Cho \(\dfrac{a}{b}\) = \(\dfrac{b}{c}\) = \(\dfrac{c}{d}\) chung minh rang \(\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\) = \(\dfrac{a}{d}\)
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Cho \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\) chung minh rang \(\dfrac{a^3+b^3+c^3}{a^3+c^3+d^3}=\dfrac{a}{d}\)
Ta co :\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)=>\(\left(\dfrac{a}{b}\right)^3=\left(\dfrac{b}{c}\right)^3=\left(\dfrac{c}{d}\right)^3\)
=> \(\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\) (1)
Mặt khác:\(\dfrac{a^3}{b^3}=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}\) (2)
Tu (1) va (2)
=> \(\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\dfrac{a}{d}\) (dpcm)
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7 tháng 11 2021 lúc 14:13
Cho \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\) Chứng minh rằng \(\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\dfrac{a}{d}\)
Đặt \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\b=ck\\c=dk\end{matrix}\right.\)
Ta có: \(\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\dfrac{b^3k^3+c^3k^3+d^3k^3}{b^3+c^3+d^3}=k^3\)
\(\dfrac{a}{d}=\dfrac{bk}{d}=\dfrac{ck^2}{d}=\dfrac{dk^3}{d}=k^3\)
Do đó: \(\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\dfrac{a}{d}\)
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Cho \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\). Chứng minh rằng :
\(\dfrac{a}{b}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=(\dfrac{a+b-c}{b+c-d})^3\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b-c}{b+c-d}\Rightarrow\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{\left(a+b-c\right)^3}{\left(b+c-d\right)^3}=\left(\dfrac{a+b-c}{b+c-d}\right)^3\left(1\right)\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\left(2\right)\)
Từ \(\left(1\right),\left(2\right)\Rightarrow\dfrac{a}{b}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\left(\dfrac{a+b-c}{b+c-d}\right)^3\)
Vậy \(\dfrac{a}{b}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\left(\dfrac{a+b-c}{b+c-d}\right)^3\left(dpcm\right)\)
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6 tháng 11 2021 lúc 20:31
Cho b^2 = ac ; c^2 = bd với b, c, d ≠ 0; b+c ≠ 0; b^3+c^3≠ d^3 3. Chứng minh rằng:
a) \(\dfrac{a^3+b^3-c^3}{b^3+c^3-d^3}=\left(\dfrac{a+b-c}{b+c-d}\right)^3\)
b) \(\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\dfrac{a}{d}\)
Cho \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)chứng minh rằng : \(\dfrac{a^3}{b^3}=\dfrac{a}{d}\)
Cho\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\) với b+c+d khác 0.
Chứng minh:\(\dfrac{a^3+b^3+c^3}{b^3+c^3-d^3}=\left(\dfrac{a+d-c}{b+c-d}\right)^3\)
Cho dfrac{a}{b}dfrac{c}{d} . Chứng minh :
a, dfrac{a^3+b^3}{c^3+d^3} dfrac{a^3-b^3}{c^3-d^3}
b, dfrac{(a+b)^3}{(c+d)^3}dfrac{a^3+b^3}{c^3+d^3}
c, dfrac{(a-b)^3}{(c-d)^3}dfrac{3a^2+2b^2}{3c^2+2d^2}
d, dfrac{4a^4+5b^4}{4c^4+5d^4}dfrac{a^2b^2}{c^2d^2}
e, dfrac{a^{10}+b^{10}}{(a+b)^{10}} dfrac{c^{10}+d^{10}}{(c+d)^{10}}
Đọc tiếp
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\) . Chứng minh :
a, \(\dfrac{a^3+b^3}{c^3+d^3} = \dfrac{a^3-b^3}{c^3-d^3}\)
b, \(\dfrac{(a+b)^3}{(c+d)^3}=\dfrac{a^3+b^3}{c^3+d^3}\)
c, \(\dfrac{(a-b)^3}{(c-d)^3}=\dfrac{3a^2+2b^2}{3c^2+2d^2}\)
d, \(\dfrac{4a^4+5b^4}{4c^4+5d^4}=\dfrac{a^2b^2}{c^2d^2}\)
e, \(\dfrac{a^{10}+b^{10}}{(a+b)^{10}} = \dfrac{c^{10}+d^{10}}{(c+d)^{10}}\)
a/
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^3}{b^3}=\frac{c^3}{d^3}\)
Áp dụng tỉ lệ thức ta có:
\(\frac{a^3}{b^3}=\frac{c^3}{d^3}\Rightarrow\frac{a^3}{c^3}=\frac{b^3}{d^3}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a^3}{c^3}=\frac{b^3}{d^3}=\frac{a^3+b^3}{c^3+d^3}=\frac{a^3-b^3}{c^3-d^3}\)
Vậy \(\frac{a^3+b^3}{c^3+d^3}=\frac{a^3-b^3}{c^3-d^3}\)
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Cho \(\dfrac{a}{b}=\dfrac{c}{d}\)chứng minh rằng :\(\dfrac{a^3+c^3+e^3}{b^3+d^3+f^3}=\dfrac{ace}{bdf}\)
a,Tìm x,y,z biết: dfrac{y+z+1}{x}dfrac{x+z+2}{y}dfrac{x+y-3}{z}dfrac{1}{x+y+z}b,Cho dfrac{a}{b}dfrac{b}{c}dfrac{c}{d}. Chứng minh rằng: (dfrac{a+b+c}{b+c+d})3dfrac{a}{d}c,Cho dfrac{a}{b}dfrac{c}{d}. Chứng minh rằng: dfrac{5a+3b}{5c+3d}dfrac{5a-3b}{5c-3d}d,Cho dfrac{3x-2y}{4}dfrac{2z-4x}{3}dfrac{4y-3z}{2}.Chứng minh rằng: dfrac{x}{2}dfrac{y}{3}dfrac{z}{4}
Đọc tiếp
a,Tìm x,y,z biết: \(\dfrac{y+z+1}{x}\)=\(\dfrac{x+z+2}{y}\)=\(\dfrac{x+y-3}{z}\)=\(\dfrac{1}{x+y+z}\)
b,Cho \(\dfrac{a}{b}\)=\(\dfrac{b}{c}\)=\(\dfrac{c}{d}\). Chứng minh rằng: (\(\dfrac{a+b+c}{b+c+d}\))3=\(\dfrac{a}{d}\)
c,Cho \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\). Chứng minh rằng: \(\dfrac{5a+3b}{5c+3d}\)=\(\dfrac{5a-3b}{5c-3d}\)
d,Cho \(\dfrac{3x-2y}{4}\)=\(\dfrac{2z-4x}{3}\)=\(\dfrac{4y-3z}{2}\).Chứng minh rằng: \(\dfrac{x}{2}\)=\(\dfrac{y}{3}\)=\(\dfrac{z}{4}\)
b/ \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
\(\Rightarrow\left(\dfrac{a}{b}\right)^3=\dfrac{a}{d}\left(1\right)\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
=> \(\left(\dfrac{a}{b}\right)^3=\left(\dfrac{a+b+c}{c+d+b}\right)^3\) (2)Từ (1) và (2)=>đpcm
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