a: 12+18+24+36+42+48

=30+60+90

=90+90

=180

b: 327-16+216-27+500

=(327-27)+(216-16)+500

=300+200+500

=1000

c: \(385\cdot485+386\cdot515\)

\(=385\cdot485+385\cdot515+515\)

\(=385\cdot1000+515=385000+515=385515\)

d: \(16\cdot48+8\cdot48+16\cdot28\)

\(=16\cdot\left(48+28\right)+8\cdot2\cdot24\)

\(=16\cdot\left(76+24\right)=16\cdot100=1600\)

e: \(2\cdot4\cdot8\cdot50\cdot25\cdot125=4\cdot25\cdot8\cdot125\cdot2\cdot50\)

\(=100\cdot100\cdot1000=10000000\)

g: \(17\cdot8+51\cdot4=17\cdot4\cdot\left(2+3\right)=17\cdot4\cdot5=17\cdot20=340\)

`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

2: \(x^2-2xy+y^2-2x+2y\)

\(=\left(x-y\right)^2-2\left(x-y\right)\)

=(x-y)(x-y-2)

3: \(3x^2-2x-5\)

\(=3x^2-5x+3x-5\)

=x(3x-5)+(3x-5)

=(3x-5)(x+1)

4: \(16-x^2+4xy-4y^2\)

\(=16-\left(x^2-4xy+4y^2\right)\)

\(=4^2-\left(x-2y\right)^2\)

=(4-x+2y)(4+x-2y)

5: \(x^2-2x+1-y^2\)

\(=\left(x-1\right)^2-y^2\)

=(x-1-y)(x-1+y)

6: \(x^2+8x+15\)

\(=x^2+3x+5x+15\)

=x(x+3)+5(x+3)

=(x+3)(x+5)

7: \(\left(x^2+6x+8\right)\left(x^2+14x+48\right)-9\)

=(x+2)(x+4)(x+6)(x+8)-9

\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)-9\)

\(=\left(x^2+10x\right)^2+40\left(x^2+10x\right)+384-9\)

\(=\left(x^2+10x\right)^2+15\left(x^2+10x\right)+25\left(x^2+10x\right)+375\)

\(=\left(x^2+10x+25\right)\left(x^2+10x+15\right)=\left(x+5\right)^2\cdot\left(x^2+10x+15\right)\)

8: \(\left(x^2-8x+15\right)\left(x^2-16x+60\right)-24x^2\)

\(=\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)-24x^2\)

\(=\left(x^2-13x+30\right)\left(x^2-11x+30\right)-24x^2\)

\(=\left(x^2+30\right)^2-24x\left(x^2+30\right)+143x^2-24x^2\)

\(=\left(x^2+30\right)^2-24x\left(x^2+30\right)+119x^2\)

\(=\left(x^2-7x+30\right)\left(x^2-17x+30\right)\)

\(=\left(x^2-7x+30\right)\left(x-2\right)\left(x-15\right)\)

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