Bài 1:

Tổng của 2 số là

  \(36\times2=72\) 

Số lớn là

  \(72-17=55\) 

Bài 2:

a) \(4567+y\div34=10987\) 

                 \(y\div34=10987-4567\) 

                 \(y\div34=6420\) 

                         \(y=6420\times34\) 

                         \(y=218280\) 

b) \(\dfrac{4}{3}+\dfrac{1}{2}\div y=2\) 

             \(\dfrac{1}{2}\div y=2-\dfrac{4}{3}\) 

              \(\dfrac{1}{2}\div y=\dfrac{2}{3}\) 

                      \(y=\dfrac{1}{2}\div\dfrac{2}{3}\) 

                      \(y=\dfrac{3}{4}\) 

Bài 3:

a) \(\dfrac{2}{5}\times\dfrac{2}{5}+\dfrac{9}{8}\div3=\dfrac{4}{25}+\dfrac{9}{8}\times\dfrac{1}{3}=\dfrac{4}{25}+\dfrac{3}{8}=\dfrac{107}{200}\) 

b) \(2-\left(\dfrac{1}{7}\times4+\dfrac{5}{21}\right)=2-\left(\dfrac{4}{7}+\dfrac{5}{21}\right)=2-\dfrac{17}{21}=\dfrac{25}{21}\)

Bài 1 : Gọi a là số lớn, b là số bé, theo đề bài ta có :

(a+b):2=36⇒a+b=72

mà b=17

Nên a=72-17=55

Bài 2 :

a) 4567+y:34=10987

⇒ y:34=10987-4567

⇒ y:34=6420

⇒ y=6420x34

⇒ y=218280

b) \(\dfrac{4}{3}+\dfrac{1}{2}:y=2\)

\(\Rightarrow\dfrac{1}{2}:y=2-\dfrac{4}{3}\)

\(\Rightarrow\dfrac{1}{2}:y=\dfrac{2}{3}\)

\(\Rightarrow y=\dfrac{1}{2}:\dfrac{2}{3}\)

\(\Rightarrow y=\dfrac{1}{2}x\dfrac{3}{2}\)

\(\Rightarrow y=\dfrac{3}{4}\)

Bài 3 :

\(\dfrac{2}{5}x\dfrac{2}{5}+\dfrac{9}{8}:3=\dfrac{4}{25}+\dfrac{9}{8}x\dfrac{1}{3}=\dfrac{4}{25}+\dfrac{3}{8}\)

= \(\dfrac{4x8}{25x8}+\dfrac{25x3}{25x8}=\dfrac{32}{200}+\dfrac{75}{200}=\dfrac{107}{200}\)

\(2-\left(\dfrac{1}{7}x4+\dfrac{5}{21}\right)=2-\left(\dfrac{4}{7}+\dfrac{5}{21}\right)=2-\left(\dfrac{12}{21}+\dfrac{5}{21}\right)=2-\dfrac{17}{21}=\dfrac{42}{21}-\dfrac{17}{21}=\dfrac{25}{21}\)

\(\dfrac{x}{y}=\dfrac{3}{7}.\\ \Rightarrow x=\dfrac{3}{7}y.\\ x-y=16.\\\Rightarrow\dfrac{3}{7}y-y=16.\\ \Rightarrow y=-28.\\ \Rightarrow x=-12.\)

\(\dfrac{x}{1,8}=\dfrac{y}{3,2}.\\ \Rightarrow\dfrac{x}{y}=\dfrac{1,8}{3,2}=\dfrac{9}{16}.\\ \Rightarrow x=\dfrac{9}{16}y.\\ y-x=7.\\ \Rightarrow y-\dfrac{9}{16}y=7.\\ \Leftrightarrow y=16.\\ \Leftrightarrow x=9.\)

\(\dfrac{x}{5}=\dfrac{y}{8}.\\ \Rightarrow\dfrac{x}{y}=\dfrac{5}{8}.\\ \Rightarrow x=\dfrac{5}{8}y.\\ x+2y=42.\\ \Rightarrow\dfrac{5}{8}y+2y=42.\\ \Leftrightarrow y=16.\\ \Rightarrow x=10.\)

