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x11+x4+1
= x11+x10+x9-x10-x9-x8+x8+x7+x6-x7-x6-x5+x5+x4+x3-x3-x2-x+x2+x+1
= x9(x2+x+1)-x8(x2+x+1)+x6(x2+x+1)-x5(x2+x+1)+x3(x2+x+1)-x(x2+x+1)+(x2+x+1)
= (x2+x+1)(x9-x8+x6-x5+x3-x+1)
\(k\left(x\right)=\dfrac{5x^2-22x+25}{x^2-4x+4}\)
\(\Leftrightarrow k\left(x\right)=\dfrac{5x^2-20x+20-x+2-x+2+1}{x^2-4x+4}\)
\(\Leftrightarrow k\left(x\right)=\dfrac{\left(5x^2-20x+20\right)-\left(x-2\right)-\left(x-2\right)+1}{x^2-4x+4}\)
\(\Leftrightarrow k\left(x\right)=\dfrac{5\left(x^2-4x+4\right)-\left(x-2\right)-\left(x-2\right)+1}{x^2-4x+4}\)
\(\Leftrightarrow k\left(x\right)=\dfrac{5\left(x-2\right)^2-\left(x-2\right)-\left(x-2\right)+1}{\left(x-2\right)^2}\)
\(\Leftrightarrow k\left(x\right)=\dfrac{5\left(x-2\right)^2}{\left(x-2\right)^2}-\dfrac{x-2}{\left(x-2\right)^2}-\dfrac{x-2}{\left(x-2\right)^2}+\dfrac{1}{\left(x-2\right)^2}\)
\(\Leftrightarrow k\left(x\right)=5-\dfrac{1}{x-2}-\dfrac{1}{x-2}+\dfrac{1}{\left(x-2\right)^2}\)
Đặt \(y=\dfrac{1}{x-2}\)
\(\Rightarrow k\left(x\right)=5-y-y+y^2=y^2-2y+1+4=\left(y-1\right)^2+4\ge4\)
Vậy GTNN của \(k\left(x\right)=4\) khi \(y=1\Rightarrow\dfrac{1}{x-2}=1\Leftrightarrow x=3\)
\(h\left(x\right)=\dfrac{x^2-x+1}{\left(x-1\right)^2}\)
\(\Leftrightarrow h\left(x\right)=\dfrac{x^2-2x+1+x-1+1}{\left(x-1\right)^2}\)
\(\Leftrightarrow h\left(x\right)=\dfrac{\left(x-1\right)^2}{\left(x-1\right)^2}+\dfrac{x-1}{\left(x-1\right)^2}+\dfrac{1}{\left(x-1\right)^2}\)
\(\Leftrightarrow h\left(x\right)=1+\dfrac{1}{x-1}+\dfrac{1}{\left(x-1\right)^2}\)
Đặt \(y=\dfrac{1}{x-1}\)
\(\Rightarrow h\left(x\right)=1+y+y^2\)
\(\Rightarrow h\left(x\right)=y^2+y+1\)
\(\Rightarrow h\left(x\right)=y^2+2.y.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)
\(\Rightarrow h\left(x\right)=\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
=> GTNN của \(h\left(x\right)=\dfrac{3}{4}\) khi \(y+\dfrac{1}{2}=0\Leftrightarrow y=\dfrac{-1}{2}\)
\(\Leftrightarrow\dfrac{1}{x-1}=\dfrac{-1}{2}\)
\(\Leftrightarrow x=-1\)
a,(5x-2y)(x2-xy+1)=5x3-5x2+5x-2yx2+2xy2-2y
=5x3-7x2y+2xy2+5x-2y
b,(x-2)(x+2)(\(\dfrac{1}{2}\) x-5)=x2-4.\(\left(\dfrac{1}{2}x-5\right)\)
=\(\dfrac{1}{2}x^3-5x^2-2x+20\)
c,\(\left(x^2-2x+3\right)\left(\dfrac{1}{2}x-5\right)\)
=\(\dfrac{1}{2}x^3-5x^2-1x^2+10x+\dfrac{3}{2}x-15\)
=\(\dfrac{1}{2}x^3-6x^2+\dfrac{23}{2}x-15\)
d,\(\left(x^2-5\right)\left(x+3\right)+\left(x+4\right)\left(x-x^2\right)\)
=\(x^3+3x^2-5x-15+x^2-x^3+4x-4x^2\)
=\(-5x+4x-15\)
=\(-x-15\)
Chúc bạn học tốt(mỏi tay quá)
a) \(x^7+x^5+1\)
\(= (x^7 - x ) + (x^5 - x^2 ) + (x^2 + x + 1)\)
\(= x(x^3 - 1)(x^3 + 1) + x^2(x^3 - 1) + (x^2 + x + 1)\)
\(= (x^2 + x + 1)(x - 1)(x^4 + x) + x^2 (x - 1)(x^2 + x + 1) +(x^2 + x +1)\)
\(= (x^2 + x + 1)[(x^5 - x^4 + x^2 - x) + (x^3 - x^2 ) + 1]\)
\(= (x^2 + x + 1)(x^5 - x^4 + x^3 - x + 1)\)
