Xét hiệu (a^2 + 4b^2 + 4c^2)-( 4ab-4ac+8bc )

= (a^2-4ab+4b^2) + 4c^2 + (4ac-8bc)

=(a-2b)^2 + 4c^2 + 4c(a-2b)

=(a-2b+2c)^2 >=0

Vậy a^2 + 4b^2 + 4c^2 >=  4ab-4ac+8bc

hok tốt

1)

a) 4x⁴+81=4x²+36x²+81-36x²

=(2x²+9)²-36x²

=(2x²+9-6x)(2x²+9+6x)

b)

(x²+x+1)(x²+x+2)-12

=(x²+x+1)(x²+x+1+1)-12

=(x²+x+1)²+(x²+x+1)-12

=(x²+x+1)²-3(x²+x+1)+4.(x²+x+1)-12

=(x²+x+1).(x²+x+1-3)+4.(x²+x+1-3)

=(x²+x+1)(x²+x-2)+4.(x²+x-2)

=(x²+x-2)(x²+x+1+4)

=(x²+x-2)(x²+x+5)

a) 4x^4 + 81 

= 4x^4 + 2.2x^2 .9 + 81 - 36x^2 

= ( 2x^2 + 9 )^2 - 36x^2

= (2x^2 - 6x + 9 )(2x^2 + 6x + 9 )

b) Đặt x^2 + x + 1 = a thay vào ta có 

a ( a+ 1 ) - 12 = a^2 + a - 12 

         = a^2 + 4a - 3a - 12 

        = a ( a+ 4 ) - 3 ( a+  4 )

         = ( a- 3 )( a+ 4 )

Thay a = x^2 + x + 1 ta có :

 ( x^2 + x + 1 - 3 )(x^2 + x + 1 + 4 ) = (x^2 +x - 2 )(x ^2 + x + 5 )

Còn phân tích đc tiếp phân tích hộ mình nha 

1)Ta có : \(A=\frac{3}{2x-x^2-4}\Leftrightarrow A=\frac{3}{-\left(x^2-2x+1\right)-3}\)\(\Leftrightarrow A=\frac{3}{-\left(x-1\right)^2-3}\)

Vì \(-\left(x-1\right)^2\le0\)nên \(-\left(x-1\right)^2-3\le-3\)

\(\Leftrightarrow A=\frac{3}{-\left(x-1\right)^2-3}\ge\frac{3}{-3}=-1\)

Vậy \(GTLN\left(A\right)=-1\) khi \(x=1\)

2)Ta có : \(a^2+4b^2+4c^2+4ac=\left(a^2+4c^2+4ac\right)+4b^2\)

                                                          \(=\left(a+2c\right)^2+\left(2b\right)^2\)                       \(\left(1\right)\)

Vì \(\left(a-b\right)^2\ge0\)nên\(a^2+b^2-2ab\ge0\Leftrightarrow a^2+b^2\ge2ab\)dấu "=" xảy ra khi \(a=b\)

Áp dụng BĐT vào (1) ta có \(\left(a+2c\right)^2+\left(2b\right)^2\ge2.\left(2b\right).\left(a+2c\right)=4b\left(a+2c\right)\)

\(\Leftrightarrow a^2+4b^2+4c^2+4ac\ge4ab+8bc\)

Nhầm , sorry bạn nha , mk làm lại nè

a² + 4b² + 4c² ≥ 4ab - 4ac + 8bc

⇔ a² - 4ab + 4b² + 4ac - 8bc + 4c² ≥ 0

⇔ ( a - 2b)² + 4c( a - 2b) + 4c² ≥ 0

⇔ ( a - 2b + 2c)² ≥ 0 ( luôn đúng ∀abc)

\(a^2+4b^2+4c^2\ge4ab-4ac+8bc\\ \Leftrightarrow a^2+4b^2+4c^2-4ab+4ac-8bc\ge0\\ \Leftrightarrow\left(a-2b+2c\right)^2\ge0\)

