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Mình sẽ viết tổng quát A thê này nhé:
\(A=\left(xy^2\right)\left(x^5y^4\right)\left(x^9y^6\right)...\left(x^{4n-3}y^{2n}\right)\)(giả sử A có n nhân tử)
Theo đề bài ta có:
\(\dfrac{n\left(4n-3+1\right)}{2}+\dfrac{n\left(2n+2\right)}{2}=3675\)
\(\Leftrightarrow2n^2-n+n^2+n=3675\)
\(\Leftrightarrow3n^2=3675\Leftrightarrow n^2=1225\Leftrightarrow n=35\)
Bậc cao nhấ của biến x là:\(4.35-3=137\)
a) =(x-y)5+(x-y)3=(x-y)3[(x-y)2+1]
b) =33(y-2x)3:-9(y-2x)=-3(y-2x)2
c) =(x-y)2 [3(x-y)3-2(x-y)2+3]:5(x-y)2=[3(x-y)3-2(x-y)2+3]/5
Ta có (x^2 + y^2 )^3 + (z^2 – x^2 )^3 – (y^2 + z^2 )^3
= (x^2 + y^2 )^3 + (z^2 – x^2 )^3 + (-y^2 - z^2 )^3
Ta thấy x^2 + y^2 + z^2 – x^2 – y^2 – z^2 = 0
=> áp dụng nhận xét ta có: (x^2+y^2 )^3+ (z^2 -x^2 )^3 -y^2 -z^2 )^3
= 3(x^2 + y^2 ) (z^2 –x^2 ) (-y^2 – z^2 )
= 3(x^2+y^2 ) (x+z)(x-z)(y^2+z^2 )
a.100x2-(x2+25)2=(10x)2-(x2+25)2=(10x-x2-25)(10x+x2+25)=(10x-x2-25)(x+5)2
b.1+(x-y+5)2-2(x-y+5)=(x-y+4)2
\(a,100x^2-\left(x^2+25\right)\)
\(=\left(10x-x^2+25\right)\left(10x+x^2+25\right)\)
\(b,1+\left(x-y+5\right)^2-2\left(x-y+5\right)^2\)
\(=\left(1-x+y-5\right)^2\)
(x -y)3 - 1 - 3(x -y)(x - y - 1)
= (x -y)3 - 3(x -y)(x - y - 1) - 1
Đặt x - y = t, khi đó ta có:
t3 - 3t. (t - 1) - 1
= t3 - 3t2 + 3t - 1
= (t - 1)3
Thay t = x - y vào (t - 1)3 , ta có: ( x - y - 1)3
Vậy (x -y)3 - 1 - 3(x -y)(x - y - 1) = ( x - y - 1)3
\(a,2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)
\(=2x^2+2y^2+x^2+2xy+y^2+x^2-2xy+y^2=3\left(x^2+y^2\right)\)\(b,\left(5x-1\right)+2\left(1-5x\right)\left(4x+5\right)+\left(5x+4\right)\)\(=\left[\left(5x-1\right)-\left(5x+4\right)\right]^2=25\)
c)\(Q=\left(x-y\right)^3+\left(x+y\right)^3+\left(x-y\right)^3-3xy\left(x+y\right)\)
\(=x^3-3x^2y+3xy^2-y^3+x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-3xy^2-3x^2y\)
\(=x^3+y^3\)
d)\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(2P=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(2P=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(2P=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(2P=\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(2P=5^{32}-1\Rightarrow P=\dfrac{5^{32}-1}{2}\)
2. CMR:
a. \(\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)=x^5-y^5\)
Ta có: VT=\(\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)=x^5+x^4y+x^3y^2+x^2y^3+xy^4-x^4y-x^3y^2-x^2y^3-xy^4-y^5=x^5-y^5=VP\)=> đpcm.
b. \(\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)=x^5+y^5\)
Ta có: VT=\(\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)=x^5-x^4y+x^3y^2-x^2y^3+xy^4+x^4y-x^3y^2+x^2y^3-xy^4+y^5=x^5+y^5=VP\)
=> đpcm.
c. \(\left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab\)
\(\Leftrightarrow x^2+bx+ax+ab=x^2+ax+bx+ab\) (đúng)
=> đpcm.