b) tương tự
Bài 2:
Cách 1:
\(x^3-7x-6=x^3-3x^2+3x^2-9x+2x-6\)
\(=\left(x^3-3x^2\right)+\left(3x^2-9x\right)+\left(2x-6\right)\)
\(=x^2.\left(x-3\right)+3x.\left(x-3\right)+2.\left(x-3\right)\)
\(=\left(x-3\right).\left(x^2+3x+2\right)\)
\(=\left(x-3\right).\left(x^2+x+2x+2\right)\)
\(=\left(x-3\right).\left[\left(x^2+x\right)+\left(2x+2\right)\right]\)
\(=\left(x-3\right).\left[x.\left(x+1\right)+2.\left(x+1\right)\right]\)
\(=\left(x-3\right).\left(x+1\right).\left(x+2\right)\)
Cách 2:
\(x^3-7x-6=x^3+x^2-x^2-x-6x-6\)
\(=\left(x^3+x^2\right)-\left(x^2+x\right)-\left(6x+6\right)\)
\(=x^2.\left(x+1\right)-x.\left(x+1\right)-6.\left(x+1\right)\)
\(=\left(x+1\right).\left(x^2-x-6\right)\)
\(=\left(x+1\right).\left(x^2+2x-3x-6\right)\)
\(=\left(x+1\right).\left[\left(x^2+2x\right)-\left(3x+6\right)\right]\)
\(=\left(x+1\right).\left[x.\left(x+2\right)-3.\left(x+2\right)\right]\)
\(=\left(x+1\right).\left(x+2\right).\left(x-3\right)\)
Chúc bạn học tốt!!! Còn 1 cách nữa nhưng mình mỏi tay quá!!! 
a, \(x^3-9x^2+6x+16=x^3+x^2-10x^2-10x+16x+16\)
\(=\left(x^3+x^2\right)-\left(10x^2+10x\right)+\left(16x+16\right)\)
\(=x^2.\left(x+1\right)-10x.\left(x+1\right)+16.\left(x+1\right)\)
\(=\left(x+1\right).\left(x^2-10x+16\right)\)
\(=\left(x+1\right).\left(x^2-2x-8x+16\right)\)
\(=\left(x+1\right).\left[\left(x^2-2x\right)-\left(8x-16\right)\right]\)
\(=\left(x+1\right).\left[x.\left(x-2\right)-8.\left(x-2\right)\right]\)
\(=\left(x+1\right).\left(x-2\right).\left(x-8\right)\)
Chúc bạn học tốt!!!
\(a,\left(x+1\right)^2-\left(x-1\right)^2-3\left(x+1\right)\left(x-1\right)\)
\(=x^2+2x+1-\left(x^2-2x+1\right)-3\left(x^2-1\right)\)
\(=x^2+2x+1-x^2+2x-1-3x^2+2=-3x^2+4x+2\)\(b,5\left(x+2\right)\left(x-2\right)-\left(2x-3\right)^2-x^2+17\)
\(=5\left(x^2-4\right)-\left(4x^2-12x+9\right)-x^2+17\)
\(=5x^2-20-4x^2+12x-9-x^2+17=12x-12\)
c)(x2+x)2-2(x2+x)-15
đặt x2+x=a ta có
a2-2a-15
=a2+3a-5a-15
=(a2+3a)-(5a+15)
=a(a+3)-5(a+3)
=(a+3)(a-5)
thay a=x2+x
(x2+x+3)(x2+x-5)
a, Vì x2 ≥ 0 , 2y2 ≥ 0 với mọi x,y
=>x2+2y2+ 1 ≥ 1
=>Phân thức trên luôn có nghĩa
a).
\(x^5+x+1=\left(x^5+x^4+x^3\right)-\left(x^4+x^3+x^2\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^3-x^2\right)\)
b).\(x^8+x^7+1=\left(x^8+x^7+x^6\right)-\left(x^6+x^5+x^4\right)+\left(x^5+x^4+x^3\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)
d).
\(x^7+x^5+1=\left(x^7+x^6+x^5\right)-\left(x^6+x^5+x^4\right)+\left(x^5+x^4+x^3\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^5-x^4+x^3-x+1\right)\)
e).
\(x^8+x^4+1=x^8+2x^4+1-x^4\\ =\left(x^4+1\right)^2-\left(x^2\right)^2\\ =\left(x^4+x^2+1\right)\left(x^4-x^2+1\right)\\ =\left(x^4-x^2+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)\)
c).
\(x^5-x^4-1=x^5-x^3-x^2-\left(x^4-x^2-x\right)+x^3-x-1\\ \left(x^3-x-1\right)\left(x^2-x+1\right)\)