$\left[{}\begin{matrix}\\\end{matrix}\right.\dfrac{-5}{12} \dfrac{6}{11}$ ] [ $\dfrac{7}{18} \dfrac{5}{11} \dfrac{5}{12}$]

E2

★彡℘é✿ทợท彡★ 2

21 tháng 1 2022

$\left[{}\begin{matrix}\\\end{matrix}\right.\dfrac{-5}{12}+\dfrac{6}{11}$ ] +[ $\dfrac{7}{18}+\dfrac{5}{11}+\dfrac{5}{12}$]

#Toán lớp 6

1

NL

Nguyễn Lê Phước Thịnh

21 tháng 1 2022

$=\dfrac{-5}{12}+\dfrac{5}{12}+\dfrac{6}{11}+\dfrac{5}{11}+\dfrac{7}{18}=1+\dfrac{7}{18}=\dfrac{25}{18}$

Đúng(0)

H

Học24

8 tháng 1 2018

Giải hệ phương trình: a) $\left\{{}\begin{matrix}\dfrac{5}{x}+\dfrac{6}{y}=9\\\dfrac{2}{x}-\dfrac{6}{y}=7\end{matrix}\right.$ c) $\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{7}{y}=21\\-\dfrac{2}{x}-\dfrac{5}{y}=-11\end{matrix}\right.$ b) $\left\{{}\begin{matrix}\dfrac{5}{x}+\dfrac{1}{y}=14\\\dfrac{8}{x}-\dfrac{1}{y}=-8\end{matrix}\right.$ d) $\left\{{}\begin{matrix}\dfrac{9}{x}+\dfrac{2}{y}=22\\\dfrac{5}{x}-\dfrac{2}{y}=13\end{matrix}\right.$ e)...

Đọc tiếp

#Toán lớp 9

1

T

thuongnguyen

8 tháng 1 2018

a) ĐK xác định : x≠0;y≠0

ta có : $\left\{{}\begin{matrix}\dfrac{5}{x}+\dfrac{6}{y}=9\\\dfrac{2}{x}-\dfrac{6}{y}=7\end{matrix}\right.$ <=> $\left\{{}\begin{matrix}\dfrac{7}{x}=16\\\dfrac{2}{x}-\dfrac{6}{y}=7\end{matrix}\right.< =>\left\{{}\begin{matrix}x=\dfrac{7}{16}\\y=-\dfrac{42}{17}\end{matrix}\right.$

Vậy S = {($\dfrac{7}{16};-\dfrac{42}{17}$)}

b) Đk xác định : x≠0;y≠0

ta có : $\left\{{}\begin{matrix}\dfrac{5}{x}+\dfrac{1}{y}=14\\\dfrac{8}{x}-\dfrac{1}{y}=-8\end{matrix}\right.< =>\left\{{}\begin{matrix}\dfrac{13}{x}=6\\\dfrac{5}{x}+\dfrac{1}{y}=14\end{matrix}\right.$ <=> $\left\{{}\begin{matrix}x=\dfrac{13}{6}\\y=\dfrac{13}{152}\end{matrix}\right.$

Vậy S={($\dfrac{13}{6};\dfrac{13}{152}$)}

c) ĐK xác định : x≠0;y≠0

ta có : $\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{7}{y}=21\\-\dfrac{2}{x}-\dfrac{5}{y}=-11\end{matrix}\right.$ <=> $\left\{{}\begin{matrix}\dfrac{2}{y}=10\\\dfrac{2}{x}+\dfrac{7}{y}=21\end{matrix}\right.< =>\left\{{}\begin{matrix}y=\dfrac{1}{5}\\x=-\dfrac{1}{7}\end{matrix}\right.$

