CMR nếu \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)(a,b,c,d khác 0). CMR \(\frac{a}{b}=\frac{c}{d}\)hoặc \(\frac{a}{b}=\frac{d}{c}\)
cho \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)với a,c,b,d khác 0,c khác +-d. CMR \(\frac{a}{b}=\frac{c}{d}hoặc,\frac{a}{b}=\frac{d}{c}\)
Ta có: \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\Rightarrow\frac{a^2+b^2}{ab}=\frac{c^2+d^2}{cd}\)
=> \(\frac{a^2}{ab}+\frac{b^2}{ab}=\frac{c^2}{cd}+\frac{d^2}{cd}\)
=> \(\frac{a}{b}+\frac{b}{a}=\frac{c}{d}+\frac{d}{c}\)
Mình chỉ làm được tới khúc này
Ta có: \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}=\frac{2ab}{2cd}=\frac{a^2+b^2+2ab}{c^2+d^2+2cd}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\left(\frac{a+b}{c+d}\right)^2\left(1\right)\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{2ab}{2cd}=\frac{a^2+b^2-2ab}{c^2+d^2-2cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\left(\frac{a-b}{c-d}\right)^2\left(2\right)\)
Từ (1) và (2) suy ra:
\(\left(\frac{a+b}{c+d}\right)^2=\left(\frac{a-b}{c-d}\right)^2\)
Trường hợp 1: \(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{\left(a+b\right)+\left(a-b\right)}{\left(c+d\right)+\left(c-d\right)}=\frac{2a}{2c}=\frac{a}{c}\left(3\right)\)
\(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{\left(a+b\right)-\left(a-b\right)}{\left(c+d\right)-\left(c-d\right)}=\frac{2b}{2d}=\frac{b}{d}\left(4\right)\)
Từ (3) và (4) suy ra \(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}\)
Trường hợp 2: \(\frac{a+b}{c+d}=\frac{-\left(a-b\right)}{c-d}=\frac{b-a}{c-d}=\frac{\left(a+b\right)+\left(b-a\right)}{\left(c+d\right)+\left(c-d\right)}=\frac{2b}{2c}=\frac{b}{c}\left(5\right)\)
\(\frac{a+b}{c+d}=\frac{b-a}{c-d}=\frac{\left(a+b\right)-\left(b-a\right)}{\left(c+d\right)-\left(c-d\right)}=\frac{2a}{2d}=\frac{a}{d}\left(6\right)\)
Từ (5) và (6) suy ra \(\frac{b}{c}=\frac{a}{d}\Rightarrow\frac{a}{b}=\frac{d}{c}\)
cho \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\) a,b,c,d khác 0
CMR: \(\frac{a}{b}=\frac{c}{d}\) hoặc \(\frac{a}{b}=\frac{d}{c}\)
1.Biết : \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)với a ,b ,c ,d khác 0
CMR: \(\frac{a}{b}=\frac{c}{d}ho\text{ặc}\frac{a}{b}=\frac{b}{c}\)
Áp dụng tính chất của dãy tỉ số bằng nhau
a^2+b^2/c^2+d^2 = a^2/c^2 = b^2 / d^2
=>a/c = b/d
=>a/b = c/d
Chúc bạn học tốt nha
dat k ; ta co a= bk , c=dk , roi tu thay vao ma rut gon nhe
Ta có \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}=\frac{2ab}{2cd}\)
\(\Rightarrow\frac{a^2+b^2}{c^2+d^2}=\frac{2ab}{2cd}=\frac{a^2+b^2+2ab}{c^2+d^2+2cd}=\frac{a^2+b^2-2ab}{c^2+d^2-2cd}\)
\(\Rightarrow\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
\(\Rightarrow\frac{a-b}{c-d}=\frac{a+b}{c+d}=\frac{a-b-a-b}{c-d-c-d}=\frac{a-b+a+b}{c-d+c+d}\)
\(\Rightarrow\frac{2b}{2d}=\frac{2a}{2c}\Rightarrow\frac{a}{b}=\frac{c}{d}\)
Biết \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\) với a,b,c,d \(\ne\)0.CMR \(\frac{a}{b}=\frac{c}{d}\)hoặc \(\frac{a}{b}=\frac{d}{c}\)
1, Cho \(\frac{a}{b}=\frac{c}{d}\)( b,c,d khác 0; c+đ khác 0). CMR:
\(\frac{ab}{cd}=\frac{\left(a+b\right)^2}{\left(c+\text{d}\right)^2}\)
a/b=c/d
=>a/c=b/d=a+b/c+d
=>a/b.c/d=(a+b)^2/(c+d)^2
=>ab/cd=(a+b)^2/(c+d)^2
Vay......
a/b=c/d
=> a/c=b/d=a+b/c+d
=> a/b.c/d=(a+b)^2/(c+d)^2
=> ab/cd=(a+b)^2/(c+d)^2
# Hok_tốt nha
Cho\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
CMR:\(\frac{a}{b}=\frac{c}{d}\)hoặc \(\frac{a}{b}=\frac{d}{c}\)
Cho a,b,c,d>0, ab+bc+cd+da=3. CMR \(\frac{a}{b^2+c^2+d^2}+\frac{b}{c^2+d^2+a^2}+\frac{c}{d^2+a^2+b^2}+\frac{d}{a^2+b^2+c^2}>\frac{4}{a+b+c+d}\)
4.Biết : \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)Cmr:
\(\frac{a}{b}=\frac{c}{d}\) hoặc \(\frac{a}{b}=\frac{d}{c}\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
=> \(\left(a^2+b^2\right)cd=ab\left(c^2+d^2\right)\)
\(\Leftrightarrow a^2cd+b^2cd-abc^2-abd^2=0\)
\(\Leftrightarrow a^2cd-abc^2+b^2cd-abd^2=0\)
\(\Leftrightarrow ac\left(ad-bc\right)+bd\left(bc-ad\right)=0\)
\(\Leftrightarrow ac\left(ad-bc\right)-bd\left(ad-bc\right)=0\)
\(\Leftrightarrow\left(ad-bc\right)\left(ac-bd\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}ad-bc=0\\ac-bd=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}ad=bc\\ac=bd\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{a}{b}=\frac{c}{d}\\\frac{a}{b}=\frac{d}{c}\end{cases}}\) (DPCM)
CMR: nếu\(\frac{a}{b}=\frac{c}{d}thì\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)