a) CMR: Nếu\(\frac{a}{b}=\frac{c}{d}\)thì\(\frac{a}{b}=\frac{a+c}{b+d}\)
b) Cho\(\frac{a+b}{a-b}=\frac{c+a}{c-a}\). CMR: a2 = bc
Bài 1 : a) CMR : nếu \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\)thì \(a=b=c\)
b) CMR : nếu \(\frac{a}{b}=\frac{c}{d}=\frac{p}{q}\)thì \(\frac{ma+nc+ep}{mb+nd+eq}=\frac{a}{b}=\frac{c}{d}=\frac{p}{q}\)
TA CÓ \(\frac{a}{b}=\frac{c}{d}=\frac{p}{q}=\frac{am}{bm}=\frac{nc}{nd}=\frac{ep}{eq}\)
ÁP DỤNG TÍNH CHẤT DÃY TỈ SỐ BẰNG NHAU TA CÓ
\(\frac{a}{b}=\frac{c}{d}=\frac{p}{q}=\frac{ma}{mb}=\frac{nc}{nd}=\frac{ep}{eq}=\frac{ma+nc+ep}{mb+nd+eq}\)(ĐPCM)
ADTC dãy tỉ số bằng nhau ta có \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1\)
\(\Rightarrow\hept{\begin{cases}a=b\cdot1=b\\b=c\cdot1=c\\c=a\cdot1=a\end{cases}\Leftrightarrow a=b=c}\)
CMR : nếu \(\frac{a+b}{a-b}=\frac{c+d}{c-d}thì\frac{a}{b}=\frac{c}{d}\)
Cách 1:Đặt \(\frac{a}{b}=\frac{c}{d}=k;\Rightarrow a=bk,c=dk\Leftrightarrow\)
\(\frac{a}{b}=\frac{bk}{b}=k\left(1\right)\)
\(\frac{c}{d}=\frac{dk}{d}=k\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\frac{a}{b}=\frac{c}{d}\)
Cách 2:Đặt: a/b = c/d = k => a = bk, c = dk
Ta có:
a + b/a - b = bk + b/bk - b = b(k+1)/ b(k-1) = k+1/k-1 (1)
c + d/c- d = dk +d/ dk - d = d(k+1)/d(k-1) = k+1/k-1 (2)
Từ (1) và (2) => a+b/a-b = c+d/c-d
Cho hai số hữu tỉ\(\frac{a}{b}\)và\(\frac{c}{d}\)(b>0,d>0).CMR
a,Nếu \(\frac{a}{b}< \frac{c}{d}\)thì ad<bc
b,Nếu ad<bc thì\(\frac{a}{b}< \frac{c}{d}\)
Cho a,b,c,d thỏa mãn $\frac{a}{b}$ =$\frac{b}{c}$ =$\frac{c}{d}$ =$\frac{d}{a}$
CMR:($\frac{2019b+2020c-2021d}{2019c+2020d-2021e}$)^3=$\frac{a^2}{bc}$
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{a}=\dfrac{a+b+c+d}{a+b+c+d}=1\\ \Rightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=d\\d=a\end{matrix}\right.\Rightarrow a=b=c=d\\ \Rightarrow VT=\left(\dfrac{2019a+2020a-2021a}{2019a+2020a-2021a}\right)^3=1^3=1=\dfrac{a^2}{a\cdot a}=VP\)
CMR nếu \(\frac{a}{b}=\frac{c}{d}\)thì \(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\)
a, Cho a,b>0 , CMR: \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
b. Cho a,b,c,d > 0. CMR: \(\frac{a-d}{d+b}+\frac{d-b}{b+c}+\frac{b-c}{c+a}+\frac{c-a}{a+d}\ge0\)
a/ Biến đổi tương đương:
\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\Leftrightarrow\frac{a+b}{ab}\ge\frac{4}{a+b}\)
\(\Leftrightarrow\left(a+b\right)^2\ge4ab\Leftrightarrow a^2+2ab+b^2\ge4ab\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\Leftrightarrow\left(a-b\right)^2\ge0\) (luôn đúng)
Vậy BĐT được chứng minh
b/ \(VT=\frac{a-d}{b+d}+1+\frac{d-b}{b+c}+1+\frac{b-c}{a+c}+1+\frac{c-a}{a+d}+1-4\)
\(VT=\frac{a+b}{b+d}+\frac{c+d}{b+c}+\frac{a+b}{a+c}+\frac{c+d}{a+d}-4\)
\(VT=\left(a+b\right)\left(\frac{1}{b+d}+\frac{1}{a+c}\right)+\left(c+d\right)\left(\frac{1}{b+c}+\frac{1}{a+d}\right)-4\)
\(\Rightarrow VT\ge\left(a+b\right).\frac{4}{b+d+a+c}+\left(c+d\right).\frac{4}{b+c+a+d}-4\)
\(\Rightarrow VT\ge\frac{4}{\left(a+b+c+d\right)}\left(a+b+c+d\right)-4=4-4=0\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=d\)
1) CMR: nếu \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{d}{e}\) thì a\(\frac{a}{e}=\left(\frac{a-b+c-d}{b-c+d-e}\right)^4\)
a) Cho \(\frac{a}{b}\)= \(\frac{b}{c}\)=\(\frac{c}{d}\)
CMR:(\(\frac{a+b+c}{b+c+d}^{ }\))\(^3\)=\(\frac{a}{d}\)
b) Cho \(\frac{a1}{a2}\)=\(\frac{a2}{a3}\)=\(\frac{a3}{a4}\)=...=\(\frac{a2008}{a2009}\)
CMR:\(\frac{a1}{a2009}\)=(\(\frac{a1+a2+a3+...+a2008}{a2+a3+a4+...+a2009}\))\(^{2008}\)
c) Cho \(\frac{a}{2003}\)=\(\frac{b}{2004}\)=\(\frac{c}{2005}\)
CMR: 4(a-b)(b-c)=(c-a)\(^{^2}\)
Cho a,b,c,d thuộc Z (b>0,d>0).CMR nếu \(\frac{a}{b}<\frac{c}{d}\) thì\(\frac{a}{b}<\frac{a+b}{c+d}<\frac{c}{d}\)