Tìm MIN :

a) \(9x^2-4x+11=\left(3x\right)^2-2.3x.\frac{4}{6}+\frac{4}{9}-\frac{95}{9}\)

\(=\left(3x-\frac{4}{6}\right)^2-\frac{95}{9}\ge\frac{95}{9}\)

Dấu "=" xảy ra \(\Leftrightarrow x=?\)

\(2x-x^2-10=-\left(x^2-2x+1\right)+9=-\left(x-1\right)^2+9\ge0\)

a) \(9x^2-6x+11=\left(3x\right)^2-2.3x+1+10=\left(3x-1\right)^2+10>0\forall x\)

b) \(3x^2-12x+81=3.\left(x^2-4x+9\right)=3.\left(x-2\right)^2+15>0\forall x\)

c) \(5x^2-5x+4=5.\left(x^2-x+\dfrac{4}{5}\right)=5.\left(x^2-x+\dfrac{1}{4}+\dfrac{11}{20}\right)=5.\left(x-\dfrac{1}{2}\right)^2+\dfrac{11}{4}>0\forall x\)

d) \(2x^2-2x+9=2.\left(x^2-x+\dfrac{9}{2}\right)=2.\left(x-\dfrac{1}{2}\right)^2+\dfrac{17}{2}>0\forall x\)

a) = (3x-1)^2+10

Do (3x-1)^2>=0 với mọi x

--> (3x-1)^2+10>0 với mọi x

a) \(9x^2-6x+11=\left(3x-1\right)^2+10\ge10>0\)

b) \(3x^2-12x+81=3\left(x-2\right)^2+69\ge69>0\)

c) \(5x^2-5x+4=5\left(x-\dfrac{1}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}>0\)

d) \(2x^2-2x+9=2\left(x-\dfrac{1}{2}\right)^2+\dfrac{17}{2}\ge\dfrac{17}{2}>0\)

MK KO BT MK MỚI HO C LỚP 6

AI HỌC LỚP 6 CHO MK XIN

\(1,=x^6+27\\ 2,=8x^3+1\\ 3,=x^6+8\\ 4,=27x^3+8\)

1. (x² + 3)(x⁴ - 3x² + 9)

= x⁶ + 27

2. (2x + 1)(4x² - 2x + 1)

= 8x³ + 1

3. (x² + 2)(x⁴ - 2x² + 4)

= x⁶ + 8

4. (3x + 2)(9x² - 6x + 4)

= 27x³ + 8

1: \(\left(x^2+3\right)\left(x^4-3x^2+9\right)=x^6+27\)

2: \(\left(2x+1\right)\left(4x^2-2x+1\right)=8x^3+1\)

3: \(\left(x^2+2\right)\left(x^4-2x^2+4\right)=x^6+8\)

4: \(\left(3x+2\right)\left(9x^2-6x+4\right)=27x^3+8\)

`@` `\text {Ans}`

`\downarrow`

Gửi c!

Bài 1: 

a) \(3x^2\left(2x^3-x+5\right)-6x^5-3x^3+10x^2\)

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

\(=10x^2+10x^2\)

\(=20x^2\)

b) \(-2x\left(x^3-3x^2-x+11\right)-2x^4+3x^3+2x^2-22x\)

\(=-2x^4+6x^3+2x^2-22x-2x^4+3x^3+2x^2-22x\)

\(=-4x^4+9x^3+4x^2-44x\)

4:

a: =>1/4x^2-1/4x^2+2x=-14

=>2x=-14

=>x=-7

b: =>2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20

=>3x^2-12x-2=3x^2-17x+20

=>5x=22

=>x=22/5

1. Đề bài sai, các biểu thức này chỉ có giá trị lớn nhất, không có giá trị nhỏ nhất

2.

\(A=\left(2x\right)^3-3^3-\left(8x^3+2\right)\)

\(=8x^3-27-8x^3-2\)

\(=-29\) 

\(B=x^3+9x^2+27x+27-\left(x^3+9x^2+27x+243\right)\)

\(=27-243=-216\)

 sửa đề lại thành tìm Max nhé1, vì mấy ý này ko có min

\(1,=>D=-\left(x^2-4x-3\right)=-\left(x^2-2.2x+4-7\right)\)

\(=-[\left(x-2\right)^2-7]=-\left(x-2\right)^2+7\le7\)

dấu"=" xảy ra<=>x=2

2, \(E=-2\left(x^2-x+\dfrac{5}{2}\right)=-2[x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{9}{4}]\)

\(=-2[\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}]\le-\dfrac{9}{2}\) dấu"=" xảy ra<=>x=1/2

3, \(F=-\left(x^2+4x-20\right)=-\left(x^2+2.2x+4-24\right)\)

\(=-[\left(x+2\right)^2-24]\le24\) dấu"=" xảy ra<=>x=-2

Bài 1: 

a) Ta có: \(D=-x^2+4x+3\)

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

\(=-\left(x^2-4x+4-7\right)\)

\(=-\left(x-2\right)^2+7\le7\forall x\)

Dấu '=' xảy ra khi x=2

c) Ta có: \(F=-x^2-4x+20\)

\(=-\left(x^2+4x-20\right)\)

\(=-\left(x^2+4x+4-24\right)\)

\(=-\left(x+2\right)^2+24\le24\forall x\)

Dấu '=' xảy ra khi x=-2

Bài 2 : phân tích các đa thức sau thành nhân tử

a, x3 - 2x2 + x

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

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

b, x2 - 2x - y2 + 1

\(=x^2-2x+1-y^2\)

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

\(=\left(x-1-y\right)\left(x-1+y\right)\)

vt mũ hộ mk đuy bạn :

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

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

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

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

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

\(b,x^2-2x-y^2+1\)

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

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

\(=\left(x-1+y\right)\left(x-1-y\right)\)

Bài 1 :

a) \(x\left(x^2+5\right)\)

\(=x^3+5x\)

\(1,\\ a,A=4x^2\left(-3x^2+1\right)+6x^2\left(2x^2-1\right)+x^2\\ A=-12x^4+4x^2+12x^2-6x^2+x^2=-x^2=-\left(-1\right)^2=-1\\ b,B=x^2\left(-2y^3-2y^2+1\right)-2y^2\left(x^2y+x^2\right)\\ B=-2x^2y^3-2x^2y^2+x^2-2x^2y^3-2x^2y^2\\ B=-4x^2y^3-4x^2y^2+x^2\\ B=-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^3-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^2+\left(0,5\right)^2\\ B=\dfrac{1}{8}-\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{8}\)

\(2,\\ a,\Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ b,\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3=8=-2^3\\ \Leftrightarrow x=2\\ c,\Leftrightarrow4x^2\left(4x-2\right)-x^3+8x^2=15\\ \Leftrightarrow16x^3-8x^2-x^3+8x^2=15\\ \Leftrightarrow15x^3=15\\ \Leftrightarrow x^3=1\Leftrightarrow x=1\)

\(P=x\left(2x+1\right)-x^2\left(x+2\right)+x^3-x+3\\ P=2x^2+x-x^3-2x^2+x^3-x+3\\ P=3\left(đfcm\right)\)