bài 1.Cho M bằng 1/2 x 3/4 x5/6 x....x 99/100 N bằng 2/3 x 4/5 x 6/7 x.....x100/101 a/So sánh M và N b/ tính M x N c/So sánh M và 1/2
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1
b;
B=1+ (7-5) + (11-9) + ...+(101-99)
B=1+2+2+..+2
B=1+25.2=51
2.
a.
ĐK : x+2 >=0 => x>=-2
\(\left|x+2\right|-x=2\\ \Rightarrow\left|x+2\right|=2+x\\ \Rightarrow\left[{}\begin{matrix}x+2=x+2\\x+2=-x-2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}0x=0\\2x=-4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}0x=0\\x=-2\end{matrix}\right.\)
Vậy x=-2
a) Ta có:
(n-1)/n < n/(n+1)
vì (n-1).(n+1)=n2-1 < n2
=>
1/2 < 2/3
3/4 < 4/5
....
99/100 < 100/101
Vậy A < B
b). Ta lại có:
A.B = 1/2 . 2/3 . 3/4 . 4/5 .... . 99/100 . 100/101 = 1/100
Mà A<B => A.A<A.B=1/100
=> A2 < 1/100
=> A < 1/10<1
3 x 25 x 8 + 4 x 6 x 37 + 2 x 38 x 12
= (3 x 8) x 25 + (4 x 6) x 37 + (2 x 12) x 38
= 24 x 25 + 24 x 37 + 24 x 38
= 24 x (25 + 37 + 38)
= 24 x 100
= 2400
3 x 25 x 8 + 4 x 6 x 37 + 2 x 38 x 12
= (3 x 8) x 25 + (4 x 6) x 37 + (2 x 12) x 38
= 24 x 25 + 24 x 37 + 24 x 38
= 24 x (25 + 37 + 38)
= 24 x 100
= 2400
Bài 2:
=(1-2-3+4)+(5-6-7+8)+...+(97-98-99+100)
=0+0+...+0
=0
Bài 3:
\(7A=\dfrac{7^{11}+7}{7^{11}+1}=1+\dfrac{6}{7^{11}+1}\)
\(7B=\dfrac{7^{12}+7}{7^{12}+1}=1+\dfrac{6}{7^{12}+1}\)
mà \(\dfrac{6}{7^{11}+1}>\dfrac{6}{7^{12}+1}\)
nên A>B
Tính nhanh mỗi biểu thức sau:
a, 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20
= (0 + 20) + (1 + 19) + (2 + 18) + (3 + 17) + (4 + 16) + (5 + 15) + (6 + 14) + (7 + 13) + (8 + 12) + (9 + 11) + 10
= 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 10
= 20 x 10 + 10
= 200 + 10
= 210
b, 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x (4 x 9 - 36)
= 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x (36 - 36)
= 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 0
= A x 0
= 0
c, (81 - 7 x 9 - 18) : (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)
= (81 - 63 - 18) : (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)
= (18 - 18) : (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)
= 0 :(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)
= 0 : A
= 0
d, (6 x 5 + 7 - 37) x (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)
= (30 + 7 - 37) x (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)
= (37 - 37) x (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)
= 0 x (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)
= 0 x A
= 0
e, (11 x 9 - 100 + 1) : (1 x 2 x 3 x 4 x ... x 10)
= (99 - 100 + 1) : (1 x 2 x 3 x 4 x ... x 10)
= (99 + 1 - 100) : (1 x 2 x 3 x 4 x ... x 10)
= (100 - 100) : (1 x 2 x 3 x 4 x ... x 10)
= 0 : (1 x 2 x 3 x 4 x ... x 10)
= 0 : A
= 0
g, (m : 1 - m x 1) : (m x 2008 + m x 2008)
= (m - m) : (m x 2008 + m x 2008)
= 0 : (m x 2008 + m x 2008)
= 0 : A
= 0
h, (2 + 4 + 6 + 8 + m x n) x (324 x 3 - 972)
= (2 + 4 + 6 + 8 + m x n) x (972 - 972)
= (2 + 4 + 6 + 8 + m x n) x 0
= A x 0
= 0
l, (1 + 2 + 3 + ... + 99) x (13 x 15 - 12 x 15 - 15)
= (1 + 2 + 3 + ... + 99) x (15 x (13 - 12 - 1))
= (1 + 2 + 3 + ... + 99) x (15 x 0)
= (1 + 2 + 3 + ... + 99) x 0
= A x 0
= 0
i, (0 x 1 x 2 x...x 99 x 100) : (2 + 4 + 6 +...+ 98)
= 0 x : (2 + 4 + 6 +...+ 98)
= 0 x A
= 0
k, (0 + 1 + 2 +...+ 97 + 99) x (45 x 3 - 45 x 2 - 45)
= (0 + 1 + 2 +...+ 97 + 99) x (45 x (3 - 2 - 4))
= (0 + 1 + 2 +...+ 97 + 99) x (45 x 0)
= (0 + 1 + 2 +...+ 97 + 99) x 0
= A x 0
= 0
\(M=\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}.....\dfrac{99}{100}\)
\(N=\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}.....\dfrac{100}{101}\)
\(M.N=\left(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}.....\dfrac{99}{100}\right).\left(\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}.....\dfrac{100}{101}\right)\)
\(M.N=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}.\dfrac{5}{6}.\dfrac{6}{7}.....\dfrac{99}{100}.\dfrac{100}{101}\)
\(M.N=\dfrac{1.2.3.4.5.6.....100}{2.3.4.5.6.....101}\)
\(M.N=\dfrac{1}{101}\)
Dấu nhân trên THCS lak dấu chấm còn Tiểu Học lak chữ "x" đó bn!
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Dấu chấm là nhân
a) \(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\) \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)
b) \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\) \(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}=1-\frac{1}{99}=\frac{98}{99}\)
c) Đặt \(C=\frac{4}{5.7}+\frac{4}{7.9}+....+\frac{4}{59.61}\)
\(\Rightarrow\frac{1}{2}C=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{59}-\frac{1}{61}\)
\(\Rightarrow\frac{1}{2}C=\frac{1}{5}-\frac{1}{61}=\frac{56}{305}\)
\(\Rightarrow C=\frac{56}{305}:\frac{1}{2}=\frac{112}{305}\)
CHÚC BẠN HỌC TỐT NHA! ĐÚNG THÌ NHA!