-5^2-5.3^2/5^3+5^2.3^2
Tính :
\(A=\dfrac{-5^2-5.3^2}{5^3+5^2.3^2}\)
Giups mk nha các bn !
\(A=\dfrac{-\left(5^2+5.3^2\right)}{5^2\left(5+3^2\right)}\)
\(A=\dfrac{-\left(5\left(5+3^2\right)\right)}{5^2\left(5+3^2\right)}\)
\(A=\dfrac{-5}{5^2}=-\dfrac{1}{5}\)
CHÚC BẠN HỌC TỐT.........
Tìm số nguyên x , nếu biết
1. 7.4^x=7.4^3
2.3/2.5^x =3/2.5^12
3 . 2^x=2.2^8
4. 5.3^x=7.3^5-2.3^5
1) \(7.4^x=7.4^3\Leftrightarrow4^x=4^3;x=3\)
2) \(\frac{3}{2.5^x}=\frac{3}{2.5^{12}}\Leftrightarrow5^x=5^{12};x=12\)
\(2^x=2.2^8=2^9;x=9\)
4) \(5.3^x=7.3^5-2.3^5\Leftrightarrow5.3^x=3^5.\left(7-2\right)\)
\(\Leftrightarrow3^5.x=3^5.5;x=5\)
Cho 3 phân số:
\(\frac{-5^2-5.3^2}{5^3+5^2.3^2};\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}};\frac{2929-101}{2.1919+404}\)
Rút gọn rồi quy đòng 3 ps trên
A=[2.315.812-5.32.(95)2 ] :\(\dfrac{1+2+3+....+1996}{998}-1817\)
\(A=\left[2\cdot3^{15}\cdot3^8-5\cdot3^2\cdot3^{10}\right]\cdot\dfrac{998}{1993006}-1817\)
\(=\left[3^{23}\cdot2-5\cdot3^{12}\right]\cdot\dfrac{998}{1993006}-1817\)
\(=3^{12}\cdot\left[3^{11}\cdot2-5\right]\cdot\dfrac{998}{1993006}-1817\)
\(=\dfrac{1}{1997}\cdot3^{12}\cdot354289-1817\)
\(\simeq94281458.14\)
Cho 3 phân số :\(\frac{-5^2-5.3^2}{5^3+5^2.3^2}\);\(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\);\(\frac{2929-101}{2.1919+404}\)rút gọn rồi qui đồng các phân số đó.
Tìm x biết
1)\(3^{x+2}+4\cdot3^{x+1}=7.3^6\)
2)\(5.3^{x+6}=2.3^5+3.3^5\)
\(5.3^x=5.3^4\)
\(5.3^x=7.3^5-2:3^5\)
5.3x = 5.34
=> x = 4
5.34 = 7.35 - 2.35
= 35 . (7 - 2)
= 5. 35
=> 3x = 35
=> x = 5
a) Ta có: \(5\cdot3^x=5\cdot3^4\)
\(\Leftrightarrow3^x=3^4\)
hay x=4
bài 1
A=\(\frac{-5^2-5.3^2}{5^3+5^2.3^2}\)
B=\(\frac{7^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
C=\(\frac{2929-101}{2.1999-404}\)
giúp mik với mik sẽ tích và kb cho
2Tìm X:
a.(19x+2.5^2):14=(13-8)^2-4^2
b.2.3^x=10.3^12+8.27^4
c.{x^2-[6^2-(8^2-9.7)^3-7.5]^3-5.3}^3=1
d.60-3(x-2)=51
e.4x-20=2^5:2^2
{ x2 - [ 62 - ( 82 - 9.7)3 - 7.5]3 - 5.3 }3 = 1
{ x2 + [ 36 - (64 - 63)3 - 35]3 - 15}3 = 1
[ x2 - ( 36 - 13 - 35 ) - 15 ]3 = 1
[ x2 - ( 36 - 1 - 35 ) - 15]3 = 1
[ x2 - ( 35 - 35 ) - 15]3 = 1
[ x2 - 0 - 15]3 = 1
( x2 - 15 )3 = 1
<=> ( x2 - 15)3 = 13
=> x2 - 15 = 1
<=> x2 = 16
=> x = 4