\(\left(x+3\right)^3-3\cdot\left(3x+1\right)^2+\left(2x+1\right)\cdot\left(4x^2-2x+1\right)=54\)

\(\Leftrightarrow x^3+9x^2+27x+27-3\cdot\left(9x^2+6x+1\right)+8x^3-4x^2+2x+4x^2-2x+1=54\)

\(\Leftrightarrow x^3+9x^2+27x+27-27x^2-18x-3+8x^3-4x^2+2x+4x^2-2x+1=54\)

\(\Leftrightarrow9x^3-18x^2+9x-29=0\)

\(\Leftrightarrow x=2,208024627\)

\(C=\frac{2x^2-3x-9}{x^2-9}=0\)

\(\Leftrightarrow2x^2-3x-9=0\)

Ta có : \(\Delta=\left(-3\right)^2-4.2.\left(-9\right)=81>0\)

\(\Rightarrow x1=\frac{3-\sqrt{81}}{2.2}=\frac{3-9}{4}=-\frac{3}{2}\)

\(\Rightarrow x2=\frac{3+\sqrt{81}}{2.2}=\frac{3+9}{4}=3\)

\(2x^2+10xy+14y^2+2x+2y+2=0\)

\(\Leftrightarrow\left(x^2+4y^2+1+2x+4xy+4y\right)+\left(x^2+6xy+9y^2\right)+\left(y^2-2y+1\right)=0\)

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

Vì \(\hept{\begin{cases}\left(x+2y+1\right)^2\ge0;\forall x,y\\\left(x+3y\right)^2\ge0;\forall x,y\\\left(y-1\right)^2\ge0;\forall x,y\end{cases}}\)

\(\Rightarrow\left(x+2y+1\right)^2+\left(x+3y\right)^2+\left(y-1\right)^2\ge0;\forall x,y\)

Do đó :\(\left(x+2y+1\right)^2+\left(x+3y\right)^2+\left(y-1\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(x+2y+1\right)^2=0\\\left(x+3y\right)^2=0\\\left(y-1\right)^2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y=1\\x=-3\\y=1\end{cases}}\)

Vậy x=-3 và y=1 

Kiến thức bổ sung 

\(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ca\)

\(\Leftrightarrow4x^2+20xy+28y^2+4x+4y+4=0\)

\(\Leftrightarrow\left(4x^2+4x+20xy+25y^2+10y+1\right)+\left(3y^2-6y+3\right)=0\)

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

\(\Leftrightarrow\hept{\begin{cases}2x+5y+1=0\\y-1=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-3\\y=1\end{cases}}\)

3x(x² - 4) = 0

Mà 3 khác 0 

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2;2\end{cases}}\)

bạn đăng vừa thôi nhé chứ đăng nhiều thế này ít người khiên trì giải hết lắm bạn nên đăng từng bài cho đỡ dài

1a) -3x²(2x³ - 2x + 1/3) = -6x⁵ + 6x³ - x²

b) (x⁴ + 2x³ - 2/3).(-3x⁴) = -3x⁸ - 6x⁷ + 2x⁴

c) (x + 3)(x - 4) = x² - 4x + 3x - 12 = x² - x - 12

d)(x - 4)(x² + 4x + 16) = (x - 4)(x² + 4x + 4²) = x³ - 64

e) 4(x - 1/2)(x + 1/2)(4x² + 1) =4(x² - 1/4)(4x²  + 1) = 4(4x⁴ + x² - x² - 1/4) = 4(4x⁴ - 1/4) = 16x⁴ - 1

B2. a) (2 - x)(x² + 2x + 4) + x(x - 3)(x + 4) - x² + 24 = 0

=> 8 - x³ + x(x² + 4x - 3x - 12) - x² + 24 = 0

=> 8 - x³ + x³ + x² - 12x - x² + 24 = 0

=> -12x + 32 = 0

=> -12x = -32

=> x = -32 : (-12) = 8/3

b) (x/2 + 3)(5 - 6x) + (12x - 2)(x/4 + 3) = 0

=> 5x/2 - 3x² + 15 - 18x + 3x² + 36x - x/2 - 6 = 0

=> 20x + 9 = 0

=> 20x = -9

=> x = -9/20

a) \(2x^2+y^2+6=4\left(x-y\right)\)

\(\Leftrightarrow2x^2+y^2+6-4x+4y=0\)

\(\Leftrightarrow\left(2x^2-4x+2\right)+\left(y^2+4y+4\right)=0\)

\(\Leftrightarrow2\left(x-1\right)^2+\left(y+2\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}\)

b/ x²(y + 2) + 1 = y²

<=> x²(y + 2) + 1 = (y + 2)(y - 2) + 4

<=> (y + 2)(x² + 2 - y) = 3

Làm tiếp nhé

\(P=\frac{2x^5-x^4-2x+1}{4x^2-1}+\frac{8x^2-4x+2}{8x^3+1}\)

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

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

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

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

\(=\frac{x^4+1}{2x+1}\)

bạn ơi tìm các giá trị của x sau khi bạn đã rút gọn í cái đề mk đăng lên là dậy đó tìm x khi P = 6 đó!