tim cap so x,y biet:
\(\dfrac{1+3.y}{12}\)\(=\dfrac{1+5.y}{5.x}\)\(=\dfrac{1+7.y}{4.x}\)
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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 = \(\dfrac{245}{64}\)
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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 = \(\dfrac{15}{13}\)
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\(\dfrac{12}{5}\) - 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 = \(\dfrac{23}{28}\)
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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 \(=1\dfrac{1}{4}\)
\(a,2\dfrac{2}{5}:y\times1\dfrac{3}{4}=\dfrac{7}{8}\\ \dfrac{12}{5}:y\times\dfrac{7}{4}=\dfrac{7}{8}\\ \dfrac{12}{5}:y=\dfrac{7}{8}:\dfrac{7}{4}\\ \dfrac{12}{5}:y=\dfrac{1}{2}\\ y=\dfrac{12}{5}:\dfrac{1}{2}=\dfrac{24}{5}\\ b,3\dfrac{2}{5}:y:1\dfrac{1}{4}=2\dfrac{3}{5}\\ \dfrac{17}{5}:y:\dfrac{5}{4}=\dfrac{13}{5}\\ y:\dfrac{5}{4}=\dfrac{17}{5}:\dfrac{13}{5}\\ y:\dfrac{5}{4}=\dfrac{17}{13}\\ y=\dfrac{17}{13}\times\dfrac{5}{4}=\dfrac{85}{52}\)
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\(c,\dfrac{12}{5}-2\dfrac{2}{5}\times y=1\dfrac{1}{4}\\ \dfrac{12}{5}-\dfrac{12}{5}\times y=\dfrac{5}{4}\\ \dfrac{12}{5}\times y=\dfrac{12}{5}-\dfrac{5}{4}\\ \dfrac{12}{5}\times y=\dfrac{23}{20}\\ y=\dfrac{23}{20}:\dfrac{12}{5}\\ y=\dfrac{23}{48}\)
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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 = \(\dfrac{23}{48}\)
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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 = \(\dfrac{4}{5}\)
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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
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Xem thêm câu trả lời
Giải các hệ phương trình sau
f.{ (2x - y) (x + 3y) = 4
{ (5x + y) (x + 3y) = 24
g.{ \(\dfrac{8x-5y-3}{7}+\dfrac{11y-4x-7}{5}=12\)
{ \(\dfrac{9x+4y-13}{5}+\dfrac{3\left(x-2\right)}{4}=15\)
h.{\(\dfrac{1}{x}+\dfrac{1}{y}=2\)
{\(\dfrac{3}{x}-\dfrac{4}{y}=-1\)
h) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=2\\\dfrac{3}{x}-\dfrac{4}{y}=-1\end{matrix}\right.\)\(\left(1\right)\)\(\left(đk:x,y\ne0\right)\)
Đặt \(a=\dfrac{1}{x},b=\dfrac{1}{y}\)
\(\left(1\right)\Leftrightarrow\) \(\left\{{}\begin{matrix}a+b=2\\3a-4b=-1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3a+3b=6\\3a-4b=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\7b=7\end{matrix}\right.\)\(\Leftrightarrow a=b=1\)
Thay a,b:
\(\Leftrightarrow\dfrac{1}{x}=\dfrac{1}{y}=1\Leftrightarrow x=y=1\left(tm\right)\)
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4 tháng 1 2022 lúc 14:32
Tìm x, y, z, t ∈ Z biết:a, dfrac{5}{x}dfrac{-10}{12} b, dfrac{4}{-6}dfrac{x+3}{9} c, dfrac{x-1}{25}dfrac{4}{x-1} d, dfrac{x+1}{y}dfrac{-3}{5}e, dfrac{-12}{6}dfrac{x}{5}dfrac{-y}{3}dfrac{Z}{-17}dfrac{-t}{-9}h, dfrac{-24}{-6}dfrac{x}{3}dfrac{4}{y^2}dfrac{Z^3}{-2}
