cho \(\dfrac{a}{b}< \dfrac{c}{d}\) trong đó b,d dương. Chứng minh rằng:
a) a.d < b.c b)\(\dfrac{a}{b}< \dfrac{a+c}{b+d}< \dfrac{c}{d}\)
Cho các số hữu tỉ \(\dfrac{a}{b}\)và\(\dfrac{c}{d}\) với mẫu dương, trong đó \(\dfrac{a}{b}\)<\(\dfrac{c}{d}\). Chứng minh rằng:
A) ad<bc
B) \(\dfrac{a}{b}\)<\(\dfrac{a+c}{b+d}\)< \(\dfrac{c}{d}\)
a) \(\dfrac{a}{b}< \dfrac{c}{d}\Rightarrow ad< bc\)
b) Tham khảo:https://olm.vn/hoi-dap/tim-kiem?q=cho+c%C3%A1c+s%E1%BB%91+h%E1%BB%AFu+t%E1%BB%89+a/b+v%C3%A0+c/d+v%E1%BB%9Bi+m%E1%BA%ABu+d%C6%B0%C6%A1ng+,+trong+%C4%91%C3%B3+a/b+%3Cc/d+.+c/m+r%E1%BA%B1ng+a)+a.d+%3Cb.c+b)+a/b+%3C+(a+c)/(b+d)%3Cc/d+&id=174343
a) Ta có: \(\left\{{}\begin{matrix}\dfrac{a}{b}< \dfrac{c}{d}\\b,d>0\end{matrix}\right.\)
\(\Rightarrow\dfrac{a}{b}.bd< \dfrac{c}{d}.bd\Rightarrow ad< bc\)
b) Ta có: \(ad< bc\Rightarrow ad+ab< bc+ab\)
\(\Rightarrow a\left(b+d\right)< b\left(a+c\right)\Rightarrow\dfrac{a}{b}< \dfrac{a+c}{b+d}\left(1\right)\)(do \(b,d>0\))
\(bc>ad\Rightarrow bc+cd>ad+cd\)
\(\Rightarrow c\left(b+d\right)>d\left(a+c\right)\Rightarrow\dfrac{c}{d}>\dfrac{a+c}{b+d}\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow\dfrac{a}{b}< \dfrac{a+c}{b+d}< \dfrac{c}{d}\)
cho đẳng thức a.d=b.c tỉ lệ thức nào sau đây sai ( a, b , c , d khác 0 ) :
A \(\dfrac{a}{b}\) = \(\dfrac{c}{d}\) B \(\dfrac{d}{b}\) = \(\dfrac{c}{a}\) C \(\dfrac{b}{d}\) = \(\dfrac{c}{a}\) D \(\dfrac{a}{c}\) = \(\dfrac{b}{d}\)
giúp mình đi nha mn =(
C. \(\dfrac{b}{d}=\dfrac{c}{a}\)
Chúc bạn học tốt!!
