Câu đặc biệt :

\(\left(3x-2\right)\left(x+1\right)^2\left(3x+8\right)=-16\)

\(\Leftrightarrow9x^4+36x^3+29x^2-14x-16=-16\)

\(\Leftrightarrow9x^4+36x^3+29x^2-14x=0\)

\(\Leftrightarrow x\left(9x^3+36x^2+29x-14\right)=0\)

\(\Leftrightarrow x\left[\left(9x^3+18x^2-7x\right)+\left(18x^2+36x-14\right)\right]=0\)

\(\Leftrightarrow x\left[x\left(9x^2+18x-7\right)+2\left(9x^2+18x-7\right)\right]=0\)

\(\Leftrightarrow x\left(x+2\right)\left(9x^2+18x-7\right)=0\)

\(\Leftrightarrow x\left(x+2\right)\left[\left(9x^2+21x\right)-\left(3x+7\right)\right]=0\)

\(\Leftrightarrow x\left(x+2\right)\left[3x\left(3x+7\right)-\left(3x+7\right)\right]=0\)

\(\Leftrightarrow x\left(x+2\right)\left(3x-1\right)\left(3x+7\right)=0\)

<=> x = 0 hoặc x + 2 = 0 hoặc 3x - 1 = 0 hoặc 3x + 7 = 0

<=> x = 0 hoặc x = - 2 hoặc x = 1/3 hoặc x = 7/3

Vậy phương trình có tập nghiệm là : \(S=\left\{0;\frac{1}{3};\frac{7}{3};-2\right\}\)

Câu 2:

a) Ta có: \(2x^2+3x+1>0\)

\(\Leftrightarrow\frac{2x^2+3x+1}{3}>\frac{0}{3}\)

\(\Leftrightarrow\frac{2}{3}x^2+x+\frac{1}{3}>0\)

=> đpcm

b) Ta có: \(4x-1< 0\)

\(\Leftrightarrow0-\left(4x-1\right)>0\)

\(\Leftrightarrow1-4x>0\)

=> đpcm

c) Ta có: \(\frac{3x-2}{4}+2\frac{1}{2}>0\)

\(\Leftrightarrow\frac{3x-2}{4}+\frac{10}{4}>0\)

\(\Leftrightarrow\frac{3x+8}{4}>0\)

\(\Rightarrow3x+8>0\)

=> đpcm

\(2x^2+3\left(x^2-1\right)=5x^2+5x\)            

\(2x^2+3x^2-3=5x^2+5x\)   

\(5x^2-3=5x^2+5x\)                 

\(-3=5x\)          

\(x=-\frac{3}{5}\)      

\(2x^2+3\left(x-1\right)\left(x+1\right)=5x\left(x+1\right)\)

\(\Leftrightarrow2x^2+3\left(x^2-1\right)-5x^2-5x=0\)

\(\Leftrightarrow2x^2+3x^2-5x^2-3-5x=0\)

\(\Leftrightarrow-5x=3\)

\(\Leftrightarrow x=\frac{-3}{5}\)

\(x^3-5x^2+8x-4\)

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

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

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

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

              Bài làm :

Ta có :

\(x^3-5x^2+8x-4\)

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

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

\(=x\left(x-2\right)^2-\left(x-2\right)^2.1\)

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

Sửa đề :

(x - 2)² - 16 = 0

=> (x - 2)² = 16

=> (x - 2)² = (\(\pm\)4)²

=> \(\orbr{\begin{cases}x-2=4\\x-2=-4\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=-2\end{cases}}\)

(x - 2)² - x² + 4 = 0

=> x² - 4x + 4 - x² + 4 = 0

=> (x² - x²) - 4x + (4 + 4) = 0

=> -4x + 8 = 0

=> -4x = -8

=> x = 2

(2x + 3)² - (\(\frac{1}{3}-2x\))² = -2/3x + 5

=> \(\left(2x+3\right)\left(2x+3\right)-\left(\frac{1}{3}-2x\right)\left(\frac{1}{3}-2x\right)=-\frac{2}{3}x+5\)

