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Bài 1:
Áp dụng t.c của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\\ =\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a^3}{b^3}=\dfrac{a.b.c}{b.c.d}=\dfrac{a}{d}\left(dpcm\right)\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\) (*)
a) Từ (*)suy ra:
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\dfrac{b^2.k^2+b^2}{d^2.k^2+d^2}=\dfrac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}\)\(=\dfrac{b^2}{d^2}\) (1)
\(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\dfrac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\dfrac{b^2.\left(k+1\right)^2}{d^2.\left(k+1\right)^2}=\dfrac{b^2}{d^2}\)(2)
Từ (1) và (2) suy ra: \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\) (đpcm)
b) Tương tự câu a nhé bạn!
ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a^6}{b^6}=\dfrac{c^6}{d^6}=\dfrac{3a^6}{3b^6}\)
Áp dụng tính chất dãy tỉ sốbằng nhau ta có:
\(\dfrac{a^6}{b^6}=\dfrac{c^6}{d^6}=\dfrac{3a^6}{3b^6}=\dfrac{a^6+c^6}{b^6+d^6}=\dfrac{\left(a+c\right)^6}{\left(b+d\right)^6}\)
=\(\dfrac{c^6+3a^6}{d^6+3b^6}\)
\(\Rightarrow\dfrac{3a^6+c^6}{3b^6+d^6}=\dfrac{\left(a+c\right)^6}{\left(b+d\right)^6}\) (ĐPCM)
Đặt
\(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\left(a+2c\right)\left(b+d\right)=\left(bk+2dk\right)\left(b+d\right)\)
\(=bk\left(b+d\right)+2dk\left(b+d\right)\)
\(=b^2k+bdk+2bdk+2d^2k\)
\(=b^2k+3bdk+2d^2k\)
\(\Rightarrow\left(a+c\right)\left(b+2d\right)=\left(bk+dk\right)\left(b+2d\right)\)
\(=bk\left(b+2d\right)+dk\left(b+2d\right)\)
\(=b^2k+2bdk+bdk+2d^2k\)
\(=b^2k+3bdk+2d^2k\)
\(VT=VP\)\(\Rightarrowđpcm\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Từ đó, ta được:\(\dfrac{\left(a+c\right)^3}{\left(b+d\right)^3}=\dfrac{\left(bk+dk\right)^3}{\left(b+d\right)^3}=\dfrac{\left[k\left(b+d\right)\right]^3}{\left(b+d\right)^3}=\dfrac{k^3.\left(b+d\right)^3}{\left(b+d\right)^3}=k^3\left(1\right)\) \(\dfrac{\left(a-c\right)^3}{\left(b-d\right)^3}=\dfrac{\left(bk-dk\right)^3}{\left(b-d\right)^3}=\dfrac{\left[k\left(b-d\right)\right]^3}{\left(b-d\right)^3}=\dfrac{k^3.\left(b-d\right)^3}{\left(b-d\right)^3}=k^3\left(2\right)\)
Từ (1) và (2) suy ra: \(\dfrac{\left(a+c\right)^3}{\left(b+d\right)^3}=\dfrac{\left(a-c\right)^3}{\left(b-d\right)^3}\)
4.a
\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\\ \Leftrightarrow\left(3x-y\right).4=3\left(x+y\right)\\ \Rightarrow12x-4y=3x+3y\\ \Rightarrow12x-3x=4y+3y\\ \Rightarrow9x=7y\\ \Rightarrow\dfrac{x}{y}=\dfrac{7}{9}\)
Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\)
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{2c}{2d}=\dfrac{a+2c}{b+2d}\)
\(\Rightarrow\dfrac{a+c}{b+d}=\dfrac{a+2c}{b+2d}\)
\(\Rightarrow\left(a+2c\right)\left(b+d\right)=\left(a+c\right)\left(b+2d\right)\left(đpcm\right)\)
Vậy...
Vì \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)
\(\Rightarrow\left[{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\) (!)
Thay (!) vào đề bài:
VT = \(c\left(k+2\right).d\left(k+1\right)\left(1\right)\)
\(VP=c\left(k+1\right).d\left(k+2\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow VT=VP\)
hay \(\left(a+2c\right)\left(b+d\right)=\left(a+c\right)\left(b+2d\right)\).
bạn sửa hộ mik \(\left(\dfrac{a^2+b^2}{c^2+d^2}\right)^2\) thành\(\dfrac{a^2+b^2}{c^2+d^2}\)nha!!
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Xét vế trái , ta có :
(a + b + c + d)(a - b - c + d)
= (bk + b + dk + d)(bk - b - dk + d)
= [(b.(k + 1) + d(k + 1)].[b.(k - 1) - d.(k - 1)]
= (b + d)(k + 1).(b - d)(k - 1)
Xét vế phải , ta có :
(a - b + c - d)(a + b - c - d)
= (bk - b + dk - d)(bk + b - dk - d)
= [b.(k - 1) + d.(k - 1)].[b(k + 1) - d(k + 1)]
= (b + d)((k - 1).(b - d)(k + 1)
Ta thấy sau khi xét hai vế , đều có thừa số chung
Vậy ....
Đoàn Đức Hiếu giúp mình