`a) x(x + 5)(x – 5) – (x + 2)(x^2 – 2x + 4) = 3`
`<=>x(x^2-25)-(x^3-8)=3`
`<=>x^3-25x-x^3+8=3`
`<=>-25x=-5`
`<=>x=1/5`
`b) (x – 3)^3 – (x – 3)(x^2 + 3x + 9) + 9(x + 1)^2 = 15`
`<=>x^3-9x^2+27x-27-(x^3-27)+9(x^2+2x+1)=15`
`<=>-9x^2+27x+9x^2+18x+9=15`
`<=>45x+9=15`
`<=>45x=6`
`<=>x=6/45=2/15`

`c) (x+5)(x^2 –5x +25) – (x – 7) = x^3`
`<=>x^3-125-x+7=x^3`
`<=>x^3-x-118=x^3`
`<=>-x-118=0`
`<=>-x=118<=>x=-118`
`d) (x+2)(x^2 – 2x + 4) – x(x^2 + 2) = 4 `
`<=>x^3+8-x^3-2x=4`
`<=>8-2x=4`
`<=>2x=4<=>x=2`

$ a/ 12x(x – 5) – 3x(4x - 10) = 120$

`<=>12x^2-60x-12x^2+30x=120`

`<=>-30x=120`

`<=>x=-4`

Vậy `x=-4`

$b/ 9x(x + 4) – 5x(3x + 2) = 112 - 2x(3x + 1)$

`<=>9x^2+36x-15x^2-10x=112-6x^2-2x`

`<=>-6x^2+26x=112-6x^2-2x`

`<=>28x=112`

`<=>x=4`

Vậy `x=4`

$c/ 3x(1 – x) - 5x(3x + 7) = 154 + 9x(5 – 2x)$

`<=>3x-3x^2-15x^2-35x=154+45x-18x^2`

`<=>-32x-18x^2=154+45x-18x^2`

`<=>77x=-154`

`<=>x=-2`

Vậy `x=-2`

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

\(=-5x\left(2x^2-x-12\right)-\left(x^2-9\right)+15x^2+20x-6x-8\)

\(=-10x^3+5x^2+60x-x^2+9+15x^2+20x-6x-8\)

\(=-10x^3+19x^2+74x+1\)

\(b,\left(4x-1\right)x\left(3x+1\right)-5x^2.x\left(x-3\right)-\left(x-4\right)x\left(x-5\right)\)\(-7\left(x^3-2x^2+x-1\right)\)

\(=\left(4x^2-x\right)\left(3x+1\right)-5x^4-15x^3-\left(x^2-4x\right)\left(x-5\right)\)\(-7x^3+14x^2-7x+7\)

\(=12x^3+x^2-x-5x^4-15x^3-x^3+9x^2+20x\)\(-7x^3+14x^2-7x+7\)

\(=-5x^4-11x^3+24x^2+12x+7\)

\(c,\left(5x-7\right)\left(x-9\right)-\left(3-x\right)\left(2-5x\right)-2x\left(x-4\right)\)

\(=5x^2-52x+63-6+17x-5x^2-2x^2+8x\)

\(=-2x^2-27x+57\)

\(d,\left(5x-4\right)\left(x+5\right)-\left(x+1\right)\left(x^2-6\right)-5x+19\)

\(=5x^2+21x-20-x^3-x^2+6x+6-5x+19\)

\(=-x^3+4x^2+22x+5\)

\(e,\left(9x^2-5\right)\left(x-3\right)-3x^2\left(3x+9\right)-\left(x-5\right)\left(x+4\right)-9x^3\)

\(=9x^3-27x^2-5x+15-9x^3-27x^2-x^2+x+20-9x^3\)

\(=-9x^3-55x^2+4x+35\)

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

\(=x^2-2x+1-x^2-4x-4\)

\(=-6x-3\)