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Tìm x, y, z, t ∈ Z biết:
a, \(\dfrac{5}{x}=\dfrac{-10}{12}\) b, \(\dfrac{4}{-6}=\dfrac{x+3}{9}\) c, \(\dfrac{x-1}{25}=\dfrac{4}{x-1}\) d, \(\dfrac{x+1}{y}=\dfrac{-3}{5}\)
e, \(\dfrac{-12}{6}=\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{Z}{-17}=\dfrac{-t}{-9}\)
h, \(\dfrac{-24}{-6}=\dfrac{x}{3}=\dfrac{4}{y^2}=\dfrac{Z^3}{-2}\)
a) \(\dfrac{5}{x}=\dfrac{-10}{12}.\Rightarrow x=-6.\)
b) \(\dfrac{4}{-6}=\dfrac{x+3}{9}.\Rightarrow x+3=-6.\Leftrightarrow x=-9.\)
c) \(\dfrac{x-1}{25}=\dfrac{4}{x-1}.\left(đk:x\ne1\right).\Leftrightarrow\dfrac{x-1}{25}-\dfrac{4}{x-1}=0.\)
\(\Leftrightarrow\dfrac{x^2-2x+1-100}{25\left(x-1\right)}=0.\Leftrightarrow x^2-2x-99=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=11.\\x=-9.\end{matrix}\right.\) \(\left(TM\right).\)
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25 tháng 7 2023 lúc 10:18
Bài 1: Tìm x; y ϵ ℤ
a) 2x - ysqrt{6} 5 + (x + 1)sqrt{6}
b) 5x + y - (2x -1)sqrt{7} ysqrt{7} + 2
Bài 2: So sánh M và N
M dfrac{dfrac{3}{4}+dfrac{3}{5}+dfrac{3}{7}-dfrac{3}{11}}{dfrac{6}{4}+dfrac{6}{5}+dfrac{6}{7}-dfrac{6}{11}}
N dfrac{dfrac{2}{3}+dfrac{2}{5}-dfrac{2}{7}-dfrac{2}{11}}{dfrac{6}{2}+dfrac{6}{5}-dfrac{6}{7}-dfrac{6}{11}}
Bài 3: Chứng minh:
dfrac{1}{2!}+dfrac{1}{3!}+dfrac{1}{4!}+...+dfrac{1}{2023!} 1
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Bài 1: Tìm x; y ϵ \(ℤ\)
a) 2x - y\(\sqrt{6}\) = 5 + (x + 1)\(\sqrt{6}\)
b) 5x + y - (2x -1)\(\sqrt{7}\) = y\(\sqrt{7}\) + 2
Bài 2: So sánh M và N
M = \(\dfrac{\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{11}}{\dfrac{6}{4}+\dfrac{6}{5}+\dfrac{6}{7}-\dfrac{6}{11}}\)
N = \(\dfrac{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}{\dfrac{6}{2}+\dfrac{6}{5}-\dfrac{6}{7}-\dfrac{6}{11}}\)
Bài 3: Chứng minh:
\(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}< 1\)
Bài 3 :
\(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}\)
\(\dfrac{1}{2!}=\dfrac{1}{2.1}=1-\dfrac{1}{2}< 1\)
\(\dfrac{1}{3!}=\dfrac{1}{3.2.1}=1-\dfrac{1}{2}-\dfrac{1}{3}< 1\)
\(\dfrac{1}{4!}=\dfrac{1}{4.3.2.1}< \dfrac{1}{3!}< \dfrac{1}{2!}< 1\)
.....
\(\)\(\dfrac{1}{2023!}=\dfrac{1}{2023.2022....2.1}< \dfrac{1}{2022!}< ...< \dfrac{1}{2!}< 1\)
\(\Rightarrow\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}< 1\)
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bài 1: giải các hệ phương trình
1)dfrac{1}{x}+dfrac{1}{y}dfrac{1}{2}
x+y9
2) dfrac{2x+1}{4}-dfrac{y-2}{3}dfrac{1}{12}
dfrac{x+5}{2}-dfrac{y+7}{3}-4
3)2|x|-y3
|x|+y3
4)2left(x+yright)+sqrt{x+1}4
left(x+yright)-3sqrt{x+1}-5
5) dfrac{7}{2x+y}+dfrac{4}{2x-y}74
dfrac{3}{2x+y}+dfrac{2}{2x-y}32
6)dfrac{1}{x}+dfrac{3}{2y+1}2
dfrac{2}{x}+dfrac{4}{2y+1}2
7) dfrac{1}{x}+dfrac{1}{y}2
dfrac{3}{x}-dfrac{1}{y}2
8)dfrac{1}{x+2}+dfrac{3}{2y-1}4
dfrac{4}{x+2}-dfrac{1}{2y-1}3
9)dfrac{4}{x+y}
+dfra...
