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

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

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

= \(\left(x^2+6x+8\right)\left(x^2+5x+8\right)\)

= \(\left(x^2+2x+4x+8\right)\left(x^2+5x+8\right)\)

= \(\left(x+2\right)\left(x+4\right)\left(x^2+5x+8\right)\)

a) x¹² + 4 = x¹² + 4x⁶ + 4 - 4x⁶ = (x⁶ + 2)² - (2x³)² 

= (x⁶ - 2x³ + 2)(x⁶ + 2x³ + 2)

b) 4x⁸ + 1 = 4x⁸ + 4x⁴  + 1 - 4x⁴ = (2x⁴ + 1)² - (2x²)² 

= (2x⁴ + 2x² + 1)(2x⁴ - 2x²  + 1)

c) x⁷ + x⁵ - 1 = x⁷ - x + x⁵ + x² - (x² - x  + 1) = x(x⁶ - 1) + x²(x³ + 1) - (x² - x + 1)

= x(x³ - 1)(x³ + 1) + x²(x + 1)(x² - x + 1) - (x2 - x + 1)

= (x⁴ - x)(x + 1)(x² - x + 1) + (x³ + x²)(x² - x + 1) - (x² - x + 1)

= (x⁵ + x⁴ - x² - x + x³ + x² - 1)(x² -x + 1)

= (x⁵ + x⁴ + x³ - x - 1)(x² - x + 1)

d) x⁷ + x⁵ + 1 = x⁷ - x + x⁵ - x² + (x² + x + 1)

= x(x³ - 1)((x³ + 1) + x²(x³ - 1) + (x² + x + 1)

= (x⁴ + x)(x  - 1)(x² + x + 1) + x²(x - 1)((x² + x + 1) + (x² + x + 1)

= (x² + x + 1)(x⁵ - x⁴ + x² - x + x³ - x² + 1)

= (x² + x + 1)(x⁵ - x⁴ + x³ - x + 1)

\(Pt\Leftrightarrow\left(x^3+4x^2+3x\right)+3\left(x+2-\sqrt[3]{2x^2+9x+8}\right)=0.\)

\(\Leftrightarrow\left(x^3+4x^2+3x\right)+3.\frac{\left(x+2\right)^3-2x^2-9x-8}{\left(x+2\right)^2+\left(x+2\right)\sqrt[3]{2x^2+9x+8}+\sqrt[3]{\left(2x^2+9x+8\right)^2}}=0\)

\(\Leftrightarrow\left(x^3+4x^2+3x\right)+3.\frac{x^3+4x^2+3x}{MS}=0\Leftrightarrow\left(x^3+4x^2+3x\right)\left(1+\frac{3}{MS}\right)=0\)

Dễ thấy MS >0 \(\Rightarrow PT\Leftrightarrow x^3+4x^2+x=0\Leftrightarrow x\left(x^2+4x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x^2+4x+3=0\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-3\end{cases}}\end{cases}}\)

\(\Rightarrow Pt\Leftrightarrow x^3+4x^2+3x=0\Leftrightarrow x\left(x^2+4x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x^2+4x+3=0\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-3\end{cases}}\end{cases}}\)\(\Rightarrow PT\Leftrightarrow x^3+4x^2+3x=0\)<=>\(x\in\left\{-3;-1;0\right\}\)

3x + 4x + 8x - 2( 32 + 8 ) = 30

15x - 2.40 = 30

15x - 80 = 30

15x = 30 + 80

15x = 110

x = 110 : 15

x = 22/3

3x+4x+8x - 2.(32+8) = 30

( 3 + 4 + 8 )x - 2.40 = 30

15x - 80= 30

15x  = 30 + 80 = 110 

x = 110 : 15 = \(\frac{22}{3}\)

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

\(x^2+5x+4=x^2+4x+x+4=x\left(x+4\right)+\left(x+4\right)=\left(x+1\right)\left(x+4\right)\)

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

\(x^4+4=\left(x^2\right)^2+2.x^2.2+2^2-4x^2=\left(x^2+2\right)^2-4x^2=\left(x^2+2\right)^2-\left(2x^2\right)=\left(x^2+2+2x\right)\left(x^2-2-2x\right)\)

Nhần mình ấn lộn bài sory m.n

8 + 9 = 17 

học tốt #

bằng 17

8 + 9 = 17

Mà bn đổi nick kiểu j zậy

\(\dfrac{11}{8};\dfrac{-7}{-8};\dfrac{-1}{8};\dfrac{-7}{8};\dfrac{-15}{8}\)

số dương luôn lớn hơn số âm

so sánh cùng mẫu thì do tử là số âm nên tử nào lớn hơn thì lớn hơn thôi em