\(\dfrac{x}{5}=\dfrac{y}{7}.\\ \Rightarrow\dfrac{x}{y}=\dfrac{5}{7}.\\ \Rightarrow x=\dfrac{5}{7}y.\\ x.y=35.\\ \Rightarrow\dfrac{5}{7}y.y=35.\\ \Leftrightarrow y^2=49.\\ \Leftrightarrow u=\pm7.\\ \Rightarrow x=\pm5.\)

Bài 1: 

a: \(M=\dfrac{2^{12}\cdot3^{10}+2^3\cdot2^9\cdot3^9\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)

\(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{11}\cdot3^{11}\cdot\left(2\cdot3-1\right)}\)

\(=\dfrac{2^{12}\cdot3^{10}\cdot6}{2^{11}\cdot3^{11}\cdot5}=\dfrac{2}{3}\cdot\dfrac{6}{5}=\dfrac{12}{15}=\dfrac{4}{5}\)

b: \(N=\left(\dfrac{-3}{4}+\dfrac{5}{13}\right)\cdot\dfrac{7}{2}-\left(\dfrac{9}{4}+\dfrac{8}{13}\right)\cdot\dfrac{7}{2}\)

\(=\dfrac{7}{2}\left(-\dfrac{3}{4}+\dfrac{5}{13}-\dfrac{9}{4}-\dfrac{8}{13}\right)\)

\(=\dfrac{7}{2}\cdot\left(-3-\dfrac{3}{13}\right)=\dfrac{7}{2}\cdot\dfrac{-42}{13}=\dfrac{-147}{13}\)

Giải:

a) \(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\) 

\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\) 

\(\Rightarrow5x\in\left\{0;\pm5;10\right\}\) 

\(\Rightarrow x\in\left\{0;\pm1;2\right\}\) 

b) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\) 

\(\Rightarrow\dfrac{4}{x-6}=\dfrac{-12}{18}\) 

\(\Rightarrow-12.\left(x-6\right)=4.18\) 

\(\Rightarrow-12x+72=72\) 

\(\Rightarrow-12x=72-72\) 

\(\Rightarrow-12x=0\) 

\(\Rightarrow x=0:-12\) 

\(\Rightarrow x=0\) 

\(\Rightarrow\dfrac{y}{24}=\dfrac{-12}{18}\) 

\(\Rightarrow y=\dfrac{-12.24}{18}=-16\) 

c) \(\dfrac{x+46}{20}=x.\dfrac{2}{5}\) 

\(\dfrac{x+46}{20}=\dfrac{2x}{5}\) 

\(\Rightarrow5.\left(x+46\right)=2x.20\) 

\(\Rightarrow5x+230=40x\) 

\(\Rightarrow5x-40x=-230\) 

\(\Rightarrow-35x=-230\) 

\(\Rightarrow x=-230:-35\) 

\(\Rightarrow x=\dfrac{46}{7}\) 

Chúc bạn học tốt!

\(a.\)

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{-14}{7}=-2\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-2\right)\cdot2=-4\\y=\left(-2\right)\cdot5=-10\end{matrix}\right.\)

\(b.\)

\(\dfrac{x}{7}=\dfrac{y}{5}=\dfrac{x-y}{7-5}=\dfrac{8}{2}=4\)

\(\Rightarrow\left\{{}\begin{matrix}x=4\cdot7=28\\y=5\cdot4=20\end{matrix}\right.\)

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{8}\)

\(\Rightarrow\dfrac{2x}{6}=\dfrac{y}{5}=\dfrac{z}{8}=\dfrac{2x+y-z}{6+5-8}=\dfrac{12}{3}=4\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{4\cdot6}{2}=12\\y=4\cdot5=20\\z=8\cdot4=32\end{matrix}\right.\)

x/3=y/5=z/8

áp dung dãy tỉ số bằng nhau

=

>x/3=y/5=z/8=2.x+y-z/2.3+5-8=12/3=4

x=12,y=20,z=32

1,a/ Theo t,c dãy tỉ số bằng nhau ta có :

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{-14}{7}=2\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=-2\\\dfrac{y}{5}=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=-10\end{matrix}\right.\)

Vậy ...

b, Theo t,c dãy tỉ số bằng nhau ta có :

\(\dfrac{x}{7}=\dfrac{y}{5}=\dfrac{x-y}{7-5}=\dfrac{8}{2}=4\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{7}=4\\\dfrac{y}{5}=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=28\\y=20\end{matrix}\right.\)

Vậy ...