Vậy S={($-\dfrac{1}{7};\dfrac{1}{5}$)}

d) ĐK xác định : x≠0;y≠0

ta có : $\left\{{}\begin{matrix}\dfrac{9}{x}+\dfrac{2}{y}=22\\\dfrac{5}{x}-\dfrac{2}{y}=13\end{matrix}\right.$ <=> $\left\{{}\begin{matrix}\dfrac{14}{x}=35\\\dfrac{5}{x}-\dfrac{2}{y}=13\end{matrix}\right.< =>\left\{{}\begin{matrix}x=\dfrac{2}{5}\\y=-4\end{matrix}\right.$

Vậy S={(0,4;-4)}

e) ĐKXĐ : x≠0;y≠0

ta có : $\left\{{}\begin{matrix}\dfrac{3}{x}+\dfrac{5}{y}=10\\-\dfrac{3}{x}-\dfrac{7}{y}=8\end{matrix}\right.$ <=> $\left\{{}\begin{matrix}-\dfrac{2}{y}=18\\\dfrac{3}{x}+\dfrac{5}{y}=10\end{matrix}\right.< =>\left\{{}\begin{matrix}y=-\dfrac{1}{9}\\x=\dfrac{3}{55}\end{matrix}\right.$ 'Vậy....

Đúng(0)

NT

Nguyễn Thành

7 tháng 10 2021

giải hệ...

Đọc tiếp

#Toán lớp 9

2

LL

Lấp La Lấp Lánh

7 tháng 10 2021

9) $\left\{{}\begin{matrix}\dfrac{7}{2x+y}+\dfrac{4}{2x-y}=74\\\dfrac{3}{2x+y}+\dfrac{2}{2x-y}=32\end{matrix}\right.$$\Leftrightarrow\left\{{}\begin{matrix}\dfrac{21}{2x+y}+\dfrac{12}{2x-y}=222\\\dfrac{21}{2x+y}+\dfrac{14}{2x-y}=224\end{matrix}\right.$

$\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{2x-y}=2\\\dfrac{7}{2x+y}+\dfrac{4}{2x-y}=74\end{matrix}\right.$$\Leftrightarrow\left\{{}\begin{matrix}2x+y=\dfrac{1}{10}\\2x-y=1\end{matrix}\right.$

$\Leftrightarrow\left\{{}\begin{matrix}-2y=\dfrac{9}{10}\\2x+y=\dfrac{1}{10}\end{matrix}\right.$$\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{9}{20}\\x=\dfrac{11}{40}\end{matrix}\right.$

10) $\left\{{}\begin{matrix}x=2y-1\\2x-y=5\end{matrix}\right.$$\Leftrightarrow\left\{{}\begin{matrix}2x-4y=-2\\2x-y=5\end{matrix}\right.$

$\Leftrightarrow\left\{{}\begin{matrix}x=2y-1\\3y=7\end{matrix}\right.$$\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{3}\\y=\dfrac{7}{3}\end{matrix}\right.$

11) $\left\{{}\begin{matrix}3x-6=0\\2y-x=4\end{matrix}\right.$$\Leftrightarrow\left\{{}\begin{matrix}3x=6\\y=\dfrac{x+4}{2}\end{matrix}\right.$$\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.$

12) $\left\{{}\begin{matrix}2x+y=5\\x+7y=9\end{matrix}\right.$$\Leftrightarrow\left\{{}\begin{matrix}2x+y=5\\2x+14y=18\end{matrix}\right.$

$\Leftrightarrow\left\{{}\begin{matrix}2x+y=5\\13y=13\end{matrix}\right.$$\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.$

Đúng(1)

LL

Lấp La Lấp Lánh

7 tháng 10 2021

13) $\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{4}{y}=2\\\dfrac{4}{x}-\dfrac{5}{y}=3\end{matrix}\right.$(ĐKXĐ: $x,y\ne0$)

$\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x}-\dfrac{16}{y}=8\\\dfrac{12}{x}-\dfrac{15}{y}=9\end{matrix}\right.$$\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{4}{y}=2\\\dfrac{1}{y}=1\end{matrix}\right.$

$\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\y=1\left(tm\right)\end{matrix}\right.$

14) $\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.$(ĐKXĐ: $x,y\ne0$)

