Toán

(p) đi qua A(-1;2) 

=> 2 = (m - 2).(-1)² 

<=> m - 2 = 2

<=> m = 4

Vậy m = 4 thì (p) đi qua A(-1 ; 2) 

Gọi số học sinh nam x ; số học sinh nữ y 

Theo đề ra => HPT : \(\left\{{}\begin{matrix}x+y=24\\2x+3y=62\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=14\end{matrix}\right.\)

\(a)8/12; 3/12\)

\(b)63/72; 64/72\)

\(c)32/48; 42/48\) nhé

\(a.\dfrac{2}{3}=\dfrac{4}{12};\dfrac{1}{4}=\dfrac{3}{12}\\ b.\dfrac{7}{8}=\dfrac{63}{72};\dfrac{8}{9}=\dfrac{64}{72}\\ c.\dfrac{4}{6}=\dfrac{32}{48};\dfrac{7}{8}=\dfrac{42}{48}\)

\(a,\dfrac{2}{3}và\dfrac{1}{4}=\dfrac{2.4}{3.4}=\dfrac{8}{12};\dfrac{1}{4}=\dfrac{1.3}{4.3}=\dfrac{3}{12}\)

\(b,\dfrac{7}{8}và\dfrac{8}{9}=\dfrac{7}{8}=\dfrac{7.9}{8.9}=\dfrac{63}{72};\dfrac{8}{9}=\dfrac{8.8}{9.8}=\dfrac{64}{72}\)

\(c,\dfrac{4}{6}và\dfrac{7}{8}=\dfrac{4}{6}=\dfrac{4.4}{6.4}=\dfrac{16}{24};\dfrac{7}{8}=\dfrac{7.3}{8.3}=\dfrac{21}{24}\)

\(a,\dfrac{1}{2}\\ b,\dfrac{1}{3}\\ c,\dfrac{1}{2}\\ d,\dfrac{1}{2}\)

\(a,\dfrac{18}{36}=\dfrac{1}{2}\)

\(b,\dfrac{7}{21}=\dfrac{1}{3}\)

\(c,\dfrac{108}{204}=\dfrac{1}{2}\)

\(d,\dfrac{45}{90}=\dfrac{1}{2}\)

1/ 2   1/3       1/2       1/2

a.MD=ME

b.xét  ∆AME và ∆AMD có

AM là chung

MD=ME(theo câu a)

vì BA=BC => AD=AE

=> ∆AME = ∆AMD(c.c.c)

\(\dfrac{7}{9}=\dfrac{7.5}{9.5}=\dfrac{35}{45};\dfrac{19}{45}\)

\(\dfrac{7}{9}và\dfrac{19}{45}=\dfrac{35}{45}và\dfrac{19}{45}\)

\(f\left(x\right)=\left(25-x^2\right)\left(x^2-4x+3\right).\)

Ta có: \(25-x^2=0.\Leftrightarrow x^2=25.\Leftrightarrow\left[{}\begin{matrix}x=5.\\x=-5.\end{matrix}\right.\)

\(x^2-4x+3=0.\Leftrightarrow\left(x-3\right)\left(x-1\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=3.\\x=1.\end{matrix}\right.\)

Bảng xét dấu:

\(x\)                       \(-\infty\)          -5               1               3            5          \(+\infty\)

\(25-x^2\)                      -       0       +       |        +      |      +     0      -

\(x^2-4x+3\)               +       |       +      0       -       0      +      |     +

\(f\left(x\right)\)                          -       0       +      0       -        0      +     0      -

Vậy \(f\left(x\right)>0.\Leftrightarrow x\in\left(-5;1\right)\cup\left(3;5\right).\\ f\left(x\right)< 0.\Leftrightarrow x\in\left(-\infty;-5\right)\cup\left(1;3\right)\cup\left(5;+\infty\right).\\ f\left(x\right)=0.\Leftrightarrow x\in\left\{-5;1;3;5\right\}.\)

6kg 3g = …6003… g

6003 g

6kg 3g =6003g

\(5/2; 1/3 =>15/6; 2/6\)

\(6/7; 3/8 => 48/56; 18/48\) nhé

\(\dfrac{25}{10}=\dfrac{5\times3}{2\times3}=\dfrac{15}{6};\dfrac{25}{75}=\dfrac{1\times2}{3\times2}=\dfrac{2}{6}\\ \dfrac{48}{56}=\dfrac{6\times8}{7\times8}=\dfrac{48}{56};\dfrac{18}{48}=\dfrac{3\times7}{8\times7}=\dfrac{21}{56}\)

a.25/10=5/2=5 x 3/2 x 3=15/6

25/75=1/3=1 x 2/3 x 2=2/6

b.48/56=6/7=6 x 8/7 x 8=48/56

18/48=3/8=3 x 7/8 x 7=21/56

a.AB=AC - BC=5 - 3=2 cm

b.AB = CD

a.AB=AC - BC=5 - 3=2 cm

b.AB = CD