Bài 1 : So sánh 2300 và 3200
Bài 2 : Tìm x , biết :
a) 23x+1 = 32x
b) 3x+2 = 273x-1
Bài 3 : Rút gọn
a) 273 - 184 / 316 - 24
b) 219 . 273 + 15 . 49 . 94 / 69 . 210 + 1210
Bài 4 : Cho : 12 + 22 + 32 + 102 = 385
Tính : 22 + 42 + 62 +.....+ 202
Bài 5 : Tìm x , biết :
a) (4/5)2x+7 = 256/625
b) (4x-3)4 = (4x-3)2
c) (2x+3)2 + (3x -2) 4 = 0
Bạn hãy đăng từng bài để tiện trao đổi. Yên tâm mình sẽ giúp bạn.
B1: \(2^{300}=\left(2^3\right)^{100}=8^{100}\)
\(3^{200}=\left(3^2\right)^{100}=9^{100}\)
\(8^{100}< 9^{100}\Rightarrow2^{300}< 3^{200}\)
Ta có : \(2^{300}=2^{3.100}=\left(2^3\right)^{100}=8^{100}\)
\(3^{200}=3^{2.100}=\left(3^2\right)^{100}=9^{100}\)
Vì : \(8^{100}< 9^{100}\)
\(\Rightarrow2^{300}< 3^{200}\)
B2.
a) \(2^{3x+1}=32^x\)
\(\Leftrightarrow2^{3x}.2=2^{5x}\)
\(\Leftrightarrow2^{5x}:2^{3x}=2\)
\(\Leftrightarrow2^{2x}=2\)
\(\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)
b) \(3^{x+2}=27^{3x-1}\)
\(\Leftrightarrow3^x.9=3^{9x}:27\)
\(\Leftrightarrow3^x=3^{9x}:3^5\)
\(\Leftrightarrow3^5=3^{8x}\) \(\Leftrightarrow8x=5\Rightarrow x=\frac{5}{8}\)
Bài 1:
\(2^{300}=\left(2^3\right)^{100}=8^{100}< 9^{100}=\left(3^2\right)^{100}=3^{200}\)
Vậy \(2^{300}< 3^{200}\)
Bài 2:
a.
\(2^{3x+1}=32^x\)
\(2^{3x+1}=\left(2^5\right)^x\)
\(2^{3x+1}=2^{5x}\)
\(3x+1=5x\)
\(3x-5x=-1\)
\(-2x=-1\)
\(x=\frac{-1}{-2}\)
\(x=\frac{1}{2}\)
b.
\(3^{2+x}=27^{3x-1}\)
\(3^{x+2}=\left(3^3\right)^{3x-1}\)
\(3^{x+2}=3^{3\times\left(3x-1\right)}\)
\(x+2=3\times\left(3x-1\right)\)
\(x+2=9x-3\)
\(9x-x=3+2\)
\(8x=5\)
\(x=\frac{5}{8}\)
Bài 4:
\(2^2+4^2+6^2+...+18^2+20^2\)
\(=\left(1\times2\right)^2+\left(2\times2\right)^2+\left(3\times2\right)^2+...+\left(9\times2\right)^2+\left(10\times2\right)^2\)
\(=1^2\times2^2+2^2\times2^2+3^2\times2^2+...+9^2\times2^2+10^2\times2^2\)
\(=2^2\times\left(1^2+2^2+3^3+...+9^2+10^2\right)\)
\(=4\times385\)
\(=1540\)
Bài 5:
a.
\(\left(\frac{4}{5}\right)^{2x+7}=\frac{256}{625}\)
\(\left(\frac{4}{5}\right)^{2x+7}=\left(\frac{4}{5}\right)^4\)
\(2x+7=4\)
\(2x=4-7\)
\(2x=-3\)
\(x=-\frac{3}{2}\)
b.
\(\left(4x-3\right)^4=\left(4x-3\right)^2\)
TH1:
\(4x-3=0\)
\(4x=3\)
\(x=\frac{3}{4}\)
TH2:
\(4x-3=1\)
\(4x=1+3\)
\(4x=4\)
\(x=\frac{4}{4}\)
\(x=1\)
TH3:
\(4x-3=-1\)
\(4x=-1+3\)
\(4x=2\)
\(x=\frac{2}{4}\)
\(x=\frac{1}{2}\)
Vậy \(x\in\left\{\frac{1}{2};\frac{3}{4};1\right\}\)
Bài 1 : So sánh 2300 và 3200
ta có : 2\(^{300}\) =(2\(^3\))\(^{100}\) =8\(^{100}\)
3\(^{200}\)=(3\(^2\))\(^{100}\) =9\(^{100}\)
vi 8 \(^{100}\)<9\(^{100}\) nen 2\(^{300}\) < 3\(^{200}\)
