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\(\dfrac{a+b}{b+c}=\dfrac{c+d}{d+a}\)
=>(a+b)(d+a)=(c+d)(b+c)
\(\Leftrightarrow ad+a^2+bd+ab=cb+c^2+db+dc\)
\(\Leftrightarrow ad+a^2+ab=cb+c^2+dc\)
\(\Leftrightarrow d\left(a-c\right)+\left(a+c\right)\left(a-c\right)+b\left(a-c\right)=0\)
\(\Leftrightarrow\left(a-c\right)\left(a+b+c+d\right)=0\)
=>a=c hoặc a+b+c+d=0(đpcm)
Cho \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\) với ( với a, b, c, d khác 0, và c \(\ne\pm d\) ). Chứng minh rằng hoặc \(\dfrac{a}{b}=\dfrac{c}{d}\) hoặc \(\dfrac{a}{b}=\dfrac{d}{c}\) ?
Bài 1: Nhân chéo
Bài 2:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
\(\Rightarrow\left(\dfrac{a}{b}\right)^3=\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}\)
\(\Rightarrow\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)
\(\Rightarrowđpcm\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a+b+c}{a+b-c}=\dfrac{a-b+c}{a-b-c}\)
\(=\dfrac{a+b+c-a+b-c}{a+b-c-a+b+c}\)
\(=\dfrac{\left(a-a\right)+\left(b+b\right)+\left(c-c\right)}{\left(a-a\right)+\left(b+b\right)+\left(c-c\right)}\)
\(=\dfrac{2b}{2b}=1\)
\(\Rightarrow a+b+c=a+b-c\)
\(\Rightarrow c=-c\)
\(\Rightarrow c+c=0\)
\(\Rightarrow2c=0\Rightarrow c=0\)
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a.b.c}{b.c.d}=\dfrac{a}{d}\left(1\right)\)
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\Rightarrow\left(\dfrac{a}{b}\right)^3=\left(\dfrac{b}{c}\right)^3=\left(\dfrac{c}{d}\right)^3\)
\(=\left(\dfrac{a+b+c}{b+c+d}\right)^3\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) ta có:
\(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)
Ta có :
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\)
\(\Leftrightarrow\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}\)\(\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{ab}{cd}\)
\(\Leftrightarrow\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{ab}{cd}\)
\(\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
\(\Leftrightarrow\dfrac{a}{b}=\dfrac{c}{d}\left(đpcm\right)\)
a: Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{k}{k-1}\)
\(\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{k}{k-1}\)
Do đó: \(\dfrac{a}{a-b}=\dfrac{c}{c-d}\)
b: Đặt a/b=c/d=k
=>a=bk; c=dk
\(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bk+b}{dk+d}\right)^2=\dfrac{b^2}{d^2}\)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2}{d^2}\)
DO đó: \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
(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)
\(\Leftrightarrow a^2-\left(b+c\right)^2=a^2-\left(b-c\right)^2\)
=>(b+c)^2=(b-c)^2
=>b+c=b-c hoặc b+c=c-b
=>b=0(loại) hoặc c=0
\(\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{a+b}{c}\)
\(\Rightarrow\dfrac{b+c}{a}+1=\dfrac{c+a}{b}+1=\dfrac{a+b}{c}+1\)
\(\Rightarrow\dfrac{a+b+c}{a}=\dfrac{a+b+c}{b}=\dfrac{a+b+c}{c}\)
Từ \(\dfrac{a+b+c}{a}=\dfrac{a+b+c}{b}\Leftrightarrow a\left(a+b+c\right)=b\left(a+b+c\right)\)
\(\Rightarrow\left(a-b\right)\left(a+b+c\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=b\\a+b+c=0\end{matrix}\right.\left(đpcm\right)\)