$\Leftrightarrow\left\{{}\begin{matrix}\dfrac{8}{x}+\dfrac{8}{y}=\dfrac{2}{3}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.$$\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{7}{y}=\dfrac{1}{3}\end{matrix}\right.$$\Leftrightarrow\left\{{}\begin{matrix}x=28\left(tm\right)\\y=21\left(tm\right)\end{matrix}\right.$

15) $\left\{{}\begin{matrix}2\sqrt{x-1}-\sqrt{y-1}=1\\\sqrt{x-1}+\sqrt{y-1}=2\end{matrix}\right.$(ĐKXĐ: $x\ge1,y\ge1$)

$\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-1}=3\\\sqrt{x-1}+\sqrt{y-1}=2\end{matrix}\right.$$\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{y-1}=1\end{matrix}\right.$

$\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-1=1\end{matrix}\right.$$\Leftrightarrow x=y=2\left(tm\right)$

Đúng(1)

L

Lizy

27 tháng 1

$\left\{{}\begin{matrix}\dfrac{1}{\left|x+3\right|}+\dfrac{4}{\left|y\right|-2}=\dfrac{11}{6}\\\dfrac{5}{\left|x+3\right|}+\dfrac{2}{\left|y\right|-2}=\dfrac{11}{6}\end{matrix}\right.$

#Toán lớp 9

1

GN

GV Nguyễn Trần Thành Đạt

Giáo viên

27 tháng 1

$Đặt:\left\{{}\begin{matrix}a=\dfrac{1}{\left|x+3\right|}\left(ĐK:x\ne-3\right)\\b=\dfrac{1}{\left|y\right|-2}\left(ĐK:y\ne\pm2\right)\end{matrix}\right.\\ Có:\left\{{}\begin{matrix}\dfrac{1}{\left|x+3\right|}+\dfrac{4}{\left|y\right|-2}=\dfrac{11}{6}\\\dfrac{5}{\left|x+3\right|}+\dfrac{2}{\left|y\right|-2}=\dfrac{11}{6}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+4b=\dfrac{11}{6}\\5a+2b=\dfrac{11}{6}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a+4b=\dfrac{11}{6}\\10a+4b=\dfrac{22}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-9a=-\dfrac{11}{6}\\a+4b=\dfrac{11}{6}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{11}{54}\\b=\dfrac{\dfrac{11}{6}-\dfrac{11}{54}}{4}=\dfrac{11}{27}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{\left|x+3\right|}=a=\dfrac{11}{54}\\\dfrac{1}{\left|y\right|-2}=b=\dfrac{11}{27}\end{matrix}\right.$

$\Leftrightarrow\left\{{}\begin{matrix}11\left|x+3\right|=54\\11\left(\left|y\right|-2\right)=27\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left|x+3\right|=\dfrac{54}{11}\\\left|y\right|=\dfrac{27}{11}+2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x+3=\dfrac{54}{11}\\x+3=\dfrac{-54}{11}\end{matrix}\right.\\\left[{}\begin{matrix}y=\dfrac{27}{11}+2\\y=-\left(\dfrac{27}{11}+2\right)\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=\dfrac{21}{11}\left(TM\right)\\x=\dfrac{-87}{11}\left(TM\right)\end{matrix}\right.\\\left[{}\begin{matrix}y=\dfrac{49}{11}\left(TM\right)\\y=-\dfrac{49}{11}\left(TM\right)\end{matrix}\right.\end{matrix}\right.\\ Vậy:\left(x;y\right)=\left\{\left(\dfrac{21}{11};\dfrac{49}{11}\right);\left(\dfrac{-87}{11};\dfrac{49}{11}\right);\left(\dfrac{21}{11};\dfrac{-49}{11}\right);\left(\dfrac{-87}{11};\dfrac{-49}{11}\right)\right\}$

Đúng(1)

NT

nguyen thi khanh nguyen

17 tháng 1 2018

Giải các hệ phương trình sau bằng cách đặt ẩn số phụ: 1) $\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.$ 2) $\left\{{}\begin{matrix}\dfrac{2}{x+2y}+\dfrac{1}{y+2x}=3\\\dfrac{4}{x+2y}-\dfrac{3}{y+2x}=1\end{matrix}\right.$ 3) $\left\{{}\begin{matrix}\dfrac{3x}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.$ 4)...

