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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
\(M=\dfrac{54.107-53}{53.107-54}=\dfrac{53\left(107-1\right)+107}{53\left(107-1\right)-1}=\dfrac{53.106-1+108}{53.106-1}=1+\dfrac{108}{53.106-1}\)
\(N=\dfrac{135.269-133}{134.269-135}=\dfrac{134\left(269-1\right)-1+270}{134\left(269-1\right)-1}=1+\dfrac{270}{134.268-1}\)
\(M-N=\dfrac{108}{53.106-1}-\dfrac{270}{134.268-1}\)
\(M-N=\dfrac{2}{2.53^2-1}-\dfrac{5}{8.67^2-1}>\dfrac{5}{10.53^2-1}-\dfrac{5}{8.67^2-1}>0\)
M>N
a) 27/82 < 26/75 ( 2025/6250 < 2132\6250)
b) -49/78 > 64/ -95 ( - 3136/7410 > -4992/7410)
c) ta có: \(A=\frac{54.107-53}{53.107}=\frac{53.107+(107-53)}{53.107+54}=\frac{53.107+54}{53.107+54}=1\)
\(B=\frac{135.269-133}{134.269+135}=\frac{134.269+\left(269-133\right)}{134.269+135}=\frac{134.269+136}{134.269+135}>1\)
\(\Rightarrow A< B\)
d) ta có: \(A=\frac{3^{10}+1}{3^9+1}=\frac{3.\left(3^9+1\right)-2}{3^9+1}=\frac{3.\left(3^9+1\right)}{3^9+1}-\frac{2}{3^9+1}=3-\frac{2}{3^9+1}\)
\(B=\frac{3^9+1}{3^8+1}=\frac{3.\left(3^8+1\right)-2}{3^8+1}=\frac{3.\left(3^8+1\right)}{3^8+1}-\frac{2}{3^8+1}=3-\frac{2}{3^8+1}\)
mà \(\frac{2}{3^9+1}< \frac{2}{3^8+1}\Rightarrow3-\frac{2}{3^9+1}< 3-\frac{2}{3^8+1}\)
=> A < B
a.Vì \(\frac{17}{19}< 1\) và \(\frac{19}{17}>1\)
nên \(\frac{17}{19}< 1< \frac{19}{17}\)
hay \(\frac{17}{19}< \frac{19}{17}\)
b) \(\frac{15}{7}=2\frac{1}{7}\) và \(\frac{25}{12}=2\frac{1}{12}\)
Vì \(2\frac{1}{7}>2\frac{1}{12}\) nên \(\frac{15}{7}>\frac{25}{12}\)
\(A=\frac{54.107-53}{53.107+54}\)
\(\Leftrightarrow A=\frac{53.107+107-53}{53.107+54}\)
\(\Leftrightarrow A=\frac{53.107+54}{53.107+54}\)
\(\Leftrightarrow A=1\)
\(B=\frac{135.269-133}{134.269+135}\)
\(\Leftrightarrow B=\frac{134.269+269-133}{134.269+135}\)
\(\Leftrightarrow B=\frac{134.269+135}{134.269+135}\)
\(\Leftrightarrow B=1\)
Vì 1 = 1 nên A =B
A= (54.107-53)/(53.107+54)
= (53+1).107-53 / 53.107+54
=53.107+107-53 / 53.107+54
=53.107+54 / 54.107 + 54
=1
B= 135.269-133 / 134.269+135
= (134+1).269-133 / 134.269+135
= 134.269+269-133 / 134.269+135
=134.269+136 / 134.269+135
=134.269+135/ 134.269+135 + 1/134.269+135
=1 + 1/134.269+135 >1=A
\(A=\frac{54.107-53}{53.107+54}=1\)
\(B=\frac{135.269-133}{134.269+135}>1\)
\(A=\frac{54.107-53}{53.107+54}<\frac{135.269-133}{134.269+135}\)
\(A=\frac{54\cdot107-53}{53\cdot107+54}=\frac{\left(53+1\right)107-53}{53\cdot107+54}=\frac{53\cdot107+107-53}{53\cdot107+54}=\frac{53\cdot107+54}{53\cdot107+54}=1\)
\(B=\frac{135\cdot268-133}{134\cdot269+135}=\frac{\left(134+1\right)\cdot268-133}{134\cdot269+135}=\frac{134\cdot268+268-133}{34\cdot269+135}=\frac{134\cdot268+135}{134\cdot269+135}=1\)
Vì 1=1 nên A=B
bài 2
a, TS= 54 . 107 -53=(53+1) .107-53=53.107+107-53=53.107+ 54
<=>
\(\frac{TS}{MS}\)=\(\frac{54.107+54}{54.107+54}\)=1
Bài 1 :
\(a)\) Gọi \(ƯCLN\left(n+1;2n+3\right)=d\)
\(\Rightarrow\)\(\hept{\begin{cases}n+1⋮d\\2n+3⋮d\end{cases}\Rightarrow\hept{\begin{cases}2\left(n+1\right)⋮d\\2n+3⋮d\end{cases}\Rightarrow}\hept{\begin{cases}2n+2⋮d\\2n+3⋮d\end{cases}}}\)
\(\Rightarrow\)\(\left(2n+2\right)-\left(2n+3\right)⋮d\)
\(\Rightarrow\)\(2n+2-2n-3⋮d\)
\(\Rightarrow\)\(\left(-1\right)⋮d\)
\(\Rightarrow\)\(d\inƯ\left(-1\right)\)
Mà \(Ư\left(-1\right)=\left\{1;-1\right\}\)
\(\Rightarrow\)\(d\in\left\{1;-1\right\}\)
Do đó :
\(ƯCLN\left(n+1;2n+3\right)=\left\{1;-1\right\}\)
Vậy \(\frac{n+1}{2n+3}\) là phân số tối giản với mọi n
Chúc bạn học tốt ~
Các câu a,b,c,d,e thì bạn sử quy đồng mẫu số hoặc tử số,mình chỉ làm câu h thôi
\(\dfrac{54.107-53}{53.107+54}\)=\(\dfrac{\left(53+1\right).107-53}{53.107+54}\)=\(\dfrac{53.107+107-53}{53.107+54}\)=\(\dfrac{53.107+54}{53.107+54}\)=1
B=\(\dfrac{135.269-133}{134.269+135}\)=\(\dfrac{\left(134+1\right).269-133}{134.269+135}\)=\(\dfrac{134.269+269-133}{134.269+135}\)=
\(\dfrac{134.269+136}{134.269+135}\)>1
=) A<B
a , \(\dfrac{2}{3}\) > \(\dfrac{1}{2}\) > \(\dfrac{1}{3}\)
b , \(\dfrac{4}{41}\) > \(\dfrac{2}{83}\) > \(\dfrac{3}{124}\) > \(\dfrac{5}{207}\)
c , \(\dfrac{24}{13}\) > \(\dfrac{16}{19}\)
d , \(\dfrac{26}{75}\) < \(\dfrac{27}{82}\)
Câu h đã có bạn làm cho nên mk ko làm nữa!