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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:
\(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}=\frac{a-b}{c-d}\Rightarrow\frac{a+b}{a-b}=\frac{c+d}{c-d}\left(đpcm\right)\)
Ta có: \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
\(\Leftrightarrow\left(a^2+b^2\right)\cdot cd=\left(c^2+d^2\right)\cdot ab\)
\(\Rightarrow a^2\cdot cd+b^2\cdot cd=c^2\cdot ab+d^2\cdot ab\)
\(\Rightarrow a^2\cdot cd+b^2\cdot cd-c^2\cdot ab-d^2\cdot ab=0\)
\(\Rightarrow\left(a^2\cdot cd-c^2\cdot ab\right)+\left(b^2\cdot cd-d^2\cdot ab\right)=0\)
\(\Rightarrow ac\cdot\left(ad-bc\right)+bd\cdot\left(bc-ad\right)=0\)
\(\Rightarrow ac\cdot\left(ad-bc\right)-bd\cdot\left(ad-bc\right)=0\)
\(\Rightarrow\left(ac-bd\right)\cdot\left(ad-bc\right)=0\)
Tự làm tiếp nhé.......
a)Gọi \(\frac{a}{b}=\frac{c}{d}=k\)
Suy ra \(\begin{cases}a=bk\\c=dk\end{cases}\)
Xét VT \(\frac{a}{a+b}=\frac{bk}{bk+b}=\frac{bk}{b\left(k+1\right)}=\frac{k}{k+1}\left(1\right)\)
Xét 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
b)Gọi \(\frac{a}{b}=\frac{c}{d}=k\)
Suy ra \(\begin{cases}a=bk\\c=dk\end{cases}\)
Xét VT \(\frac{a}{a-b}=\frac{bk}{bk-b}=\frac{bk}{b\left(k-1\right)}=\frac{k}{k-1}\left(1\right)\)
Xét 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
Ta có\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
<=> cd(a2 + b2) = ab(c2 + d2)
<=> a2cd + b2cd - abc2 - abd2 = 0
<=> (a2cd - abc2) + (b2cd - abd2) = 0
<=> ac(ad - bc) + bd(bc - ad) = 0
<=> ac(ad - bc) - bd(ad - bc) = 0
<=> (ac - bd)(ad - bc) = 0
<=> \(\orbr{\begin{cases}ac-bd=0\\ad-bc=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}ac=bd\\ad=bc\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{a}{d}=\frac{b}{c}\\\frac{a}{b}=\frac{c}{d}\end{cases}}\left(\text{đpcm}\right)\)
\(\Leftrightarrow\left(a^{2016}+b^{2016}\right).\left(c^{2016}-d^{2016}\right)=\left(a^{2016}-b^{2016}\right).\left(c^{2016}+d^{2016}\right)\)
\(\Leftrightarrow ac^{2016}-ad^{2016}+bc^{2016}-bd^{2016}=ac^{2016}+ad^{2016}-bc^{2016}-bd^{2016}\)
\(\Leftrightarrow-\left(ad^{2016}-bc^{2016}\right)=ad^{2016}-bc^{2016}\)
nếu \(-\left(ad^{2016}-bc^{2016}\right)=ad^{2016}-bc^{2016}=0\)
\(\Rightarrow ad^{2016}-bc^{2016}=0\Rightarrow ad=bc\Rightarrow\frac{a}{b}=\frac{c}{d}\left(1\right)\)
nếu \(\text{}-\left(ad^{2016}-bc^{2016}\right)=ad^{2016}-bc^{2016}\ne0\Rightarrow ad=-bc\Rightarrow\frac{a}{b}=-\frac{c}{d}\left(2\right)\)
từ (1) và (2) => đpcm
a. Ta có: a+b/b = (a+b):b = a:b + b:b = a/b +1
Ta có: c+d/d = (c+d):d = c:d + d:d = c/d +1
Ta có: a/b = c/d
1 = 1
=> a+b/b = c+d/d
Áp dụng cách tương tự cho các câu khác!
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\)
\(\Rightarrow\left(\frac{a}{c}\right)^3=\left(\frac{a+b}{c+d}\right)^3\)
\(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^3}{c^3}=\frac{b^3}{d^3}\)
\(\frac{a^3}{c^3}=\frac{b^3}{d^3^.}=\frac{a^3-b^3}{c^3-d^3}\)
Vậy \(\left(\frac{a+b}{c+d}\right)^3=\frac{a^3-b^3}{c^3-d^3}\)
a) Ta có: \(\frac{a}{b}=\frac{c}{d}.\)
\(\Rightarrow\frac{a}{b}+1=\frac{c}{d}+1\)
\(\Rightarrow\frac{a}{b}+\frac{b}{b}=\frac{c}{d}+\frac{d}{d}\)
\(\Rightarrow\frac{a+b}{b}=\frac{c+d}{d}\left(đpcm1\right).\)
b) Ta có: \(\frac{a}{b}=\frac{c}{d}.\)
\(\Rightarrow\frac{a}{b}-1=\frac{c}{d}-1\)
\(\Rightarrow\frac{a}{b}-\frac{b}{b}=\frac{c}{d}-\frac{d}{d}\)
\(\Rightarrow\frac{a-b}{b}=\frac{c-d}{d}\left(đpcm2\right).\)
c) Ta có: \(\frac{a}{b}=\frac{c}{d}.\)
\(\Rightarrow\frac{b}{a}=\frac{d}{c}.\)
\(\Rightarrow\frac{b}{a}+1=\frac{d}{c}+1\)
\(\Rightarrow\frac{b}{a}+\frac{a}{a}=\frac{d}{c}+\frac{c}{c}\)
\(\Rightarrow\frac{b+a}{a}=\frac{d+c}{c}\left(đpcm3\right).\)
d) Ta có: \(\frac{a}{b}=\frac{c}{d}.\)
\(\Rightarrow\frac{b}{a}=\frac{d}{c}.\)
\(\Rightarrow\frac{b}{a}-1=\frac{d}{c}-1\)
\(\Rightarrow\frac{b}{a}-\frac{a}{a}=\frac{d}{c}-\frac{c}{c}\)
\(\Rightarrow\frac{b-a}{a}=\frac{d-c}{c}\left(đpcm4\right).\)
Còn 2 câu kia tí nữa mình làm sau nhé.
Chúc bạn học tốt!
=>\(\frac{a}{a}+\frac{b}{a}=\frac{c}{c}+\frac{d}{c}\)
=>\(1+\frac{b}{a}=1+\frac{d}{c}\)
=> \(\frac{b}{a}=\frac{d}{c}\)=> \(\frac{a}{b}=\frac{c}{d}\)
Còn phần trừ bạn làm tương tự