tìm x,y biết
\(\dfrac{\left(x-5\right)^2}{3}=\)\(\dfrac{7-y^2}{2}\)
Những câu hỏi liên quan
Tìm x,y biết :
a) \(\left|3.x-\dfrac{1}{2}\right|+\left|\dfrac{1}{4}.y+\dfrac{3}{5}\right|\)= 0
b)\(\left|\dfrac{3}{2}.x+\dfrac{1}{9}\right|+\left|\dfrac{5}{7}.y-\dfrac{1}{2}\right|\le0\)
a) \(\left|3x-\dfrac{1}{2}\right|+\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|=0\)
Do \(\left|3x-\dfrac{1}{2}\right|,\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|\ge0\forall x,y\)
\(\Rightarrow\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\\dfrac{1}{4}y+\dfrac{3}{5}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-\dfrac{12}{5}\end{matrix}\right.\)
b) \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|+\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\le0\)
Do \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|,\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\ge0\forall x,y\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{9}=0\\\dfrac{5}{7}y-\dfrac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{27}\\y=\dfrac{7}{10}\end{matrix}\right.\)
Đúng 1
Bình luận (0)
a,Tìm x,y,z biết/: \(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{4}=\dfrac{z}{5}\) và \(x^2-y^2=-16\)
b, Tìm x biết: \(\left|2x+3\right|=x+2\)
a) \(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{x^2-y^2}{4-9}=\dfrac{-16}{-5}=\dfrac{16}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=4.\dfrac{16}{5}\\y^2=9.\dfrac{16}{5}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\pm\left(2.\dfrac{4}{\sqrt[]{5}}\right)=\pm\dfrac{8\sqrt[]{5}}{5}\\y=\pm\left(3.\dfrac{4}{\sqrt[]{5}}\right)=\pm\dfrac{12\sqrt[]{5}}{5}\end{matrix}\right.\)
\(\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow z=\dfrac{5}{4}y=\dfrac{5}{4}.\left(\pm\dfrac{12\sqrt[]{5}}{5}\right)=\pm3\sqrt[]{5}\)
b) \(\left|2x+3\right|=x+2\)
\(\Rightarrow\left[{}\begin{matrix}2x+3=x+2\\2x+3=-x-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\3x=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\3x=-\dfrac{5}{3}\end{matrix}\right.\)
Đúng 1
Bình luận (0)
Đính chính
Dòng cuối \(3x=-\dfrac{5}{3}\rightarrow x=-\dfrac{5}{3}\)
Đúng 0
Bình luận (0)
Tìm x, y biết :
\(\left(x+y-2\right)^2+7=\dfrac{14}{\left|y-1\right|+\left|y-3\right|}\)
Ta có: \(\left(x+y-2\right)^2+7\ge7\Rightarrow\dfrac{14}{\left|y-1\right|+\left|y-3\right|}\ge7\)
\(\Rightarrow\left|y-1\right|+\left|y-3\right|\le2\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\left|y-1\right|=0\\\left|y-3\right|=2\end{matrix}\right.\\\left\{{}\begin{matrix}\left|y-1\right|=2\\\left|y-3\right|=0\end{matrix}\right.\\\left|y-1\right|=\left|y-3\right|=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=1\\y=3\\y=2\end{matrix}\right.\Rightarrow}\left[{}\begin{matrix}x=1\\x=-1\\x=0\end{matrix}\right.\)
Đúng 0
Bình luận (0)
Tìm y biết:
a) 2y . (y - dfrac{1}{7}) 0
b) -dfrac{2}{5}+dfrac{2}{3}y+dfrac{1}{6}ydfrac{-4}{15}
Tìm các số x thỏa mãn:
a) x.left(x-dfrac{4}{7}right)0
b)left(x-dfrac{2}{5}x^2right) 0
c)left(x-dfrac{2}{5}right).left(x+dfrac{3}{7}right)0
d)left(x-dfrac{2}{5}right).left(x+dfrac{3}{7}right).left(x+dfrac{3}{4}right)0
(* dấu . là dấu nhân nha bạn ^-^)
Đọc tiếp
Tìm y biết:
a) 2y . (y - \(\dfrac{1}{7}\)) = 0
b) \(-\dfrac{2}{5}+\dfrac{2}{3}y+\dfrac{1}{6}y=\dfrac{-4}{15}\)
Tìm các số x thỏa mãn:
a) \(x.\left(x-\dfrac{4}{7}\right)>0\)
b)\(\left(x-\dfrac{2}{5}x^2\right)< 0\)
c)\(\left(x-\dfrac{2}{5}\right).\left(x+\dfrac{3}{7}\right)>0\)
d)\(\left(x-\dfrac{2}{5}\right).\left(x+\dfrac{3}{7}\right).\left(x+\dfrac{3}{4}\right)>0\)
(* dấu . là dấu nhân nha bạn ^-^)
Bài 1:
a, \(2y.\left(y-\dfrac{1}{7}\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}2y=0\\y-\dfrac{1}{7}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=0\\y=\dfrac{1}{7}\end{matrix}\right.\)
Vậy \(y\in\left\{0;\dfrac{1}{7}\right\}\)
b, \(\dfrac{-2}{5}+\dfrac{2}{3}y+\dfrac{1}{6}y=\dfrac{-4}{15}\)
\(\Rightarrow\dfrac{5}{6}y=\dfrac{-4}{15}+\dfrac{2}{5}\)
\(\Rightarrow\dfrac{5}{6}y=\dfrac{2}{15}\)
\(\Rightarrow y=\dfrac{4}{25}\)
Vậy \(y=\dfrac{4}{25}\)
Chúc bạn học tốt!!!
