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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:
\(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}=t\)
Ta có : \(\left\{{}\begin{matrix}\left(\dfrac{a+b+c}{b+c+d}\right)^3=t^3\\\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}=t^3\end{matrix}\right.\)
Ta có đpcm
Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{b^3k^3+b^3}{d^3k^3+d^3}=\dfrac{b^3}{d^3}\)
\(\dfrac{\left(a+b\right)^3}{\left(c+d\right)^3}=\dfrac{\left(bk+b\right)^3}{\left(dk+d\right)^3}=\dfrac{b^3}{d^3}\)
Do đó: \(\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{\left(a+b\right)^3}{\left(c+d\right)^3}\)
ta có \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
\(\Rightarrow\dfrac{a^3}{b^3}=\dfrac{\left(a+b+c\right)^3}{\left(b+c+d\right)^3}\)
\(\Leftrightarrow\dfrac{a}{b}\cdot\dfrac{b}{c}\cdot\dfrac{c}{d}=\dfrac{\left(a+b+c\right)^3}{\left(b+c+d\right)^3}\)
\(\Leftrightarrow\dfrac{a}{d}=\dfrac{\left(a+b+c\right)^3}{\left(b+c+d\right)^3}\left(đpcm\right)\)
Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\)
\(\Rightarrow a=bk;b=ck;c=dk\) (1)
Thay (1) vào đề bài:
\(VT=\left(\frac{bk+ck+dk}{ck+dk+d}\right)^3=\left[\frac{k\left(c+d\right)+bk}{k\left(c+d\right)+d}\right]^3=\left(\frac{bk}{d}\right)^3=\frac{bk}{d}\)
\(VP=\frac{bk}{d}\)
\(\Rightarrow VT=VP\)
hay \(\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a}{d}\rightarrowđpcm.\)
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: \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)
\(\Rightarrow\dfrac{\left(a+b\right)^3}{\left(c+d\right)^3}=\dfrac{a^3}{c^3}=\dfrac{b^3}{d^3}\)(1)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a^3}{c^3}=\dfrac{b^3}{d^3}=\dfrac{a^3+b^3}{c^3+d^3}\)(2)
Từ (1) và (2) \(\Rightarrow\) đpcm
đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
a) \(\dfrac{a-b}{a}=\dfrac{c-d}{c}\)
\(\dfrac{a-b}{a}=\dfrac{bk-b}{bk}=\dfrac{b\left(k-1\right)}{bk}=\dfrac{k-1}{k}\left(1\right)\)
\(\dfrac{c-d}{c}=\dfrac{dk-d}{dk}=\dfrac{d\left(k-1\right)}{dk}=\dfrac{k-1}{k}\left(2\right)\)
từ \(\left(1\right),\left(2\right)\Rightarrow\dfrac{a-b}{a}=\dfrac{c-d}{c}\)
b) \(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\)
\(\dfrac{ab}{cd}=\dfrac{bk.b}{dk.d}=\dfrac{b^2.k}{d^2,k}=\dfrac{b^2}{d^2}\)(3)
\(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{\left(bk\right)^2-b^2}{\left(dk\right)^2-d^2}=\dfrac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\dfrac{b^2}{d^2}\)(4)
từ (3) (4) \(\Rightarrow\)......
c) \(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{a^2+b^2}{c^2+d^2}\)
\(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\dfrac{b^2}{d^2}\) (5)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\dfrac{b^2}{d^2}\left(6\right)\)
từ (5) (6)\(\Rightarrow\)...............
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk,d=ck\)
a) \(\dfrac{a^2-b^2}{ab}=\dfrac{b^2k^2-b^2}{bk.b}=\dfrac{b^2\left(k^2-1\right)}{b^2.k}=\dfrac{k^2-1}{k}\) (1)
\(\dfrac{c^2-d^2}{cd}=\dfrac{d^2k^2-d^2}{dk.d}=\dfrac{d^2\left(k^2-1\right)}{d^2k}=\dfrac{k^2-1}{k}\) (2)
Tử (1) và (2) \(\Rightarrow\dfrac{a^2-b^2}{ab}=\dfrac{c^2-d^2}{cd}\)
b) \(\dfrac{\left(a+b\right)^2}{a^2+b^2}=\dfrac{\left(bk+b\right)^2}{b^2k^2+b^2}=\dfrac{\left[b\left(k+1\right)\right]^2}{b^2\left(k^2+1\right)}\)
\(=\dfrac{b^2\left(k+1\right)^2}{b^2\left(k^2+1\right)}=\dfrac{\left(k+1\right)^2}{k^2+1}\) (1)
\(\dfrac{\left(c+d\right)^2}{c^2+d^2}=\dfrac{\left(dk+d\right)^2}{d^2k^2+d^2}=\dfrac{\left[d\left(k+1\right)\right]^2}{d^2\left(k^2+1\right)}\)
\(=\dfrac{d^2\left(k+1\right)^2}{d^2\left(k^2+1\right)}=\dfrac{\left(k+1\right)^2}{k^2+1}\) (2)
Từ (1) và (2) \(\Rightarrow\dfrac{\left(a+b\right)^2}{a^2+b^2}=\dfrac{\left(c+d\right)^2}{c^2+d^2}\)
Chúc bạn học tốt ♥v♥
a: a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{k}{k-1}\)
\(\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{k}{k-1}=\dfrac{a}{a-b}\)
b: \(\dfrac{a}{b}=\dfrac{bk}{b}=k\)
\(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=k=\dfrac{a}{b}\)
c \(\dfrac{a}{3a+b}=\dfrac{bk}{3bk+b}=\dfrac{k}{3k+1}\)
\(\dfrac{c}{3c+d}=\dfrac{dk}{3dk+d}=\dfrac{k}{3k+1}=\dfrac{a}{3a+b}\)
d: \(\dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=k^2\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=k^2=\dfrac{ac}{bd}\)
Ta có: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{a+b+c}{b+c+d}\)
\(\dfrac{b}{c}=\dfrac{a+b+c}{b+c+d}\)
\(\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
\(\Rightarrow\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}.\dfrac{a+b+c}{b+c+d}.\dfrac{a+b+c}{b+c+d}\)
\(\Rightarrow\dfrac{a}{d}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\rightarrowđpcm\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\left(1\right)\)
Từ \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
Ta xét tích: \(\dfrac{a}{b}.\dfrac{a}{b}.\dfrac{a}{b}=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left(\dfrac{a+b+c}{b+c+d}\right)=\dfrac{a}{d}\left(dpcm\right)\)