F = 1 + 3 mu 1 + 3mu 2 + 3 mu 3 + ... + 3 mu 100
Mong cac ban lam giup mimh
nho cac ban giup minh nhe
1.tim x thuoc N biet:
a)(2x+1)mu 3=125 b)(x-5)mu 4=(x-5)mu 6 c)2 mu x-15=17 d)(7x-11)mu 3=2 mu 5. 5mu 2+200
2.viet cac tich sau hoac thuong duoi dang luy thua cua mot so:
a)2 mu 5 . 8 mu 4 b)25.125 c)25 mu 5:25 mu 7
3.viet cac tich, thuong sau duoi dang luy thua:
a) 2 mu 10:8 mu 3 b)12 mu 7:6 mu 7 c)5 mu 8:25 mu 2
4.tinh gia tri cac bieu thuc sau:
a mu3 . a mu 9 (a mu 5)mu7 (a mu 6)mu 4. a mu 12 4.5 mu 2-2.3 mu 2
Bài 1 :
a) (2x + 1)3 = 125
=> (2x + 1)3 = 53
=> 2x + 1 = 5
=> 2x = 5 - 1
=> 2x = 4
=> x = 2
b) (x - 5)4 = (x - 5)6
Với hai mũ khác nhau , ta chỉ có thể tìm được giá trị biểu thức bằng 1 hoặc 0 (giá trị của chúng bằng nhau)
+) (x - 5)4 = (x - 5)6 = 0
=> (x - 5)4 = 0
=> (x - 5)4 = 04
=> x - 5 = 0 => x = 0 + 5 = 5
+) (x - 5)4 = (x- 5)6 = 1
=> (x - 5)4 = 1
=> (x - 5)4 = 14
=> x - 5 = 1
=> x = 1 + 5
=> x = 6
Bài 4 :
a3 . a9 = a3 + 9 = a12
(a5)7.(a6)4 .a12 = a35 . a24 . a12 = a35 + 24 + 12 = a71
4.52 - 2.32 = 4.25 - 2.9
= 100 - 18
= 82
mong cac ban giup, minh can gap lam,tuy minh trinh bay hoi xau nhung mong cac ban giup
3.viet cac tich, thuong sau duoi dang luy thua:
a) \(\dfrac{2^{10}}{8^3}\)
\(=\dfrac{2^{10}}{\left(2^3\right)^3}\)
\(=\dfrac{2^{10}}{2^9}\)
\(=2^1\)
Tinh tong sau hop li :
A=2+2 mu 2 + 2 mu 3 + ............+2 mu 100
B=1+3 mu 2 + 3 mu 3 + .......... + 3 mu 2009
C = 4+ 4 mu 2 + 4 mu 3 + ...............+ 4 mu n
Giup minh voi sory may minh ko co dau minh dang can gap lam !!!
Lần sau viết cái đề rõ rõ ra nhs!!!
a) \(A=2+2^2+2^3+................+2^{100}\)
\(\Rightarrow2A=2^2+2^3+2^4+................+2^{100}+2^{101}\)
\(\Rightarrow2A-A=\left(2^2+2^3+..............+2^{100}+2^{101}\right)-\left(2+2^2+............+2^{100}\right)\)
\(\Rightarrow A=2^{101}-2\)
b) \(B=1+3+3^2+..................+3^{2009}\)
\(\Rightarrow3B=3+3^2+3^3+..................+3^{2009}+3^{2010}\)
\(\Rightarrow3B-B=\left(3+3^2+...............+3^{2010}\right)-\left(1+3+3^2+.............+3^{2009}\right)\)
\(\Rightarrow2B=3^{2010}-1\)
\(\Rightarrow B=\dfrac{3^{2010}-1}{2}\)
c) \(C=4+4^2+4^3+................+4^n\)
\(\Rightarrow4C=4^2+4^3+.................+4^n+4^{n+1}\)
\(\Rightarrow4C-C=\left(4^2+4^3+.............+4^n+4^{n+1}\right)-\left(4+4^2+............+4^n\right)\)
\(\Rightarrow3C=4^{n+1}-4\)
\(\Rightarrow C=\dfrac{4^{n+1}-4}{3}\)
\(A=2+2^2+2^3+...+2^{100}\)
\(2A=2^2+2^3+2^4+...+2^{101}\)
\(\Rightarrow2A-A=2^{101}-2\)
\(\Rightarrow A=2^{101}-2\)
Vậy \(A=2^{101}-2\).
\(B=1+3^2+3^3+...+3^{2009}\)
\(3B=3+3^3+3^4+...+3^{2010}\)
\(\Rightarrow3B-B=3^{2010}-7\)
\(\Rightarrow2B=3^{2010}-7\)
\(\Rightarrow B=\dfrac{3^{2010}-7}{2}\)
Vậy \(B=\dfrac{3^{2010}-7}{2}\).
\(C=4+4^2+4^3+...+4^n\)
\(4C=4^2+4^3+4^4+...+4^{n+1}\)
\(\Rightarrow4C-C=4^{n+1}-4\)
\(\Rightarrow3C=4^{n+1}-4\)
\(\Rightarrow C=\dfrac{4^{n+1}-4}{3}\)
Vậy \(C=\dfrac{4^{n+1}-4}{3}\).
tong S =3 mu 0 cong 3mu 2cong 3mu 4 cong 3 mu 6 cong 3 mu 8 cong .......... cong 3 mu 2006 cong 3 mu 2008 tan cung bang chu so nao . vi sao
cho s =1+3 mu n1+ 3 mu2+ 3mu 3+3 mu 4+......3 mu 30
tìm chữ số tận cùng
cac ban oi giup minh voi
1.tim a,b thuoc Z,biet:a.(2b-3)=-6
2.cho x,y thuoc Z thoa man x mu 2 +y mu 2 chia het cho 3.chung to x va y chia het cho 3.
C=1+3+3mu 2+3 mu3+....+3 mu 11
\(C=1+3+3^2+3^3+...+3^{11}\)
\(\Rightarrow C=3^0+3^1+3^2+3^3+...+3^{11}\)
\(\Rightarrow3C=3^1+3^2+3^3+3^4+...+3^{12}\)
\(\Rightarrow3C-C=\left(3^1+3^2+3^3+3^4+...+3^{12}\right)-\left(3^0+3^1+3^2+3^3+...+3^{11}\right)\)
\(\Rightarrow2C=3^{12}-3^0\)
\(\Rightarrow C=\frac{3^{12}-1}{2}\)
Tìm x thuộc N biết
a, 2 mũ x . 7 = 56
b, 3 mu x : 9 = 27
c, ( 3 . x 1 ) mu 3 . 4 mu 6 = 4 mu 9
Dấu chấm là dấu nhân nha. phan c minh viet mu 3 o ben tren dau ngoac nha .cac ban dung hieu la 1 mu 3
a. 2x . 7 = 56
2x = 56 : 7
2x = 8
\(\Rightarrow\)23 = 8
Vậy x = 8
a,
2^x.7=56
2^x=56:7
2^x=8
2^x=2^3
=>x=3
b,
3^x:9=27
3^x=27.9
3^x=243
3^x=3^5
=>x=5
c, mình không hiểu
giải nhanh hộ tớ tớ cần gấp
9[3mu x[ 243
25[5mu x [ 625
9[3 mu n [27
25 [ 5mu h [ 3152
2 mu n + 4.2 mu n =5.2 mũ 5
2 mu n /4=16
6.2 mu n+3.2 mu n =9.2 mu n
3 mu n / 3 mu 2 = 243