=> \(2x\left(2x+3\right)+3\left(2x+3\right)-\frac{1}{3}\left(\frac{1}{3}-2x\right)+2x\left(\frac{1}{3}-2x\right)=-\frac{2}{3}x+5\)

=> \(4x^2+6x+6x+9-\frac{1}{9}+\frac{2}{3}x+\frac{2}{3}x-4x^2=-\frac{2}{3}x+5\)

=> \(\left(4x^2-4x^2\right)+\left(6x+6x+\frac{2}{3}x+\frac{2}{3}x\right)+\left(9-\frac{1}{9}\right)+\frac{2}{3}x-5=0\)

=> \(\frac{40}{3}x+\frac{80}{9}+\frac{2}{3}x-5=0\)

=> \(\frac{40}{3}x+\frac{2}{3}x+\frac{80}{9}-5=0\)

=> 14x + 80/9 - 5 = 0

=> x = -5/18

Câu a thiếu kết quả để tìm

Câu b)

(x + 2)(x² - 2x + 4x² - 2x + 4) - x(x² + 2) = 15

=> (x + 2)(x² - 4x + 4x² + 4) - x³ + 2x = 15

=> (x + 2)(5x² - 4x + 4) - x³ + 2x = 15

=> x(5x² - 4x + 4) + 2(5x² - 4x + 4) - x³ + 2x = 15

=> 5x³ - 4x² + 4x + 10x² - 8x + 8 - x³ + 2x = 15

=> (5x³ - x³) + (-4x² + 10x²) + (4x - 8x + 2x) + 8 = 15

=> 4x³ + 6x² - 2x + 8 = 15

=> 2(2x³ + 3x² - x + 4) = 15

=> (2x³ + 3x² - x + 4) = 15/2 => vô nghiệm

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

b) \(\left(\frac{1}{3}x+2y\right)\left(\frac{1}{9}x^2-\frac{2}{3}xy+4y^2\right)=\left(\frac{1}{3}x\right)^3+\left(2y\right)^3=\frac{1}{27}x^3+8y^3\)

c) \(\left(x-3y\right)\left(x^2+3xy+9y^2\right)=x^3-\left(3y\right)^3=x^3-27y^3\)

d) \(\left(x^2-\frac{1}{3}\right)\left(x^4+\frac{1}{2}x^2+\frac{1}{9}\right)=\left(x^2\right)^3-\left(\frac{1}{3}\right)^3=x^6-\frac{1}{9}\)

( x + 4 )( x² - 4x + 16 ) = x³ + 4³ = x³ + 64

( 1/3x + 2y )( 1/9x² - 2/3xy + 4y² ) = ( 1/3x )³ - ( 2y )³ = 1/27x³ - 8y³

( x - 3y )( x² + 3xy + 9y² ) = x³ - ( 3y )³ = x³ - 27y³

( x² - 1/3 )( x⁴ + 1/3x² + 1/9 ) = ( x² )³ - ( 1/3 )³ = x⁶ - 1/27

HĐT số 6 + 7 bạn nhé ^^

Làm mỗi ý đầu !! Mấy ý kia tự làm nha !

1) Biến đổi vế trái , ta có :

\(x^2+xy+y^2+1\)

\(\Leftrightarrow x^2+xy+\frac{1}{4}y^2+\frac{3}{4}y^2+1\)

\(\Leftrightarrow\left(x+\frac{1}{2}y\right)^2+\frac{3}{4}y^2+1>0\left(đpcm\right)\)

x² + xy + y² + 1

 \(=\left[x^2+2\cdot x\cdot\frac{y}{2}+\left(\frac{y}{2}\right)^2\right]+\frac{3y^2}{4}+1\)

\(=\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}+1\ge1>0\forall x,y\left(đpcm\right)\)

\(4x-x^2\)

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

\(=-\left(x-2\right)^2+4\le4\forall x\)

\(-x^2+4x-10\)

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

\(=-\left(x-2\right)^2-6\le-6< 0\forall x\left(đpcm\right)\)