Đọc tiếp

#Toán lớp 9

6

CW

Cold Wind

17 tháng 1 2018

hỏi trước tí, bạn biết giải cái hệ này chứ?

$\left\{{}\begin{matrix}2x+y=3\\2x-3y=1\end{matrix}\right.$

Đúng(0)

NT

nguyen thi khanh nguyen

17 tháng 1 2018

VAG

Đúng(0)

TT

Thầy Tùng Dương

VIP

28 tháng 1 2021 - olm

Câu hỏi hay

Giải các hệ phương...

Đọc tiếp

#Toán lớp 9

31

NT

Nguyễn Thị Kim Thoa 1977, GV Địa–Ho...

3 tháng 2 2021

Phương trình đâu bạn ?

Đúng(0)

VD

Vũ Diệu Linh

4 tháng 2 2021

$x = 144,$ $y = 36$ .

Đúng(0)

H

Học24

8 tháng 1 2018

Giải hệ phương trình : a) $\left\{{}\begin{matrix}4\dfrac{1}{x}+\dfrac{1}{y}=12\\\dfrac{1}{x}+\dfrac{1}{y}=-3\end{matrix}\right.$ b) $\left\{{}\begin{matrix}5\dfrac{1}{x}+2\dfrac{1}{y}=6\\2\dfrac{1}{x}-\dfrac{1}{y}=3\end{matrix}\right.$ c) $\left\{{}\begin{matrix}3\dfrac{1}{x}-6\dfrac{1}{y}=2\\\dfrac{1}{x}-\dfrac{1}{y}=5\end{matrix}\right.$ d) $\left\{{}\begin{matrix}\dfrac{1}{x}-4\dfrac{1}{y}=5\\2\dfrac{1}{x}-3\dfrac{1}{y}=1\end{matrix}\right.$ e)...

Đọc tiếp

#Toán lớp 9

1

T

thuongnguyen

8 tháng 1 2018

$a.\left\{{}\begin{matrix}4\dfrac{1}{x}+\dfrac{1}{y}=12\\\dfrac{1}{x}+\dfrac{1}{y}=-3\end{matrix}\right.$ (1)

ĐK xác định : x≠0 ; y≠0

Đặt ẩn phụ : a = $\dfrac{1}{x}$ ; b = $\dfrac{1}{y}$

Thay vào (1) ta được :

$\left\{{}\begin{matrix}4a+b=12\\a+b=-3\end{matrix}\right.$ <=> $\left\{{}\begin{matrix}3a=15\\a+b=-3\end{matrix}\right.< =>\left\{{}\begin{matrix}a=5\\b=-8\end{matrix}\right.$ <=> $\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=-\dfrac{1}{8}\end{matrix}\right.$

Vậy S = {($\dfrac{1}{5};-\dfrac{1}{8}$)}

$b.\left\{{}\begin{matrix}5\dfrac{1}{x}+2\dfrac{1}{y}=6\\2\dfrac{1}{x}-\dfrac{1}{y}=3\end{matrix}\right.$ (2)

ĐK xác định : x≠0 ; y≠0

Đặt ẩn phụ : a = 1/x ; b = 1/y

Thay vào (2) ta được : $\left\{{}\begin{matrix}5a+2b=6\\2a-b=3\end{matrix}\right.$ <=> $\left\{{}\begin{matrix}5a+2b=6\\4a-2b=6\end{matrix}\right.< =>\left\{{}\begin{matrix}9a=12\\2a-b=3\end{matrix}\right.$ <=> $\left\{{}\begin{matrix}a=\dfrac{4}{3}\\b=-\dfrac{1}{3}\end{matrix}\right.$ <=> $\left\{{}\begin{matrix}x=\dfrac{3}{4}\\y=-3\end{matrix}\right.$

