Bài 1: 

a: \(\left(\dfrac{1}{3}x+2\right)\left(3x-6\right)\)

\(=x^2-3x+6x-12\)

\(=x^2+3x-12\)

b: \(\left(x+3\right)\left(x^2-3x+9\right)=x^3+27\)

c: \(\left(-2xy+3\right)\left(xy+1\right)\)

\(=-2x^2y^2-2xy+3xy+3\)

\(=-2x^2y^2+xy+3\)

d: \(x\left(xy-1\right)\left(xy+1\right)\)

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

\(=x^3y^2-x\)

Bài 2: 

a: Ta có: \(M=\left(3x+2\right)\left(9x^2-6x+4\right)\)

\(=27x^3+8\)

\(=27\cdot\dfrac{1}{27}+8=9\)

b: Ta có: \(N=\left(5x-2y\right)\left(25x^2+10xy+4y^2\right)\)

\(=125x^3-8y^3\)

\(=125\cdot\dfrac{1}{125}-8\cdot\dfrac{1}{8}\)

=0

nhầm xíu '-'

Tìm min:

$F=3x^2+x-2=3(x^2+\frac{x}{3})-2$

$=3[x^2+\frac{x}{3}+(\frac{1}{6})^2]-\frac{25}{12}$

$=3(x+\frac{1}{6})^2-\frac{25}{12}\geq \frac{-25}{12}$

Vậy $F_{\min}=\frac{-25}{12}$. Giá trị này đạt tại $x+\frac{1}{6}=0$
$\Leftrightarrow x=\frac{-1}{6}$

Tìm min

$G=4x^2+2x-1=(2x)^2+2.2x.\frac{1}{2}+(\frac{1}{2})^2-\frac{5}{4}$

$=(2x+\frac{1}{2})^2-\frac{5}{4}\geq 0-\frac{5}{4}=\frac{-5}{4}$ (do $(2x+\frac{1}{2})^2\geq 0$ với mọi $x$)

Vậy $G_{\min}=\frac{-5}{4}$. Giá trị này đạt tại $2x+\frac{1}{2}=0$

$\Leftrightarrow x=\frac{-1}{4}$

1/

a/ \(D=2x\left(10x^2-5x-2\right)-5x\left(4x^2-2x-1\right)\)

\(D=2x\left[10\left(x^2-\frac{1}{2}x-\frac{1}{5}\right)\right]-5x\left[4\left(x^2-\frac{1}{2}x-\frac{1}{4}\right)\right]\)

\(D=20x\left(x^2-\frac{1}{2}x-\frac{1}{5}\right)-20x\left(x^2-\frac{1}{2}x-\frac{1}{4}\right)\)

\(D=20x^3-10x^2-4x-20x^3+10x^2+5x\)

\(D=x\)

b/ Mình xin sửa lại đề:

Tính giá trị biểu thức \(E\left(x\right)=x^5-13x^4+13x^3-13x^2+13x+2012\)

Tại x = 12

\(E\left(x\right)=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x-1\right)x+2012\)

\(E\left(x\right)=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2-x+2012\)

\(E\left(x\right)=2012-x\)

\(E\left(x\right)=2000\)

2/

a/ \(2x\left(x-5\right)-x\left(3+2x\right)=26\)

<=> \(2x^2-10x-3x-2x^2=26\)

<=> \(-13x=26\)

<=> \(x=-2\)

b/ Bạn vui lòng coi lại đề.

3a/ Ta có \(D=x\left(5x-3\right)-x^2\left(x-1\right)+x\left(x^2-6x\right)-10+3x\)

\(D=5x^2-3x-x^3+x^2+x^3-6x^2-10+3x\)

\(D=-10\)

Vậy giá trị của D không phụ thuộc vào x (đpcm)

Giúp mik vs^^

a: 3x-5>15-x

=>4x>20

hay x>5

b: \(3\left(x-2\right)\left(x+2\right)< 3x^2+x\)

=>3x²+x>3x²-12

=>x>-12

a: 3x-5>15-x

=>3x+x>15+5

=>4x>20

=>x>5

b: \(3\left(x-2\right)\left(x+2\right)<3x^2+x\)

=>\(3\left(x^2-4\right)<3x^2+x\)

=>\(3x^2-12-3x^2-x<0\)

=>-x-12<0

=>x+12>0

=>x>-12

c: \(\left(2x+1\right)^2+3x\left(1-x\right)\le\left(x+2\right)^2\)

=>\(4x^2+4x+1+3x-3x^2\le x^2+4x+4\)

=>\(x^2+7x+1\le x^2+4x+4\)

=>7x+1<=4x+4

=>7x-4x<=4-1

=>3x<=3

=>x<=1

d: \(\frac{5x-20}{3}-\frac{2x^2+x}{2}>\frac{x\left(1-3x\right)}{3}-\frac{5x}{4}\)

=>\(\frac{4\left(5x-20\right)-6\left(2x^2+x\right)}{12}>\frac{4x\left(1-3x\right)-15x}{12}\)

=>\(4\left(5x-20\right)-6\left(2x^2+x\right)>4x\left(1-3x\right)-15x\)

=>\(20x-80-12x^2-6x>4x-12x^2-15x\)

=>14x-80>-11x

=>25x>80

=>\(x>\frac{80}{25}=\frac{16}{5}\)

e: 4-2x<=3x-6

=>-2x-3x<=-6-4

=>-5x<=-10

=>x>=2

f: \(\left(x+4\right)\left(5x-1\right)>5x^2+16x+2\)

=>\(5x^2-x+20x-4>5x^2+16x+2\)

=>19x-4>16x+2

=>3x>6

=>x>2

g: \(x\left(2x-1\right)-8<5-2x\left(1-x\right)\)

=>\(2x^2-x-8<5-2x+2x^2\)

=>-x-8<-2x+5

=>-x+2x<5+8

=>x<13

h: \(\frac{3x-1}{4}-\frac{3\left(x-2\right)}{8}-1>\frac{5-3x}{2}\)

=>\(\frac{2\left(3x-1\right)}{8}-\frac{3\left(x-2\right)}{8}-\frac88>\frac{4\left(5-3x\right)}{8}\)

=>2(3x-1)-3(x-2)-8>4(5-3x)

=>6x-2-3x+6-8>20-12x

=>3x-4>20-12x

=>15x>24

=>\(x>\frac{24}{15}\)

=>x>1,6

\(7x^2\left(x^2-5x+2\right)-5x\left(x^3-7x^2+3x\right)\)

\(=7x^4-35x^3+14x^2-5x^4+35x^3-15x^2\)

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

Thay: \(x=-\frac{1}{2}\) vào được

\(2.\left(-\frac{1}{2}\right)^4-\left(-\frac{1}{2}\right)^2\)

\(=2.\frac{1}{16}-\frac{1}{4}\)

\(=\frac{1}{8}-\frac{1}{4}=\frac{1}{8}-\frac{2}{8}=-\frac{1}{8}\)

P/s: Ko chắc

cacs bạn giúp mik vs mik đang cần gấp

\(2x^2-3x=2.(-1)^2-3.(-1)=2-(-3)=5\)
\(5x^2-3x-16=5.2^2-3.2-16=20-6-16=-2\)
\(5x-7y+10=5.\frac{1}{5}\)\(-7.\frac{1}{7}\)\(+10=1-1+10=10\)
\(2x-3y^2+4z^3=2.2+3.(-1)^2+4(-1)=4+3-4=3\)
Học tốt!

\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)

\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)

Bài 4:

a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)

\(\Leftrightarrow6x-9-2x+4=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

hay \(x=\dfrac{13}{3}\)

c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

hay x=1