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\(=x\left(\frac{x^2}{4}+x+1\right)=x\left(\frac{x}{2}+1\right)^2\)
\(\left(x+5\right)^2-3\left(x+5\right)\)
\(=\left(x+5\right)\left(x+5-3\right)\)
\(=\left(x+5\right)\left(x+2\right)\)
\(2x\left(x-3\right)-\left(x-3\right)^2\)
\(=\left(x-3\right)\left(2x-x+3\right)\)
\(=\left(x-3\right)\left(x+3\right)\)
25n(n-1)-50(n-1) luôn chia hết cho 150 với mọi n là số nguyên
giúp mình chứng minh nha . Cám ơn mấy bạn
1. \(xy\left(a^2+2b^2\right)-ab\left(2x^2+y^2\right)\)
\(=xya^2+2xyb^2-2abx^2-aby^2\)
\(=xya^2-aby^2-2abx^2+2xyb^2\)
\(=ay\left(ax-by\right)-2bx\left(ax-by\right)\)
\(=\left(ay-2bx\right)\left(ax-by\right)\)
2. \(xy\left(a^2+2b^2\right)+ab\left(2x^2+y^2\right)\)
\(=xya^2+2xyb^2+2abx^2+aby^2\)
\(=xya^2+aby^2+2abx^2+2xyb^2\)
\(=ay\left(ax+by\right)+2bx\left(ax+by\right)\)
\(=\left(ay+2bx\right)\left(ax+by\right)\)
Ta có: x^7 + x^5 + 1 = x^7 - x + x^5 - x^2 + x^2 + x + 1
=x(x^6 - 1) + x^2(x^3 - 1) + (x^2 +x +1)
=x(x^3 -1)(x^3 +1) +x^2(x^3-1) + (x^2 + x + 1)
=x(x-1)(x^2 + x +1)(x^3 +1) + x^2(x-1)(x^2 +x +1) +(x^2 +x +1)
=(x^2 +x +1)[x(x-1)(x^3 +1) +x^2(x-1) +1]
=(x^2 +x +1)[ x^5 - x^4 + x^3 - x + 1]
a) 3x3-2x2+2 chia x+1= 3x2-5x+5 dư -3 b) -3 chia hết x+1 vậy chon x =2
1)
a) \(-7x\left(3x-2\right)\)
\(=-21x^2+14x\)
b) \(87^2+26.87+13^2\)
\(=87^2+2.87.13+13^2\)
\(=\left(87+13\right)^2\)
\(=100^2\)
\(=10000\)
2)
a) \(x^2-25\)
\(=x^2-5^2\)
\(=\left(x-5\right)\left(x+5\right)\)
b) \(3x\left(x+5\right)-2x-10=0\)
\(\Leftrightarrow3x\left(x+5\right)-\left(2x-10\right)=0\)
\(\Leftrightarrow3x\left(x+5\right)-2\left(x-5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\3x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\3x=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy..........
3)
a) \(A:B=\left(3x^3-2x^2+2\right):\left(x+1\right)\)
Vậy \(\left(3x^3-2x^2+2\right):\left(x+1\right)=\left(3x^2-5x-5\right)+7\)
b)
Để \(A⋮B\Rightarrow7⋮\left(x+1\right)\)
\(\Rightarrow\left(x+1\right)\in U\left(7\right)=\left\{-1;1-7;7\right\}\)
Vì x là số nguyên nên x=0 ; x=6 thì \(A⋮B\)
\(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
\(x^7+x^5+1\)
\(=x^7+x^6+x^5-x^6-x^5-x^4+x^5+x^4+x^3-x^3-x^2-x+x^2+x+1\)
\(=x^5\left(x^2+x+1\right)-x^4\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^5-x^4+x^3-x+1\right)\left(x^2+x+1\right)\)
Thank you bạn nhiều nha!