Đúng 0
Bình luận (0)
Bài 1:
a, \(2y\left(y-\dfrac{1}{7}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2y=0\\y-\dfrac{1}{7}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=0\\y=\dfrac{1}{7}\end{matrix}\right.\)
Vậy...
b, \(\dfrac{-2}{5}+\dfrac{2}{3}y+\dfrac{1}{6}y=\dfrac{-4}{15}\)
\(\Rightarrow\dfrac{5}{6}y=\dfrac{2}{15}\)
\(\Rightarrow y=\dfrac{4}{25}\)
Vậy...
Bài 2:
a, \(x\left(x-\dfrac{4}{7}\right)>0\)
\(\Rightarrow\left\{{}\begin{matrix}x>0\\x-\dfrac{4}{7}>0\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x< 0\\x-\dfrac{4}{7}< 0\end{matrix}\right.\)
\(\Rightarrow x>\dfrac{4}{7}\left(x\ne0\right)\) hoặc \(x< \dfrac{4}{7}\left(x\ne0\right)\)
Vậy...
Các phần còn lại tương tự nhé
Đúng 0
Bình luận (0)
a, \(x.\left(x-\dfrac{4}{7}\right)>0\)
\(\Rightarrow\left\{{}\begin{matrix}x>0\\x-\dfrac{4}{7}>0\end{matrix}\right.\) hay \(\left\{{}\begin{matrix}x< 0\\x-\dfrac{4}{7}< 0\end{matrix}\right.\)
+, Xét \(\left\{{}\begin{matrix}x>0\\x-\dfrac{4}{7}>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>0\\x>\dfrac{4}{7}\end{matrix}\right.\Rightarrow x>\dfrac{4}{7}\)(1)
+, Xét \(\left\{{}\begin{matrix}x< 0\\x-\dfrac{4}{7}< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< 0\\x< \dfrac{4}{7}\end{matrix}\right.\Rightarrow x< 0\)(2)
Vậy \(x>\dfrac{4}{7}\) và \(x< 0\)
b, \(\left(x-\dfrac{2}{5}x^2\right)< 0\)
\(\Rightarrow x.\left(1-\dfrac{2}{5}x\right)< 0\)
\(\Rightarrow\left\{{}\begin{matrix}x>0\\1-\dfrac{2}{5}x< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>0\\\dfrac{2}{5}x>1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>0\\x>2,5\end{matrix}\right.\)
\(\Rightarrow x>2,5\)
Vậy \(x>2,5\)
c, \(\left(x-\dfrac{2}{5}\right).\left(x+\dfrac{3}{7}\right)>0\)
\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{2}{5}>0\\x+\dfrac{3}{7}>0\end{matrix}\right.\) hay \(\left\{{}\begin{matrix}x-\dfrac{2}{5}< 0\\x+\dfrac{3}{7}< 0\end{matrix}\right.\)
+, Xét \(\left\{{}\begin{matrix}x-\dfrac{2}{5}>0\\x+\dfrac{3}{7}>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>\dfrac{2}{5}\\x>\dfrac{-3}{7}\end{matrix}\right.\) \(\Rightarrow x>\dfrac{2}{5}\)
+, Xét \(\left\{{}\begin{matrix}x-\dfrac{2}{5}< 0\\x+\dfrac{3}{7}< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< \dfrac{2}{5}\\x< \dfrac{-3}{7}\end{matrix}\right.\) \(\Rightarrow x< \dfrac{-3}{7}\)
Vậy \(x>\dfrac{2}{5};x< \dfrac{-3}{7}\)
Chúc bạn học tốt!!!
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
Giải các hệ phương trình sau:
1, left{{}begin{matrix}left|x-1right|+dfrac{2}{y}2-left|x-1right|+dfrac{4}{y}1end{matrix}right.
2, left{{}begin{matrix}2left|x-1right|-dfrac{5}{y-1}-3left|x-1right|+dfrac{2}{y-1}3end{matrix}right.
