Giải hệ phương trình sau:
\(\left\{{}\begin{matrix}\dfrac{8}{3}a+\dfrac{b}{3}=0,1\\\dfrac{8}{3}.3b+\dfrac{b}{3}=0,1\end{matrix}\right.\)
Giải các hệ phương trình sau bằng phương pháp thế:
a)\(\left\{{}\begin{matrix}3x-2y=11\\4x-5y=3\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\dfrac{x}{2}-\dfrac{y}{3}=1\\5x-8y=3\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}3x+5y=1\\2x-y=-8\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}\dfrac{x}{y}=\dfrac{2}{3}\\x+y-10=0\end{matrix}\right.\)
a: \(\left\{{}\begin{matrix}3x-2y=11\\4x-5y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x=11+2y\\4x-5y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y+\dfrac{11}{3}\\4\left(\dfrac{2}{3}y+\dfrac{11}{3}\right)-5y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y+\dfrac{11}{3}\\\dfrac{8}{3}y+\dfrac{44}{3}-5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}y+\dfrac{11}{3}\\-\dfrac{7}{3}y=3-\dfrac{44}{3}=-\dfrac{35}{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=5\\x=\dfrac{2}{3}\cdot5+\dfrac{11}{3}=\dfrac{10}{3}+\dfrac{11}{3}=\dfrac{21}{3}=7\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}\dfrac{x}{2}-\dfrac{y}{3}=1\\5x-8y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}+1\\5x-8y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}y+2\\5\left(\dfrac{2}{3}y+2\right)-8y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y+2\\\dfrac{10}{3}y+10-8y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{14}{3}y=3-10=-7\\x=\dfrac{2}{3}y+2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=7:\dfrac{14}{3}=7\cdot\dfrac{3}{14}=\dfrac{3}{2}\\x=\dfrac{2}{3}\cdot\dfrac{3}{2}+2=3\end{matrix}\right.\)
c: \(\left\{{}\begin{matrix}3x+5y=1\\2x-y=-8\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=2x+8\\3x+5\left(2x+8\right)=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2x+8\\3x+10x+40=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=2x+8\\13x=-39\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-3\\y=2\cdot\left(-3\right)+8=8-6=2\end{matrix}\right.\)
d: \(\left\{{}\begin{matrix}\dfrac{x}{y}=\dfrac{2}{3}\\x+y-10=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y\\x+y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{3}y+y=10\\x=\dfrac{2}{3}y\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{5}{3}y=10\\x=\dfrac{2}{3}y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=6\\x=\dfrac{2}{3}\cdot6=4\end{matrix}\right.\)
a,\(\left\{{}\begin{matrix}\dfrac{y}{2}-\dfrac{x+y}{5}=0,1\\\dfrac{y}{5}-\dfrac{x-y}{2}=0,1\end{matrix}\right.\)
b,\(\left\{{}\begin{matrix}x+y=140\\x-\dfrac{x}{8}=y+\dfrac{x}{8}\end{matrix}\right.\)
Giải hệ phương trình sau:
a. \(\left\{{}\begin{matrix}\dfrac{5}{\sqrt{x-2}}+\sqrt{3-y}=8\\\dfrac{2}{\sqrt{x-2}}+3\sqrt{3-y}=11\end{matrix}\right.\)
b. \(\left\{{}\begin{matrix}\dfrac{5}{\sqrt{x}-2}+\sqrt{3-y}=8\\\dfrac{2}{\sqrt{x}-2}+3\sqrt{3-y}=11\end{matrix}\right.\)
c. \(\left\{{}\begin{matrix}3\sqrt{2x-1}+\dfrac{4}{2-\sqrt{y}}=10\\5\sqrt{2x-1}-\dfrac{8}{2-\sqrt{y}}=2\end{matrix}\right.\)
Giải các hệ phương trình sau bằng phương pháp cộng đại số:
a) \(\left\{{}\begin{matrix}\sqrt{2}x-y=3\\x+\sqrt{2}y=\sqrt{2}\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\dfrac{x}{2}-2y=\dfrac{3}{4}\\2x+\dfrac{y}{3}=-\dfrac{1}{3}\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\dfrac{2x-3y}{4}-\dfrac{x+y-1}{5}=2x-y-1\\\dfrac{x+y-1}{3}+\dfrac{4x-y-2}{4}=\dfrac{2x-y-3}{6}\end{matrix}\right.\)
a) Ta có: \(\left\{{}\begin{matrix}\sqrt{2}x-y=3\\x+\sqrt{2}y=\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2}x-y=3\\\sqrt{2}x+2y=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-3y=1\\x+\sqrt{2}y=\sqrt{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{3}\\x=\sqrt{2}-\sqrt{2}y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{3}\\x=\sqrt{2}-\sqrt{2}\cdot\dfrac{-1}{3}=\dfrac{4\sqrt{2}}{3}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{4\sqrt{2}}{3}\\y=-\dfrac{1}{3}\end{matrix}\right.\)
