Cho \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}\) \(\left(a,b,c,d\ne0;a+b+c+d\ne0\right)\)
Tính: \(M=\frac{3a-2b}{c+d}+\frac{3b-2c}{d+a}+\frac{3c-2d}{a+b}+\frac{3d-2a}{b+c}\)
cho dãy tỉ số bằng nhau
\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}\)
\(=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)
tính giá trị biểu thức \(M=\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}\)
\(\left(a,b,c,d\ne0;a+b+c+d\ne0;a+b\ne0;b+c\ne0;c+d\ne0;d+a\ne0\right)\)
1) Cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\)
Chứng minh \(\frac{a}{b}=\frac{a-c}{b-d}\left(b,d\ne0\right)\)
2) Cho \(\frac{a}{b}=\frac{c}{d}\)
Chứng minh \(\frac{a+b}{a-b}=\frac{c+d}{c-d}\left(a-b\ne0;c-d\ne0\right)\)
1) \(\frac{a}{b}=\frac{c}{d}=\frac{a-c}{b-d}\)
-->\(\frac{a}{b}=\frac{a-c}{b-d}\left(đpcm\right)\)
2) ta có \(\frac{a}{b}=\frac{c}{d}\)
đặt a=kb và c=kd
\(\frac{a+b}{a-b}=\frac{kb+b}{kb-b}=\frac{b\left(k+1\right)}{b\left(k-1\right)}=\frac{k+1}{k-1}\left(1\right)\)
\(\frac{c+d}{c-d}=\frac{kd+d}{kd-d}=\frac{d\left(k+1\right)}{d\left(k-1\right)}=\frac{k+1}{k-1}\left(2\right)\)
từ (1) và (2) --> \(\frac{a+b}{a-b}=\frac{c+d}{c-d}\left(đpcm\right)\)
chứng minh Từ \(\frac{a}{b}=\frac{c}{d}\left(\left(a-b\right)\ne0,\left(c-d\right)\ne0\right)\Rightarrow\frac{a+b}{a-b}=\frac{c+d}{c-d}\)
Cho \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}\)\(\left(a+b+c+d\ne0\right)\)Tìm M = \(\frac{2a-b}{c+d}=\frac{2b-c}{d+a}=\frac{2c-d}{a+b}=\frac{2d-a}{b+c}\)
Ta có : \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}\) (đề bài)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{b+c+d+a}=1\)
\(\Rightarrow\begin{cases}\frac{a}{b}=1\\\frac{b}{c}=1\\\frac{c}{d}=1\\\frac{d}{a}=1\end{cases}\Rightarrow\begin{cases}a=b\\b=c\\c=d\\d=a\end{cases}\)
\(\Rightarrow a=b=c=d\)
Thay \(b=a\) ; \(c=a\) ; \(d=a\) vào biểu thức \(M=\frac{2a-b}{c+d}=\frac{2b-c}{d+a}=\frac{2c-d}{a+b}=\frac{2d-a}{b+c}\) ta có :
\(M=\frac{2a-a}{a+a}=\frac{2a-a}{a+a}=\frac{2a-a}{a+a}=\frac{2a-a}{a+a}\)
\(M=\frac{1a}{2a}=\frac{1a}{2a}=\frac{1a}{2a}=\frac{1a}{2a}=\frac{1}{2}\)
Vậy \(M=\frac{1}{2}\)
Cho a = b = c và \(c=\frac{bd}{b-d}\left(b\ne0;d\ne0\right)\)
Chứng minh \(\frac{a}{b}=\frac{c}{d}\)
Ta có :
\(c=\frac{bd}{b-d}\)
\(\Rightarrow b-d=\frac{bd}{c}\left(c\ne0\right)\)
\(a=b+c\Rightarrow c=a-b\)
\(\Rightarrow c=\frac{bd}{b-d}=a-b\)
\(\Rightarrow bd=\left(a-b\right).\left(b-d\right)\)
\(\Rightarrow ab-ad-b^2+bd=bd\)
\(\Rightarrow a\left(b-d\right)-b^2=0\)
\(\Rightarrow a.\frac{bd}{c}-b^2=0\)
\(\Rightarrow\frac{ad}{c}-b=0\)
\(\Rightarrow\frac{ad-bc}{c}=0\)
\(\Rightarrow ad-bc=0\)
\(\Rightarrow ad=bc\)
\(\Rightarrow\frac{a}{b}=\frac{c}{d}\left(đpcm\right)\)
Chúc bạn học tốt !!!
