Tính hợp lí:
\(\frac{54\cdot107-53}{53\cdot107-54}\)
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\(\frac{200}{201}+\frac{201}{202}>\frac{200}{201+202}+\frac{201}{201+202}=\frac{200+201}{201+202}\)
\(A=\frac{54.107-53}{53.107+54}=\frac{\left(53+1\right)107-53}{53.107+54}=\frac{53.107+107-53}{53.107+54}=\frac{53.107+54}{53.107+54}=1\)
\(B=\frac{135.269-133}{134.269+135}=\frac{\left(134+1\right)269-133}{134.269+135}=\frac{134.269+269-133}{134.269+135}=\frac{134.269+136}{134.269+135}>1\)
Vậy A<B
câu a là
200/201+201/202>200/202+201/202>1
200+201/201+201<1
=>200/201+201/202>1>200+201/201+202
a) Xét \(\frac{1999.2000-2}{1998.1999+3997}=\frac{1999.\left(1998+2\right)-2}{1998.1999+3997}=\frac{1999.1998+1999.2-2}{1998.1999+3997}=\frac{1999.1998+3996}{1998.1998+3997}\)
=> A < B
\(3)1579-(53+1579)-(53) \)
\(=1579-53-1579+53\)
\(=(1579-1579)-(-53+53)\)
\(=0-0\)
\(=0\)
\(4)2001-15+54-2001-54\)
\(=(2001-2001)+(54-54)-15\)
\(=0+0-15\)
\(=0-15\)
\(=-15\)
Bạn nên sửa đề thành: \(\frac{54\cdot107-53}{53\cdot107+54}\)
Thì \(=\frac{54\cdot\left(54+53\right)-53}{53\left(53+54\right)+54}=\frac{54^2+53\left(53+1\right)-53}{53^2+54\cdot\left(54-1\right)+54}=\frac{54^2+53^2}{53^2+54^2}=1\)
Chứ nguyên đề thì tính hợp lý sao?
a: \(\dfrac{1\cdot5\cdot6+2\cdot10\cdot12}{1\cdot3\cdot52\cdot6\cdot10}=\dfrac{5\cdot6\left(1+2\cdot2\cdot2\right)}{3\cdot52\cdot6\cdot10}\)
\(=\dfrac{1}{2}\cdot\dfrac{9}{156}\)
\(=\dfrac{9}{312}=\dfrac{3}{104}\)
Đề sai chắc lun.
\(\frac{54.107-53}{53.107-54}=\frac{\left(53+1\right).107-53}{53.107-54}=\frac{53.107+107-53}{53.107-54}=\frac{53.107+54}{53.107-54}=1-1=0\)