`a) x^3 - 9x^2 + 14x = 0`

\(\Rightarrow\) `x^3 - 7x^2 - 2x^2 + 14x = 0`

\(\Rightarrow\) `(x^3 - 2x^2) - (7x^2 - 14x) =0`

\(\Rightarrow\) `x^2.(x - 2) - 7x.(x - 2) =0`

\(\Rightarrow\) `(x^2 - 7x)(x-2)=0`

\(\Rightarrow\) `x.(x-7)(x-2)=0`

\(\Rightarrow\left[\begin{array}{l}x=0\\ x-7=0\\ x-2=0\end{array}\right.\) \(\Rightarrow\left[\begin{array}{l}x=0\\ x=0+7\\ x=0+2\end{array}\right.\) \(\Rightarrow\left[\begin{array}{l}x=0\\ x=7\\ x=2\end{array}\right.\)

Vậy \(x\in\left\lbrace0;7;2\right\rbrace\)

`3.x^3 - 5x^2 + 8x - 4 = 0`

\(\Rightarrow\) `x^3 - x^2 - 4x^2 + 4x + 4x - 4 =0`

\(\Rightarrow\) `x^2 . (x-1) - 4x(x-1) + 4.(x-1) =0`

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

\(\Rightarrow\) `(x-2)^2(x-1)=0`

\(\Rightarrow\left[\begin{array}{l}x-2=0\\ x-1=0\end{array}\right.\) \(\Rightarrow\left[\begin{array}{l}x=2\\ x=1\end{array}\right.\)

Vậy \(x\in\left\lbrace2;1\right\rbrace\)

Bài làm

16x - 14x - x = 2

x ( 16 - 14 - 1 ) = 2

x . 1 = 2

x       = 2

Vậy x = 2

# Học tốt #

\(16\cdot X-14\cdot X-X=2\)

=> \(\left(16-14\right)\cdot X-X=2\)

=>\(2\cdot X-X=2\)

=> \(X=2\)

#CHÚC BẠN HỌC TỐT

giúp mik với

a) x\(\in\){1;2;3;4;6;12}
b) x\(\in\){4;8;16}

x\(\varepsilon\)(-\(\frac{1}{2}\);\(\frac{1}{2}\))

\(E=-16x^2+3x-3=-\left(4x-\frac{3}{8}\right)^2-\frac{183}{64}\le\frac{-183}{64}\)

 Vậy \(MaxE=\frac{-183}{64}\) khi \(x=\frac{3}{32}\)

Bạn xem lại đề phần \(F\) nhé.

\(G=-3x^2-9x+2=-3\left(x^2+3x-\frac{2}{3}\right)=-3[x^2+2x.\frac{3}{2}+\left(\frac{3}{2}\right)^2]+\frac{35}{4}\)

\(=-3\left(x+\frac{3}{2}\right)^2+\frac{35}{4}\le\frac{35}{4}\forall x\)

Vậy \(MaxG=\frac{35}{4}\) khi: \(\left(x+\frac{3}{2}\right)^2=0\Rightarrow x=\frac{-3}{2}\)

\(H=-7x^2+14x-3=-7\left(x^2-2x+\frac{3}{7}\right)=-7\left(x^2-2x+1\right)+4=-7\left(x-1\right)^2+4\le4\forall x\)

Vậy \(MaxH=4\) khi: \(\left(x-1\right)^2=0\Rightarrow x=1\)

-14

chắc k ?? vừa ghi Violympic kêu sai