Bài 1:

a)-x^2+4x-5

=-(x²-4x+5)<0 với mọi x

=>-x^2+4x-5<0 với mọi x

b)x^4+3x^2+3

\(=\left(x^2+\frac{3}{2}\right)^2+\frac{3}{4}>0\)với mọi x

=>x^4+3x^2+3>0 với mọi x

c) bn xét từng th ra

Bài 2:

a)9x^2-6x-3=0

=>3(3x²-2x-1)=0

=>3x²-2x-1=0

=>3x²+x-3x-1=0

=>x(3x+1)-(3x+1)=0

=>(x-1)(3x+1)=0

b)x^3+9x^2+27x+19=0

=>(x+1)(x²+8x+19) (dùng pp nhẩm nghiệm rồi mò ra)

c)x(x-5)(x+5)-(x+2)(x^2-2x+4)=3

=>x³-25x-x³-8=3

=>-25x-8=3

=>-25x=1

=>x=-11/25

mk sửa 1 tí ở dấu => thứ 2 từ dưới lên là

=>-25x=11

(x+2)(x+3)-(x-2)(x+5)=0

=> x²+5x+6-x²-3x+10=0

=>2x+16=0 

 =>2x=-16

=>x=-8

a) Ta có: \(6x\left(x-5\right)+3x\left(7-2x\right)=18\)

\(\Leftrightarrow6x^2-30x+21x-6x^2=18\)

\(\Leftrightarrow-9x=18\)

hay x=-2

Vậy: S={-2}

b) Ta có: \(2x\left(3x+1\right)+\left(4-2x\right)\cdot3x=7\)

\(\Leftrightarrow6x^2+2x+12x-6x^2=7\)

\(\Leftrightarrow14x=7\)

hay \(x=\dfrac{1}{2}\)

Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)

c) Ta có: \(0.5x\left(0.4-4x\right)+\left(2x+5\right)\cdot x=-6.5\)

\(\Leftrightarrow0.2x-2x^2+2x^2+5x=-6.5\)

\(\Leftrightarrow5.2x=-6.5\)

hay \(x=-\dfrac{5}{4}\)

Vậy: \(S=\left\{-\dfrac{5}{4}\right\}\)

d) Ta có: \(\left(x+3\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)=6\)

\(\Leftrightarrow x^2+5x+6-\left(x^2+3x-10\right)=6\)

\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)

\(\Leftrightarrow2x+16=6\)

\(\Leftrightarrow2x=-10\)

hay x=-5

Vậy: S={-5}

e) Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)

\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)

\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)

\(\Leftrightarrow14x=0\)

hay x=0

Vậy: S={0}

a: =>x^2-25-x^2-3x=10

=>-3x=35

=>x=-35/3

b: =>4x^2-9-4(x^2+4x+4)=5

=>4x^2-9-4x^2-16x-16-5=0

=>-16x-30=0

=>x=-15/8

c: =>9x^2+45x-9x^2+4=7

=>45x=3

=>x=1/15

d: =>x^3+3x^2+3x+1-x^3-3x^2+5x=8

=>8x=7

=>x=7/8

\(a,\left(x+2\right)^2+\left(x+3\right)^2-2\left(x-2\right)\left(x-3\right)=19\\ \Leftrightarrow x^2+4x+4+x^2+6x+9-2x^2+10x-12=19\\ \Leftrightarrow20x=20\\ \Leftrightarrow x=1\\ b,\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-5\right)=15\\ \Leftrightarrow x^3+8-x^3+5x=15\\ \Leftrightarrow5x=7\\ \Leftrightarrow x=\dfrac{7}{5}\\ c,\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=17\\ \Leftrightarrow x^3-3x^2+3x+1+8-x^3+3x^2+6x=17\\ \Leftrightarrow9x=8\\ \Leftrightarrow x=\dfrac{8}{9}\)

a. (x + 2)² + (x + 3)² - 2(x - 2)(x - 3) = 19

<=> (x² + 4x + 4) + (x² + 6x + 9) - (2x + 4)(x - 3) = 19

<=> x² + 4x + 4 + x² + 6x + 9 - 2x² + 6x - 4x + 12 = 19

<=> x² + x² - 2x² + 4x + 6x + 6x - 4x + 9 + 4 + 12 - 19 = 0

<=> 12x + 6 = 0

<=> 6(2x + 1) = 0

<=> 2x + 1 = 0

<=> 2x = -1

<=> x = \(\dfrac{-1}{2}\)  

b. (x + 2)(x² - 2x + 4) - x(x² - 5) = 15

<=> x³ - 2x² + 4x + 2x² - 4x + 8 - x³ + 5x - 15 = 0

<=> x³ - x³ - 2x² + 2x² + 4x - 4x + 5x + 8 - 15 = 0

<=> 5x - 7 = 0

<=> 5x = 7

<=> x = \(\dfrac{7}{5}\)

c: \(-x^2+2x-2=-\left(x-1\right)^2-1\le-1\forall x\)

\(\Leftrightarrow V\ge-1\forall x\)

Dấu '=' xảy ra khi x=1