K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

\(y\div\frac{6}{5}=\frac{7}{6}+2\)

\(y\div\frac{6}{5}=\frac{7}{6}+\frac{12}{6}\)

\(y\div\frac{6}{5}=\frac{19}{6}\)

\(y=\frac{19}{6}\times\frac{6}{5}\)

\(y=\frac{19}{5}\)

17 tháng 9 2023

\(a,y+\dfrac{2}{3}=\dfrac{5}{2}\)

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

\(y=\dfrac{15}{6}-\dfrac{4}{6}\)

\(y=\dfrac{11}{6}\)

\(b,3\dfrac{4}{5}-y=\dfrac{18}{5}\)

\(y=3\dfrac{4}{5}-\dfrac{18}{5}\)

\(y=\dfrac{19}{5}-\dfrac{18}{5}\)

\(y=\dfrac{1}{5}\)

\(c,y-4\dfrac{5}{6}=2\dfrac{1}{6}+\dfrac{5}{6}\)

\(y-\dfrac{29}{6}=\dfrac{13}{6}+\dfrac{5}{6}\)

\(y-\dfrac{29}{6}=\dfrac{18}{6}\)

\(y=\dfrac{18}{6}+\dfrac{29}{6}\)

\(y=\dfrac{47}{6}\)

30 tháng 5 2021

1)\(\left(x+1\right).\left(y-2\right)=0\)                                       \(\left(x,y\inℤ\right)\)

\(\Rightarrow\orbr{\begin{cases}x+1=0\\y-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\y=2\end{cases}}\)

2)\(\left(x-5\right).\left(y-7\right)=1\)

x-51-1
y-71-1
x64
y86

3)\(\left(x+4\right).\left(y-2\right)=2\)

x+412-1-2
y-221-2-1
x-3-2-5-6
y4301

4)\(\left(x-4\right).\left(y+3\right)=-3\)

x-41-13-3
y+3-33-11
x5371
y-60-4-2

5)\(\left(x+3\right).\left(y-6\right)=-4\)

x+3-11-442-2
y-64-41-1-22
x-4-2-71-1-5
y1027548

6)\(\left(x-8\right).\left(y+7\right)=5\)

x-815-1-5
y+751-5-1
x91373
y-2-6-12-8

7)\(\left(x+7\right).\left(y-3\right)=-6\)

x+7-11-66-22-33
y-36-61-13-32-2
x-8-6-13-1-9-5-10-4
y9-3426051

8)\(\left(x-6\right).\left(y+2\right)=7\)

x-617-1-7
y+271-7-1
x7135-1
y5-1-9-3

ok :)

25 tháng 7 2023

Bài 3 :

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

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

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

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

.....

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

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

25 tháng 7 2023

Bạn xem lại đề 2, phần mẫu của N

11 tháng 12 2022

a:=>x+1=0 và y-2=0

=>x=-1 và y=2

b: \(\Leftrightarrow\left(x-5;y-7\right)\in\left\{\left(1;1\right);\left(-1;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(6;8\right);\left(4;6\right)\right\}\)

c: (x+4)(y-2)=2

=>\(\left(x+4;y-2\right)\in\left\{\left(1;2\right);\left(2;1\right);\left(-1;-2\right);\left(-2;-1\right)\right\}\)

hay \(\left(x,y\right)\in\left\{\left(-3;4\right);\left(-2;3\right);\left(-5;0\right);\left(-6;1\right)\right\}\)

f: =>(x-12)(y-6)=-2

=>\(\left(x-12;y-6\right)\in\left\{\left(1;-2\right);\left(-2;1\right);\left(-1;2\right);\left(2;-1\right)\right\}\)

hay \(\left(x,y\right)\in\left\{\left(13;4\right);\left(10;7\right);\left(11;8\right);\left(14;5\right)\right\}\)

21 tháng 7 2015

cầu giúp đỡ ,mik còn nhiều lắm T_T

7/6 - 3/5 : 6 

= 7/6 - 3/5 x 1/6 

= 7/6 - 1/10 

= 70/60 - 6/60 

= 64/60 

= 16/15 

2/7 x 3 - y = 5/14 

6/7 - y = 5/14 

y = 6/7 - 5/14 

y = 12/14 - 5/14 

y = 7/14

29 tháng 12 2021

\(\left\{{}\begin{matrix}\dfrac{7}{\sqrt{x}-7}-\dfrac{4}{\sqrt{y}+6}=\dfrac{5}{3}.\\\dfrac{5}{\sqrt{x}-7}+\dfrac{3}{\sqrt{y}+6}=2\dfrac{1}{6}.\end{matrix}\right.\) \(\left(x,y\ge0;x\ne49\right).\)

\(\Leftrightarrow\left\{{}\begin{matrix}7\dfrac{1}{\sqrt{x}-7}-4\dfrac{1}{\sqrt{y}+6}=\dfrac{5}{3}.\\5\dfrac{1}{\sqrt{x}-7}+3\dfrac{1}{\sqrt{y}+6}=\dfrac{13}{6}.\end{matrix}\right.\)

Đặt \(\dfrac{1}{\sqrt[]{x}-7}=a\)\(\dfrac{1}{\sqrt[]{y}+6}=b\left(a,b\ne0\right).\)

\(\Rightarrow\left\{{}\begin{matrix}7a-4b=\dfrac{5}{3}.\\5a+3b=\dfrac{13}{6}.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{3}.\\b=\dfrac{1}{6}.\end{matrix}\right.\) \(\left(TM\right).\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x}-7}=\dfrac{1}{3}.\\\dfrac{1}{\sqrt{y}+6}=\dfrac{1}{6}.\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}-7=3.\\\sqrt{y}+6=6.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=10.\\\sqrt{y}=0.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=100\left(TM\right).\\y=0\left(TM\right).\end{matrix}\right.\)

Vậy hệ phương trình có nghiệm duy nhất là: \(\left(x;y\right)=\left(100;0\right).\)