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_Minh ngụy_
a) ( 1000-13) . ( 1000-23) . ( 1000-33) ...( 1000 -503)
\(=\left(1000-1^3\right)\cdot\left(1000-2^3\right)\cdot...\cdot\left(1000-10^3\right)\cdot.....\cdot\left(1000-50^3\right)\)
\(=\left(1000-1^3\right)\cdot\left(100-2^3\right)\cdot...\cdot\left(1000-1000\right)\cdot...\cdot\left(1000-50^3\right)\)
\(=\left(1000-1^3\right)\cdot\left(1000-2^3\right)\cdot......\cdot0\cdot......\left(1000-50^3\right)\)
\(=0\)
b) (1/125-1/13) . (1/125-1/23).( 1/125-1/33)...( 1/125-1/253)
\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\cdot\left(\frac{1}{125}-\frac{1}{2^3}\right)\cdot...\cdot\left(\frac{1}{125}-\frac{1}{5^3}\right)\cdot...\cdot\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\cdot\left(\frac{1}{125}-\frac{1}{2^3}\right)\cdot...\cdot\left(\frac{1}{125}-\frac{1}{125}\right)\cdot...\cdot\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\cdot\left(\frac{1}{125}-\frac{1}{2^3}\right)\cdot....\cdot0\cdot...\cdot\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=0\)
\(\left(2x+1\right)^2=25\)
\(\Rightarrow2x+1\in\left\{-5;5\right\}\)
\(\Rightarrow2x\in\left\{-6;4\right\}\)
\(\Rightarrow x\in\left\{-3;2\right\}\)
Vậy..
\(\left(x-1\right)^3=-125\)
\(\left(x-1\right)^3=-5^3\)
\(x-1=-5\)
\(x=-4\)
Vậy...
\(7^{x+2}.2.7^{x-1}=345\)
\(7^x.\left(7^2+\dfrac{2}{7}\right)=345\)
\(7x=7\)
\(x=1\)
Vậy...
đơn giản thôi bạn
3 (x + 25)
= 3x + 3*25 (bạn nhân 3 với từng số trong ngoặc) (*là nhân nhé)
= 3x + 75
a, 5\(^{-1}\)25\(^n\)=125
\(\frac{1}{5}\)25\(^n\)=125
25\(^n\)=625
n=2
b, 3\(^{-1}\)3\(^n\)+6. 3\(^{n-1}\)=7.3\(^6\)
3\(^{n-1}\)+6. 3\(^{n-1}\)=7.3\(^6\)
3\(^{n-1}\)(6+1)=7.3\(^6\)=3\(^{n-1}\).7
3\(^6\)=3\(^{n-1}\)
6=n-1
n=7
c,3\(^4\)<1/9.27^n <3^10
3^4<1/(3^2) .3^(3n)<3^10
3^4<3^(3n-2)<3^10
3n-2 chia 3 dư 1 nên 3n-2 = 7
n=3
d,25<5^n:5<625
5^2<5^(n-1)<5^4
n-1=3 nên n=4
P/s : mk làm từng phần một
\(\left(\frac{2}{3}\right)^x=\frac{8}{27}=\left(\frac{2}{3}\right)^3\)
=> x = 3
Vậy,........
2)
\(\left(\frac{5}{6}x+\frac{1}{2}\right)^2=\frac{9}{16}=\left(\frac{3}{4}\right)^2\)
\(\frac{5}{6}x+\frac{1}{2}=\frac{3}{4}\)
\(\frac{5}{6}x=\frac{1}{4}\)
\(x=\frac{3}{10}\)
thank!
\(A=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)\left(\frac{1}{125}-\frac{1}{3^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)\\ A=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)\left(\frac{1}{125}-\frac{1}{3^3}\right)\left(\frac{1}{125}-\frac{1}{4^3}\right)\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)\\ A=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)\left(\frac{1}{125}-\frac{1}{3^3}\right)\left(\frac{1}{125}-\frac{1}{4^3}\right)\left(\frac{1}{125}-\frac{1}{125}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)\\ A=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)\left(\frac{1}{125}-\frac{1}{3^3}\right)\left(\frac{1}{125}-\frac{1}{4^3}\right)\cdot0\cdot...\left(\frac{1}{125}-\frac{1}{25^3}\right)\\ A=0\)