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\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)A=54-53/53+54=1/107=2/214
B=135-133/134+135=2/169
tự so sánh tiếp
a , Ta có : \(1-\frac{54}{59}=\frac{5}{59}\) \(=\frac{50}{590}\) ; \(1-\frac{541}{591}=\frac{50}{591}\)
Vì \(\frac{50}{590}>\frac{50}{591}\)nên \(\frac{54}{59}< \frac{541}{591}\)
\(a)\) Ta có công thức :
\(\frac{a}{b}< \frac{a+c}{b+c}\)\(\left(\frac{a}{b}< 1;a,b,c\inℕ^∗\right)\)
Áp dụng vào ta có :
\(A=\frac{10^{11}-1}{10^{12}-1}< \frac{10^{11}-1+11}{10^{12}-1+11}=\frac{10^{11}+10}{10^{12}+10}=\frac{10\left(10^{10}+1\right)}{10\left(10^{11}+1\right)}=\frac{10^{10}+1}{10^{11}+1}=B\)
\(\Rightarrow\)\(A< B\)
Vậy \(A< B\)
Chúc bạn học tốt ~
\(\frac{200}{201}+\frac{201}{202}>\frac{200}{201+202}+\frac{201}{201+202}=\frac{200+201}{201+202}\)
\(A=\frac{54.107-53}{53.107+54}=\frac{\left(53+1\right)107-53}{53.107+54}=\frac{53.107+107-53}{53.107+54}=\frac{53.107+54}{53.107+54}=1\)
\(B=\frac{135.269-133}{134.269+135}=\frac{\left(134+1\right)269-133}{134.269+135}=\frac{134.269+269-133}{134.269+135}=\frac{134.269+136}{134.269+135}>1\)
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
bang nhau
Giai:
A=1.3.5.7...97.99=\(\frac{\left(1.3.5...97.99\right).\left(2.4.6...100\right)}{2.4.6...100}\)
=\(\frac{1.2.3.4...99.100}{\left(1.2\right).\left(2.2\right)...\left(2.50\right)}\)
=\(\frac{\left(1.2.3...50\right).\left(51.52...99.100\right)}{\left(1.2.3...49.50\right).2^{50}}\)
=\(\frac{51.52...99.100}{2.2...2.2}\)
=\(\frac{51}{2}.\frac{52}{2}.\frac{53}{2}...\frac{100}{2}\)
mà B=\(\frac{51}{2}.\frac{52}{2}.\frac{53}{2}...\frac{100}{2}\)
Nên A=B
Vậy A=B
\(1.3.5.7...97.99=\frac{100!}{2.4.6.8...100}\)
\(=\frac{1.2.3.4...100}{1.2.2.2.3.2...50.2}\)
\(=\frac{51.52.53...100}{2}\)
Vậy \(A=B\)
hdhiq
nam or nữ vậy