Thay 11= 10+1 ta có

x¹⁰-(10+1)x⁹+(10+1)x⁸-(10+1)x⁷+(10+1)x⁶-(10+1)x⁵+(10+1)x⁴-(10+1)x³+(10+1)x²-(10+1)x+2

= x¹⁰-(x+1)x⁹+(x+1)x⁸-(x+1)x⁷+(x+1)x⁶-(x+1)x⁵+(x+1)x⁴-(x+1)x³+(x+1)x²-(x+1)x+2

= x¹⁰-x¹⁰-x⁹+x⁹+x⁸-x⁸-x⁷+x⁷+x⁶-x⁶-x⁵+x⁵+x⁴-x⁴-x³+x³+x²-x²-x+2

= -x+2 

Thay x=10 vào bt

= -10+2

= -8

Câu 1:

 \(x^4+5x^3-12x^2+5x+1=x^4+7x^3+x^2-2x^3-14x^2-x+x^2+7x+1\)

\(=\left(x^4+7x^3+x^2\right)-\left(2x^3+14x^2+x\right)+\left(x^2+7x+1\right)\)

\(=x^2\left(x^2+7x+1\right)-2x\left(x^2+7x+1\right)+\left(x^2+7x+1\right)\)

\(=\left(x^2-2x+1\right)\left(x^2+7x+1\right)\)

\(=\left(x-1\right)^2\left(x^2+7x+1\right)\)

Câu 2:

\(\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)-24x^2=x^4-24x^3+203x^2-720x+900-24x^2\)

\(=x^4-24x^3+179x^2-720x+900\)

\(=\left(x^4-7x^3+30x^2\right)-\left(17x^3-119x^2+510x\right)+\left(30x^2-210x+900\right)\)

\(=x^2\left(x^2-7x+30\right)-17x\left(x^2-7x+30\right)+30\left(x^2-7x+30\right)\)

\(=\left(x^2-17x+30\right)\left(x^2-7x+30\right)\)

\(=\left(x^2-2x-15x+30\right)\left(x^2-7x+30\right)\)

\(=\left[x\left(x-2\right)-15\left(x-2\right)\right]\left(x^2-7x+30\right)\)

\(=\left(x-15\right)\left(x-2\right)\left(x^2-7x+30\right)\)

Câu 3:

\(2x^3+11x^2+3x-36=\left(2x^3+14x^2+24x\right)-\left(3x^2+21x+36\right)\)

\(=2x\left(x^2+7x+12\right)-3\left(x^2+7x+12\right)\)

\(=\left(2x-3\right)\left(x^2+7x+12\right)\)

\(=\left(2x-3\right)\left(x^2+3x+4x+12\right)\)

\(=\left(2x-3\right)\left[x\left(x+3\right)+4\left(x+3\right)\right]\)

\(=\left(2x-3\right)\left(x+3\right)\left(x+4\right)\)

a: A⋮B

=>\(4x^2-6x+a\) ⋮x-3

=>\(4x^2-12x+6x-18+a+18\) ⋮ x-3

=>a+18=0

=>a=-18

b: A⋮B

=>\(2x^3-7x^2-11x+a-8\) ⋮\(2x^2+3x+4\)

=>\(2x^3+3x^2+4x-10x^2-15x-20+a+12\) ⋮\(2x_{}^2+3x+4\)

=>a+12=0

=>a=-12

a: \(\Leftrightarrow\left(x+12-3x\right)\left(x+12+3x\right)=0\)

=>(-2x+12)(4x+12)=0

=>x=-3 hoặc x=6

b: \(\Leftrightarrow20x^3-15x^2+45x-45=0\)

=>\(x\simeq0.93\)

d: =>-4x+28+11x=-x+3x+15

=>7x+28=2x+15

=>5x=-13

=>x=-13/5

e: \(\Leftrightarrow4x^3-12x+x=4x^3-3x+5\)

=>-9x=-3x+5

=>-6x=5

=>x=-5/6

a: Sửa đề: B=|2x+1|+|2x+3|

Ta có; B=|2x+1|+|2x+3|

=|2x+3|+|-2x-1|

=>B>=|2x+3-2x-1|=2∀x

Dấu '=' xảy ra khi (2x+1)(2x+3)<=0

=>\(-\frac32\le x\le-\frac12\)

b: ĐKXĐ: x>=1/2

\(\sqrt{2x-1}\ge0\forall x\) thỏa mãn ĐKXĐ

=>\(3\sqrt{2x-1}\ge0\forall x\) thỏa mãn ĐKXĐ

=>\(3\sqrt{2x-1}+\frac34\ge\frac34\forall x\) thỏa mãn ĐKXĐ

=>C>=3/4∀x thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi 2x-1=0

=>2x=1

=>x=1/2

c: \(2\left(x-3\right)^2\ge0\forall x;\frac{7}{11}\left|3y+7\right|\ge0\forall y\)

=>\(2\left(x-3\right)^2+\frac{7}{11}\left|3y+7\right|\ge0\forall x,y\)

=>\(-2\left(x-3\right)^2-\frac{7}{11}\left|3y+7\right|\le0\forall x,y\)

=>\(-2\left(x-3\right)^2-\frac{7}{11}\left|3y+7\right|-2011\le-2011\forall x,y\)

dấu '=' xảy ra khi x-3=0 và 3y+7=0

=>x=3 và y=-7/3

c​âu c:x^4-2x^3-x^2+x^3-2x^2-x+5x^2-10x-5=x^2(x^2-2x-1)+x(x^2-2x-1)+5(x^2-2x-1)=(x^2-2x-1)(x^2+x+5)

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