$\Rightarrow$ ⇒b-1=1/2*b+1

$\Rightarrow$⇒b-1/2*b=1+1

$\Rightarrow$⇒1/2*b=2

$\Rightarrow$⇒b=2:1/2

$\Rightarrow$⇒b=4

Vậy a=4-1=3

Vậy có 3 con ngựa và 4 đứa trẻ.

a) Ta có:

\(\widehat{MAN}=\widehat{MAB}+\widehat{BAC}+\widehat{NAC}\)

=> \(\widehat{MAN}=60+60+60\)

=>\(\widehat{MAN}=180\)

=> MAN là góc bẹt

=> M, A, N thẳng hàng

b) Ta có:

+) \(\widehat{MAC}=\widehat{MAB}+\widehat{BAC}\)

=> \(\widehat{MAC}=60+60=120\) (1)

+) \(\widehat{NAB}=\widehat{NAC}+\widehat{BAC}\)

=> \(\widehat{NAB}=60+60=120\) (2)

Từ (1) và (2) => \(\widehat{MAC}=\widehat{NAB}\)

Xét tam giác MAC và BAN có:

MA=AB

\(\widehat{MAC}=\widehat{NAB}\) (cmt)

NA=AC

=> Tam giác MAB = tam giác BAN ( c-g-c)

=> BN=CM (2 cạnh tương ứng)

c) Ta có: Tam giác MAB = tam giác BAN (câu b)

=> \(\widehat{MCA}=\widehat{NBA}\)( 2 góc tương ứng)

Mặt khác ta có:

\(\widehat{NOC}=180-\left[\left(\widehat{MCA}+\widehat{ACN}\right)+\left(\widehat{ANC}-\widehat{BAN}\right)\right]\)

=> \(\widehat{NOC}=180-\left[\left(\widehat{MCA}+60\right)+\left(60-\widehat{BNA}\right)\right]\)

=> \(\widehat{NOC}=180-\widehat{MCA}+60+60-\widehat{BAN}\)

\(\Rightarrow\widehat{NOC}=180-\left(60+60\right)\)

=>\(\widehat{NOC}=120\)

Ta lại có:

\(\widehat{NOC}+\widehat{BOC}=180\) (kề bù)

=> \(120+\widehat{BOC}=180\)

=> \(\widehat{BOC}=60\)

Find the maximum value of

a) A=\(4x-x^2+3\)

b) B= \(x-x^2\)

c) C= \(2x-2x^2-5\)

d) D = \(4x-x^2-5\)

e) E = \(-x^2-6x-10\)

f) F = \(8x-x^2-20\)

g) G = \(4x-2x^2+3\)

h) H = \(12x-12^2+2\)

i) I = \(-x^2+7x+12\)

k) K= \(-3x^2-9x-25\)

Find the maximum value of:

a) \(-3x^2-9x-25\)

b) \(x-x^2\)

c) \(-x^2+7x+12\)