=> B= (x-1)(x^2-x+1).2(x+1)3(x^2+x+1)

=> B= 6(x-1)(x^2+x+1).(x+1)(x^2-x+1)

=>B =6(x^3-1)(x^3+1)

=> B 6x^6-6

Bài 4: 

b: \(=x^2z\left(-1+3-7\right)=-5x^2z=-5\cdot\left(-1\right)^2\cdot\left(-2\right)=10\)

c: \(=xy^2\left(5+0.5-3\right)=2.5xy^2=2.5\cdot2\cdot1^2=5\)

a.ta có

\(\left(x+3\right)^3+\left(x^2+1\right)\left(x-2\right)=x^3+9x^2+27x+27+x^3-2x^2+x-2\)

\(=2x^3+7x^2+28x+25\)

b.\(\left(2x-1\right)^2-\left(2-x\right)^3=4x^2-4x+1+x^3-6x^2+12x-8\)

\(=x^3-2x^2+8x-7\)

a) (x + 3)³ + (x² + 1)(x - 2)

= x³ + 9x² + 27x + 27 + x³ - 2x + x - 2

= x³ + x³ + 9x² + 27x - 2x + x + 27 - 2

= 2x³ + 9x² + 26x + 25

b) (2x - 1)² - (2 - x)³

= 4x² - 4x + 1 - ( 8 - 12x + 6x² - x³)

= 4x² - 4x + 1 - 8 + 12x - 6x² + x³  

= x³ + 4x² - 6x² + 12x - 4x + 1 -  8

= x³ - 2x² + 8x - 7

\(a,B=4\sqrt{x=1}-3\sqrt{x+1}+2\)\(\sqrt{x+1}+\sqrt{x+1}\)

\(=4\sqrt{x+1}\)

\(b,\)đưa về \(\sqrt{x+1}=4\Rightarrow x=15\)

a, Với \(x\ge-1\)

\(\Rightarrow B=4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}\)

\(=4\sqrt{x+1}\)

b, Ta có B = 16 hay 

\(4\sqrt{x+1}=16\Leftrightarrow\sqrt{x+1}=4\)bình phương 2 vế ta được 

\(\Leftrightarrow x+1=16\Leftrightarrow x=15\)

a) B = 4√x+1                                   b) x = 15

a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

b: Ta có: \(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)

\(=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)

\(=\dfrac{2}{x+\sqrt{x}+1}\)

c: Ta có: \(x+\sqrt{x}+1>0\forall x\) thỏa mãn ĐKXĐ

\(\Leftrightarrow\dfrac{2}{x+\sqrt{x}+1}>0\forall x\)

a) x(x + 1) - 2x(x - 2) = x² + x - 2x² + 4x = -x² + 5x

b) -3x(x - 1) + (x - 1)(x + 1) = -3x² + 3x + x² - 1 = -2x² + 3x  - 1

c) (3x - 2)(3x + 2) - (x - 1)(x + 2) = 9x² - 4 - x² - x + 2 

= 8x² - x - 2

a, x(x+1) - 2x(x -2 )

= x² +x  - 2x² + 4x  = -x² + 5x

b, -3x( x - 1 ) + ( x -1 ) ( x+1 )

= -3x² + 3x + x² -1

= -2x² + 3x -1

c, ( 3x-2 ) ( 3x + 2 ) - ( x -1 ) ( x +2 )

= 9x² - 4 - ( x² + 2x -x -2 ) 

= 9x² -4 - x² -2x + x + 2

= 8x² -x -2

*Sxl

Trả lời:

a) x ( x + 1 ) - 2x ( x - 2 ) = x² + x - 2x² + 4x = - x² + 5x

b) - 3x ( x - 1 ) + ( x - 1 ) ( x + 1 ) = - 3x² + 3x + x² - 1 = - 2x² + 3x - 1

c) ( 3x - 2 ) ( 3x + 2 ) - ( x - 1 ) ( x + 2 ) = 9x² - 4 - ( x² + 2x - x - 2 ) = 9x² - 4 - x² - 2x + x + 2 = 8x² - x - 2

Bài 5:

\(x^3=18+3\sqrt[3]{\left(9+4\sqrt{5}\right)\left(9-4\sqrt{5}\right)}\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)\\ \Leftrightarrow x^3=18+3x\sqrt[3]{1}\\ \Leftrightarrow x^3-3x=18\\ y^3=6+3\sqrt[3]{\left(3-2\sqrt{2}\right)\left(3+2\sqrt{2}\right)}\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)\\ \Leftrightarrow y^3=6+3y\sqrt[3]{1}\\ \Leftrightarrow y^3-3y=6\\ P=x^3+y^3-3\left(x+y\right)+1993\\ P=\left(x^3-3x\right)+\left(y^3-3y\right)+1993\\ P=18+6+1993=2017\)

x3=18+33√(9+4√5)(9−4√5)(3√9+4√5+3√9−4√5)⇔x3=18+3x3√1⇔x3−3x=18y3=6+33√(3−2√2)(3+2√2)(3√3+2√2+3√3−2√2)⇔y3=6+3y3√1⇔y3−3y=6P=x3+y3−3(x+y)+1993P=(x3−3x)+(y3−3y)+1993P=18+6+1993=2017

a: \(=9-4\sqrt{5}\cdot\dfrac{1}{\sqrt{5}}=9-4=5\)

b:  \(=\sqrt{5}-2-\dfrac{1}{2}\cdot2\sqrt{5}=-2\)