Luôn đúng với \(\forall x\in R\)

a² + 4b² + 4c² ≥ 4ab - 4ac + 8bc

⇔ a² - 4ab + 4b² - 4ac + 8bc + 4c² ≥ 0

⇔ ( a - 2b)² - 4c( a - 2b) + 4c² ≥ 0

⇔ ( a - 2b - 2c)² ≥ 0 ( luôn đúng ∀a,b,c )

\(a^2+4b^2+4c^2\ge4ab-4ac+8bc\)

\(\Leftrightarrow\left(a-2b+2c\right)^2\ge0\forall a,b,c\)

\("="\Leftrightarrow b=\dfrac{a}{2}+c\)

cac giup minh di minh sap phai nop roi

a²+4b²⁺4c²>= 4ab-4ac+8bc

a²+4b²⁺4c² - 4ab +4ac-8bc

(a² - 4ab+4b²)+4c²+(4ac-8bc>=0)

suy ra (a-2b²)+2.2c.(a-2b)+(2c)²

(a-2b+2c)²>=0

dau = xảy ra khi va chỉ khi a+2c=2b

a²+4b²+4c²>= 4ab-4ac+8bc(dpcm)

ban giai day du cho minh di. minh lam de nop ma

Bài 1 : 

a) \(x^2+y^2\)

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

\(\Leftrightarrow\left(x+y\right)^2-2xy=\left(-3\right)^2-2.\left(-28\right)=65\)

b) \(x^3+y^3\)

\(\Leftrightarrow\left(x+y\right)\left(x^2-xy+y^2\right)\)

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

\(\Leftrightarrow\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]=\left(-3\right)\left[\left(-3\right)^2-3.\left(-28\right)\right]=-279\)

c) \(x^4+y^4\)

\(\Leftrightarrow\left(x+y\right)^4-4x^3y-4xy^3-6x^2y^2=\left(-3\right)^4-4\left(-28\right).65-6\left(-28\right)^2=2657\)

Bài 3:

Có:    \(x^3+y^3+z^3=\left(x+y\right)^3-3xy\left(x+y\right)+z^3\)

=>     \(x^3+y^3+z^3=\left(-z\right)^3-3xy.-z+z^3\)

=>     \(x^3+y^3+z^3=-z^3+z^3+3xyz=3xyz\)

=> TA CÓ ĐPCM.

VẬY      \(x+y+z=0\Rightarrow x^3+y^3+z^3=3xyz\)

Bài 2 :

a) Giả sử  \(a^2+b^2+c^2+d^2\ge ab+ac+ad\)

\(\Leftrightarrow a^2+b^2+c^2+d^2-ab-ac-ad\ge0\)

\(\Leftrightarrow4a^2+4b^2+4c^2+4d^2-4ab-4ac-4ad\ge0\)

\(\Leftrightarrow\left(a-2b\right)^2+\left(a-2c\right)^2+\left(a-2d^2\right)\ge0\)( luôn đúng )

\(\RightarrowĐPCM\)

b) Sửa đề : \(a^2+4b^2+4c^2\ge2ab-2ac+4bc\)

Ta có : \(\left(a+2c\right)^2\ge0\Leftrightarrow a^2+4c^2\ge-4ac\left(1\right)\)

Áp dụng BĐT Cô - si ta có :

\(\hept{\begin{cases}a^2+4b^2\ge4ab\left(2\right)\\4b^2+4c^2\ge8bc\left(3\right)\end{cases}}\)

(1) + (2) + (3) 

\(\Leftrightarrow2a^2+8b^2+8c^2\ge4ab-4ac+8bc\)

\(\Leftrightarrow2\left(a^2+4b^2+4c^2\right)\ge4\left(ab-ac+2bc\right)\)

\(\Leftrightarrow a^2+4b^2+4c^2\ge2ab-2ac+4bc\)