Không biê

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

\(A=-3x^2+6x-7=-3\left(x^2-2x+1-1\right)-7\)

\(=-3\left(x-1\right)^2-4\le-4\)Dấu ''='' xảy ra khi x = 1

\(B=-2x^2+5x+1=-2\left(x^2-\dfrac{5}{2}x\right)+1\)

\(=-2\left(x^2-2.\dfrac{5}{4}x+\dfrac{25}{16}-\dfrac{25}{16}\right)+1\)

\(=-2\left(x-\dfrac{5}{4}\right)^2+\dfrac{33}{8}\le\dfrac{33}{8}\)Dấu ''='' xảy ra khi x = 5/4

C;D chỉ có GTNN thôi bạn nhé \(C=2x^2-8x+13=2\left(x^2-4x+4-4\right)+13\)

\(=2\left(x-2\right)^2+5\ge5\)Dấu ''='' xảy ra khi x = 2

\(D=x^2-3x+5=x^2-2.\dfrac{3}{2}x+\dfrac{9}{4}-\dfrac{9}{4}+5\)

\(=\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\)Dấu ''='' xảy ra khi x = 3/2 

d: Ta có: \(D=x^2-3x+5\)

\(=x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{11}{4}\)

\(=\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{3}{2}\)

Biểu thức C với D là tìm GTNN.

a: \(x^2-5x+10\)

\(=x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}+\dfrac{15}{4}\)

\(=\left(x-\dfrac{5}{2}\right)^2+\dfrac{15}{4}>0\forall x\)

b: \(2x^2+8x+15\)

\(=2\left(x^2+4x+\dfrac{15}{2}\right)\)

\(=2\left(x^2+4x+4+\dfrac{7}{2}\right)\)

\(=2\left(x+2\right)^2+7>0\forall x\)

a) \(A=x^2-4x+1=\left(x-2\right)^2-3\ge-3\)

\(minA=-3\Leftrightarrow x=2\)

b) \(B=-x^2-8x+5=-\left(x+4\right)^2+21\le21\)

\(maxB=21\Leftrightarrow x=-4\)

c) \(C=2x^2-8x+19=2\left(x-2\right)^2+11\ge11\)

\(minC=11\Leftrightarrow x=2\)

d) \(D=-3x^2-6x+1=-3\left(x+1\right)^2+4\le4\)

\(maxD=4\Leftrightarrow x=-1\)

a) A = (x-2)^2 - 3 >= -3

--> A nhỏ nhất bằng -3

 <=> x = 2

b) B = -(x+4)^2 + 21 <= 21

--> B lớn nhất bằng 21

<=> x = -4

a/ \(x^2-2.4x+16+y^2+2y+1+z^2=16\Leftrightarrow\left(x-4\right)^2+\left(y+1\right)^2+z^2=16\)

\(\Rightarrow\left\{{}\begin{matrix}I\left(4;-1;0\right)\\R=\sqrt{16}=4\end{matrix}\right.\)

b/ \(x^2+y^2+z^2+2x-y+5z-\dfrac{2}{3}=0\Leftrightarrow x^2+2x+1+y^2-2.\dfrac{1}{2}y+\dfrac{1}{4}+z^2+2.\dfrac{5}{2}z+\dfrac{25}{4}=\dfrac{2}{3}+1+\dfrac{1}{4}+\dfrac{25}{4}\)

\(\Leftrightarrow\left(x+1\right)^2+\left(y-\dfrac{1}{2}\right)^2+\left(z+\dfrac{5}{2}\right)^2=\dfrac{49}{6}\) \(\Rightarrow\left\{{}\begin{matrix}I\left(-1;\dfrac{1}{2};-\dfrac{5}{2}\right)\\R=\dfrac{7}{\sqrt{6}}\end{matrix}\right.\)

P/s: câu c bạn tự làm nốt ạ!

