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AD t/c DTS = nhau ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{a+c}{b+d}\left(dpcm\right)\)
Theo bài ra:
\(\dfrac{a}{b+c}=\dfrac{b}{a+c}=\dfrac{c}{a+b};a\ne b\ne c;a,b,c\ne0\)
\(P=\dfrac{b+c}{a}+\dfrac{a+c}{b}+\dfrac{a+b}{c}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{b+c}=\dfrac{b}{a+c}=\dfrac{c}{a+b}=\dfrac{a+b+c}{b+c+a+c+a+b}=\dfrac{a+b+c}{2a+2b+2c}=\dfrac{a+b+c}{2\left(a+b+c\right)}=\dfrac{1}{2}\)
\(hay:\dfrac{a}{b+c}=\dfrac{1}{2}\Rightarrow a=\dfrac{b+c}{2}\)
Thay \(a=\dfrac{b+c}{2}\) vào \(P\), ta có:
\(P=\dfrac{b+c}{\dfrac{b+c}{2}}+\dfrac{b+c+c}{b}+\dfrac{b+c+b}{c}\\ P=\dfrac{2\left(b+c\right)}{b+c}+\dfrac{2c+b}{b}+\dfrac{2b+c}{c}\\ P=2+\dfrac{2c}{b}+\dfrac{b}{b}+\dfrac{2b}{c}+\dfrac{c}{c}\\ P=2+\dfrac{2c}{b}+1+\dfrac{2b}{c}+1\\ P=\left(2+1+1\right)+\dfrac{2c}{b}+\dfrac{2b}{c}\\ P=4+\dfrac{2c}{b}+\dfrac{2b}{c}\\ P=4+\dfrac{2c+2b}{b+c}\\ P=4+\dfrac{2\left(b+c\right)}{b+c}\\ P=4+2\\ P=6\)
Vậy: \(P=6\)
Theo dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=\dfrac{a+b+c}{b+c+a}=1\) (vì \(a+b+c\ne0\))
\(\Rightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\Rightarrow a=b=c=\pm1\)
a, ta có :
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a}{2c}=\dfrac{2b}{2d}\)
áp dụng tính chất dă y tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a}{2c}=\dfrac{2b}{2d}=\dfrac{a+2b}{c+2d}=\dfrac{2a-b}{2c-d}\)
\(\Rightarrow\dfrac{a+2b}{c+2d}=\dfrac{2a-b}{2c-d}\Rightarrow\dfrac{a+2b}{2a-b}=\dfrac{c+2d}{2c-d}\) (ĐPCM)
b, ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{3c}{3d}\)
áp dụng tính chất dă tỉ số bằng nhau ta có :
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{3c}{3d}=\dfrac{a+3c}{b+3d}=\dfrac{a-c}{b-d}\)
\(\Rightarrow\dfrac{a+3c}{b+3d}=\dfrac{a-c}{b-d}\)
\(\Rightarrow\left(a+3c\right)\left(b-d\right)=\left(b+3d\right)\left(a-c\right)\) (ĐPCM)
Bài 1:
\(3^{-1}.3^n+4.3^n=13.3^5\)
\(\Rightarrow3^{n-1}+4.3.3^{n-1}=13.3^5\)
\(\Rightarrow3^{n-1}\left(1+4.3\right)=13.3^5\)
\(\Rightarrow3^{n-1}.13=13.3^5\)
\(\Rightarrow3^{n-1}=3^5\)
\(\Rightarrow n-1=5\)
\(\Rightarrow n=6\)
Vậy n = 6
Bài 2a: Câu hỏi của Nguyễn Trọng Phúc - Toán lớp 7 | Học trực tuyến
a/ Đặt :
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có :
\(VT=\dfrac{a-b}{a+b}=\dfrac{bk-b}{bk+b}=\dfrac{b\left(k-1\right)}{b\left(k+1\right)}=\dfrac{k-1}{k+1}\)\(\left(1\right)\)
\(VP=\dfrac{c-d}{c+d}=\dfrac{dk-d}{dk+d}=\dfrac{d\left(k-1\right)}{d\left(k+1\right)}=\dfrac{k-1}{k+1}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)
b/ Đặt :
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(VT=\dfrac{2a+5b}{3a-4b}=\dfrac{2bk+5b}{3bk-4b}=\dfrac{b\left(2k+5\right)}{b\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\left(1\right)\)
\(VP=\dfrac{2c+5d}{3c-4d}=\dfrac{2dk+5d}{3dk-4d}=\dfrac{d\left(2k+5\right)}{d\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)
a) Từ \(\dfrac{a}{b}=\dfrac{c}{d}\) \(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)
Từ \(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\) \(\Rightarrow\dfrac{c-d}{c+d}=\dfrac{a-b}{a+b}\)
b) Từ \(\dfrac{a}{b}=\dfrac{c}{d}\) \(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a}{2c}=\dfrac{3a}{3c}=\dfrac{4b}{4d}=\dfrac{5b}{5d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{2a}{2c}=\dfrac{3a}{3c}=\dfrac{4b}{4d}=\dfrac{5b}{5d}=\dfrac{2a+5b}{2c+5d}=\dfrac{3a-4b}{3c-4d}\)
Từ \(\dfrac{2a+5b}{2c+5d}=\dfrac{3a-4b}{3c-4d}\) \(\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)
Đặt a/b=c/d=k
=>a=bk; c=dk
\(\left(\dfrac{a+b}{c+d}\right)^3=\left(\dfrac{bk+b}{dk+d}\right)^3=\dfrac{b^3}{d^3}\)
\(\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{b^3k^3+b^3}{d^3k^3+d^3}=\dfrac{b^3}{d^3}\)
Do đó: \(\left(\dfrac{a+b}{c+d}\right)^3=\dfrac{a^3+b^3}{c^3+d^3}\)
thiếu đè
Thiếu nhé:
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\)
\(\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\)
Ta có điều phải chứng minh