21x22x23x24x25 = 63a5600. Hay tim a
21x22x23x24x25 = 63a5600. Hãy tìm a.
vế phải có 21 và 24 chia hết cho 3 nên tích trên chia cho 9.
Để 63a5600 chia hết cho 9 thìa=7
Tìm a biết :
21 x 22 x 23 x 24 x 25 = 63a5600
a=7
nho k cho minh va chuc ban hoc gioi
a+88n=11237
hay tim a
hay tim so A= zyzt biet a-2yzt=xt
a) cho 1538+3425=S. khong lam phep tinh, hay tim gia tri cua: S-1538; S-3425
b) cho 9142-2451=D. khong lam phep tinh, hay tim gia tri cua: D+2451; 9142-D
a) \(S-1538=\left(1538+3425\right)-1538=1538+3425-1538=3425\)
\(S-3425=1538+3425-3425=1538\)
b) \(D+2451=9142-2451+2451=9142\)
\(9142-D=9142-\left(9142-2451\right)=9142-9142+2451=2451\)
a) cho 1538 + 3425 = S. khong lam phep tinh , hay tim gia tri cua S - 1538 ; S - 3425
b) cho 9142 - 2451= D. khong lam phep tinh hay tim gia tri cua D +2451 ; 9142 -D
1 , 4 , 7 , 10 , ......... m
a , tim tong cua day so
b , hay neu cach tim
hay thay a=2b vao dang thuc a+b=30 va tim a;b
cho góc nhon A biet sinA=0,8 hay tim cosA,tanA,cotA
\(\sin^2\widehat{A}+\cos^2\widehat{A}=1\Leftrightarrow\cos^2\widehat{A}=1-\dfrac{16}{25}=\dfrac{9}{25}\\ \Leftrightarrow\cos\widehat{A}=\dfrac{3}{5}\\ \tan\widehat{A}=\dfrac{\sin\widehat{A}}{\cos\widehat{A}}=\dfrac{4}{5}:\dfrac{3}{5}=\dfrac{4}{3}\\ \cot\widehat{A}=\dfrac{1}{\tan\widehat{A}}=\dfrac{3}{4}\)
\(\sin A=0,8\Rightarrow A=arcsin0,8_{ }\)
\(\Rightarrow\cos A=cos\left(arcsin0,8\right)=\dfrac{3}{5}\)
tanA=tan(arcsin0,8)=4/3
cotA=1:4/3=3/4