Question

Giải hệ phương trình:

a) \(\left\{{}\begin{matrix}\dfrac{3}{4}x+\dfrac{2}{3}y=16\\\dfrac{5}{2}x-\dfrac{3}{5}y=11\end{matrix}\right.\)

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

Answer 1

b)\(\left\{{}\begin{matrix}2\sqrt{3}x-2y=2\\5\sqrt{2}x+2y=\sqrt{6}\end{matrix}\right.< =>\left\{{}\begin{matrix}x\left(2\sqrt{3}+5\sqrt{2}\right)=2+\sqrt{6}\\5\sqrt{2}x+2y=\sqrt{6}\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}x=\dfrac{2+\sqrt{6}}{2\sqrt{3}+5\sqrt{2}}=\dfrac{3\sqrt{3}+2\sqrt{2}}{19}\\5\sqrt{2}.\dfrac{3\sqrt{3}+2\sqrt{2}}{19}+2y=\sqrt{6}\end{matrix}\right.< =>\left\{{}\begin{matrix}x=\dfrac{3\sqrt{3}+2\sqrt{2}}{19}\\y=\dfrac{-10+2\sqrt{6}}{19}\end{matrix}\right.\)Vậy, ..................

Answer 2

a) \(\left\{{}\begin{matrix}\dfrac{15}{8}x+\dfrac{5}{3}y=40\\\dfrac{15}{8}x-\dfrac{9}{20}y=\dfrac{33}{4}\end{matrix}\right.< =>\left\{{}\begin{matrix}\dfrac{127}{60}y=\dfrac{127}{4}\\\dfrac{15}{8}x-\dfrac{9}{20}y=\dfrac{33}{4}\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}y=15\\\dfrac{15}{8}x-\dfrac{9}{20}.15=\dfrac{33}{4}\end{matrix}\right.< =>\left\{{}\begin{matrix}y=15\\x=8\end{matrix}\right.\)

Vậy, ..........