Các câu hỏi tương tự
Bài 1: Giải các hệ PTa) left{{}begin{matrix}dfrac{2}{x}+dfrac{3}{y-2}4dfrac{4}{x}-dfrac{1}{y-2}1end{matrix}right. b) left{{}begin{matrix}3sqrt{x}+2sqrt{y}162sqrt{x}-3sqrt{y}-11end{matrix}right. c) left{{}begin{matrix}dfrac{1}{2}left(x+2right)left(y+1right)dfrac{1}{2}xy+5dfrac{1}{3}left(x-3right)left(y-5right)dfrac{1}{3}xy-dfrac{4}{3}end{matrix}right.
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Bài 1: Giải các hệ PT
a) \(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y-2}=4\\\dfrac{4}{x}-\dfrac{1}{y-2}=1\end{matrix}\right.\) b) \(\left\{{}\begin{matrix}3\sqrt{x}+2\sqrt{y}=16\\2\sqrt{x}-3\sqrt{y}=-11\end{matrix}\right.\) c) \(\left\{{}\begin{matrix}\dfrac{1}{2}\left(x+2\right)\left(y+1\right)=\dfrac{1}{2}xy+5\\\dfrac{1}{3}\left(x-3\right)\left(y-5\right)=\dfrac{1}{3}xy-\dfrac{4}{3}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{1}{x+2}+2\sqrt{y-3}=7\\\dfrac{2}{x+2}-3\sqrt{y-3}=-7\end{matrix}\right.\)
GIẢI HPT
\(\left\{{}\begin{matrix}\dfrac{7}{\sqrt{x}-7}-\dfrac{4}{\sqrt{y}+6}=\dfrac{5}{3}\\\dfrac{5}{\sqrt{x}-7}+\dfrac{3}{\sqrt{y}+6}=2\dfrac{1}{6}\end{matrix}\right.\)
21 tháng 12 2022 lúc 21:12
\(Ghpt:\left\{{}\begin{matrix}\sqrt{3x}\left(1+\dfrac{1}{x+y}\right)=2\\\sqrt{7y}\left(1-\dfrac{1}{x+y}\right)=4\sqrt{2}\end{matrix}\right.\)
giải các hệ phương trìnhleft{{}begin{matrix}dfrac{2x+1}{4}-dfrac{y-2}{3}dfrac{1}{12}dfrac{x+5}{2}dfrac{y+7}{3}-4end{matrix}right.b2.Asqrt{3+sqrt{5}}+sqrt{7-3sqrt{5}}-sqrt{2}B3. Tìm ĐKXĐdfrac{1}{xsqrt{x}+1}-dfrac{2}{sqrt{x}+1}b4. so sánh A với 1Adfrac{sqrt{x}}{x-sqrt{x}+1}b5.tínha,sin47+2sin38-cos43-cos52b, Cdfrac{2sin^2x-1}{sin x-cos x}
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giải các hệ phương trình
\(\left\{{}\begin{matrix}\dfrac{2x+1}{4}-\dfrac{y-2}{3}=\dfrac{1}{12}\\\dfrac{x+5}{2}=\dfrac{y+7}{3}-4\end{matrix}\right.\)
b2.
\(A=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)
B3. Tìm ĐKXĐ
\(\dfrac{1}{x\sqrt{x}+1}-\dfrac{2}{\sqrt{x}+1}\)
b4. so sánh A với 1
A=\(\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\)
b5.tính
a,\(\sin47+2\sin38-\cos43-\cos52\)
b, \(C=\dfrac{2\sin^2x-1}{\sin x-\cos x}\)
\(\left\{{}\begin{matrix}\dfrac{1}{\left|x-1\right|}+\dfrac{3}{\sqrt{y-2}}=4\\\dfrac{5}{\left|x-1\right|}-\dfrac{2}{\sqrt{y-2}}=3\end{matrix}\right.\)
17 tháng 1 2022 lúc 20:22
Giải hệ phương trình:
a,\(\left\{{}\begin{matrix}\sqrt{x+y}\left(\sqrt{y}+1\right)=\sqrt{x^2+y^2}+2\\x\sqrt{y-1}+y\sqrt{x-1}=\dfrac{x^2+4y-4}{2}\end{matrix}\right.\)
b, \(\left\{{}\begin{matrix}x^3+2y^2=x^2y+2xy\\2\sqrt{x^2-2y-1}+\sqrt[3]{y^3-14}=x-2\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{3}{\left|x+2\right|}+\dfrac{1}{\sqrt{y-2}}=4\\\dfrac{2}{\left|x+2\right|}-\dfrac{1}{\sqrt{y-2}}=1\end{matrix}\right.\)
giải hệ: \(\left\{{}\begin{matrix}\dfrac{1}{x-y}+\dfrac{1}{x+y}=2\\\dfrac{2}{x+y}+\dfrac{3}{x+y}=5\end{matrix}\right.\)
giải pt: \(\sqrt{x^2-4x+7}=\sqrt{x+1}\)