}=\frac{3}{5}: \frac{4}{7}=\frac{21}{20}$
Bài 2:
Chiều dài hình chữ nhật là:
$\frac{3}{5}: \frac{3}{4}=\frac{4}{5}$ (m)
Chu vi hình chữ nhật:
$2\times (\frac{3}{4}+\frac{4}{5})=\frac{31}{10}$ (m)
\(a,y+\dfrac{2}{3}=\dfrac{5}{2}\)
\(y=\dfrac{5}{2}-\dfrac{2}{3}\)
\(y=\dfrac{15}{6}-\dfrac{4}{6}\)
\(y=\dfrac{11}{6}\)
\(b,3\dfrac{4}{5}-y=\dfrac{18}{5}\)
\(y=3\dfrac{4}{5}-\dfrac{18}{5}\)
\(y=\dfrac{19}{5}-\dfrac{18}{5}\)
\(y=\dfrac{1}{5}\)
\(c,y-4\dfrac{5}{6}=2\dfrac{1}{6}+\dfrac{5}{6}\)
\(y-\dfrac{29}{6}=\dfrac{13}{6}+\dfrac{5}{6}\)
\(y-\dfrac{29}{6}=\dfrac{18}{6}\)
\(y=\dfrac{18}{6}+\dfrac{29}{6}\)
\(y=\dfrac{47}{6}\)
a,Ta có:
\(\dfrac{x}{y}=\dfrac{7}{4}=\dfrac{x}{7}=\dfrac{y}{4}\)
ÁP dụng tcdtsbn , ta có:
\(\dfrac{x}{7}=\dfrac{y}{4}=\dfrac{x+y}{7+4}=\dfrac{33}{11}=3\)
\(\Rightarrow\left\{{}\begin{matrix}x=21\\y=12\end{matrix}\right.\)
b,
\(\Rightarrow3.\left(x-1\right)=-24\)
\(\Rightarrow x-1=-8\)
\(\Rightarrow x=-7\)
A)\(\dfrac{x}{y}=\dfrac{7}{4}\Rightarrow\dfrac{x}{7}=\dfrac{y}{4}\)
Áp dụng t/c dtsbn ta có:
\(\dfrac{x}{7}=\dfrac{y}{4}=\dfrac{x+y}{7+4}=\dfrac{33}{11}=3\)
\(\dfrac{x}{7}=3\Rightarrow x=21\\ \dfrac{y}{4}=3\Rightarrow y=12\)
B) \(3\left(x-1\right)+5=-19\\ \Rightarrow3\left(x-1\right)=-24\\ \Rightarrow x-1=-8\\ \Rightarrow x=-7\)