Vậy S = {($\dfrac{3}{4};-3$ )}

c) $\left\{{}\begin{matrix}3\dfrac{1}{x}-6\dfrac{1}{y}=2\\\dfrac{1}{x}-\dfrac{1}{y}=5\end{matrix}\right.$

ĐK xác định : x≠0 ; y ≠0

Áp dụng quy tác cộng đại số ta có :

$\left\{{}\begin{matrix}3\dfrac{1}{x}-6\dfrac{1}{y}=2\\\dfrac{1}{x}-\dfrac{1}{y}=5\end{matrix}\right.< =>\left\{{}\begin{matrix}3\dfrac{1}{x}-6\dfrac{1}{y}=2\\3\dfrac{1}{x}-3\dfrac{1}{y}=15\end{matrix}\right.$ <=> $\left\{{}\begin{matrix}-3\dfrac{1}{y}=-13\\\dfrac{1}{x}-\dfrac{1}{y}=5\end{matrix}\right.< =>\left\{{}\begin{matrix}y=\dfrac{3}{13}\\x=\dfrac{3}{28}\end{matrix}\right.$

Vậy S = {($\dfrac{3}{28};\dfrac{3}{13}$)}

d) $\left\{{}\begin{matrix}\dfrac{1}{x}-4\dfrac{1}{y}=5\\2\dfrac{1}{x}-3\dfrac{1}{y}=1\end{matrix}\right.$

ĐK xác định : x≠0 ; y≠0

áp dụng quy tắc cộng đại số ta có :

$\left\{{}\begin{matrix}\dfrac{1}{x}-4\dfrac{1}{y}=5\\2\dfrac{1}{x}-3\dfrac{1}{y}=1\end{matrix}\right.< =>\left\{{}\begin{matrix}2\dfrac{1}{x}-8\dfrac{1}{y}=10\\2\dfrac{1}{x}-3\dfrac{1}{y}=1\end{matrix}\right.$ <=> $\left\{{}\begin{matrix}-5\dfrac{1}{y}=9\\\dfrac{1}{x}-4\dfrac{1}{y}=5\end{matrix}\right.< =>\left\{{}\begin{matrix}y=-\dfrac{5}{9}\\x=-\dfrac{5}{11}\end{matrix}\right.$

Vậy S = {($-\dfrac{5}{11};-\dfrac{5}{9}$)}

e) ĐK xác định x≠0 ; y≠0

$\left\{{}\begin{matrix}\dfrac{1}{x}-3\dfrac{1}{y}=4\\6\dfrac{1}{x}-\dfrac{1}{y}=2\end{matrix}\right.< =>\left\{{}\begin{matrix}\dfrac{1}{x}-3\dfrac{1}{y}=4\\18\dfrac{1}{x}-3\dfrac{1}{y}=6\end{matrix}\right.$ <=> $\left\{{}\begin{matrix}-17\dfrac{1}{x}=-2\\\dfrac{1}{x}-3\dfrac{1}{y}=4\end{matrix}\right.$ <=>$\left\{{}\begin{matrix}x=\dfrac{17}{2}\\y=-\dfrac{17}{22}\end{matrix}\right.$

Vậy S={($\dfrac{17}{2};-\dfrac{17}{22}$)}

Đúng(0)

PT

Phạm Thị Thu Trang

6 tháng 12 2018

Câu 1:Thực hiện phép tính(tính một cách hợp lí nếu có thể): a)$\dfrac{1}{2}-\dfrac{-3}{6}=+\dfrac{5}{3}-\dfrac{9}{12}$ b)$\begin{matrix}&\left(\dfrac{-2}{3}\right)\end{matrix}.\dfrac{3}{11}+\left(\dfrac{-16}{9}\right):\dfrac{11}{3}$ c)\(\begin{matrix}&\left(\dfrac{2}{3}\right)^0\end{matrix}-\sqrt{9+}\left(-\dfrac{^{...

Đọc tiếp

#Toán lớp 7

1