Cho \(\frac{a}{b}=\frac{c}{d}\)CMR \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
Cho : \(\frac{a}{b}=\frac{c}{d}CMR:\)\(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}v\text{à}\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)
\(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\)
\(\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}=\frac{a^2-b^2}{c^2-d^2}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\left(\frac{a+b}{c+d}\right)^2=\frac{ab}{cd}\)
Vậy \(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)và \(\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)
Cho \(\frac{a}{b}=\frac{c}{d}\). CMR:
a) \(\frac{a^2-b^2}{c^2-d^2}=\frac{ab}{cd}\)
b) \(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{ab}{cd}\)
a) áp dụng tính chất của dãy tỉ số bằng nhau ta có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-c^2}{b^2-d^2}\)
Do \(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{ab}{cd}\)=> đpcm
b) áp dụng tính chất của dãy tỉ số bằng nhau ta có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{ab}{cd}=\left(\frac{a-c}{b-d}\right)^2\)=> đpcm
Cho a,b,c,d>0, ab+bc+cd+da=3. CMR \(\frac{a}{b^2+c^2+d^2}+\frac{b}{c^2+d^2+a^2}+\frac{c}{d^2+a^2+b^2}+\frac{d}{a^2+b^2+c^2}>\frac{4}{a+b+c+d}\)
cho \(\frac{a}{b}< \frac{c}{d}\). CMR\(\frac{a}{b}< \frac{ab+cd}{b^2+d^2}< \frac{c}{d}\)
Cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}.CMR:\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)
Từ giả thiết: \(\frac{a}{b}=\frac{c}{d}\)=>ad=bc (1)
Ta có: ab(c2-d2)=abc2-abd2=acbc-adbd (2)
cd(a2-b2)=a2cd-b2cd=acad-bcbd (3)
Từ (1) ,(2),(3)=> ab(c2-d2)=cd(a2-b2)=>\(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\) (đpcm)
cho tỉ lệ thức : \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}cmr\frac{a}{b}=\frac{c}{d}\)
(a² + b²) / (c² + d²) = ab/cd
<=> (a² + b²)cd = ab(c² + d²)
<=> a²cd + b²cd = abc² + abd²
<=> a²cd - abc² - abd² + b²cd = 0
<=> ac(ad - bc) - bd(ad - bc) = 0
<=> (ac - bd)(ad - bc) = 0
<=> ac - bd = 0 hoặc ad - bc = 0
<=> ac = bd hoặc ad = bc
<=> a/b = d/c hoặc a/b = c/d (đpcm)
Ta có : \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}=\frac{2ab}{2cd}=\frac{a^2+2ab+b^2}{c^2+2cd+d^2}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{ab}{cd}\)
\(\Rightarrow\frac{\left(a+b\right)\left(a+b\right)}{\left(c+d\right)\left(c+d\right)}=\frac{ab}{cd}\)
\(\Rightarrow\frac{c\left(a+b\right)}{a\left(c+d\right)}=\frac{b\left(c+d\right)}{d\left(a+b\right)}=\frac{ca+cb}{ac+ad}=\frac{bc+db}{da+db}=\frac{ca-bd}{ca-bd}=1\)
\(\Rightarrow ca+cb=ac+ad\Rightarrow cb=ad\Rightarrow\frac{a}{b}=\frac{c}{d}\)
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 \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
CMR \(\left[{}\begin{matrix}\frac{a}{b}=\frac{c}{d}\\\frac{a}{b}=\frac{d}{c}\end{matrix}\right.\)
Cho:
\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}.CMR:\frac{a}{b}=\frac{c}{d}\)
(a^2+b^2)/(c^2+d^2)=a^2/c^2=b^2/d^2(theo t/c dãy tỉ số bằng nhau)
a^2/c^2=ab/cd<=>a/c=b/d<=>a/b=c/d(đpcm)
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 \(\frac{a}{b}=\frac{d}{c}\)
em gửi bài qua fb của thầy nhé thầy HD giải cho, tìm fb của thầy qua sđt: 0975705122
Ta có :
\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}=\frac{2ab}{2cd}=\frac{a^2+b^2+2ab}{c^2+d^2+2cd}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\left(\frac{a+b}{c+d}\right)^2\)( 1 )
\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}=\frac{2ab}{2cd}=\frac{a^2+b^2-2ab}{c^2+d^2-2cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\left(\frac{a-b}{c-d}\right)^2\)( 2 )
Từ ( 1 ) và ( 2 ) suy ra : \(\left(\frac{a+b}{c+d}\right)^2=\left(\frac{a-b}{c-d}\right)^2\)
TH1 : \(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{\left(a+b\right)+\left(a-b\right)}{\left(c+d\right)+\left(c-d\right)}=\frac{2a}{2c}=\frac{a}{c}\)( 3 )
TH2 : \(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{\left(a+b\right)-\left(a-b\right)}{\left(c+d\right)-\left(c-d\right)}=\frac{2b}{2d}=\frac{b}{d}\)( 4 )
Từ ( 3 ) và ( 4 ) suy ra : \(\frac{a}{c}=\frac{b}{d}\)hay \(\frac{a}{b}=\frac{c}{d}\)
TH2 : \(\frac{a+b}{c+d}=\frac{b-a}{d-c}=\frac{2b}{2c}=\frac{b}{c}\)( 5 )
\(\frac{a+b}{c+d}=\frac{b-a}{d-c}=\frac{2a}{2d}=\frac{a}{d}\)( 6 )
Từ ( 5 ) và ( 6 ) suy ra : \(\frac{b}{c}=\frac{a}{d}\)hay \(\frac{a}{b}=\frac{d}{c}\)
Vậy nếu \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)thì \(\orbr{\begin{cases}\frac{a}{b}=\frac{c}{d}\\\frac{a}{b}=\frac{d}{c}\end{cases}}\)