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a)13x3x32,27+67,63x39
=39x32,27+67,63x39
=39x(32,27+67,63)
=39x100
=3900
b,= 1- [ 1/2 x 1/3 x1/4 x..... x 1/100 ]
=1/2 x 2/3 x 3/4 x .......x 99/100
= 1x2x3x......x99 / 2x3x4x...... x100 [ rút gọn ]
= 1/100
a) \(\left(2x+1\right)^3=125\)
\(\Rightarrow\left(2x+1\right)^3=5^3\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow2x=5-1\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=4:2\)
\(\Rightarrow x=2\)
Vậy x = 2
b) \(\left(x-5\right)^4=\left(x-5\right)^6\)
\(\Rightarrow\left(x-5\right)^4-\left(x-5\right)^6=0\)
\(\Rightarrow\left(x-5\right)^4\left[1-\left(x-5\right)^2\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-5\right)^4=0\\1-\left(x-5\right)^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^2=1\end{cases}}\)
TH 1 : \(\left(x-5\right)^4=0\Rightarrow x-5=0\Rightarrow x=5\)
TH 2 : \(\left(x-5\right)^2=1\Rightarrow\orbr{\begin{cases}x-5=1\\x-5=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=4\end{cases}}\)
Vậy \(x\in\left\{5;6;4\right\}\)
c) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\Rightarrow\left(2x-15\right)^3\left[\left(2x-15\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2=1\end{cases}}\)
TH 1 : \(\left(2x-15\right)^3=0\Rightarrow2x-15=0\Rightarrow2x=15\Rightarrow x=\frac{15}{2}\)
TH 2 : \(\left(2x-15\right)^2=1\Rightarrow\orbr{\begin{cases}2x-15=1\\2x-15=-1\end{cases}}\Rightarrow\orbr{\begin{cases}2x=16\\2x=14\end{cases}}\Rightarrow\orbr{\begin{cases}x=8\\x=7\end{cases}}\)
Vậy \(x\in\left\{\frac{15}{2};8;7\right\}\)
_Chúc bạn học tốt_
Bài 2:
\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2003}{2004}\)
\(=\frac{1}{2004}\)
A)\(a=3\frac{3}{4}\)
suy ra \(\frac{\left(3\frac{3}{4}-1,75\right):0,01}{\left(1-0,75\right)x80}\)
= 10
b) nếu A = 41,25 thì ta có \(\frac{\left(a-1,75\right):0,01}{\left(1-0,75\right)x80}=41,25\)
suy ra ( a -1,75) : 0,01: ( 1-0,75) x 80 = 41.25
( a - 1, 75 ) : 0,01 : 0,25 = 41,25 : 80
( a - 1, 75 ) : 0,01 : 0,25 = 33/64
a - 1,75 = 33 / 64 x 0,25 x 0,01
a - 1,75 = 33/25600
a = 33/ 25600+ 1,75
a = 1,751289063
=
b) Ta có:
\(A=\frac{\left(a-1,75\right)\div0,01}{\left(1-0,75\right)\times80}=41,25\)
\(\frac{\left(a-1,75\right)\div0,01}{0,25\times80}=41,25\)
\(\left(a-1,75\right)\div0,01\div20=41,25\)
\(\left(a-1,75\right)\div0,01=41,25\times20\)
\(\left(a-1,75\right)\div0,01=825\)
\(\left(a-1,75\right)=825\times0,01\)
\(a-1,75=8,25\)
\(a=8,25+1,75\)
\(a=10\)
Vậy a phải bằng 10 thì biểu thức A có giá trị là 41,25
Gợi ý: Các biểu thức mũ chẵn đều không âm.
\(a^{2n}+b^{2n}\le0\Leftrightarrow a^{2n}+b^{2n}=0\Leftrightarrow a=b=0\)
a,\(\left(x-\frac{2}{5}\right)^{2010}+\left(y+\frac{3}{7}\right)^{468}\)< \(0\)
Vì \(\left(x-\frac{2}{5}\right)^{2010}\);\(\left(y+\frac{3}{7}\right)^{468}\)đều > \(0\)
=> \(\left(x-\frac{2}{5}\right)^{2010}=0\)
\(\left(y+\frac{3}{7}\right)^{468}=0\)
=> \(\left(x-\frac{2}{5}\right)^{2010}=0^{2010}\)
\(\left(y+\frac{3}{7}\right)^{468}=0^{468}\)
=> \(x-\frac{2}{5}=0\)
\(y-\frac{3}{7}=0\)
=> \(x=\frac{2}{5}\)
\(y=\frac{3}{7}\)
Vậy \(x=\frac{2}{5}\)\(y=\frac{3}{7}\)
\(a\odot b=bx\left(a-5\right)-12\)
\(6\odot\left(15\odot8\right)=6\odot\left(8x\left(15-5\right)-12\right)\)
\(=6\odot68\)
\(=68x\left(6-5\right)-12\)
\(=56\)