3, left{{}begin{matrix}dfrac{1}{x-5}+dfrac{6}{sqrt{y}-2}2dfrac{2}{x-5}-dfrac{1}{sqrt{y}-2}-9end{matrix}right.
4, 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}}dfrac{13}{6}end{matrix}right.
Tìm đi...
Đọc tiếp
Giải các hệ phương trình sau:
1, \(\left\{{}\begin{matrix}\left|x-1\right|+\dfrac{2}{y}=2\\-\left|x-1\right|+\dfrac{4}{y}=1\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}2\left|x-1\right|-\dfrac{5}{y-1}=-3\\\left|x-1\right|+\dfrac{2}{y-1}=3\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}\dfrac{1}{x-5}+\dfrac{6}{\sqrt{y}-2}=2\\\dfrac{2}{x-5}-\dfrac{1}{\sqrt{y}-2}=-9\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}\dfrac{7}{\sqrt{x+7}}-\dfrac{4}{\sqrt{y-6}}=\dfrac{5}{3}\\\dfrac{5}{\sqrt{x+7}}+\dfrac{3}{\sqrt{y-6}}=\dfrac{13}{6}\end{matrix}\right.\)
Tìm điều kiện giúp mình nhé!
1: \(\left\{{}\begin{matrix}\left|x-1\right|+\dfrac{2}{y}=2\\-\left|x-1\right|+\dfrac{4}{y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6}{y}=3\\\left|x-1\right|=2-\dfrac{2}{y}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=2\\\left|x-1\right|=2-\dfrac{2}{2}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x\in\left\{2;0\right\}\end{matrix}\right.\)
2: \(\left\{{}\begin{matrix}2\left|x-1\right|-\dfrac{5}{y-1}=-3\\\left|x-1\right|+\dfrac{2}{y-1}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\left|x-1\right|-\dfrac{5}{y-1}=-3\\2\left|x-1\right|+\dfrac{4}{y-1}=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{9}{y-1}=-9\\\left|x-1\right|+\dfrac{2}{y-1}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=2\\\left|x-1\right|=3-\dfrac{2}{2}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x\in\left\{3;-1\right\}\end{matrix}\right.\)
3: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x-5}+\dfrac{12}{\sqrt{y}-2}=4\\\dfrac{2}{x-5}-\dfrac{1}{\sqrt{y}-2}=-9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{13}{\sqrt{y}-2}=13\\\dfrac{1}{x-5}=2-\dfrac{6}{\sqrt{y}-2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=9\\\dfrac{1}{x-5}=2-\dfrac{6}{3-2}=2-\dfrac{6}{1}=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=9\\x-5=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{19}{4}\\y=9\end{matrix}\right.\)
Đúng 0
Bình luận (0)
Tìm x,y,z biết :1) x:y:z3:5:left(-2right) và 5x-y+3z-162) dfrac{x}{2}dfrac{y}{-3};dfrac{z}{3}dfrac{y}{4} và x+y+z5,23) 2x3y;7z5y và 3x-7y+5z304) 3x4y5z và x-left(y+zright)-215) dfrac{x-1}{2}dfrac{y-2}{3}dfrac{z-3}{4} và 2x+3y-z50
Đọc tiếp
Tìm x,y,z biết :
1) \(x:y:z=3:5:\left(-2\right)\) và \(5x-y+3z=-16\)
2) \(\dfrac{x}{2}=\dfrac{y}{-3};\dfrac{z}{3}=\dfrac{y}{4}\) và \(x+y+z=5,2\)
3) \(2x=3y;7z=5y\) và \(3x-7y+5z=30\)
4) \(3x=4y=5z\) và \(x-\left(y+z\right)=-21\)
5) \(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\) và \(2x+3y-z=50\)
Giải hệ phương trình:
a)left{{}begin{matrix}dfrac{3x+2}{x-1}-dfrac{3y-1}{y+2}0dfrac{2}{x-1}+dfrac{3}{y+2}1end{matrix}right.
b)left{{}begin{matrix}dfrac{4x-5}{x+1}+dfrac{2y-3}{y-5}8dfrac{3}{x+1}-dfrac{2}{y-5}-1end{matrix}right.
c)left{{}begin{matrix}dfrac{x+y-2}{x+1}+dfrac{3-x}{y+1}dfrac{5}{4}dfrac{3left(x+y-2right)}{x+1}-dfrac{5-x+2y}{y+1}dfrac{3}{4}end{matrix}right.
d)left{{}begin{matrix}dfrac{x-y+1}{x-3}+dfrac{x+1}{y-3}dfrac{-7}{2}dfrac{2left(x-y+1right)}{x-3}-dfrac{x+y-2}{y-3}-dfrac{9}{2}...
