a)

Cho A(X) = 0

 -18+2x =0

2x = 18

x = 9

Vậy nghiệm của đa thức A(x) là9

b)

CHo B(x) = 0

(x+1)(x-2) =0

TH1)

x+1= 0

x = -1

TH2)

x-2 =0

x = 2

Vậy nghiệm của đa thức B(x) = -1 hoặc 2

a) choA(x) = 0

\(=>-18+2x=0\)

\(=>2x=18=>x=9\)

b) cho B(x) = 0

\(=>\left(x+1\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)

a) A(x) = -18+2x

   A(x) = 0

-18 + 2x = 0

         2x = 0 + (-18)

         2x = -18

           x = -18 :2

           x = -9

Vậy nghiệm của A(x) là: x=-9.

b) B(x) = (x+1)(x-2)

    B(x) = 0

(x+1)(x-2) = 0

TH1: (x+1) = 0

          x = -1

TH2: (x-2) = 0

          x = 2 

Vậy nghiệm của B(x) là: x ∈ {-1;2}.

a) \(6x^2-2x=2x\left(3x-1\right)\)

\(2x\left(3x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}2x=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy \(S=\left\{0;\dfrac{1}{3}\right\}\)

b) \(x^2+5x+6=x^2+2x+3x+6=x\left(x+2\right)+3\left(x+2\right)=\left(x+3\right)\left(x+2\right)\)

\(\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)

Vậy \(S=\left\{-3;-2\right\}\)

\(\text{Đ}K\text{X}\text{Đ}\)

\(\frac{x-3}{x+1}\)

\(x-3\ne0\Leftrightarrow x\ne3\)

\(p=\sqrt{x^2-2xa+a^2}+\sqrt{x^2-2xb+b^2}\)

\(=\sqrt{\left(x-a\right)^2}+\sqrt{\left(x-b\right)^2}\)

\(=\left|x-a\right|+\left|x-b\right|\)

\(=\left|x-a\right|+\left|b-x\right|\ge\left|x-a+b-x\right|=\left|b-a\right|\)

Dấu \(=\)khi \(\left(x-a\right)\left(b-x\right)\ge0\).

a) \(12+x+\left(-5\right)=-18-2x\)

\(\Rightarrow12+x-5=-18-2x\)

\(\Rightarrow x+7+18+2x=0\)

\(\Rightarrow3x=-25\)

\(\Rightarrow x=-\dfrac{25}{3}\) 

b) \(\left(-14\right)-x+\left(-15\right)=-10+\left(4-2x\right)\)

\(\Rightarrow-14-x-15=-10+4-2x\)

\(\Rightarrow-x-29=-2x-6\)

\(\Rightarrow-x+2x=-6+29\)

\(\Rightarrow x=23\)

c) \(x-\left(-19\right)-\left(-11\right)=-\left(3x+40\right)\)

\(\Rightarrow x+19+11=-3x-40\)

\(\Rightarrow x+30=-3x-40\)

\(\Rightarrow x+3x=-40-30\)

\(\Rightarrow4x=-70\)

\(\Rightarrow x=-\dfrac{35}{2}\)

\(a,\Rightarrow\left[{}\begin{matrix}x-1=2x\\1-x=2x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\\ b,\Rightarrow\left[{}\begin{matrix}x+x-2=2\left(x\ge2\right)\\x+2-x=2\left(0\le x< 2\right)\\-x+2-x=2\left(x< 0\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\left(x\ge2\right)\left(tm\right)\\x=0\left(0\le x< 2\right)\left(tm\right)\\x=0\left(x< 0\right)\left(ktm\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)

a: Ta có: \(\left|x-1\right|=2x\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=2x\left(x\ge1\right)\\x-1=-2x\left(x< 1\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\left(loại\right)\\x=\dfrac{1}{3}\left(nhận\right)\end{matrix}\right.\)

a, Ta có \(\left(a-3+15\right)x^2y^3=4x^2y^3\Rightarrow a+12=4\Leftrightarrow a=-8\)

b, \(10ax^6y^6=-20x^6y^6\Rightarrow10a=-20\Leftrightarrow a=-2\)

a, Ta có ( a − 3 + 15 ) x 2 y 3 = 4 x 2 y 3 ⇒ a + 12 = 4 ⇔ a = − 8 b, 10 a x 6 y 6 = − 20 x 6 y 6 ⇒ 10 a = − 20 ⇔ a = − 2