2/a, Theo t,c dãy tỉ số bằng nhau ta có :

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{x+y+z}{2+5+7}=\dfrac{56}{14}=4\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=4\\\dfrac{y}{5}=4\\\dfrac{z}{7}=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=10\\z=28\end{matrix}\right.\)

Vậy ...

b/ \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{8}\)

\(\Leftrightarrow\dfrac{2x}{6}=\dfrac{y}{5}=\dfrac{z}{8}\)

Theo t,c dãy tỉ số bằng nhau ta có :

\(\dfrac{2x}{6}=\dfrac{y}{5}=\dfrac{z}{8}=\dfrac{2x+y-z}{6+5-8}=\dfrac{12}{3}=4\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2x}{6}=4\\\dfrac{y}{5}=4\\\dfrac{z}{8}=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=24\\y=20\\z=32\end{matrix}\right.\)

Vậy ..

Bài Giải:

Bài 1:

a) Theo đề bài, ta có:

\(\dfrac{x}{2}=\dfrac{y}{5}\)và x+y=-4

Áp dụng tính chất của dãy tỉ số bằng nhau

Ta có: \(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{-14}{7}=-2\)

Suy ra: x = 2 . (-2) =-4

y = 5 . (-2) =-10

Vậy: x = -4 và y = -10

Mấy câu sau cậu cứ dựa vào bài trên để giải nhé!

Tick cho Phong nhé:>

Yêu nhiều>3

#Phong_419

Bài 1
a) \(\dfrac{x}{2}=\dfrac{y}{5}\)
\(\text{Mà }x+y=-14\)
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{-14}{7}=-2\)
\(\Rightarrow\dfrac{x}{2}=-2\Rightarrow x=2.\left(-2\right)=-4\)
\(\Rightarrow\dfrac{y}{5}=-2\Rightarrow y=5.\left(-2\right)=-10\)
\(\text{Vậy }\left\{{}\begin{matrix}x=-4\\y=-10\end{matrix}\right.\)
b) \(\dfrac{x}{7}=\dfrac{y}{5}\)
\(\text{Mà }x-y=8\)
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{7}=\dfrac{y}{5}=\dfrac{x-y}{7-5}=\dfrac{8}{2}=4\)
\(\Rightarrow\dfrac{x}{7}=4\Rightarrow x=4.7=28\)
\(\Rightarrow\dfrac{y}{5}=4\Rightarrow y=4.5=20\)
\(\text{Vậy }\left\{{}\begin{matrix}x=28\\y=20\end{matrix}\right.\)
Bài 2:
a) \(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}\)
\(\text{Mà }x+y+z=56\)
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{x+y+z}{2+5+7}=\dfrac{56}{14}=4\)
\(\Rightarrow\dfrac{x}{2}=4\Rightarrow x=4.2=8\)
\(\Rightarrow\dfrac{y}{5}=4\Rightarrow y=4.5=20\)
\(\Rightarrow\dfrac{z}{7}=4\Rightarrow z=4.7=28\)
\(\text{Vậy }\left\{{}\begin{matrix}x=8\\y=20\\z=28\end{matrix}\right.\)
b) \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{8}\)
\(\Rightarrow\dfrac{2x}{6}=\dfrac{y}{5}=\dfrac{z}{8}\)
\(\text{Mà }2x+y-z=12\)
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\dfrac{2x}{6}=\dfrac{y}{5}=\dfrac{z}{8}=\dfrac{2x+y-z}{6+5-8}=\dfrac{12}{3}=4\)
\(\Rightarrow\dfrac{2x}{6}=4\Rightarrow2x=4.6=24\Rightarrow x=24:2=12\)
\(\Rightarrow\dfrac{y}{5}=4\Rightarrow y=4.5=20\)
\(\Rightarrow\dfrac{z}{8}=4\Rightarrow z=4.8=32\)
\(\text{Vậy }\left\{{}\begin{matrix}x=12\\y=20\\z=32\end{matrix}\right.\)