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Đặt a/b=c/d=k
=>a=bk; c=dk
(a+2c)(b+2023d)
=(bk+2dk)(b+2023d)
=k(b+2d)(b+2023d)
=(bk+2023kd)(b+2d)
=(a+2023c)(b+2d)
a/b=c/d=k
=> a=bk, c=dk
thế vào các biểu thức đó rồi sử dụng phân phối
1) Ta có:
\(\dfrac{a}{a+b}\)=\(\dfrac{c}{c+d}\)
=>a.(c+d) = c.(a+b)
a.c+a.d = a.c+b.d
Do đó a.d=b.d
=>\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)( đpcm)
Câu 2:
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{3a+2c}{3b+2d}=\dfrac{3bk+2dk}{3b+2d}=k\)
\(\dfrac{-5a+3c}{-5b+3d}=\dfrac{-5bk+3dk}{-5b+3d}=k\)
=>\(\dfrac{3a+2c}{3b+2d}=\dfrac{-5a+3c}{-5b+3d}\)
b: \(\dfrac{a^2}{b^2}=\dfrac{b^2k^2}{b^2}=k^2\)
\(\dfrac{2c^2-ac}{2d^2-bd}=\dfrac{c\left(2c-a\right)}{d\left(2d-b\right)}=\dfrac{dk}{d}\cdot\dfrac{2dk-bk}{2d-b}=k^2\)
=>\(\dfrac{a^2}{b^2}=\dfrac{2c^2-ac}{2d^2-bd}\)
Theo bài ra ta có :
\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)
\(\Rightarrow\frac{2a+b+c+d}{a}-1=\frac{a+2b+c+d}{b}-1=\frac{a+b+2c+d}{c}-1=\frac{a+b+c+2d}{d}-1\)
\(\Rightarrow\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
Nếu a + b + c + d = 0
\(\Rightarrow\frac{0}{a}=\frac{0}{b}=\frac{0}{c}=\frac{0}{d}\)
\(\Rightarrow\orbr{\begin{cases}a=b=c=d\\a\ne b\ne c\ne d\end{cases}}\)(loại)
Nếu a + b + c + d \(\ne\)0
=> \(\frac{1}{a}=\frac{1}{b}=\frac{1}{c}=\frac{1}{d}\)
=> a = b = c = d (đpcm)
giả sử \(\frac{a}{b}=\frac{c}{d}\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{2a}{2c}=\frac{2b}{2d}\)
áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\frac{a}{c}=\frac{b}{d}=\frac{2a+b}{2c+d}=\frac{a-2b}{c-2d}\)
\(=>\frac{a}{c}=\frac{b}{d}=\frac{2a+b}{a-2b}=\frac{2c+d}{c-2d}\)
vậy \(\frac{2a+b}{a-2b}=\frac{2c+d}{c-2d}=>\frac{a}{b}=\frac{c}{d}\left(dpcm\right)\)
p/s: ko chắc lắm mong là ko sai =]
\(\frac{a}{b}=\frac{c}{d}\Leftrightarrow ad=bc\)
\(\Rightarrow ad+ad+bc=bc+ad+bc\)
\(\Rightarrow2ad+bc=2bc+ad\)
\(\Rightarrow ab+2ad+bc+2cd=ab+2bc+ad+2cd\)
\(\Rightarrow a\left(b+2d\right)+c\left(b+2d\right)=b\left(a+2c\right)+d\left(a+2c\right)\)
\(\Rightarrow\left(a+c\right)\left(b+2d\right)=\left(a+2c\right)\left(b+d\right)\rightarrowđpcm\)
DỄ MÀ
(a+2c)(b+d)=ab+ad+2bc+2cd
(a+c)(b+2d)=ab+2ad+bc+2cd
Vì a/b=c/d nên ad=bc
suy ra đpcm
Đặt a/b=c/d=k
=>a=bk; c=dk
(a+2c)(b+2023d)
=(bk+2dk)(b+2023d)
=k(b+2d)(b+2023d)
=(bk+2023kd)(b+2d)
=(a+2023c)(b+2d)