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45*[15+1]-17/45*15+28
45*15+45-17/45*15+28
45*15+28/45*15+28
=1
k nha
\(\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{4}\right)x\left(1-\frac{1}{5}\right)x\left(1-\frac{1}{6}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}\)
\(=\frac{1.2.3.4.5}{2.3.4.5.6}=\frac{1}{6}\)
\(\left(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}\right):\left(\frac{1}{6}+\frac{1}{10}-\frac{1}{15}\right)\)
=\(\frac{1}{3}+\frac{1}{5}\)
=\(\frac{5}{3}\)
a,
\(\frac{326}{325}-\frac{1}{325}=1\) \(\frac{325}{324}-\frac{1}{324}=1\)
Vì \(\frac{1}{325}< \frac{1}{324}\)nên \(\frac{326}{325}< \frac{325}{324}\)
b,
( 11 x 9 - 900 x 0,1 - 8 ) x ( 56,7 x 0,5 + 56,7 x 9,5 )
= ( 99 - 90 - 8 ) x [ 56,7 x ( 0,5 + 9,5 ) ]
= ( 9 - 8 ) x ( 56,7 x 10 )
= 1 x 567 = 567
\(\frac{x}{5}=\frac{20}{x}\)
Ta có :
x . x = 20 . 5
x2 = 100
=> x = 10 hoặc x = -10
a) \(\frac{5}{12}+\frac{5}{6}-\frac{3}{4}=\frac{5}{12}+\frac{10}{12}-\frac{9}{12}=\frac{5+10-9}{12}=\frac{6}{12}=\frac{1}{2}.\)
b) \(1-\left(\frac{1}{5}+\frac{1}{2}\right)=1-\left(\frac{2}{10}+\frac{5}{10}\right)=1-\frac{7}{10}=\frac{3}{10}\)
Chúc bn học tốt !!!
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_
=53,4-3,56
=49,84