Giải hệ phương trình
a,\(\dfrac{3}{x-1}+\dfrac{1}{y+2}=4 Và\dfrac{2}{x-1}-\dfrac{1}{y+2}=1\)
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giải các hệ phương trìnha dfrac{5}{x-1}+dfrac{1}{y-1}10dfrac{1}{x-1}-dfrac{3}{y-1}18b dfrac{5}{x+y-3}-dfrac{2}{x-y+1}8dfrac{3}{x+y-3}+dfrac{1}{x-y+1}dfrac{3}{2}c sqrt{x-1}-3sqrt{y+2}22sqrt{x-1}+5sqrt{y+2}15d 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}e 7x^2+13y-395x^2-11y33f 2left(x-1right)^2-3y^375left(x-1right)^2+6y^34
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giải các hệ phương trình
a \(\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\)
\(\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\)
b \(\dfrac{5}{x+y-3}-\dfrac{2}{x-y+1}=8\)
\(\dfrac{3}{x+y-3}+\dfrac{1}{x-y+1}=\dfrac{3}{2}\)
c \(\sqrt{x-1}-3\sqrt{y+2}=2\)
\(2\sqrt{x-1}+5\sqrt{y+2}=15\)
d \(\dfrac{7}{\sqrt{x-7}}-\dfrac{4}{\sqrt{y+6}}=\dfrac{5}{3}\)
\(\dfrac{5}{\sqrt{x-7}}+\dfrac{3}{\sqrt{y+6}}=\dfrac{13}{6}\)
e \(7x^2+13y=-39\)
\(5x^2-11y=33\)
f \(2\left(x-1\right)^2-3y^3=7\)
\(5\left(x-1\right)^2+6y^3=4\)
a) Ta có: \(\left\{{}\begin{matrix}\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\\\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\\\dfrac{5}{x-1}-\dfrac{15}{y-1}=90\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{16}{y-1}=-80\\\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y-1=\dfrac{-1}{5}\\\dfrac{1}{x-1}=18+\dfrac{3}{y-1}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{4}{5}\\x-1=\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=\dfrac{4}{5}\end{matrix}\right.\)
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3 tháng 12 2021 lúc 20:17
Giải hệ phương trìnha)left{{}begin{matrix}x+ydfrac{x-3}{2}x+2ydfrac{2-4y}{15}end{matrix}right. b)left{{}begin{matrix}dfrac{1}{x}+dfrac{1}{y}-1dfrac{3}{x}-dfrac{2}{y}7end{matrix}right.c)left{{}begin{matrix}sqrt{x+3}-2sqrt{y+1}22sqrt{x+3}+sqrt{y+1}4end{matrix}right. d)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}2dfrac{1}{9}end{matrix}right.
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Giải hệ phương trình
a)\(\left\{{}\begin{matrix}x+y=\dfrac{x-3}{2}\\x+2y=\dfrac{2-4y}{15}\end{matrix}\right.\) b)\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=-1\\\dfrac{3}{x}-\dfrac{2}{y}=7\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}\sqrt{x+3}-2\sqrt{y+1}=2\\2\sqrt{x+3}+\sqrt{y+1}=4\end{matrix}\right.\) d)\(\left\{{}\begin{matrix}\dfrac{7}{\sqrt{x}-7}-\dfrac{4}{\sqrt{y}+6}=\dfrac{5}{3}\\\dfrac{5}{\sqrt{x}-7}+\dfrac{3}{\sqrt{y}+6}=2\dfrac{1}{9}\end{matrix}\right.\)
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}+\dfrac{3}{y}=-3\\\dfrac{3}{x}-\dfrac{2}{y}=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{y}=-10\\\dfrac{1}{x}+\dfrac{1}{y}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{2}\\x=1\end{matrix}\right.\)
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giải các hệ phương trình
a)\(\left\{{}\begin{matrix}x^2+y^2=1\\x^3+y^3=1\end{matrix}\right.\) b) \(\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{1}{y}=\dfrac{5}{12}\\x^2+y^2=1\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}x^2-xy+y^2=3\\2x^2-xy+3y^2=12\end{matrix}\right.\)
Đề bài: Giải các phương trìnha) dfrac{1}{x} - dfrac{2}{x+1} dfrac{3}{x^2+x}b) dfrac{x+2}{x-2} - dfrac{1}{x} dfrac{2}{xleft(x-2right)}c) dfrac{x-2}{x+2} - dfrac{3}{x-2} dfrac{2left(x-11right)}{x^2-4}
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Đề bài: Giải các phương trình
a) \(\dfrac{1}{x}\) - \(\dfrac{2}{x+1}\) = \(\dfrac{3}{x^2+x}\)
b) \(\dfrac{x+2}{x-2}\) - \(\dfrac{1}{x}\) = \(\dfrac{2}{x\left(x-2\right)}\)
c) \(\dfrac{x-2}{x+2}\) - \(\dfrac{3}{x-2}\) = \(\dfrac{2\left(x-11\right)}{x^2-4}\)
a) ĐKXĐ: \(x\notin\left\{0;-1\right\}\)
Ta có: \(\dfrac{1}{x}-\dfrac{2}{x+1}=\dfrac{3}{x^2+x}\)
\(\Leftrightarrow\dfrac{x+1}{x\left(x+1\right)}-\dfrac{2x}{x\left(x+1\right)}=\dfrac{3}{x\left(x+1\right)}\)
Suy ra: \(-x+1=3\)
\(\Leftrightarrow-x=2\)
hay x=-2(thỏa ĐK)
Vậy: S={-2}
