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Sửa đề \(M=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)
Ta có: \(a^3+b^3+c^3=3ab\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Rightarrow\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{cases}}\)
TH1: a+b+c=0
=> \(\hept{\begin{cases}a=-\left(b+c\right)\\b=-\left(a+c\right)\\c=-\left(a+b\right)\end{cases}}\)
Thay vào M ta được M=\(\left(1-\frac{b+c}{b}\right)\left(1-\frac{a+c}{c}\right)\left(1-\frac{a+b}{a}\right)\)
\(\Rightarrow M=\frac{-c}{b}\cdot\frac{-a}{c}\cdot\frac{-b}{a}=-1\)
TH2: \(a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Rightarrow M=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
a) Ta có: (a + b + c + d)(a - b - c +d )=( (a + d) + (b + c) )( (a + d) - (b + c) )
=(a + d )2 - (b +c )2 (1)
(a - b + c - d)(a + b - c - d)=(a - d)2 - (b - c)2 (2)
Từ (1) và (2) => a2 + 2ad + d2 - b2 - 2bc - c2=a2 - 2ad + d2 - b2 + 2bc - c2
4ad=4bc => ad=bc <=> \(\frac{a}{c}=\frac{b}{d}\) (đpcm)
\(b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c}\)
\(c^2=bd\Rightarrow\frac{b}{c}=\frac{c}{d}\)
\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)
\(\Rightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\left(\frac{a}{b}\right)\left(\frac{b}{c}\right)\left(\frac{c}{d}\right)\)
\(\Rightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a}{d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau có :
\(\Rightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\)
Mà \(\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a}{d}\)
\(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{d}\)
Vậy ...
a2 + b2 + (a + b)2 = c2 + d2 + (c +d)2 => 2.(a2 + b2) + 2ab = 2.(c2 + d2) + 2cd
=> a2 + b2 + ab = c2 + d2 + cd (1)
+) a4 + b4 + (a + b)4 = (a2 + b2)2 - 2a2.b2 + (a + b)4 = [(a2 + b2)2 - a2.b2] + [(a + b)4 - a2.b2]
= (a2 + b2 - ab). (a2 + b2 + ab) + [(a + b)2 - ab].[(a+ b)2 + ab]
= (a2 + b2 - ab). (a2 + b2 + ab) + (a2 + b2 + ab). (a2 + b2 + 3ab) = (a2 + b2 + ab). [(a2 + b2 - ab) + (a2 + b2 + 3ab)]
= 2.(a2 + b2 + ab).(a2 + b2 + ab) = 2.(a2 + b2 + ab)2 (2)
Tương tự: c4 + d4 + (c+d)4 = 2. (c2 + d2 + cd)2 (3)
Từ (1)(2)(3) => đpcm
Lời giải:
$a+b=c+d$
$(a+b)^2=(c+d)^2\Rightarrow a^2+b^2+2ab=c^2+d^2+2cd$
$\Rightarrow ab=cd\Rightarrow \frac{a}{d}=\frac{c}{b}$.
Đặt $\frac{a}{d}=\frac{c}{b}=k$
$\Rightarrow a=dk; c=bk$. Khi đó:
$a+b=c+d$
$\Leftrightarrow dk+b=bk+d$
$\Leftrightarrow k(d-b)=d-b$
$\Leftrightarrow (d-b)(k-1)=0$
$\Rightarrow d=b$ hoặc $k=1$.
Nếu $b=d$ thì do $ab=cd\Rightarrow a=c$.
$\Rightarrow b^{2013}=d^{2013}; a^{2013}=c^{2013}$
$\Rightarrow a^{2013}+b^{2013}=c^{2013}+d^{2013}$
Nếu $k=1\Rightarrow a=d; b=c$
$\Rightarrow a^{2013}=d^{2013}; b^{2013}=c^{2013}$
$\Rightarrow a^{2013}+b^{2013}=c^{2013}+d^{2013}$
Lời giải:
$a+b=c+d$
$(a+b)^2=(c+d)^2\Rightarrow a^2+b^2+2ab=c^2+d^2+2cd$
$\Rightarrow ab=cd\Rightarrow \frac{a}{d}=\frac{c}{b}$.
Đặt $\frac{a}{d}=\frac{c}{b}=k$
$\Rightarrow a=dk; c=bk$. Khi đó:
$a+b=c+d$
$\Leftrightarrow dk+b=bk+d$
$\Leftrightarrow k(d-b)=d-b$
$\Leftrightarrow (d-b)(k-1)=0$
$\Rightarrow d=b$ hoặc $k=1$.
Nếu $b=d$ thì do $ab=cd\Rightarrow a=c$.
$\Rightarrow b^{2013}=d^{2013}; a^{2013}=c^{2013}$
$\Rightarrow a^{2013}+b^{2013}=c^{2013}+d^{2013}$
Nếu $k=1\Rightarrow a=d; b=c$
$\Rightarrow a^{2013}=d^{2013}; b^{2013}=c^{2013}$
$\Rightarrow a^{2013}+b^{2013}=c^{2013}+d^{2013}$
a,
x=2005=> 2006=x+1 . Thay vào biểu thức A có:
\(A=x^{20}-\left(x+1\right)x^{19}+\left(x+1\right)x^{18}-\left(x+1\right)x^{17}+....+\left(x+1\right)x^2-\left(x+1\right)x+\left(x+1\right)\)A=\(x^{20}-x^{20}+x^{19}-x^{19}+x^{18}-x^{18}+...+x^3+x^2-x^2-x+x+1\)
A=1
b,
B=\(x^5-\left(x+1\right)x^4+\left(x+2\right)x^3-\left(2x+1\right)x^2+\left(x-1\right)x\)
B=\(x^5-x^5-x^4+x^4+2x^3-2x^3-x^2+x^2-x\)
B=x=14
Ta có:
\(a^2+b^2=c^2+d^2\)
\(\Leftrightarrow a^2-c^2=d^2-b^2\)
\(\Leftrightarrow\left(a-c\right)\left(a+c\right)=\left(d-b\right)\left(d+b\right)\)
Mà \(a+b=c+d\Leftrightarrow a-c=d-b\)
\(\Leftrightarrow\left(a-c\right)\left(a+c\right)=\left(a-c\right)\left(d+b\right)\)
TH1: \(a-c\ne0\)
\(\Rightarrow a+c=d+b\Leftrightarrow a-b=d-c\left(1\right)\)
Lại có: \(a+b=c+d\left(2\right)\)
Cộng (1) và (2) theo vế ta có: \(2a=2d\Leftrightarrow a=d\)\(\Rightarrow b=c\)
\(\Rightarrow a^{2006}=d^{2006}\); \(b^{2006}=c^{2006}\)
\(\Rightarrow a^{2006}+b^{2006}=c^{2006}+d^{2006}\)(*)
TH2: \(a-c=0\)
\(\Rightarrow a=c\)\(\Rightarrow b=d\)
\(\Rightarrow a^{2006}=c^{2006};b^{2006}=d^{2006}\)
\(\Rightarrow a^{2006}+b^{2006}=c^{2006}+d^{2006}\)(**)
Từ (*) và (**) \(\Rightarrow a^{2006}+b^{2006}=c^{2006}+d^{2006}\)