Bài 3:

Ta có: \(2n^2+n-7⋮n-2\)

\(\Leftrightarrow2n^2-4n+5n-10+3⋮n-2\)

\(\Leftrightarrow n-2\in\left\{1;-1;3;-3\right\}\)

hay \(n\in\left\{3;1;5;-1\right\}\)

`a,`

`P(x)=M(x)+N(x)`

`P(x)=`\(\left(5x^4+8x^2-9x^3-12x-6\right)+\left(-5x^2+9x^3-5x^4+12x-8\right)\)

`P(x)= 5x^4+8x^2-9x^3-12x-6-5x^2+9x^3-5x^4+12x-8`

`P(x)=(5x^4-5x^4)+(-9x^3+9x^3)+(8x^2-5x^2)+(-12x+12x)+(-6-8)`

`P(x)=3x^2-14`

`b,`

`M(x)=N(x)+Q(x)`

`-> Q(x)=M(x)-N(x)`

`-> Q(x)=(5x^4+8x^2-9x^3-12x-6)-(-5x^2+9x^3-5x^4+12x-8)`

`Q(x)=5x^4+8x^2-9x^3-12x-6+5x^2-9x^3+5x^4-12x+8`

`Q(x)=(5x^4+5x^4)+(-9x^3-9x^3)+(8x^2+5x^2)+(-12x-12x)+(-6+8)`

`Q(x)=10x^4-18x^3+13x^2-24x+2`

a) \(x^2-x+x=4\)

\(x^2=4\)

\(x=\pm2\)

b) \(3x\left(x-5\right)-2\left(x-5\right)=0\)

\(\left(x-5\right)\left(3x-2\right)=0\)

\(\left[{}\begin{matrix}x=5\\x=\dfrac{2}{3}\end{matrix}\right.\)

c) Ta có: \(a+b+c=5-3-2=0\)

\(\left[{}\begin{matrix}x=1\\x=\dfrac{c}{a}=\dfrac{-2}{5}\end{matrix}\right.\)

d) Đặt \(x^2=t\left(t\ge0\right)\) . Lúc đó phương trình trở thành :

\(t^2-11t+18=0\)

\(\left[{}\begin{matrix}t=9\left(tmđk\right)\\t=2\left(tmđk\right)\end{matrix}\right.\)

\(t=9\rightarrow x^2=9\rightarrow x=\pm3\)

\(t=2\rightarrow x^2=2\rightarrow x=\pm\sqrt{2}\)

\(3^{8x+4}=81^{x+3}\)

\(3^{8x+4}=\left(3^4\right)^{x+3}\)

\(3^{8x+4}=3^{4x+12}\)

\(\Rightarrow8x+4=4x+12\)

\(\Rightarrow8x-4x=12-4\)

\(\Rightarrow4x=8\Rightarrow x=2\)

2

  3^{8.x + 4} = 81^{x + 3}

  3^{8.x + 4} = (3⁴)^{x + 3}

  3^{8.x + 4} = 3^{4.x + 12}

  8.x + 4 = 4.x + 12

8.x - 4.x = 12 - 4

        4.x = 8

           x = 8 : 4

           x = 2

3⁵ˣ⁺⁴ = 81ˣ⁺³

3⁵ˣ⁺⁴ = (3⁴)ˣ⁺³

3⁵ˣ⁺⁴ = 3⁴ˣ⁺¹²

5x + 4 = 4x +12

5x - 4x =12 - 4

x = 8

    `x^3 + 81x - 170 = 0`

`<=>x^3 - 2x^2 + 2x^2 - 4x + 85x - 170 = 0`

`<=> x^2 ( x - 2 ) + 2x ( x - 2 ) + 85 ( x - 2 ) = 0`

`<=> ( x - 2 ) ( x^2 + 2x + 85 ) = 0`

`<=> ( x - 2 ) [ ( x + 1 )^2 + 84 ] = 0`

  Mà `( x + 1 )^2 + 84 > 0 AA x`

   `=> x - 2 = 0`

`<=> x = 2`

Vậy `S = { 2 }`

a:Ta có: \(x\left(x-1\right)+x=4\)

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

\(\Leftrightarrow x^2=4\)

hay \(x\in\left\{2;-2\right\}\)

b: Ta có: \(3x\left(x-5\right)-2x+10=0\)

\(\Leftrightarrow\left(x-5\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{2}{3}\end{matrix}\right.\)

c: Ta có: \(5x^2-3x-2=0\)

\(\Leftrightarrow5x^2-5x+2x-2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{5}\end{matrix}\right.\)

d: Ta có: \(x^4-11x^2+18=0\)

\(\Leftrightarrow x^4-9x^2-2x^2+18=0\)

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

\(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)

a) x(x-1)+x=4

⇔x²=4⇔\(x=\pm2\)

b)3x(x-5)-2x+10=0

⇔3x(x-5)-2(x-5)=0

⇔(x-5)(3x-1)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{1}{3}\end{matrix}\right.\)

c)5x²-3x-2=0

⇔ 5x(x-1)+2(x-1)=0

⇔ (x-1)(5x+2)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{5}\end{matrix}\right.\)

d)x⁴-11x²+18=0

⇔ x²(x²-2)-9(x²-2)=0

⇔ (x²-2)(x²-9)=0

\(\Leftrightarrow\left[{}\begin{matrix}x^2=2\\x^2=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm\sqrt{2}\\x=\pm3\end{matrix}\right.\)