Cho các số hữu tỉ \(\dfrac{a}{b}\) và \(\dfrac{c}{d}\) với mẫu dương, trong đó \(\dfrac{a}{b}< \dfrac{c}{d}\) . Chứng minh rằng :
\(\dfrac{a}{b}< \dfrac{a+c}{b+d}< \dfrac{c}{d}\)
`a/b<(a+c)/(b+d)`
`<=>a(b+d)<b(a+c)`
`<=>ab+ad<ad<bc`
`<=>ad<bc`
`<=>a/b<c/d`(theo giả thiết)
`(a+c)/(b+d)<c/d`
`<=>d(a+c)<c(b+d)`
`<=>ad+cd<bc+dc`
`<=>ad<bc`
`<=>a/b<c/d`(theo giả thiết)`
`=>a/b<(a+c)/(b+d)<c/d`
Giúp cái ạ, giúp đi ạ, GẤP
1, tìm a sao cho
\(\dfrac{-5}{12}< \dfrac{a}{10}< \dfrac{1}{4}\)
2, cho \(\dfrac{a}{b}\dfrac{ }{ }\) và \(\dfrac{c}{d}\) ( b; d \(\ne\) 0 )
Chứng tỏ:
a, Nếu \(\dfrac{a}{b}< \dfrac{c}{d}\) thì a.d < b.c
b, Nếu a.d < b.c thì \(\dfrac{a}{b}< \dfrac{c}{d}\)
ở bài 1 câu a là \(\dfrac{2}{5}\) ạ, mình nhầm :>>
2 . a ) nếu \(\dfrac{a}{b}< \dfrac{c}{d}\)thì a.d < b.c
\(\dfrac{a}{b}\cdot\dfrac{c}{d}=\dfrac{ac}{bd}\Rightarrow\dfrac{a}{bd}< \dfrac{c}{bd}\Rightarrow a< c\)
vì a<c => a.d < b.c
=> đcpm
b) ko ghi lại đề
vì a.d<c.d => \(\dfrac{a}{bd}< \dfrac{c}{bd}\Rightarrow\dfrac{a}{b}< \dfrac{c}{d}\)( bn suy luận ngược với a nhé )
Cho tỉ lệ thức \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\). Chứng minh rằng \(\dfrac{a.d}{c.d}=\dfrac{a^2-b^2}{b^2-d^2}\)và \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
Đẳng thức đầu tiên sai:
Ví dụ: \(a=1;b=2;c=3;d=6\) thì \(\dfrac{a}{b}=\dfrac{c}{d}\)
Nhưng \(\dfrac{a.d}{c.d}\ne\dfrac{a^2-b^2}{b^2-d^2}\)
Với đẳng thức thứ 2:
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)
\(\Rightarrow\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
Cho tỉ lệ thức: \(\dfrac{a}{b}=\dfrac{c}{d}\left(b,d\ne0\right)\). Chứng minh rằng:
a) \(\dfrac{a}{a-b}=\dfrac{c}{c-d}\)
a: Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{k}{k-1}\)
\(\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{k}{k-1}\)
Do đó: \(\dfrac{a}{a-b}=\dfrac{c}{c-d}\)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\) . Chứng minh rằng
Với a2=b.c thì \(\dfrac{a+b}{a-b}+\dfrac{c+a}{c-a}\)
Có \(\dfrac{a}{b}=\dfrac{c}{d}=>ad=bc\) => a2 = ad => a=d
Xét \(\dfrac{a+b}{a-b}=\dfrac{c+a}{c-a}\)
<=> (a+b)(c-a) = (a-b)(c+a)
<=> (a+b)(c-d) = (a-b)(c+d)
<=> ac - ad + bc - bd = ac + ad -bc -bd
<=> 2bc = 2ad (luôn đúng) => đpcm
Cho các số thực a,b,c,d,e thỏa mãn \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}\)chứng minh rằng: \(\left(\dfrac{2019b+2020c-2021d}{2019c+2020d-2021e}\right)=\dfrac{a^2}{b.c}\)
Sửa: CMR: \(\left(\dfrac{2019b+2020c-2021d}{2019c+2020d-2021e}\right)^3=\dfrac{a^2}{bc}\)
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}=\dfrac{2019b+2020c-2021d}{2019c+2020d-2021e}\\ \Rightarrow\left(\dfrac{a}{b}\right)^3=\left(\dfrac{2019b+2020c-2021d}{2019c+2020d-2021e}\right)^3\left(1\right)\\ \dfrac{a}{b}=\dfrac{b}{c}=k\Rightarrow a=bk;b=ck\Rightarrow a=ck^2\\ \Rightarrow\dfrac{a^2}{bc}=\dfrac{c^2k^4}{ck\cdot c}=k^3=\left(\dfrac{a}{b}\right)^3\left(2\right)\\ \left(1\right)\left(2\right)\RightarrowĐpcm\)
cho 4 số dương a, b, c, d. chứng minh \(\dfrac{a+b}{b+c+d}+\dfrac{b+c}{c+d+a}+\dfrac{c+d}{d+a+b}+\dfrac{d+a}{a+b+c}\ge\dfrac{8}{3}\)