`@` `\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

a) Cách 1: Khai triển HĐT rút gọn được 3 x 2  + 6x + 7 = 0

Vì (3( x 2  + 2x + 1) + 4 < 0 với mọi x nên giải được  x ∈ ∅

Cách 2. Chuyển vế đưa về ( x   +   3 ) 3 =  ( x   - 1 ) 3  Û x + 3 = x - 1

Từ đó tìm được x ∈ ∅

b) Đặt  x 2  = t với t ≥ 0 ta được  t 2  + t - 2 = 0

Giải ra ta được t = 1 (TM) hoặc t = -2 (KTM)

Từ đó tìm được x = ± 1

c) Biến đổi được 

d) Biến đổi về dạng x(x - 2) (x - 4) = 0. Tìm được x ∈ {0; 2; 4}

`@` `\text {Ans}`

`\downarrow`

`a)`

Thu gọn:

`P(x)=`\(5x^4 + 3x^2 - 3x^5 + 2x - x^2 - 4 +2x^5\)

`= (-3x^5 + 2x^5) + 5x^4 + (3x^2 - x^2) + 2x - 4`

`= -x^5 + 5x^4 + 2x^2 + 2x - 4`

`Q(x) =`\(x^5 - 4x^4 + 7x - 2 + x^2 - x^3 + 3x^4 - 2x^2\)

`= x^5 + (-4x^4 + 3x^4) - x^3 + (x^2 - 2x^2) + 7x - 2`

`= x^5 - x^4 - x^3 - x^2 + 7x - 2`

`@` Tổng:

`P(x)+Q(x)=`\((-x^5 + 5x^4 + 2x^2 + 2x - 4) + (x^5 - x^4 - x^3 - x^2 + 7x - 2)\)

`= -x^5 + 5x^4 + 2x^2 + 2x - 4 + x^5 - x^4 - x^3 - x^2 + 7x - 2`

`= (-x^5 + x^5) - x^3 + (5x^4 - x^4) + (2x^2 - x^2) + (2x + 7x) + (-4-2)`

`= 4x^4 - x^3 + x^2 + 9x - 6`

`@` Hiệu:

`P(x) - Q(x) =`\((-x^5 + 5x^4 + 2x^2 + 2x - 4) - (x^5 - x^4 - x^3 - x^2 + 7x - 2)\)

`= -x^5 + 5x^4 + 2x^2 + 2x - 4 - x^5 + x^4 + x^3 + x^2 - 7x + 2`

`= (-x^5 - x^5) + (5x^4 + x^4) + x^3 + (2x^2 + x^2) + (2x - 7x) + (-4+2)`

`= -2x^5 + 6x^4 + x^3 + 3x^2 - 5x - 2`

`b)`

`@` Thu gọn:

\(H (x) = ( 3x^5 - 2x^3 + 8x + 9) - ( 3x^5 - x^4 + 1 - x^2 + 7x)\)

`= 3x^5 - 2x^3 + 8x + 9 - 3x^5 + x^4 - 1 + x^2 - 7x`

`= (3x^5 - 3x^5) + x^4 - 2x^3 - x^2 + (8x + 7x) + (9+1)`

`= x^4 - 2x^3 - x^2 + 15x + 10`

\(R( x) = x^4 + 7x^3 - 4 - 4x ( x^2 + 1) + 6x\)

`= x^4 + 7x^3 - 4 - 4x^3 - 4x + 6x`

`= x^4 + (7x^3 - 4x^3) + (-4x + 6x) - 4`

`= x^4 + 3x^3 + 2x - 4`

`@` Tổng:

`H(x)+R(x)=` \((x^4 - 2x^3 - x^2 + 15x + 10)+(x^4 + 3x^3 + 2x - 4)\)

`= x^4 - 2x^3 - x^2 + 15x + 10+x^4 + 3x^3 + 2x - 4`

`= (x^4 + x^4) + (-2x^3 + 3x^3) - x^2 + (15x + 2x) + (10-4)`

`= 2x^4 + x^3 - x^2 + 17x + 6`

`@` Hiệu: 

`H(x) - R(x) =`\((x^4 - 2x^3 - x^2 + 15x + 10)-(x^4 + 3x^3 + 2x - 4)\)

`=x^4 - 2x^3 - x^2 + 15x + 10-x^4 - 3x^3 - 2x + 4`

`= (x^4 - x^4) + (-2x^3 - 3x^3) - x^2 + (15x - 2x) + (10+4)`

`= -5x^3 - x^2 + 13x + 14`

`@` `\text {# Kaizuu lv u.}`