1) (3x-5/9) mu 200 + (3y+0,4/3) mu 2008 =0
tim cap x;y thoa man
2) (2x-1) mu 2k + (y-(1/3))mu 4k = 0
ai truoc thi mk like
( 3x -1) mu 5 = 32
9 x - 1 ) mu 2 = ( 2x - 3 ) mu 2
a. (3x-1)5=32 b. (9x-1)2=(2x-3)2
=>(3x-1)5=23 =>9x-1=2x-3
=>3x-1=2 =>9x-1-2x+3=0
=>3x=3 =>7x+2=0
=>x=1 =>7x=-2
=>x=-2/7
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) \(\dfrac{2^{10}}{8^3}\)
\(=\dfrac{2^{10}}{\left(2^3\right)^3}\)
\(=\dfrac{2^{10}}{2^9}\)
\(=2^1\)
Tìm x biet.
a)x mu 2 -5=0
b)(2x-1)mu 2-(3x+1)mu 2=0
c)4 phần 9×x mũ 2=-4x-9
a) \(x^2-5=0\)
\(\Leftrightarrow x^2=5\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=5\\x=-5\end{array}\right.\)
b) \(\left(2x-1\right)^2-\left(3x+1\right)^2=0\)
\(\Leftrightarrow\left(2x-1+3x+1\right)\left(2x-1-3x-1\right)=0\)
\(\Leftrightarrow5x\left(-x-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\-x-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=-2\end{array}\right.\)
c) \(\frac{4}{9}\cdot x^2=-4x-9\)
\(\Leftrightarrow\left(\frac{2}{3}x\right)^2+4x+9=0\)
\(\Leftrightarrow\left(\frac{2}{3}x+3\right)^2=0\)
\(\Leftrightarrow\)\(\frac{2}{3}x+3=0\Leftrightarrow x=-\frac{9}{2}\)
tính tổng
B= 7-7 mu 4 + 7 mu 4 -........+7 mu 301
A = 1 + 5 mũ 2 + 5 mu 4 + 5 mu 6 +.....+5 mu 200
tính
A= 1/7+1/7mu 2 + 1/7 mu 3+......+1/7mu 100
B=-4/5+4/5 mu 2 - 4/5 mu 3 + ....+4/5mu 200
tính A=25 mũ 8 + 25 mũ 4 + 25 mu 20 +......+25 mu 4 +1 / 25 mu 20 + 25 mu 28 + 25 mu 26 +.....= 25 mu 2 +1
7 × 3 mu x + 20 × 3 mu x = 3 mu 25
So sanh :
a, 9 mu 5 va 27 mu 3
b, 3 mu 200 va 2 mu 300
c, 31 mu 11 va 17 mu 14
d, 199 mu 20 va 2003 mu 15
e, 2 mu 1993 va 7 mu 714
a, 9^5>27^3
b,3^200>2^300
c, 32^11<17^14
a) \3x=9\=\4-x\+0
b)(x-5)16 =(x-5)12
c)(2x-5)(5-x mu 2)(x mu 2 +1=0
d)5 mu x + 5 mu x+2 =650
a, (1/2)mu x =64
b, x mu 3 +27 =0
c, 3 mu 2 /2 mu n =4
d, 625 / 5 mu n = 5
e, 27 mu n . 3 mu n = 3 mu 2
so sanh
a, 3 mu 200 va 2 mu 300
b, 2 mu 225 va 3 mu 150
Minh can gap giup minh voi:
a) \3x=9\=\4-x\+0
b)(x-5)16 =(x-5)12
c)(2x-5)(5-x mu 2)(x mu 2 +1=0
d)5 mu x + 5 mu x+2 =650
bai 1; cho tong M =126 +213+x. Tim x de M chia het cho 3
bai 2; chung to rang tong ; A= 2 + 2 mu 3 + 2 mu 4 + 2 mu 5 + 2 mu 6 +2 mu 7 +2 mu 9 + 2 mu 10 + 2 mu 12 chia het cho 5
Có : 126 chia hết cho 3, 213 chia hết cho 3
Để được M chia hết cho 3 thì x phải chia hết cho 3
Hay gọi là 3k ( k thuộc N)
2.
Hình như đầu bài bài 2 sai
dung do khong sai dau