b) Ta có: \(\left\{{}\begin{matrix}\dfrac{x}{2}-2y=\dfrac{3}{4}\\2x+\dfrac{y}{3}=-\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-8y=3\\2x+\dfrac{1}{3}y=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{25}{3}y=\dfrac{10}{3}\\2x-8y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{2}{5}\\2x=3+8y=3+8\cdot\dfrac{-2}{5}=-\dfrac{1}{5}\end{matrix}\right.\)
hay \(\left\{{}\begin{matrix}x=-\dfrac{1}{10}\\y=-\dfrac{2}{5}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-\dfrac{1}{10}\\y=-\dfrac{2}{5}\end{matrix}\right.\)
c) Ta có: \(\left\{{}\begin{matrix}\dfrac{2x-3y}{4}-\dfrac{x+y-1}{5}=2x-y-1\\\dfrac{x+y-1}{3}+\dfrac{4x-y-2}{4}=\dfrac{2x-y-3}{6}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5\left(2x-3y\right)}{20}-\dfrac{4\left(x+y-1\right)}{20}=\dfrac{20\left(2x-y-1\right)}{20}\\\dfrac{4\left(x+y-1\right)}{12}+\dfrac{3\left(4x-y-2\right)}{12}=\dfrac{2\left(2x-y-3\right)}{12}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}10x-15y-4x-4y+4=40x-20y-20\\4x+4y-4+12x-3y-6=4x-2y-6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x-19y+4-40x+20y+20=0\\16x+y-10-4x+2y+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-34x+y=-24\\12x+3y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-102x+3y=-72\\12x+3y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-114x=-76\\12x+3y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\12\cdot\dfrac{2}{3}+3y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\3y=4-8=-4\end{matrix}\right.\)
hay \(\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-\dfrac{4}{3}\end{matrix}\right.\)
giải hệ phương trình
\(\left\{{}\begin{matrix}\dfrac{x+y}{xy}=\dfrac{3}{8}\\\dfrac{y+x}{yz}=\dfrac{3}{4}\\\dfrac{x+z}{xz}=\dfrac{5}{6}\end{matrix}\right.\)
Lời giải:
HPT \(\Leftrightarrow \left\{\begin{matrix}
\frac{1}{x}+\frac{1}{y}=\frac{3}{8}\\
\frac{1}{y}+\frac{1}{z}=\frac{3}{4}\\
\frac{1}{z}+\frac{1}{x}=\frac{5}{6}\end{matrix}\right.\Rightarrow 2(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=\frac{3}{8}+\frac{3}{4}+\frac{5}{6}\)
\(\Leftrightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{47}{48}\)
\(\Rightarrow \left\{\begin{matrix} \frac{1}{z}=\frac{47}{48}-\frac{3}{8}\\ \frac{1}{x}=\frac{47}{48}-\frac{3}{4}\\ \frac{1}{y}=\frac{47}{48}-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=\frac{48}{29}\\ y=\frac{48}{11}\\ z=\frac{48}{7}\end{matrix}\right.\)
giải hệ phương trình
\(\left\{{}\begin{matrix}\dfrac{8}{\sqrt{x}-3}+\dfrac{11}{|2y-1|}=5\\\dfrac{4}{\sqrt{x}-3}+\dfrac{1}{|1-2y|}=3\end{matrix}\right.\)
Đặt 1/căn x-3=a; 1/|2y-1|=b
Theo đề, ta có; 8a+11b=5 và 4a+b=3
=>a=7/9; b=-1/9
=>|2y-1|=-9(loại)
=>Hệ vô nghiệm
Giải hệ phương trình sau bằng phương pháp cộng đại số:
a) \(\left\{{}\begin{matrix}49x+7y=-1\\\dfrac{-4}{3}x-2y=\dfrac{4}{3}\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}4x+3y=13\\5x-3y=-31\end{matrix}\right.\)
a) Ta có: \(\left\{{}\begin{matrix}49x+7y=-1\\-\dfrac{4}{3}x-2y=\dfrac{4}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}98x+14y=-2\\-\dfrac{28}{3}x-14y=\dfrac{28}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{266}{3}x=\dfrac{22}{3}\\49x+7y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{133}\\49\cdot\dfrac{11}{133}+7y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{133}\\7y=-1-\dfrac{77}{19}=-\dfrac{96}{19}\end{matrix}\right.\)
hay \(\left\{{}\begin{matrix}x=\dfrac{11}{133}\\y=-\dfrac{96}{133}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{11}{133}\\y=-\dfrac{96}{133}\end{matrix}\right.\)
b) Ta có: \(\left\{{}\begin{matrix}4x+3y=13\\5x-3y=-31\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}9x=-18\\4x+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\3y=13-4x=13-4\cdot\left(-2\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\3y=21\end{matrix}\right.\)
hay \(\left\{{}\begin{matrix}x=-2\\y=7\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-2\\y=7\end{matrix}\right.\)
Giải hệ phương trình
\(\left\{{}\begin{matrix}3\left|x-1\right|+\dfrac{2}{y-1}=8\\\left|x-1\right|-\dfrac{3}{y-1}=-1\end{matrix}\right.\)
=>3|x-1|+2/y-1=8 và 3|x-1|=9/y-1=-3
=>11/y-1=11 và |x-1|-3/y-1=-1
=>y-1=1 và |x-1|=2
=>y=2 và (x-1=2 hoặc x-1=-2)
=>y=2 và (x=3 hoặc x=-1)
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....