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Giải:
Ta có:
\(a+b+c+d=0\)
\(\Leftrightarrow a+b=-c-d\)
\(\Leftrightarrow a+b=-\left(c+d\right)\)
Từ đó, suy ra:
\(\left(a+b\right)^3=-\left(c+d\right)^3\)
\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3=-\left(c^3+3c^2d+3cd^2+d^3\right)\)
\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3=-c^3-3c^2d-3cd^2-d^3\)
\(\Leftrightarrow a^3+3ab\left(a+b\right)+b^3=-c^3-3cd\left(c+d\right)-d^3\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-3cd\left(c+d\right)-3ab\left(a+b\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-3cd\left(c+d\right)+3ab\left(c+d\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(c+d\right)\left(ab-cd\right)\)
Vậy ...
\(a+b+c=0\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)=0
\(\Leftrightarrow\)\(a^3+ab^2+ac^2-a^2b-a^2c-abc+a^2b+b^3+bc^2-ab^2-\)
\(abc-b^2c+ca^2+bc^2+c^3-abc-ac^2-bc^2\)=0
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\Leftrightarrow a^3+b^3-3abc=-c^3\)
a+b+c+d=0
=>a+b=-(c+d)
=> (a+b)^3=-(c+d)^3
=> a^3+b^3+3ab(a+b)=-c^3-d^3-3cd(c+d)
=> a^3+b^3+c^3+d^3=-3ab(a+b)-3cd(c+d)
=> a^3+b^3+c^3+d^3=3ab(c+d)-3cd(c+d) ( vi a+b = - (c+d))
==> a^3 +b^3+c^3+d^3==3(c+d)(ab-cd) (dpcm)
a+b+c+d=0
=>a+b=-(c+d)
=>(a+b)^3=-(c+d)^3
=> a^3+b^3+3ab(a+b)=-c^3-d^3-3cd(c+d)
=>a^3+b^3+c^3+d^3=-3ab(a+b)-3cd(c+d)
=>a^3+b^3+c^3+d^3=3ab(c+d)-3cd(c+d) vi a+b=-(c+d)
=> a^3+b^3+c^3+d^3=3(c+d)(ab+cd)
Xem lai gium mk nha!!
Ta có : a+b+c+d=0
=>a+b=-(c+d)
=> (a+b)^3=-(c+d)^3
=> a^3+b^3+3ab(a+b)=-c^3-d^3-3cd(c+d)
=> a^3+b^3+c^3+d^3=-3ab(a+b)-3cd(c+d)
=> a^3+b^3+c^3+d^3=3ab(c+d)-3cd(c+d) ( vi a+b = - (c+d))
==> a^3 +b^^3+c^3+d^3==3(c+d)(ab-cd)
Ta có a+b+c+d=0
=> a+b=-(c+d)
=> (a+b)^3=-(c+d)^3
=> a^3+b^3+3ab(a+b) = -c^3-d^3-3cd(c+d)
=> a^3+b^3+c^3+d^3 = -3ab(a+b)-3cd(c+d)
=> a^3+b^3+c^3+d^3 = 3ab(c+d)-3cd(c+d) (vì a+b = - (c+d))
=> a^3 +b^3+c^3+d^3 = 3(c+d)(ab-cd) (đpcm)
a)
\(a^4+3>4a\)
<=> \(a^4-4a+3>0\)
<=> \(a^4-a^3+a^3-a^2+a^2-a-3a+3>0\)
<=> \(a^3\left(a-1\right)+a^2\left(a-1\right)+a\left(a-1\right)-3\left(a-1\right)\)
<=> \(\left(a-1\right)\left(a^3+a^2+a-3\right)>0\)
\(a)\)\(\left(a+b+c+d\right)\left(a-b-c+d\right)=\left(a-b+c-d\right)\left(a+b-c-d\right)\)
\(\Leftrightarrow\)\(\frac{a+b+c+d}{a-b+c-d}=\frac{a+b-c-d}{a-b-c+d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{a+b+c+d}{a-b+c-d}=\frac{a+b-c-d}{a-b-c+d}=\frac{a+b+c+d+a+b-c-d}{a-b+c-d+a-b-c+d}=\frac{2\left(a+b\right)}{2\left(a-b\right)}=\frac{a+b}{a-b}\) \(\left(1\right)\)
Lại có :
\(\frac{a+b+c+d}{a-b+c-d}=\frac{a+b-c-d}{a-b-c+d}=\frac{a+b+c+d-a-b+c+d}{a-b+c-d-a+b+c-d}=\frac{2\left(c+d\right)}{2\left(c-d\right)}=\frac{c+d}{c-d}\) \(\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) suy ra \(\frac{a+b}{a-b}=\frac{c+d}{c-d}\)\(\Leftrightarrow\)\(\frac{a+b}{c+d}=\frac{a-b}{c-d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{a+b+a-b}{c+d+c-d}=\frac{2a}{2c}=\frac{a}{c}\) \(\left(3\right)\)
Lại có :
\(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{a+b-a+b}{c+d-c+d}=\frac{2b}{2d}=\frac{b}{d}\) \(\left(4\right)\)
Từ \(\left(3\right)\) và \(\left(4\right)\) suy ra \(\frac{a}{c}=\frac{b}{d}\) ( đpcm )
Chúc bạn học tốt ~
\(b)\)\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow\)\(a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\)\(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\)\(a^2+b^2+c^2-ab-bc-ca=0\) ( vì \(a+b+c=0\) )
\(\Leftrightarrow\)\(2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\)\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\)\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\)\(\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Leftrightarrow}\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow}a=b=c}\)
Vậy ...
Chúc bạn học tốt ~
a+b+c+d=0
=>c+d=-(a+b)
\(a^3+b^3+c^3+d^3\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+\left(c+d\right)^3-3cd\left(c+d\right)\)
\(=\left(a+b\right)^3-\left(a+b\right)^3-3ab\left(a+b\right)-3cd\left(c+d\right)\)
=-3ab(a+b)-3cd(c+d)
\(=3ab\left(c+d\right)-3cd\left(c+d\right)=3\left(c+d\right)\left(ab-cd\right)\)