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giải phương trình
a, \(\dfrac{3}{2x-1}+1=\dfrac{2x-1}{2x+1}\)
b,\(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x+3}+\dfrac{4}{x^2+2x-3}=1\)
c,\(\dfrac{5}{x^2+x-6}-\dfrac{2}{x^2+4x+3}=\dfrac{-3}{2x-1}\)
d, \(\left(x^2-4\right)\left(2x+3\right)=\left(x^2-4\right)\left(x-1\right)\)
e, \(x^3+x^2+x+1=0\)
\(a,\dfrac{3}{2x-1}+1=\dfrac{2x-1}{2x+1};ĐKXĐ:x\ne\pm\dfrac{1}{2}\\ \Leftrightarrow\dfrac{3}{2x-1}-\dfrac{2x-1}{2x+1}+1=0\\ \Leftrightarrow\dfrac{3\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}-\dfrac{\left(2x-1\right)\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}+\dfrac{\left(2x-1\right)\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}=0\\ \Rightarrow3\left(2x+1\right)-\left(2x-1\right)^2+\left(2x-1\right)\left(2x+1\right)=0\\ \Leftrightarrow6x+3-\left(4x^2-4x+1\right)+\left(4x^2-1\right)=0\\ \Leftrightarrow6x+3-4x^2+4x-1+4x^2-1=0\\ \Leftrightarrow10x+1=0\\ \Leftrightarrow10x=-1\\ \Leftrightarrow x=-\dfrac{1}{10}\)
Vậy \(x\in\left\{-\dfrac{1}{10}\right\}\)
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giải phương trình
a.(2x- 1)x x^2+ 9x (1 - 2x) = 0
b. \(\dfrac{x+4}{5}\)-x -5= \(\dfrac{x+3}{3}\)- \(\dfrac{x-2}{2}\)
c.(x- 5)x (6x+ 3)= (2x-7)x (3x + 5)
d. \(\dfrac{x+4}{5}\)-2x+ 1= \(\dfrac{x}{3}\)- \(\dfrac{2-x}{6}\)
b: =>1/4x+4/5-x-5=1/3x+1-1/2x+1
=>-3/4x+1/6x=2+5-4/5=24/5
=>x=-288/35
c: =>6x^2+3x-30x-15=6x^2+10x-21x-35
=>-27x-15=-11x-35
=>-16x=-20
=>x=5/4
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Giải hệ phương trình :
\(\left\{{}\begin{matrix}x^2+x+\dfrac{1}{y}\left(1+\dfrac{1}{y}\right)=4\\x^3+\dfrac{x}{y^2}+\dfrac{x^2}{y}+\dfrac{1}{y^3}=4\end{matrix}\right.\)
câu 3: giải hệ phương trìnha) left{{}begin{matrix}5a+b5b-10a-19end{matrix}right.b) left{{}begin{matrix}dfrac{5x}{6}-ydfrac{-5}{6}dfrac{2x}{2x+y}+3ydfrac{-2}{3}end{matrix}right.c)left{{}begin{matrix}xsqrt{3}+3y12x-ysqrt{3}sqrt{3}end{matrix}right.d) left{{}begin{matrix}dfrac{1}{x}-dfrac{6}{y}dfrac{5}{x}+dfrac{6}{y}13end{matrix}right.17giúp mk vs ạ mk cần gấp
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câu 3: giải hệ phương trình
a) \(\left\{{}\begin{matrix}5a+b=5\\b-10a=-19\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\dfrac{5x}{6}-y=\dfrac{-5}{6}\\\dfrac{2x}{2x+y}+3y=\dfrac{-2}{3}\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}x\sqrt{3}+3y=1\\2x-y\sqrt{3}=\sqrt{3}\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{6}{y}\\\dfrac{5}{x}+\dfrac{6}{y}=13\end{matrix}\right.=17\)
giúp mk vs ạ mk cần gấp
a) \(\left\{{}\begin{matrix}5a+b=5\\b-10a=-19\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}5a+b=5\\15a=24\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{8}{5}\\b=-3\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{6}{y}=17\\\dfrac{5}{x}+\dfrac{6}{y}=13\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{6}{y}=17\\\dfrac{6}{x}=30\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=-\dfrac{1}{2}\end{matrix}\right.\)
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giải các hệ phương trình sau:a) left{{}begin{matrix}dfrac{1}{x}+dfrac{1}{y+z}dfrac{1}{2}dfrac{1}{y}+dfrac{1}{z+x}dfrac{1}{3}dfrac{1}{z}+dfrac{1}{x+y}dfrac{1}{4}end{matrix}right.b) left{{}begin{matrix}dfrac{x}{y}-dfrac{y}{x}dfrac{5}{6}x^2-y^25end{matrix}right.c) left{{}begin{matrix}dfrac{5}{sqrt{x-7}}+dfrac{3}{sqrt{y+6}}dfrac{13}{6}dfrac{7}{sqrt{x-7}}-dfrac{2}{sqrt{y+6}}dfrac{5}{3}end{matrix}right.
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giải các hệ phương trình sau:
a) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y+z}=\dfrac{1}{2}\\\dfrac{1}{y}+\dfrac{1}{z+x}=\dfrac{1}{3}\\\dfrac{1}{z}+\dfrac{1}{x+y}=\dfrac{1}{4}\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\dfrac{x}{y}-\dfrac{y}{x}=\dfrac{5}{6}\\x^2-y^2=5\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\dfrac{5}{\sqrt{x-7}}+\dfrac{3}{\sqrt{y+6}}=\dfrac{13}{6}\\\dfrac{7}{\sqrt{x-7}}-\dfrac{2}{\sqrt{y+6}}=\dfrac{5}{3}\end{matrix}\right.\)