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Lời giải:
Áp dụng tính chất tổng 3 góc trong 1 tam giác bằng $180^0$
Hình 1: Hình không rõ ràng. Bạn xem lại.
Hình 2: $x+x+120^0=180^0$
$2x+120^0=180^0$
$2x=60^0$
$x=60^0:2=30^0$
Hình 3:
$2y+y+90^0=180^0$
$3y=180^0-90^0=90^0$
$y=90^0:3=30^0$
Đổi 30 phút = 0,5 giờ
Quãng sông từ A đến B dài là:
\(x\) \(\times\) 0,5 + y \(\times\) 1 = 0,5\(x\) + y (km)
Kết luận Quãng đường từ A đên B dài: 0,5\(x\) + y (km)
\(5x=3y\Rightarrow x=\dfrac{3y}{5}\)
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:
\(\left(-\dfrac{3}{2};-\dfrac{5}{2}\right);\left(\dfrac{3}{2};\dfrac{5}{2}\right)\)
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.
$z=180^0-30^0-25^0=125^0$






1; (\(\dfrac{3}{4}\))2. 42 - (\(\dfrac{1}{2}\))2 : 2 - 2\(\dfrac{3}{4}\)
= \(\dfrac{3^2}{4^2}\).42 - \(\dfrac{1}{4}\) : 2 - 2\(\dfrac{3}{4}\)
= 9 - \(\dfrac{1}{8}\) - \(\dfrac{11}{4}\)
= \(\dfrac{72}{8}\) - \(\dfrac{1}{8}-\dfrac{22}{8}\)
= \(\dfrac{71}{8}\) - \(\dfrac{22}{8}\)
= \(\dfrac{49}{8}\)
2; (\(\dfrac{3}{5}\))2.52 - (2\(\dfrac{1}{4}\))3 : (\(\dfrac{3}{4}\))3 - 3
= \(\dfrac{9}{5^2}\).52 - (\(\dfrac{9}{4}\))3 : (\(\dfrac{3}{4}\))3 - 3
= 9 - (\(\dfrac{9}{4}\) : \(\dfrac{3}{4}\))3 - 3
= 9 - 33 - 3
= 9 - 27 - 3
= 9 - (27 + 3)
= 9 - 30
= - 21
3; 25.(\(\dfrac{-1}{5}\))3 + \(\dfrac{1}{5}\) - 2.(\(\dfrac{-1}{2}\))2 - \(\dfrac{1}{2}\)
= 25.\(\dfrac{-1}{125}\) + \(\dfrac{1}{5}\) - 2.\(\dfrac{1}{4}\) - \(\dfrac{1}{2}\)
= -\(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{2}\) - \(\dfrac{1}{2}\)
= 0 - \(\dfrac{1}{2}\) - \(\dfrac{1}{2}\)
= - \(\dfrac{1}{2}\) - \(\dfrac{1}{2}\)
= -1
4; 4.(\(\dfrac{1}{2}\))3 + 3.(\(\dfrac{1}{2}\))2 - 2.(\(\dfrac{1}{2}\))0
= 4.\(\dfrac{1}{8}\) + 3.\(\dfrac{1}{4}\) - 2
= \(\dfrac{1}{2}\) + \(\dfrac{3}{4}\) - 2
= \(\dfrac{5}{4}\) - 2
= - \(\dfrac{3}{4}\)
5; (- \(\dfrac{1}{2}\))2.4 + \(\dfrac{1}{3}\).32 + (\(\dfrac{1}{2020}\))0
= \(\dfrac{1}{4}\).4 + \(3\) + 1
= 1 + 3 + 1
= 4 + 1
= 5
6; 5.(\(\dfrac{-2}{5}\))2 + 2.(\(\dfrac{-2}{5}\)) + 4.(\(\dfrac{-15}{7}\))0
= 5.\(\dfrac{4}{5^2}\) - \(\dfrac{4}{5}\) + 4.1
= \(\dfrac{4}{5}\) - \(\dfrac{4}{5}\) + 4
= 0 + 4
= 4
7; 3. (\(\dfrac{-2}{3}\))2 - 2.(\(\dfrac{-2}{3}\)) + 4.(\(\dfrac{-2}{3}\))0
= \(\dfrac{4}{3}\) + \(\dfrac{4}{3}\) + 4.1
= \(\dfrac{8}{3}\) + 4
= \(\dfrac{20}{3}\)
8; 9.(\(\dfrac{2}{3}\) )2 + 2.(\(\dfrac{-9}{2}\)) + 4.(\(-\dfrac{15}{7}\))0
= 9.\(\dfrac{4}{9}\) - 9 + 4.1
= 4 - 9 + 4
= -5 + 4
= - 1
9; (\(\dfrac{2}{3}\))3 - 4.(-1\(\dfrac{3}{4}\))2 + (- \(\dfrac{2}{3}\))3
= [(\(\dfrac{2}{3}\))3 + (-\(\dfrac{2}{3}\))3] - 4.(-\(\dfrac{7}{4}\))2
= [(\(\dfrac{2}{3}\))3 - (\(\dfrac{2}{3}\))3] - \(\dfrac{49}{4}\)
= 0 - \(\dfrac{49}{4}\)
= - \(\dfrac{49}{4}\)
10; (\(\dfrac{-1}{3}\))-1 - (\(\dfrac{-6}{7}\))0 + (\(\dfrac{1}{2}\))2 : 2
= - 3 - 1 + \(\dfrac{1}{8}\)
= -4 + \(\dfrac{1}{8}\)
= - \(\dfrac{31}{8}\)
11; 25.(\(\dfrac{-1}{5}\))2 + \(\dfrac{1}{5}\) - 9.(\(\dfrac{-1}{9}\))2 + \(\dfrac{1^{20}}{9}\)
= 1 + \(\dfrac{1}{5}\) - \(\dfrac{1}{9}\) + \(\dfrac{1}{9}\)
= \(\dfrac{6}{5}\) + (-\(\dfrac{1}{9}\) + \(\dfrac{1}{9}\))
= \(\dfrac{6}{5}\) + 0
= \(\dfrac{6}{5}\)
12; (-\(\dfrac{1}{3}\))2 + (-\(\dfrac{1}{4}\))3.64 + ( - \(\dfrac{2015}{2016}\))0
= \(\dfrac{1}{9}\) - 1 + 1
= \(\dfrac{1}{9}\) + (-1 + 1)
= \(\dfrac{1}{9}\) + 0
= \(\dfrac{1}{9}\)
13; \(\dfrac{1}{3}\) - \(\dfrac{1}{3}\) : (-\(\dfrac{2}{3}\))2 + (-3)3.(7\(\dfrac{7}{9}\) - 9\(\dfrac{2}{3}\))
= \(\dfrac{1}{3}\) - \(\dfrac{3}{4}\) - 27.(\(\dfrac{70}{9}\) - \(\dfrac{29}{3}\))
= ( \(\dfrac{1}{3}\) - \(\dfrac{3}{4}\)) - 27.(-\(\dfrac{17}{9}\))
= -\(\dfrac{5}{12}\) + 51
= \(\dfrac{607}{12}\)