1+1+9
9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = ?
= 100 nha bạn tk nha
chúc bạn học giỏi
9 + 9 + 9 + 9 +9 +9 +9 +9 +9 +9 + 1 +1 + 1 + 1 + 1 + 1 + 1 + 1 +1 +1
= (9 + 1) + (9 + 1) + (9 + 1) + (9 + 1) + (9 + 1) + (9 + 1) + (9 + 1) + (9 + 1) + (9 + 1) + (9 + 1)
= 10 + 10 + 10 + 10 + 10 +10 + 10 + 10 +10 +10
= 100
Tìm x:
(1+1+1+1+1+1+...)*x=(9+9+9+9+9+9+9+9+9)+9
(100 số 1)
(1+1+1+1+1+1+...) x X = (9+9++9+9+9+9+9+9+9) + 9
100 số 1
(1 x 100) x X = (9 x9) +9
100 x X = 81 +9
100 x X = 90
X = 90 : 100
X = 0,9
Like nha!
(1+1+1+1+1+1+...)*x=(9+9+9+9+9+9+9+9+9)+9
(100 số 1 )
(1×100)*x = (9×9)+9
100*x=81+9
100*x = 90
x= 90:100
x=0,9
(1+1+1+1+1+1+...) x X = (9+9++9+9+9+9+9+9+9) + 9
100 số 1
(1 x 100) x X = (9 x9) +9
100 x X = 81 +9
100 x X = 90
X = 90 : 100
1/3 + 5/9 1/9 x 9/3
1/7 - 1/9 1/3 : 1/7
3 : 5/9 9 + 9/3
3 x 5/9 4 - 2/4
\(\dfrac{1}{3}+\dfrac{5}{9}=\dfrac{6}{18}+\dfrac{10}{18}=\dfrac{16}{18}=\dfrac{8}{9}\)
\(\dfrac{1}{7}-\dfrac{1}{9}=\dfrac{9}{63}-\dfrac{7}{63}=\dfrac{2}{63}\)
\(3:\dfrac{5}{9}=3.\dfrac{9}{5}=\dfrac{27}{5}\)
\(3.\dfrac{5}{9}=\dfrac{15}{9}=\dfrac{5}{3}\)
\(\dfrac{1}{9}.\dfrac{9}{3}=\dfrac{1}{3}\)
\(\dfrac{1}{3}:\dfrac{1}{7}=\dfrac{7}{3}\)
\(9+\dfrac{9}{3}=9+3=12\)
\(4-\dfrac{2}{4}=4-\dfrac{1}{2}=\dfrac{7}{2}\)
\(\dfrac{1}{3}\) \(+\) \(\dfrac{5}{9}\) \(=\) \(\dfrac{3}{9}\) \(+\) \(\dfrac{5}{9}\) \(=\) \(\dfrac{3+5}{9}\) \(=\) \(\dfrac{8}{9}\)
\(\dfrac{1}{7}\) \(-\) \(\dfrac{1}{9}\) \(=\) \(\dfrac{9}{63}\) \(-\) \(\dfrac{7}{63}\) \(=\) \(\dfrac{9-7}{63}\) \(=\) \(\dfrac{2}{63}\)
\(\dfrac{1}{9}\) \(\times\) \(\dfrac{9}{3}\) \(=\) \(\dfrac{1\times9}{9\times3}\) \(=\) \(\dfrac{1}{3}\)
\(\dfrac{1}{3}\) \(\div\) \(\dfrac{1}{7}\) \(=\) \(\dfrac{1}{3}\) \(\times\) \(\dfrac{7}{1}\) \(=\) \(\dfrac{1\times7}{3\times1}\) \(=\) \(\dfrac{7}{3}\)
\(3\) \(\div\) \(\dfrac{5}{9}\) \(=\) \(\dfrac{3}{1}\) \(\div\) \(\dfrac{5}{9}\) \(=\dfrac{3}{1}\times\dfrac{9}{5}=\dfrac{3\times9}{1\times5}=\dfrac{27}{5}\)
\(3\times\dfrac{5}{9}=\dfrac{3}{1}\times\dfrac{5}{9}=\dfrac{3\times5}{1\times9}=\dfrac{5}{3}\)
\(9+\dfrac{9}{3}=\dfrac{9}{1}+\dfrac{9}{3}=\dfrac{27}{3}+\dfrac{9}{3}=\dfrac{27+9}{3}=\dfrac{36}{3}=12\)
\(4\) \(-\dfrac{2}{4}=\dfrac{4}{1}-\dfrac{2}{4}=\dfrac{16}{4}-\dfrac{2}{4}=\dfrac{14}{4}=\dfrac{7}{2}\)
1.1.1.1.11..1.1.11..1..1.1..1.1.1.1.1.1+11+1+1+1+1+1+1+1+1+1+1+1+1+1+1-9+9+9+9+9.3.0
zzzz ko trả lời thì thôi chứ đừng chửi thề nhé
bấm mt k nhanh hơn chắc con đ*
9*9*9*9*9*9*9*9*9*9*9*9+1*0+1*0+1=?????????????
9*9*9*9*9*9*9*9*9*9*9*9+1*0+1*0+1
=282429536481+1+1+1
=282429536484
1+1+1+1+1+1+1+1+1+1+1+9+9+9+
9*9*9*9*1*1*1*1*0
Tính nhanh: F= (9+1)(9^2+1)(9^4+1)(9^8+1)....(9^32+1)
Lời giải:
Sử dụng công thức $(a-1)(a+1)=a^2-1$ ta có:
$8F=(9-1)(9+1)(9^2+1)(9^4+1)(9^8+1)...(9^{32}+1)$
$=(9^2-1)(9^2+1)(9^4+1)(9^8+1)...(9^{32}+1)$
$=(9^4-1)(9^4+1)(9^8+1)...(9^{32}+1)$
$=(9^8-1)(9^8+1)...(9^{32}+1)$
$=(9^{16}-1)...(9^{32}+1)=(9^{32}-1)(9^{32}+1)=9^{64}-1$
$\Rightarrow F=\frac{9^{64}-1}{8}$
So sánh
(9+1)(9^2+1)(9^4+1)(9^8+1)(9^16+1)(9^32+1) và 9^64-1
Nhờ mọi người ạ!!!!:3
Làm em tick cho^^
Ta có \(\left(9+1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
\(=\frac{1}{8}\left(9-1\right)\left(9+1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
\(=\frac{1}{8}\left(9^2-1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
cứ như thế
\(=\frac{1}{8}\left(9^{64}-1\right)< 9^{64}-1\)=>đpcm