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bài 1: giải các hệ phương trình
1)\(\dfrac{1}{x}\)+\(\dfrac{1}{y}\)=\(\dfrac{1}{2}\)
x+y=9
2) \(\dfrac{2x+1}{4}-\dfrac{y-2}{3}=\dfrac{1}{12}\)
\(\dfrac{x+5}{2}-\dfrac{y+7}{3}=-4\)
3)\(2|x|-y=3\)
\(|x|+y=3\)
4)\(2\left(x+y\right)+\sqrt{x+1}=4\)
\(\left(x+y\right)-3\sqrt{x+1}=-5\)
5) \(\dfrac{7}{2x+y}+\dfrac{4}{2x-y}=74\)
\(\dfrac{3}{2x+y}+\dfrac{2}{2x-y}=32\)
6)\(\dfrac{1}{x}+\dfrac{3}{2y+1}=2\)
\(\dfrac{2}{x}+\dfrac{4}{2y+1}=2\)
7) \(\dfrac{1}{x}+\dfrac{1}{y}=2\)
\(\dfrac{3}{x}-\dfrac{1}{y}=2\)
8)\(\dfrac{1}{x+2}+\dfrac{3}{2y-1}=4\)
\(\dfrac{4}{x+2}-\dfrac{1}{2y-1}=3\)
9)\(\dfrac{4}{x+y} +\dfrac{1}{y-1}=5\)
\(\dfrac{1}{x+y}-\dfrac{2}{y-1}=-1\)
10)\(\dfrac{7}{\sqrt{2x+3}}-\dfrac{4}{\sqrt{3}-y}=\dfrac{5}{3}\)
\(\dfrac{5}{\sqrt{2x+3}}+\dfrac{3}{\sqrt{3-y}}=\dfrac{13}{6}\)
11)\(\dfrac{3x}{x-1}-\dfrac{2}{y+2}=4\)
\(\dfrac{2x}{x-1}+\dfrac{1}{y+2}=5\)
12) \(\dfrac{7}{\sqrt{x}-7}-\dfrac{4}{\sqrt{y}+6}=\dfrac{5}{3}\)
\(\dfrac{5}{\sqrt{x}-7}+\dfrac{3}{\sqrt{y}+6}2\dfrac{1}{6}\)
13) \(3\sqrt{x-1}+2\sqrt{y}=13\)
\(2\sqrt{x-1}-\sqrt{y}=4\)
14) 6x + 6y = 5xy
\(\dfrac{4}{x}-\dfrac{3}{y}=1\)
mọi người giúp mk với 
câu 6 sai nha
sửa : \(\dfrac{1}{x}+\dfrac{3}{2y+1}=2\)
\(\dfrac{2}{x}+\dfrac{4}{2y+1}=3\)
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8 tháng 6 2021 lúc 20:08
Tìm các số nguyên x,y biết:a)dfrac{6}{2x+1}dfrac{2}{7}b) dfrac{24}{7x-3}dfrac{-4}{25}c) dfrac{4}{x-6}dfrac{y}{24}dfrac{-12}{18}d) dfrac{-1}{5}ledfrac{x}{8}ledfrac{1}{4}e) dfrac{x+46}{20}xdfrac{2}{5}f) ydfrac{5}{y}dfrac{86}{y} ( xdfrac{2}{5};ydfrac{5}{y} là các hỗn số)
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Tìm các số nguyên x,y biết:
a)\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)
b) \(\dfrac{24}{7x-3}=\dfrac{-4}{25}\)
c) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\)
d) \(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\)
e) \(\dfrac{x+46}{20}=x\dfrac{2}{5}\)
f) \(y\dfrac{5}{y}=\dfrac{86}{y}\) ( \(x\dfrac{2}{5};y\dfrac{5}{y}\) là các hỗn số)
a,\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)
⇒\(\dfrac{6}{2x+1}=\dfrac{6}{21}\)
⇒\(2x+1=21\)
\(2x=21-1\)
\(2x=20\)
⇒\(x=10\)
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Thu gọn đa thức:C=5\(x^3\)\(y^2\)+3\(x^2\)\(y^3\) \(\dfrac{1}{3}\) \(x^4\) \(y^5\) - \(\dfrac{1}{7}\)+\(\dfrac{1}{2}\) \(x^2\)\(y^3\) - 3\(x^4\) \(y^5\)+ \(\dfrac{1}{3}\)-4\(x^3\) \(y^2\)
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\(C=5x^3y^2-4x^3y^2+3x^2y^3+\dfrac{1}{2}x^2y^3+\dfrac{1}{3}x^4y^5-3x^4y^5-\dfrac{1}{7}\)
\(=x^3y^2+\dfrac{7}{2}x^2y^3-\dfrac{8}{3}x^4y^5-\dfrac{1}{7}\)
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