a: \(x\left(1-2x\right)+2x^2=14\)

=>\(x-2x^2+2x^2=14\)

=>x=14

b: \(x\left(x-5\right)+3x-15=0\)

=>\(\left(x-5\right)\left(x+3\right)=0\)

=>\(\left[{}\begin{matrix}x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

\(a,\left(8-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}8-x=0\\x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-5\end{matrix}\right.\\ b,2x\left(x+81\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x=0\\x+81=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-81\end{matrix}\right.\)

a)\(\left(8-x\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}8-x=0\\x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\\x=-5\end{matrix}\right.\)
b)\(2x\left(x+81\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x=0\\x+81=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-81\end{matrix}\right.\)

1: Ta có: \(\left(3-x\right)^2+\left(2x+1\right)^2-\left(2-x\right)^2-\left(2x+1\right)^2=0\)

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

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

\(\Leftrightarrow x=\dfrac{5}{2}\)

2: Ta có: \(\left(1-2x\right)^2-3\left(x-1\right)^2+\left(x+1\right)^2-\left(x-1\right)^2-\left(x-1\right)^2=0\)

\(\Leftrightarrow4x^2-4x+1-3x^2+6x-3+\left(x+1\right)^2-2\left(x-1\right)^2=0\)

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

\(\Leftrightarrow2x^2+4x+1-2x^2+4x-2=0\)

\(\Leftrightarrow x=\dfrac{1}{8}\)

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

\(\Rightarrow\left[{}\begin{matrix}x=5\\2x=9\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{9}{2}\end{matrix}\right.\)

b) \(2\left(3x-15\right)\left(5-x\right)=0\)

\(\Rightarrow6\left(x-5\right)\left(5-x\right)=0\Rightarrow x=5\)

(x - 5)(2x - 3²) = 0

=> \(\left[\begin{array}{} x - 5 = 0\\ 2x - 3^{2} = 0 \end{array} \right.\)=> \(\left[\begin{array}{} x = 0 - 5 = -5\\ 2x = 0 - 3^{2} = 0 - 9 = -9 => x = \dfrac{9}{2} \end{array} \right.\)

b) 2(3x - 15)(5 - x) = 0

      (6x - 30)(5 - x) = 0

=> \(\begin{cases} 6x - 30 = 0\\ 5 - x =0 \end{cases} \) => \(\begin{cases} 6x = 0 + 30 = 30 => x = 5\\ x = 0 + 5 = 5 \end{cases} \)

\(a)\left(x-2\right)\left(x^2+2x-3\right)\ge0.\)

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

Ta có: \(x-2=0.\Leftrightarrow x=2.\\ x^2+2x-3=0.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=-3.\end{matrix}\right.\)

Bảng xét dấu:

x                   \(-\infty\)       -3       1       2     \(+\infty\)

\(x-2\)                    -      |    -   |   -   0   +

\(x^2+2x-3\)         +     0    -   0  +   |    +

\(f\left(x\right)\)                     -     0    +  0   -  0   +

Vậy \(f\left(x\right)\ge0.\Leftrightarrow x\in\left[-3;1\right]\cup[2;+\infty).\)

\(b)\dfrac{x^2-9}{-x+5}< 0.\)

Đặt \(g\left(x\right)=\dfrac{x^2-9}{-x+5}.\)

Ta có: \(x^2-9=0.\Leftrightarrow\left[{}\begin{matrix}x=3.\\x=-3.\end{matrix}\right.\)

\(-x+5=0.\Leftrightarrow x=5.\)

Bảng xét dấu:

x            \(-\infty\)      -3       3        5       \(+\infty\)

\(x^2-9\)            +   0   -   0   +   |    +

\(-x+5\)          +    |   +   |    +  0    -

\(g\left(x\right)\)              +    0   -   0   +  ||    -

Vậy \(g\left(x\right)< 0.\Leftrightarrow x\in\left(-3;3\right)\cup\left(5;+\infty\right).\)

Đây nè bạn

b) (x+1)^3-x(x-2)^2+x-1=0

 ⇔x^3+3x^2+3x+1-(x^3-4x^2+4x)=0

⇔ x^3+3x^2+3x+1-x^3+4x^2-4x+x-1=0

⇔7x^2-2=0

⇔7x^2=2

⇔7x^2=-2⇔x=-3

⇔7x^2=2⇔x=-căn 5

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

\(B=n-2\)

\(A\text{⋮}B\) ⇒ \(\left(x^3-2x+n\right)\text{⋮}\left(n-2\right)\)

⇒ \(\left[\left(x^3-2x^2\right)+\left(2x^2-4x\right)+\left(2x-4\right)+\left(n+4\right)\right]\text{⋮}\left(n-2\right)\)

⇒ \(\left[x^2\left(x-2\right)+2x\left(x-2\right)+2\left(x-2\right)+\left(n+4\right)\right]\text{⋮}\left(n-2\right)\)

⇒ \(\left[\left(x-2\right)\left(x^2+2x+2\right)+\left(n+4\right)\right]\text{⋮}\left(x-2\right)\)

Vì \(\left(x-2\right)\left(x^2+2x+2\right)\text{⋮}\left(n-2\right)\)

Để \(A\text{⋮}B\)

⇒ \(n+4=0\)

⇒ \(n=-4\)

a: f(-2)=4+3=7

f(-1)=2+3=5

f(0)=3

f(1/2)=-1+3=2

f(-1/2)=1+3=4

b: g(-1)=1-1=0

f(0)=0-1=-1