Khó quá, ai giúp em đi ạ

\(a,=4x^2+3xy-y^2+4xy-4x^2=7xy-y^2\\ b,=x^2-9-x^3+3x+x^2-3=-x^3+2x^2+3x-12\\ c,=-2x^2+12x-18+5x^2+4x-1=3x^2+16x-19\\ d,=8x^3+1-3x^3+6x^2=5x^3+6x^2+1\\ e,=\left(3x^2+4x+15x+20\right):\left(3x+4\right)\\ =\left(3x+4\right)\left(x+5\right):\left(3x+4\right)\\ =x+5\\ f,=\left(x^3+4x^2-3x+3x^2+12x-9+3x+3\right):\left(x^2+4x-3\right)\\ =\left[\left(x^2+4x-3\right)\left(x+3\right)+3x+3\right]:\left(x^2+4x-3\right)\\ =x+3\left(dư.3x+3\right)\)

\(\left(-3x-2\right)^2+\left(3x+5\right)\left(5-3x\right)=-7\)

\(\Leftrightarrow9x^2+12x+4+15x-9x^2+25-15x=-7\)

\(\Leftrightarrow12x+36=0\Leftrightarrow x=-3\)

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

\(\Leftrightarrow x^3+2x^2+2x+2x^2+4x+4-x\left(x^2-16x+64\right)=16x^2-9\)

\(\Leftrightarrow x^3+4x^2+6x+4-x^3+16x^2-64=16x^2-9\)

\(\Leftrightarrow4x^2+6x-51=0\)

\(\cdot\Delta=6^2-4.4.\left(-51\right)=852\)

Vậy pt có 2 nghiệm phân biệt

\(x_1=\frac{-6+\sqrt{852}}{8}\);\(x_2=\frac{-6-\sqrt{852}}{8}\)

giúp mình bài ni với :3x^2(x+1)-5x(x+1)^2+4(x+1)

a) \(2x\left(x+1\right)-x^2\left(x+2\right)+x^3-x+4=0\)

\(\Leftrightarrow2x^2+2x-x^3-2x^2+x^3-x+4=0\)

\(\Leftrightarrow x+4=0\)

\(\Leftrightarrow x=-4\)

Vậy ...

b) \(4x\left(3x+2\right)-6x\left(2x+5\right)+21\left(x-1\right)=0\)

\(\Leftrightarrow12x^2+8x-12x^2-30x+21x-21=0\)

\(\Leftrightarrow-x-21=0\)

\(\Leftrightarrow x=-21\)

Vậy ...

c) \(5x\left(12x+7\right)-3x\left(2x-5\right)=-100\)

\(\Leftrightarrow60x^2+35x-6x^2+15x+100=0\)

\(\Leftrightarrow54x^2+50x+100=0\)

\(\Leftrightarrow54\left(x^2+\frac{25}{27}x+\frac{625}{2916}\right)+\frac{290975}{2916}=0\)

\(\Leftrightarrow54\left(x+\frac{25}{54}\right)^2+\frac{290975}{2916}=0\left(ktm\right)\)

Vậy phương trình vô nghiệm.

d) \(x\left(x-1\right)-x^2+2x=5\)

\(\Leftrightarrow x^2-x-x^2+2x-5=0\)

\(\Leftrightarrow x-5=0\)

\(\Leftrightarrow x=5\)

Vậy ...

e) \(2x^3\left(2x-3\right)-x^2\left(4x^2-6x+2\right)=0\)

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

\(\Leftrightarrow-2x^2=0\)

\(\Leftrightarrow x=0\)

Vậy ...

Dài ... quá :))

A(x) = x³ - 2x + 3x² - 3/2x + x⁴ - x³ + 5x - 7 - 0,7x² + 2x⁴ - 3/4

       = (x³ - x³) + (-2x - 3/2x + 5x) + (3x² - 0,7x²) + (x⁴ + 2x⁴) + (-7 - 3/4)

       = 3/2x + 2,3x² + 3x⁴ - 31/4

Sắp xếp : A(x) =  3x⁴ + 0x³ + 2,3x² + 3/2x - 31/4

b(x) = 3x⁵ - 12x³ - 6x² + 2x⁵ - 2x⁴ + 4x² + x⁵ - 2x⁴

       = (3x⁵ + x⁵ + 2x⁵)  - 12x³ + (-6x² + 4x²) + (-2x⁴ - 2x⁴)

       = 6x⁵ - 12x³ - 2x² - 4x⁴

Sắp xếp : B(x) = 6x⁵ - 4x⁴ - 12x³ - 2x²

Tính :

h(x) = a(x) + b(x)

=> h(x) = (3x⁴ + 0x³ + 2,3x²+ 3/2x - 31/4) + (6x⁵ - 4x⁴ - 12x³ - 2x²)

=> h(x) = 3x⁴ + 0x³ + 2,3x² + 3/2x - 31/4 + 6x⁵ - 4x⁴ - 12x³ - 2x²

=> h(x) = (3x⁴ - 4x⁴) + (0x³ - 12x³) + (2,3x² - 2x²) + 3/2x - 31/4 + 6x⁵

=> h(x) = -x⁴ - 12x³ + 0,3x² + 3/2x - 31/4 + 6x⁵

Còn bài trừ tương tự nhưng đổi dấu vế thứ hai thôi ...