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\(\frac{a}{b}=\frac{c}{d}\\ \Rightarrow\frac{a}{c}=\frac{b}{d}\\ \Rightarrow\frac{a^{2013}}{c^{2013}}=\frac{b^{2013}}{d^{2013}}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\\ \Rightarrow\frac{a^{2013}}{c^{2013}}=\frac{b^{2013}}{d^{2013}}=\left(\frac{a-b}{c-d}\right)^{2013}\left(1\right)\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{a^{2013}}{c^{2013}}=\frac{b^{2013}}{d^{2013}}=\frac{a^{2013}+b^{2013}}{c^{2013}+d^{2013}}\left(2\right)\)
\(\left(1\right)\left(2\right)\Rightarrow\left(\frac{a-b}{c-d}\right)^{2013}=\frac{a^{2013}+b^{2013}}{c^{2013}+d^{2013}}\)
Sửa đề: x+y+z=0
\(x^3+y^3+z^3=3xyz\)
=>\(\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=0\)
=>\(\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)=0\)
=>\(\left(x+y+z\right)\left[x^2+2xy+y^2-xz-yz+z^2-3xy\right]=0\)
=>\(\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)=0\)
=>\(\left(x+y+z\right)\left(2x^2+2y^2+2z^2-2xy-2xz-2yz\right)=0\)
=>\(\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\right]=0\)(1)
x<>y<>z
=>\(x-y< >0;y-z< >0;x-z< >0\)
=>\(\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\ne0\left(2\right)\)
Từ (1),(2) suy ra x+y+z=0
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{a}{b} = \frac{c}{d} = \frac{{a - c}}{{b - d}}\); \(\frac{a}{b} = \frac{c}{d} = \frac{{a + 2c}}{{b + 2d}}\)
Như vậy, \(\frac{{a - c}}{{b - d}} = \frac{{a + 2c}}{{b + 2d}}\) (đpcm)
Ta có: \(\frac{a+b}{b+c}=\frac{c+d}{d+a}.\)
\(\Rightarrow\frac{a+b}{c+d}=\frac{b+c}{d+a}\)
\(\Rightarrow\frac{a+b}{c+d}+1=\frac{b+c}{d+a}+1.\)
\(\Rightarrow\frac{a+b}{c+d}+\frac{c+d}{c+d}=\frac{b+c}{d+a}+\frac{d+a}{d+a}.\)
\(\Rightarrow\frac{a+b+c+d}{c+d}=\frac{b+c+d+a}{d+a}\)
Nếu \(a+b+c+d\ne0.\)
\(\Rightarrow c+d=d+a\)
\(\Rightarrow c=a\left(đpcm1\right).\)
Nếu \(a+b+c+d=0\) thì hợp với đề.
\(\Rightarrow a+b+c+d=0\left(đpcm2\right).\)
Chúc bạn học tốt!
Ta có: \(\dfrac{a+3c}{b+3d}=\dfrac{a+c}{b+d}\left(b\ne-d;b\ne-3d;b\ne0;d\ne0\right)\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
+, \(\dfrac{a+3c}{b+3d}=\dfrac{a+c}{b+d}=\dfrac{a+3c-\left(a+c\right)}{b+3d-\left(b+d\right)}=\dfrac{a+3c-a-c}{b+3d-b-d}=\dfrac{2c}{2d}=\dfrac{c}{d}\)
Khi đó: \(\dfrac{a+c}{b+d}=\dfrac{c}{d}\)
+, \(\dfrac{a+c}{b+d}=\dfrac{c}{d}=\dfrac{a+c-c}{b+d-d}=\dfrac{a}{b}\) (đpcm)
a: \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Leftrightarrow\dfrac{a}{b}-1=\dfrac{c}{d}-1\)
hay \(\dfrac{a-b}{b}=\dfrac{c-d}{d}\)
Có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
Đặt \(\frac{a}{c}=\frac{b}{d}=k\left(1\right)\\ \Rightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)
\(\frac{a-b}{c-d}=\frac{ck-dk}{c-d}=\frac{k\left(c-d\right)}{c-d}=k\left(2\right)\)
(1)(2) \(\Rightarrow\frac{a}{c}=\frac{a-b}{c-d}\)
\(\)Ta có: \(a+b+c=0 \Rightarrow b+c=-a \Rightarrow (b+c)^2=(-a)^2 \Leftrightarrow b^2+c^2+2bc=a^2 \Leftrightarrow a^2-b^2-c^2=2bc\)
Tương tự: \(b^2-c^2-a^2=2ca;c^2-a^2-b^2=2ab\)
\(P=...=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ca}+\dfrac{c^2}{2bc}=\dfrac{a^3+b^3+c^3}{2abc}=\dfrac{3abc}{2abc}=\dfrac{3}{2}\)
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Bổ đề \(a+b+c=0 \Leftrightarrow a^3+b^3+c^3\)
Ở đây ta c/m chiều thuận:
Với \(a+b+c=0 \Leftrightarrow a+b=-c \Rightarrow (a+b)^3=(-c)^3 \Leftrightarrow a^3+b^3+3ab(a+b)=-c^3 \Leftrightarrow a^3+b^3+c^3=3abc(QED)\)
\(c^2-2ac+a^2+2ab-2bc=a^2\)
\(\Rightarrow\left(a-c\right)^2+2b\left(a-c\right)=a^2\)
\(c^2-2bc+b^2+2a\left(b-c\right)=b^2\Rightarrow\left(b-c\right)^2+2a\left(b-c\right)=b^2\)
\(\Rightarrow B=\frac{\left(a-c\right)^2+2b\left(a-c\right)+\left(a-c\right)^2}{\left(b-c\right)^2+2a\left(b-c\right)+\left(b-c\right)^2}=\frac{2\left(a-c\right)\left(a-c+b\right)}{2\left(b-c\right)\left(b-c+a\right)}=\frac{a-c}{b-c}\)
Ta có
\(B=\frac{\left(x-b\right)\left(x-c\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(x-c\right)\left(x-a\right)}{\left(b-c\right)\left(b-a\right)}+\frac{\left(x-a\right)\left(x-b\right)}{\left(c-a\right)\left(c-b\right)}\)
\(=\frac{\left(x-b\right)\left(x-c\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(x-c\right)\left(x-a\right)}{\left(b-c\right)\left(c-a\right)}+\frac{\left(x-c\right)\left(x-a\right)}{\left(a-c\right)\left(b-a\right)}+\frac{\left(x-a\right)\left(x-b\right)}{\left(c-a\right)\left(c-b\right)}\)
\(=\frac{\left(x-b\right)\left(x-c\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(x-c\right)\left(x-a\right)}{\left(b-c\right)\left(c-a\right)}-\frac{\left(x-c\right)\left(x-a\right)}{\left(a-c\right)\left(a-b\right)}+\frac{\left(x-a\right)\left(x-b\right)}{\left(a-c\right)\left(c-b\right)}\)
\(=\frac{\left(x-b\right)\left(x-c\right)-\left(x-c\right)\left(x-a\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(x-c\right)\left(x-a\right)-\left(x-b\right)\left(x-a\right)}{\left(b-c\right)\left(c-a\right)}\)
\(=\frac{\left(x-c\right)\left(a-b\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(x-a\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)}\).
\(=\frac{x-c}{a-c}-\frac{x-a}{a-c}=\frac{x-c-x+a}{a-c}\)
\(=1\)