Đọc tiếp
Giải hệ phương trình:
a)\(\left\{{}\begin{matrix}\dfrac{3x+2}{x-1}-\dfrac{3y-1}{y+2}=0\\\dfrac{2}{x-1}+\dfrac{3}{y+2}=1\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\dfrac{4x-5}{x+1}+\dfrac{2y-3}{y-5}=8\\\dfrac{3}{x+1}-\dfrac{2}{y-5}=-1\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}\dfrac{x+y-2}{x+1}+\dfrac{3-x}{y+1}=\dfrac{5}{4}\\\dfrac{3\left(x+y-2\right)}{x+1}-\dfrac{5-x+2y}{y+1}=\dfrac{3}{4}\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}\dfrac{x-y+1}{x-3}+\dfrac{x+1}{y-3}=\dfrac{-7}{2}\\\dfrac{2\left(x-y+1\right)}{x-3}-\dfrac{x+y-2}{y-3}=-\dfrac{9}{2}\end{matrix}\right.\)
e)\(\left\{{}\begin{matrix}x^2-y^2+2y=1\\\left(x+y\right)^2-2x-2y=0\end{matrix}\right.\)
f)\(\left\{{}\begin{matrix}4x^2+y^2-4xy=4\\x^2+y^2-2\left(xy+8\right)=0\end{matrix}\right.\)
ĐỀ 2:
1. Tính:
A. left(-dfrac{2}{3}right)^2+left(-dfrac{7}{8}right)+left(-dfrac{11}{12}right)
B. left(dfrac{-1}{3}right)^2:dfrac{1}{6}-2.left(dfrac{-1}{2}right)^3
C. dfrac{-1}{5}-left(dfrac{1}{2}+dfrac{3}{4}right)^2:dfrac{5}{8}
D. left|dfrac{-3}{2}+1,2right|+1dfrac{2}{3}:6
2. Tìm x, biết:
a. dfrac{2}{3}x-dfrac{1}{3}xdfrac{5}{12}
b. left(x-dfrac{12}{7}right):1dfrac{1}{5}dfrac{4}{7}
c. dfrac{2}{5}+left|x+1right|dfrac{3}{4}
3. Tìm x, y biết:
a. dfrac{x}{18}dfrac{y}{15}và x - y -30
b. 7...
Đọc tiếp
ĐỀ 2:
1. Tính:
A. \(\left(-\dfrac{2}{3}\right)^2+\left(-\dfrac{7}{8}\right)+\left(-\dfrac{11}{12}\right)\)
B. \(\left(\dfrac{-1}{3}\right)^2:\dfrac{1}{6}-2.\left(\dfrac{-1}{2}\right)^3\)\
C. \(\dfrac{-1}{5}-\left(\dfrac{1}{2}+\dfrac{3}{4}\right)^2:\dfrac{5}{8}\)
D. \(\left|\dfrac{-3}{2}+1,2\right|+1\dfrac{2}{3}:6\)
2. Tìm x, biết:
a. \(\dfrac{2}{3}x-\dfrac{1}{3}x=\dfrac{5}{12}\)
b. \(\left(x-\dfrac{12}{7}\right):1\dfrac{1}{5}=\dfrac{4}{7}\)
c. \(\dfrac{2}{5}+\left|x+1\right|=\dfrac{3}{4}\)
3. Tìm x, y biết:
a. \(\dfrac{x}{18}=\dfrac{y}{15}\)và x - y = -30
b. 7x = 9x và 10x - 8x = 68
c. \(\left(x-\dfrac{1}{2}\right)^{50}+\left(y+\dfrac{1}{3}\right)^{40}=0\)
* Lm nhanh nha
Bn tách ra đi,mỏi tay lắm luôn ik,đánh máy mà.
Đúng 1
Bình luận (1)
Giải hệ phương trình:
a) left{{}begin{matrix}dfrac{5}{x}+dfrac{6}{y}9dfrac{2}{x}-dfrac{6}{y}7end{matrix}right. c) left{{}begin{matrix}dfrac{2}{x}+dfrac{7}{y}21-dfrac{2}{x}-dfrac{5}{y}-11end{matrix}right.
b) left{{}begin{matrix}dfrac{5}{x}+dfrac{1}{y}14dfrac{8}{x}-dfrac{1}{y}-8end{matrix}right. d) left{{}begin{matrix}dfrac{9}{x}+dfrac{2}{y}22dfrac{5}{x}-dfrac{2}{y}13end{matrix}right. e) left{{}begin{matrix}dfrac{3}{x}+dfrac{5}{y}10-dfrac{3}{x}-dfrac{7}{y}8end{matrix}right.
Đọc tiếp
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) \(\left\{{}\begin{matrix}\dfrac{3}{x}+\dfrac{5}{y}=10\\-\dfrac{3}{x}-\dfrac{7}{y}=8\end{matrix}\right.\)
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
Bình luận (0)