Cho \(\frac{a}{b}=\frac{c}{d}\left(a,b,c\ne0;a\ne b,c\ne d\right)\).CMR: \(\frac{a}{a-b}=\frac{c}{c-d}\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=bk;c=dk\)
\(\Rightarrow VT=\frac{bk}{bk-b}=\frac{bk}{b\left(k-1\right)}=\frac{k}{k-1}\left(1\right)\)
\(\Rightarrow VP=\frac{c}{c-d}=\frac{dk}{dk-d}=\frac{dk}{d\left(k-1\right)}=\frac{k}{k-1}\left(2\right)\)
Từ (1) và (2) =>Đpcm
cho \(\frac{a}{b}=\frac{c}{d}\left(b,c,d\ne0;c-2d\ne0\right)\)
chứng minh rằng \(\frac{\left(a-2b^4\right)}{\left(c-2d^4\right)}=\frac{a^4+2017b^4}{c^4+2017d^a}\)
cho \(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}\left(a,b,c,d\ne0\right)\)
Tính \(A=\frac{2011a-2010b}{c+d}+\frac{2011b+2010c}{a+d}+\frac{2011c-2010d}{a+b}+\frac{2011d-2010a}{b+c}\)
Áp dụng TC DTSBN ta có :
\(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}=\frac{a+b+c+d}{2b+2c+2d+2a}=\frac{a+b+c+d}{2\left(a+b+c+d\right)}=\frac{1}{2}\)
\(\Rightarrow\frac{a}{2b}=\frac{1}{2}\Rightarrow a=\frac{1}{2}.2b\Rightarrow a=b\) (1)
\(\Rightarrow\frac{b}{2c}=\frac{1}{2}\Rightarrow b=\frac{1}{2}.2c\Rightarrow b=c\) (2)
\(\Rightarrow\frac{c}{2a}=\frac{1}{2}\Rightarrow c=\frac{1}{2}.2a\Rightarrow c=a\) (3)
\(\Rightarrow\frac{d}{2a}=\frac{1}{2}\Rightarrow d=\frac{1}{2}.2a\Rightarrow d=a\) (4)
Từ (1);(2);(3):(4) \(\Rightarrow a=b=c=d\) .Thay vào A ta được :
\(A=\frac{2011a-2010a}{a+a}+\frac{2011a+2010a}{a+a}+\frac{2011a-2010a}{a+a}+\frac{2011a-2010a}{a+a}\)
\(=\frac{a}{2a}+\frac{4021a}{2a}+\frac{a}{2a}+\frac{a}{2a}=\frac{a+4021a+a+a}{2a}=\frac{4024a}{2a}=\frac{4024}{2}=2012\)
Vậy \(A=2012\)
cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\left(a,b,c,d\ne0\right),a\ne\pm b,c\ne\pm d\)
cm \(\frac{a+b}{a-d}=\frac{c+d}{c-d}\)
Lời giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=t(t\neq \pm 1)\) \(\Rightarrow a=bt;c=dt\)
Khi đó:
\(\frac{a+b}{a-b}=\frac{bt+b}{bt-b}=\frac{b(t+1)}{b(t-1)}=\frac{t+1}{t-1}\)
\(\frac{c+d}{c-d}=\frac{dt+d}{dt-d}=\frac{d(t+1)}{d(t-1)}=\frac{t+1}{t-1}\)
\(\Rightarrow \frac{a+b}{a-b}=\frac{c+d}{c-d}\) (đpcm)
Cách khác:
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau,ta có:
\(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}=\frac{a-b}{c-d}\Rightarrow\frac{a+b}{a-b}=\frac{c+d}{c-d}\left(đpcm\right)\)