\(VT=x\left(x^2+3x+2\right)=x^3+3x^2+2x\)

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

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

\(=3\)

Vậy bt trên ko phụ vào biến x

b) \(x\left(3x^2-x+5\right)-\left(2x^3+3x-16\right)-x\left(x^2-x+2\right)\)

\(=3x^3-x^2+5x-2x^3-3x+16-x^3+x^2-2x\)

\(=16\)

Vậy bt trên ko phụ vào biến x

Tks bạn nha

hepl me

ai trả lời đúng mình sẽ k nha

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

\(=y\left(x^4-y^4\right)-y\left(x^4-y^4\right)\) 

\(=0\)

mấy câu kia phá theo hằng đẳng thức rồi thu ngọn 

kết quả không chứa biến là được

học tốt

\(2\left(x-2\right)\left(x+3\right)-x^2+4=0\)

\(2\left(x^2+3x-2x-6\right)-x^2+4=0\)

\(2x^2+6x-4x-12-x^2+4=0\)

\(x^2+2x-8=0\)

\(x^2+4x-2x-8=0\)

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

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

\(\Rightarrow\orbr{\begin{cases}x+4=0\rightarrow x=\left(-4\right)\\x-2=0\rightarrow x=2\end{cases}}\)

3/ 

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

\(=2\left(x^2+2x+1\right)-3\left(x^2-2x+1\right)+\left(5x-x^2+10-2x\right)\)

\(=2x^2+4x+2-3x^2+6x-3+5x-x^2+10-2x\)

\(=-2x^2+13x+9\)

b/ \(\left(3x-1\right)^3+\left(3x-1\right)^3-6x^2+9\)

\(=2\left(3x-1\right)^3-6x^2+9\)

\(=2\left(\left(3x\right)^3-3\left(3x\right)^2\cdot1+3\cdot3x\cdot1-1\right)-6x^2+9\)

\(=2\left(27x^3-27x^2+9x-1\right)-6x^2+9\)

\(=54x^3-54x^2+18x-2-6x^2+9\)

\(=54x^3-60x^2+18x+7\)

Số hơi dài, nên dễ tính sai -,- tính mik hay cẩu thả có j sai ibbb ạ

2) 2.(x - 2).(x + 3) - x² + 4 = 0

<=> x² + 2x - 8 = 0

<=> (x - 2).(x + 4) = 0

        x - 2 = 0 hoặc x + 4 = 0

        x = 0 + 2         x = 0 - 4

        x = 2               x = -4

=> x = 2 hoặc x = -4

3) a) 2.(x + 1)² - 3.(x - 1)² + (x + 2).(5 - x)

= 2.(x² + 2x + 1) - 3.(x² - 2x + 1) + (x + 2).(5 - x)

= 2x² + 4x + 2 - 3x² + 6x - 3 + (x + 2).(5 - x)

= 2x² + 4x + 2 - 3x² + 6x - 3 + 3x - x² + 10

= (2x² - 3x² - x²) + (4x + 6x + 3x) + (2 - 3 + 10)

= -2x² + 13x + 9

b) (3x - 1)³ + (3x - 1)³ - 6x² + 9

= 2.(3x - 1)³ - 6x² + 9

= 2.(27x³ - 27x² + 9x - 1) - 6x² + 9

= 54x³ - 54x² + 18x - 2 - 6x² + 9

= 54x³ + (-54x² - 6x²) + 18x + (-2 + 9)

= 54x³ - 60x + 18x + 7

4) a) A = (x - 5).(x + 2) + 3.(x - 2).(x + 2) - (3x - 1)² + 5x²

A = (x - 5).(x + 2) + 3.(x - 2).(x + 3) - (9x² - 6x + 1) + 5x²

A = x² - 3x - 10 + 3x² - 12 - (9x² - 6x + 1) + 5x²

A = x² - 3x - 10 + 3x² - 12 - 9x² + 6x - 1 + 5x²

A = (x² + 3x² - 9x² + 5x²) + (-3x + 6x) + (-10 - 12 - 1)

A = 3x - 23 (1)

b) Thay x = 1/2 vào (1), ta có:

A = 3x - 23 = 3.(1/2) - 23

                   = 3/2 - 23

                   = -43/2

A khi x = 1/2 là -43/2

a) \(C=\left(\dfrac{x}{x^2-x-6}-\dfrac{x-1}{3x^2-4x-15}\right):\dfrac{x^4-2x^2+1}{3x^2+11x+10}\cdot\left(x^2-2x+1\right)\) (ĐK: \(x\ne-\dfrac{5}{3};x\ne3;x\ne-2;x\ne1\))

\(C=\left[\dfrac{x}{\left(x-3\right)\left(x+2\right)}-\dfrac{x-1}{\left(x-3\right)\left(3x+5\right)}\right]:\dfrac{\left(x^2-1\right)^2}{\left(3x+5\right)\left(x+2\right)}\cdot\left(x-1\right)^2\)

\(C=\left[\dfrac{x\left(3x+5\right)}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}-\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x-3\right)\left(3x+5\right)\left(x+2\right)}\right]\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x^2-1\right)^2\left(x-1\right)^2}\)

\(C=\dfrac{3x^2+5x-x^2-2x+x+2}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x^2-1\right)^2\left(x-1\right)^2}\)

\(C=\dfrac{2x^2+4x+2}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x+1\right)^2\left(x-1\right)^4}\)

\(C=\dfrac{2\left(x+1\right)^2}{\left(3x+5\right)\left(x-3\right)\left(x+2\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x+1\right)^2\left(x-1\right)^4}\)

\(C=\dfrac{2}{\left(x-1\right)^4\left(x-3\right)}\)

b) Thay x = 2003 ta có: 

\(C=\dfrac{2}{\left(2003-1\right)^4\left(2003-3\right)}=\dfrac{2}{2002^4\cdot2000}=\dfrac{1}{2002^4\cdot1000}\)

c) \(C>0\) khi: 

\(\dfrac{2}{\left(x-1\right)^4\left(x-3\right)}>0\) mà: \(\left\{{}\begin{matrix}2>0\\\left(x-1\right)^4>0\end{matrix}\right.\)

\(\Leftrightarrow x-3>0\)

\(\Leftrightarrow x>3\) (đpcm) 

1) ĐKXĐ: \(\left\{{}\begin{matrix}x^2+x+1\ge0\\x^2+1\ne0\end{matrix}\right.\)

Ta có:

+) \(x^2+x+1=\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{3}{4}\)

\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\)

+) \(x^2+1\ge1>0\forall x\)

Vậy biểu thức luôn xác định với mọi x

2) ĐKXĐ: \(\left\{{}\begin{matrix}x^2-2x+3>0\\x^2-x+1\ge0\end{matrix}\right.\)

Ta có: 

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

\(=\left(x-1\right)^2+2\ge2>0\forall x\)

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

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\)

Vậy biểu thức luôn xác định với mọi x