a ) x=0; x = -(căn bậc hai(7)*i-3)/8;x = (căn bậc hai(7)*i+3)/8;

b ) -(y-x-3)*(y+x+3)

a) \(12x^3-9x^2+3x\)

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

b) \(x^2-y^2+6x+9\)

\(=\left(x^2+6x+9\right)-y^2\)

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

\(=\left(x+y+3\right)\left(x-y+3\right)\)

1) \(f\left(x\right)=-3x^2-12x+5\)

\(\Rightarrow f\left(x\right)=-3\left(x^2+4x\right)+5\)

\(\Rightarrow f\left(x\right)=-3\left(x^2+4x+4\right)+5+12\)

\(\Rightarrow f\left(x\right)=-3\left(x+2\right)^2+17\le17\left(-3\left(x+2\right)^2\le0,\forall x\right)\)

\(\Rightarrow GTLN\left(f\left(x\right)\right)=17\left(tạix=-2\right)\)

2) \(f\left(x\right)=-8x^2+20x\)\

\(\Rightarrow f\left(x\right)=-8\left(x^2+\dfrac{5}{2}x\right)\)

\(\Rightarrow f\left(x\right)=-8\left(x^2+\dfrac{5}{2}x+\dfrac{25}{16}\right)+\dfrac{25}{2}\)

\(\Rightarrow f\left(x\right)=-8\left(x+\dfrac{5}{4}\right)^2+\dfrac{25}{2}\le\dfrac{25}{2}\left(-8\left(x+\dfrac{5}{4}\right)^2\le0,\forall x\right)\)

\(\Rightarrow GTLN\left(f\left(x\right)\right)=\dfrac{25}{2}\left(tạix=-\dfrac{5}{4}\right)\)

1) �(�)=−3�2−12�+5f(x)=−3x2−12x+5

⇒�(�)=−3(�2+4�)+5⇒f(x)=−3(x2+4x)+5

⇒�(�)=−3(�2+4�+4)+5+12⇒f(x)=−3(x2+4x+4)+5+12

⇒�(�)=−3(�+2)2+17≤17(−3(�+2)2≤0,∀�)⇒f(x)=−3(x+2)2+17≤17(−3(x+2)2≤0,∀x)

$\Rightarrow GTLN\left(f\left(x\right)\right)=17\left(tạix=-2\right)$

`M=-9x^2+6x-3`

`M=-(9x^2-6x+3)`

`M=-(9x^2-6x+1+2)`

`M=-(3x-1)^2-2`

Vì `-(3x-1)^2 <= 0 AA x`

`<=>-(3x-1)^2-2 <= -2 AA x`

  Hay `M <= -2 AA x`

Dấu "`=`" xảy ra `<=>(3x-1)^2=0<=>3x-1=0<=>x=1/3`

Vậy `GTLN` của `M` là `-2` khi `x=1/3`

\(M=-9x^2+6x-3\)

\(M=-\left(9x^2-6x+3\right)\)

\(M=-\left[\left(3x-1\right)^2+2\right]\)

\(M=-\left(3x-1\right)^2-2\)

\(\Rightarrow Max_M=-2\) khi \(3x-1=0\)

                                 \(\Leftrightarrow x=\dfrac{1}{3}\)

`-9x^2 + 6x - 3`.

`= -(3x - 1)^2 - 2`.

Vì `(3x-1)^2 >=0 => -(3x-1)^2 <=0 => -(3x-1)^2 - 2 <= -2`

Dấu bằng xảy ra `<=> 3x - 1 = 0 => x = 1/3`.

Vậy `Max_M = -2 <=> x = 1/3`.

a) Sửa đề: \(x^2+3x+1\rightarrow x^2+2x+1\)

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

b) \(x^2+y^2+2xy=\left(x+y\right)^2\)

c) \(9x^2+12x+4=\left(3x+2\right)^2\)

d) \(-4x^2-9-12x=-\left(4x^2+12x+9\right)=-\left(2x+3\right)^2\)

a) \(=\left(6x\right)^2-2.6x.1+1=\left(6x-1\right)^2\)

b) \(=5xy\left(x^2+2x+1\right)=5xy\left(x+1\right)^2\)

c) \(=\left(3x-y\right)^2-25=\left(3x-y-5\right)\left(3x-y+5\right)\)

d) \(=x\left(x+1\right)+7\left(x+1\right)=\left(x+1\right)\left(x+7\right)\)

sr mk nhầm đề

Ta có: (x - y)² ≥ 0 <=> x² - 2xy + y² ≥ 0 <=> x² + y² ≥ 2xy 
hay 2xy ≤ x² + y² , dấu " = " xảy ra <=> x = y 
tương tự: 
+) 2yz ≤ y² + z² 
+) 2xz ≤ x² + z² 

cộng 3 vế của 3 bđt trên 
--> 2xy + 2yz + 2xz ≤ 2(x² + y² + z²) 
--> xy + yz + xz ≤ x² + y² + z² 
--> xy + yz + xz + 2xy + 2yz + 2xz ≤ x² + y² + z² + 2xy + 2yz + 2xz 
--> 3(xy + yz + xz) ≤ (x + y + z)² 
--> 3(xy + yz + xz) ≤ 3² 
--> xy + yz + xz ≤ 3 

Vậy MaxP = 3 ; Dấu " = " xảy ra <=> x = y = z = 1 

:D

5.Chỉ cần bấm máy tính là xong,chúc bạn học tốt

M=-2x²-12x-13

=-2(x²+6x+6,5)

=-2(x²+6x+9-3,5)

=-2(x+3)²+7

Min M=7 tại x=-3

Min M=5 tại x=mấy vậy Linh Miu Ly Ly?

tham khảo

Dấu "=" xảy ra khi 

Dấu "=" xảy ra khi 

b:=y^2+2y+1+9x^2-12x+4

=(y+1)^2+(3x-2)^2

a:

SỬa đề: 5y^2

=y^2-10y+25+9x^2+4y^2-12xy

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