\(5x=3y\Rightarrow x=\dfrac{3y}{5}\)

Thay \(x=\dfrac{3y}{5}\) vào biểu thức \(x^2-y^2=-4\) ta có:

\(\left(\dfrac{3y}{5}\right)^2-y^2=-4\)

\(\dfrac{9y^2}{25}-y^2=-4\)

\(-\dfrac{16}{25}y^2=-4\)

\(y^2=-\dfrac{4}{\dfrac{-16}{25}}\)

\(y^2=\dfrac{25}{4}\)

\(\Rightarrow y=-\dfrac{5}{2};y=\dfrac{5}{2}\)

*) \(y=-\dfrac{5}{2}\Rightarrow x=\dfrac{3.\left(-\dfrac{5}{2}\right)}{5}=-\dfrac{3}{2}\)

*) \(y=\dfrac{5}{2}\Rightarrow x=\dfrac{3.\dfrac{5}{2}}{5}=\dfrac{3}{2}\)

Vậy ta được các cặp giá trị \(\left(x;y\right)\) thỏa mãn:

\(\left(-\dfrac{3}{2};-\dfrac{5}{2}\right);\left(\dfrac{3}{2};\dfrac{5}{2}\right)\)

ai biết gì đâu

Lời giải:
Áp dụng tính chất tổng 3 góc trong một tam giác bằng $180^0$

a.

$x=180^0-80^0-45^0=55^0$

b.

$y=180^0-30^0-90^0=60^0$

c.

$z=180^0-30^0-25^0=125^0$

a: Ta có: \(3x+\left(x-\frac{9}{20}\right)=-\frac{13}{40}\)

=>\(3x+x-\frac{9}{20}=-\frac{13}{40}\)

=>\(4x=-\frac{13}{40}+\frac{9}{20}=-\frac{13}{40}+\frac{18}{40}=\frac{5}{40}=\frac18\)

=>\(x=\frac18:4=\frac{1}{32}\)

b: \(x+\left(\frac14x-2,5\right)=-\frac{11}{20}\)

=>\(x+\frac14x-2,5=-\frac{11}{20}\)

=>\(1,25x=-0,55+2,5=1,95\)

=>\(x=\frac{1.95}{1.25}=\frac{195}{125}=\frac{39}{25}\)

c: \(\frac35x+\left(x+0,5\right)=-\frac{13}{15}\)

=>\(\frac35x+x+0,5=-\frac{13}{15}\)

=>\(\frac85x=-\frac{13}{15}-0,5=-\frac{26}{30}-\frac{15}{30}=-\frac{41}{30}\)

=>\(x=-\frac{41}{30}:\frac85=-\frac{41}{30}\cdot\frac58=\frac{-41}{6\cdot8}=-\frac{41}{48}\)

d: \(-\frac23x+\left(4x-\frac67\right)=\frac{9}{21}\)

=>\(-\frac23x+4x-\frac67=\frac37\)

=>\(\frac{10}{3}x=\frac37+\frac67=\frac97\)

=>\(x=\frac97:\frac{10}{3}=\frac97\cdot\frac{3}{10}=\frac{27}{70}\)

bài 11: câu a:

\(3x+\left(x-\frac{9}{20}\right)=-\frac{13}{40}\)

\(3x+x-\frac{9}{20}=-\frac{13}{40}\)

\(4x=-\frac{13}{40}+\frac{9}{20}\)

\(4x=-\frac{13}{40}+\frac{18}{40}\)

\(4x=\frac{5}{40}\)

\(4x=\frac18\)

\(x=\frac18:4=\frac18\cdot\frac14=\frac{1}{32}\)

b. \(x+\left(\frac14x-2,5\right)=-\frac{11}{20}\)

\(x+\frac14x-2,5=-\frac{11}{20}\)

\(\frac54x-2,5=-\frac{11}{20}\)

\(\frac54x=-\frac{11}{20}+2,5\)

\(\frac54x=\frac{39}{20}\)

\(x=\frac{39}{20}:\frac54=\frac{39}{20}\cdot\frac45=\frac{39}{25}\)

c. \(\frac35x+\left(x+0,5\right)=-\frac{13}{15}\)

\(\frac35x+x+0,5=-\frac{13}{15}\)

\(\frac85x+\frac12=-\frac{13}{15}\)

\(\frac85x=-\frac{13}{15}-\frac12\)

\(\frac85x=-\frac{41}{30}\)

\(x=-\frac{41}{30}:\frac85=-\frac{41}{30}\cdot\frac58=-\frac{41}{48}\)

\(d.-\frac23x+\left(4x-\frac67\right)=\frac{9}{21}\)

\(-\frac23x+4x-\frac67=\frac{9}{21}\)

\(\frac{10}{3}x=\frac97\)

\(x=\frac97:\frac{10}{3}=\frac97\cdot\frac{3}{10}=\frac{27}{70}\)

a)35/50

b)24/42

c)275/250

d)21/30