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\(2005^3-1=\left(2005-1\right)\left(2005^2+2005+1\right)=2004\times\left(2005^2+2005+1\right)⋮2004\left(\text{đ}pcm\right)\)
\(2005^3+125=\left(2005+5\right)\left(2005^2-2005\times5+5^2\right)=2010\times\left(2005^2-2005\times5+5^2\right)⋮2010\)
\(x^6+1=\left(x^2+1\right)\left(x^4-x^2+1\right)⋮x^2+1\left(\text{đ}pcm\right)\)
\(x^6-y^6=\left(x^2-y^2\right)\left(x^4+x^2y^2+y^2\right)=\left(x-y\right)\left(x+y\right)\left(x^4+x^2y^2+y^4\right)⋮x-y;x+y\left(\text{đ}pcm\right)\)
Câu b bài 1 :
B = x2x2 + x2x2 + x2y2 + x2y2 + x2y2 + y2y2 + y2
= ( x2x2 + x2y2 ) + ( x2x2 + x2y2 ) + ( x2y2 + y2y2 ) + y2
= x2( x2 + y2 ) + x2( x2 + y2 ) + y2( x2 + y2 ) + y2
= ( x2 + y2 ) (x2 + x2 + y2 ) + y2
= 1( x2 + 1) + y2
= x2 + y2 +1 = 2
\(x\left(x-1\right)-3x+3=0\)
<=> \(x\left(x-1\right)-3\left(x-1\right)=0\)
<=> \(\left(x-3\right)\left(x-1\right)=0\)
<=> \(\hept{\begin{cases}x-3=0\\x-1=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=3\\x=1\end{cases}}\)
\(3x\left(x-2\right)+10-5x=0\)
<=> \(3x\left(x-2\right)+5\left(2-x\right)=0\)
<=> \(3x\left(x-2\right)-5\left(x-2\right)=0\)
<=> \(\left(3x-5\right)\left(x-2\right)=0\)
<=> \(\hept{\begin{cases}3x-5=0\\x-2=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=\frac{5}{3}\\x=2\end{cases}}\)
học tốt
Bài 3:
a: \(=35^{2018}\left(35-1\right)=35^{2018}\cdot34⋮17\)
b: \(=43^{2018}\left(1+43\right)=43^{2018}\cdot44⋮11\)
1/\(A=\left(1-x^n\right)\left(1+x^n\right)+\left(2-y^n\right)\left(2+y^n\right)\)
\(=1-x^{2n}+4-y^{2n}=5-x^{2n}-y^{2n}\)
Vì \(x^{2n}\ge0;y^{2n}\ge0\Rightarrow x^{2n}+y^{2n}\ge0\Rightarrow-\left(x^{2n}+y^{2n}\right)\le0\)
\(\Rightarrow A=5-\left(x^{2n}+y^{2n}\right)=5-x^{2n}-y^{2n}\le5\)
Dấu "=" xảy ra khi x = y = 0
Vậy Amax = 5 khi x = y = 0
2/
x3+3x2+3x+1=-1
<=>(x+1)3=-1
<=>x+1=-1
<=>x=-2
3/
a, \(A=2005^3-1=\left(2005-1\right)\left(2005^2-2005.1+1^2\right)=2004\left(2005^2-2005+1\right)⋮2004\)
b, \(B=2000^3+125=2000^3+5^3=\left(2000+5\right)\left(2000^2-2000.1+1^2\right)=2010\left(\frac{2000^2}{2}-\frac{2000}{2}+\frac{1}{2}\right)⋮2010\)