a) y=10

b) y=7

a) x vô nghĩa

b) x=0,8;x=-0,8

d: Đặt x/2=y/5=k

=>x=2k; y=5k

Ta có: xy=10

nên k²=1

Trường hợp 1: k=1

=>x=2; y=5

Trường hợp 2: k=-1

=>x=-2; y=-5

\(a.\dfrac{-4}{7}-\dfrac{5}{13}\times\dfrac{-39}{25}+\dfrac{-1}{42}:\dfrac{-5}{6}\) 

\(=\dfrac{-4}{7}+\dfrac{3}{5}+\dfrac{1}{35}\) \(=\dfrac{1}{35}+\dfrac{1}{35}=\dfrac{2}{35}\) 

\(b.\dfrac{2}{9}\times\left[\dfrac{4}{5}:\left(\dfrac{1}{5}-\dfrac{2}{15}\right)+1\dfrac{2}{3}\right]-\dfrac{-5}{27}\) 

\(=\dfrac{2}{9}\times\left[\dfrac{4}{5}:\dfrac{1}{15}+\dfrac{5}{3}\right]-\dfrac{-5}{27}\) 

\(=\dfrac{2}{9}\times\left(12+\dfrac{5}{3}\right)-\dfrac{-5}{27}\) 

\(=\dfrac{2}{9}\times\dfrac{41}{3}-\dfrac{-5}{27}=\dfrac{82}{27}-\dfrac{-5}{27}=\dfrac{29}{9}\)

\(a.\)

\(\dfrac{x}{2}=\dfrac{y}{5}\)

Áp dụng tính chất dãy tỉ số bằng nhau : 

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{35}{7}=5\)

\(\Rightarrow x=5\cdot2=10\\ y=5\cdot5=25\)

\(b.\)

\(\dfrac{x+2}{y+10}=\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{x+2}{1}=\dfrac{y+10}{5}\)

\(\Leftrightarrow\dfrac{3x+6}{3}=\dfrac{y+10}{5}\)

Áp dụng tính chất dãy tỉ số bằng nhau : 

\(\Leftrightarrow\dfrac{3x+6}{3}=\dfrac{y+10}{5}=\dfrac{y+10-3x-6}{5-3}=\dfrac{2-4}{2}=-1\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x+6=-3\\y+10=-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-15\end{matrix}\right.\)

\(c.\)

\(\dfrac{x}{4}=\dfrac{y}{5}\)

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

Áp dụng tính chất dãy tỉ số bằng nhau : 

\(\dfrac{2x}{8}=\dfrac{y}{5}=\dfrac{2x-y}{8-5}=\dfrac{15}{3}=5\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=5\cdot8\\y=5\cdot5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=20\\y=25\end{matrix}\right.\)

a) Ta có: \(\dfrac{x}{2}=\dfrac{y}{5}\)

mà x+y=35

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{35}{7}=5\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{x}{2}=5\\\dfrac{y}{5}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=25\end{matrix}\right.\)

Vậy: (x,y)=(10;25)

b) Ta có: \(\dfrac{x+2}{y+10}=\dfrac{1}{5}\)

nên \(\dfrac{x+2}{1}=\dfrac{y+10}{5}\)

hay \(\dfrac{3x+6}{3}=\dfrac{y+10}{5}\)

mà y-3x=2 

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{3x+6}{3}=\dfrac{y+10}{5}=\dfrac{y-3x+10-6}{5-3}=\dfrac{2+4}{2}=3\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{3x+6}{3}=3\\\dfrac{y+10}{5}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+6=9\\y+10=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=3\\y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)

Vậy: (x,y)=(1;5)

c) Ta có: \(\dfrac{x}{4}=\dfrac{y}{5}\)

nên \(\dfrac{2x}{8}=\dfrac{y}{5}\)

mà 2x-y=15

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{2x}{8}=\dfrac{y}{5}=\dfrac{2x-y}{8-5}=\dfrac{15}{3}=5\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{x}{4}=5\\\dfrac{y}{5}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=20\\y=25\end{matrix}\right.\)

Vậy: (x,y)=(20;25)

Bài 4:

\(\dfrac{3}{4}+y:\dfrac{2}{5}=\dfrac{37}{16}\)

\(\Rightarrow y:\dfrac{2}{5}=\dfrac{37}{16}-\dfrac{3}{4}\)

\(\Rightarrow y:\dfrac{2}{5}=\dfrac{25}{16}\)

\(\Rightarrow y=\dfrac{2}{5}\cdot\dfrac{25}{16}\)

\(\Rightarrow y=\dfrac{5}{8}\)

________________

\(456+y:87=23987\)

\(\Rightarrow y:87=23987-456\)

\(\Rightarrow y:87=23531\)

\(\Rightarrow y=23531\cdot87\)

\(\Rightarrow y=2047197\)

a)\(\dfrac{4}{5}\times\dfrac{5}{8}:\dfrac{4}{5}\)

\(=\left(\dfrac{4}{5}:\dfrac{4}{5}\right)\times\dfrac{5}{8}\)

