\(\left(3x-2\right)^2-2\left(x-1\right)^2\)

\(=9x^2-12x+4-2\left(x^2-2x+1\right)\)

\(=9x^2-12x+4-2x^2+4x-2\)

\(=7x^2-16x+2\)

a: (x+1)^3-x(x-2)^2+x-1=0

=>x^3+3x^2+3x+1-x(x^2-4x+4)+x-1=0

=>x^3+3x^2+4x-x^3+4x^2-4x=0

=>7x^2=0

=>x=0

b: =>x^3-3x^2+3x-1-x^3-27+3x^2-12=2

=>3x=2+1+27+12=39+3=42

=>x=14

\(A+B=\) \((5xzy-6x^2+8xy)+(3x^2+2xyz-8xy+y^2)\)

           \(=\) \(5xzy-6x^2+8xy+3x^2+2xyz-8xy+y^2\)

           \(=5xzy+\left(-6x^2+3x^2\right)+\left(8xy-8xy\right)+2xyz+y^2\)

          \(=5xzy-3x^2+2xyz+y^2\)

Lời giải:

$(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=2$

$\Leftrightarrow (x+\sqrt{x^2+1})(x-\sqrt{x^2+1})(y+\sqrt{y^2+1})=2(x-\sqrt{x^2+1})$

$\Leftrightarrow -(y+\sqrt{y^2+1})=2(x-\sqrt{x^2+1})$

$\Leftrightarrow 2x+\sqrt{y^2+1}=2\sqrt{x^2+1}-y$

$\Rightarrow (2x+\sqrt{y^2+1})^2=(2\sqrt{x^2+1}-y)^2$
$\Leftrightarrow 4x^2+y^2+1+4x\sqrt{y^2+1}=4(x^2+1)+y^2-4y\sqrt{x^2+1}$

$\Leftrightarrow 4(x\sqrt{y^2+1})+y\sqrt{x^2+1})=3$

$\Leftrightarrow 4Q=3$

$\Leftrightarrow Q=\frac{3}{4}$ 

\(x=3\ge2\Leftrightarrow y=3+1=4\\ x=-1< 2\Leftrightarrow y=\left(-1\right)^2-2=1-2=-1\\ x=2\ge2\Leftrightarrow y=2+1=3\)

P(x) = 9x⁴ + 5x³ + 8x² - 15x³ - 4x² - x⁴ + 15 - 7x²

        = (9x⁴ - x⁴) + (5x³ - 15x³) + (8x² - 4x² - 7x²) + 15

        = 8x⁴ - 10x³ - 3x² + 15

Ta có: P(1) = 8. 1⁴ - 10. 1³ - 3. 1² + 15 = 8 - 10 - 3 + 15 = 10

          P(0) = 8. 0⁴ - 10. 0³ - 3. 0² + 15 = 0 - 0 - 0 + 15 = 15

          P(-1) = 8.(-1)⁴ - 10(-1)³ - 3(-1)² + 15 = -8 - (-10) - (-3) + 15 = 20

\(A+B=-x^2-5yz+z^2+7yz-z^2+5x^2=4x^2+2yz\)

\(A-B=-x^2-5yz+z^2+z^2-7yz-5x^2=-6x^2-12yz+2z^2\)