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

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

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

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

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

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

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

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

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

1)a)=>x²+y²+2xy-4(x²-y²-2xy)

=>x²+y²+2xy-4.x²+4y²+8xy

=>-3.x²+5y²+10xy

ai nhanh tui se k

b) Ta có: 

\(x^2+2x+2=x^2+2x+1+1=\left(x+1\right)^2+1\ge1>0\forall x\)

Suy ra đpcm.

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

= (6x² + 4x - 3x - 2)(3 - x)

= 18x² + 3x - 6 - 6x³ - x² + 2x

= 6x³ + 17x² +5x - 6

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

x² - 25 - 2xy + y² = (x² - 2xy + y²) - 25 = (x - y)² - 5² = (x - y + 5)(x - y - 5)

x⁴ + x² + 1 = x⁴ + 2x² + 1 - x² = (x² + 1)² - x² = (x² + 1 - x)(x² + 1 + x)

4. 

a. a²(a + 1) + 2a(a + 1) = (a + 1)(a² + 2a) = a(a + 1)(a + 2)

Vì a(a + 1)(a + 2) là tích của 3 số nguyên liên tiếp => a(a + 1)(a + 2) chia hết cho 2 và 3

Mà ƯCLN(2,3) = 1 => a(a + 1)(a + 2) chia hết cho 6 (đpcm)

b. x² + 2x + 2 = x² + 2x + 1 + 1 = (x + 1)² + 1

Vì (x + 1)² \(\ge\)0 với mọi x => (x + 1)² + 1 > 0 với mọi x (đpcm)

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

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

\(=x^2-x^3-2+2x+x^3+27\)

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

\(=x^2+2x+25\)

Bài 2: 

a: \(B=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{6}{3\left(x-2\right)}+\dfrac{1}{x-2}\right):\left(\dfrac{x^2-4+16-x^2}{x+2}\right)\)

\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)

\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)

\(=\dfrac{x-x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{12}=\dfrac{-1}{6\left(x-2\right)}\)

b: Thay x=1/2 vào B, ta được:

\(B=\dfrac{-1}{6\cdot\left(\dfrac{1}{2}-2\right)}=\dfrac{-1}{6\cdot\dfrac{-3}{2}}=\dfrac{1}{9}\)

Thay x=-1/2 vào B, ta được:

\(B=\dfrac{-1}{6\cdot\left(-\dfrac{1}{2}-2\right)}=-\dfrac{1}{15}\)

c: Để B=2 thì \(\dfrac{-1}{6\left(x-2\right)}=2\)

=>6(x-2)=-1/2

=>x-2=-1/12

hay x=23/12

help voi a

a: \(\left(3x+2\right)^2+4x-3x^2+2\left(5x-2\right)\left(5x+2\right)-75x^2\)

\(=9x^2+12x+4+4x-3x^2+50x^2-8-75x^2\)

\(=-19x^2+16x-4\)