a: =2x^5-15x^3-x^2-2x^5-x^3=-16x^3-x^2

b: =x^3+3x^2-2x-3x^2-9x+6

=x^3-11x+6

c: \(=\dfrac{4x^3+2x^2-6x^2-3x-2x-1+5}{2x+1}\)

\(=2x^2-3x-1+\dfrac{5}{2x+1}\)

a) \(6x^3\left(\dfrac{1}{3}x^2-\dfrac{5}{2}-\dfrac{1}{6}\right)-2x^5-x^3\)

\(=6x^3\left(\dfrac{1}{3}x^2-\dfrac{16}{6}\right)-2x^5-x^3\)

\(=2x^5-16x^3-2x^5-x^3\)

\(=-17x^3\)

b) \(\left(x+3\right)\left(x^2+3x-2\right)\)

\(=x^3+3x^2-2x+3x^2+9x-6\)

\(=x^3+6x^2+7x-6\)

c) \(\left(4x^3-4x^2-5x+4\right):\left(2x+1\right)\)

\(=2x^2+4x^3-2x-4x^2-\dfrac{5}{2}-5x+\dfrac{2}{x}+4\)

\(=4x^3-2x^2-7x+\dfrac{2}{x}+\dfrac{3}{2}\)

1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)

=>-13x=0

=>x=0

2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

=>3x=13

=>x=13/3

3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)

=>-2x^2=0

=>x=0

4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

=>-8x=6-14=-8

=>x=1

`1)2x(x-5)-(3x+2x^2)=0`

`<=>2x^2-10x-3x-2x^2=0`

`<=>-13x=0`

`<=>x=0`

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`2)x(5-2x)+2x(x-1)=13`

`<=>5x-2x^2+2x^2-2x=13`

`<=>3x=13<=>x=13/3`

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`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`

`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`

`<=>x=0`

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`4)5x(x-1)-(x+2)(5x-7)=0`

`<=>5x^2-5x-5x^2+7x-10x+14=0`

`<=>-8x=-14`

`<=>x=7/4`

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`5)6x^2-(2x-3)(3x+2)=1`

`<=>6x^2-6x^2-4x+9x+6=1`

`<=>5x=-5<=>x=-1`

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`6)2x(1-x)+5=9-2x^2`

`<=>2x-2x^2+5=9-2x^2`

`<=>2x=4<=>x=2`

\(a,x^2\left(x-2x^3\right)\)

\(=x^3-2x^5\)

\(b,\left(x-2\right)\left(x-x^2+4\right)\)

\(=x^2-x^3+4x-2x+2x^2-8\)

\(=3x^2-x^3+2x-8\)

\(c,\left(x^2-1\right)\left(x^2+2x\right)\)

\(=x^4+2x^3-x^2-2x\)

\(d,\left(2x-1\right)\left(3x+2\right)\left(3-x\right)\)

\(=\left(6x^2+4x-3x-2\right)\left(3-x\right)\)

\(=\left(6x^2+x-2\right)\left(3-x\right)\)

\(=18x^2+3x-6-6x^3-x^2+2x\)

\(=17x^2+5x-6-6x^3-x^2\)

\(e,\left(x+3\right)\left(x^2+3x-5\right)\)

\(=x^3+3x^2-5x+3x^2+9x-15\)

\(=x^3+6x^2+4x-15\)

\(f,\left(xy-2\right)\left(x^3-2x-6\right)\)

\(=x^4y-2x^2y-6xy-2x^3+4x-12\)

\(g,\left(5x^3-x^2+2x-3\right)\left(4x^2-x+2\right)\)

\(=20x^5-4x^4+8x^3-12x^2-5x^4+x^3-2x^2+3x+10x^3-2x^2+4x-6\)

\(=20x^5-9x^4+19x^3-16x^2+7x-6\)

a. x²(x−2x³)= x³-2x⁵

b. (x−2)(x−x²+4)= x²-x³+4x-2x+2x²-8= -x³+3x²+2x-8

c. (x²−1)(x²+2x)= x⁴+2x³-x²-2x

d. (2x−1)(3x+2)(3−x) = (6x²+x-2)(3-x)=18x²-6x³+3x-x²-6+2x =-6x³+17x²+5x-6

e. (x+3)(x²+3x−5)= x³+3x²-5x+3x²+9x-15= x³+6x²+4x-15

f. (xy−2)(x³−2x−6)= x⁴y-2x²y-6xy-2x³+4x+12

g. (5x³−x²+2x−3)(4x²−x+2)= 20x⁵-9x⁴+19x³-12x²+7x-6

Em làm bài 2 nha!

\(A=\frac{3-4x}{x^2+1}\Leftrightarrow Ax^2+4x+A-3=0\) (1)

+)\(A=0\Rightarrow x=\frac{3}{4}\)

+) A khác 0 thì (1) là pt bậc 2.

\(\Delta'=\left(2\right)^2-A\left(A-3\right)\ge0\Leftrightarrow4-A^2+3A\ge0\Leftrightarrow-1\le A\le4\)

Vậy...

Bài 1: (bài nào nghĩ ra thì em làm trước)

C = \(\frac{2x^2-6x+5}{\left(x-1\right)^2}\). Đặt x - 1 = y >0 thì x = y + 1 >1

Khi đó \(C=\frac{2\left(y+1\right)^2-6\left(y+1\right)+5}{y^2}=\frac{2y^2-2y+1}{y^2}\)

\(=\frac{1}{y^2}-\frac{2}{y}+2\). đặt \(\frac{1}{y}=t>0\). \(C=t^2-2t+2=\left(t-1\right)^2+1\ge1\)

Đẳng thức xảy ra khi t = 1 suy ra y = 1 suy ra x = 2

Vậy Min C = 1 khi x = 2