\(=1\times\dfrac{5}{8}=\dfrac{5}{8}\)

b)\(\dfrac{5}{6}+\left(\dfrac{1}{2}:\dfrac{3}{2}+\dfrac{4}{5}\right)\)

\(=\dfrac{5}{6}+\left(\dfrac{1}{3}+\dfrac{4}{5}\right)\)

\(=\dfrac{5}{6}+\dfrac{17}{15}\)

\(=\dfrac{59}{30}\)

Bài 2:

a) \(\dfrac{3}{4}+y:\dfrac{2}{5}=\dfrac{37}{16}\)

\(y:\dfrac{2}{5}=\dfrac{37}{16}-\dfrac{3}{4}\)

\(y:\dfrac{2}{5}=\dfrac{25}{16}\)

\(y=\dfrac{25}{16}\times\dfrac{2}{5}\)

\(y=\dfrac{5}{8}\)

b)\(456+y:87=23987\)

\(y:87=23987-456\)

\(y:87=23531\)

\(y=23531\times87\)

\(y=2047197\)

a) 4/5 x 5/8 : 4/5

= 5/8

b) 5/6 + ( 1/2 : 3/2 + 4/5)

= 5/6 + (1/3 + 4/5)

= 5/6 + 17/15

= 59/30

B4:

3/4 + y : 2/5 = 37/16

y : 2/5 = 25/16

y = 5/8.

456 + y : 87 = 23987

y : 87 = 23531

y = 2047197.

1: Ta có: \(\dfrac{x}{3}=\dfrac{y}{6}\)

mà 4x-y=42

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{3}=\dfrac{y}{6}=\dfrac{4x-y}{4\cdot3-6}=\dfrac{42}{12-6}=\dfrac{42}{6}=7\)

=>\(x=7\cdot3=21;y=6\cdot7=42\)

2: \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\)

mà x-2y+3z=33

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x-2y+3z}{2-2\cdot3+3\cdot5}=\dfrac{33}{2-6+15}=\dfrac{33}{11}=3\)

=>\(x=3\cdot2=6;y=3\cdot3=9;z=3\cdot5=15\)

3: \(\dfrac{x}{y}=\dfrac{6}{5}\)

=>\(\dfrac{x}{6}=\dfrac{y}{5}\)

mà x+y=121

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{6}=\dfrac{y}{5}=\dfrac{x+y}{6+5}=\dfrac{121}{11}=11\)

=>\(x=11\cdot6=66;y=11\cdot5=55\)

Bài 2: 

\(=\dfrac{28}{25}\cdot\dfrac{15}{7}\cdot5=\dfrac{75}{25}\cdot4=12\)

Bài 1: 

a: \(x+\dfrac{7}{8}=\dfrac{13}{2}:4=\dfrac{13}{8}\)

nên x=13/8-7/8=6/8=3/4

b: \(x:\dfrac{5}{3}=\dfrac{6}{5}-\dfrac{2}{3}=\dfrac{18-10}{15}=\dfrac{8}{15}\)

nên \(x=\dfrac{8}{15}\cdot\dfrac{5}{3}=\dfrac{8}{9}\)

Bài 3 :

\(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}\)

\(\dfrac{1}{2!}=\dfrac{1}{2.1}=1-\dfrac{1}{2}< 1\)

\(\dfrac{1}{3!}=\dfrac{1}{3.2.1}=1-\dfrac{1}{2}-\dfrac{1}{3}< 1\)

\(\dfrac{1}{4!}=\dfrac{1}{4.3.2.1}< \dfrac{1}{3!}< \dfrac{1}{2!}< 1\)

.....

\(\)\(\dfrac{1}{2023!}=\dfrac{1}{2023.2022....2.1}< \dfrac{1}{2022!}< ...< \dfrac{1}{2!}< 1\)

\(\Rightarrow\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}< 1\)

Bạn xem lại đề 2, phần mẫu của N

@Nguyễn Đức Trí: Đề bài nó như vậy mà

Bài 1:

+) \(\dfrac{7}{8}\times y=\dfrac{3}{2}+\dfrac{6}{4}=3\)

\(y=3:\dfrac{7}{8}=\dfrac{24}{7}\)

+) \(\dfrac{1}{y}\times\left(\dfrac{2}{5}+\dfrac{1}{5}\right)=\dfrac{10}{3}\)

\(\dfrac{1}{y}=\dfrac{10}{3}:\dfrac{3}{5}=\dfrac{50}{9}\)

\(y=\dfrac{9}{50}\)

Bài 2:

+) \(=\dfrac{2}{5}\times\left(\dfrac{4}{7}+\dfrac{3}{7}\right)\)

\(=\dfrac{2}{5}\times\dfrac{7}{7}=\dfrac{2}{5}\)

+) \(\dfrac{2}{9}:\dfrac{2}{3}:\dfrac{3}{9}\)

\(\dfrac{2}{9}\times\dfrac{3}{2}\times\dfrac{9}{3}=1\)

viết rứa ai mà biết