\(\Leftrightarrow a^2+b^2+c^2\ge\frac{3}{2}-\frac{3}{4}\)

\(\Leftrightarrow a^2+b^2+c^2\ge a+b+c-\frac{3}{4}\)

\(\Leftrightarrow\left(a^2-a+\frac{1}{4}\right)+\left(b^2-b+\frac{1}{4}\right)+\left(c^2-c+\frac{1}{4}\right)\ge0\)

\(\Leftrightarrow\left(a-\frac{1}{2}\right)^2+\left(b-\frac{1}{2}\right)^2+\left(c-\frac{1}{2}\right)^2\ge0\) (Luôn đúng)

Dấu '' = '' xảy ra khi: \(a=b=c=\frac{1}{2}\)

Vậy BĐT được chứng minh

\(\frac{1}{2}=\frac{a}{2a}\)

\(A=x^2-2xy+6y^2-12x+2y+45\)

\(\Rightarrow A=x^2-2xy+y^2+5y^2-12x+12y-10y+36+4+5\)

\(\Rightarrow A=\left(x^2-2xy+y^2-12x+12y+36\right)+\left(5y^2-10y+5\right)+4\)

\(\Rightarrow A=\text{[}\left(x-y\right)^2-12\left(x+y\right)+6^2\text{]}+5\left(y^2-2y+1\right)+4\)

\(\Rightarrow A=\left(x-y+6\right)^2+5\left(y-1\right)^2+4\)

Vì: (x-y+6)^2+5(y-1)^2>0

=>(x-y+6)^2+5(y-1)^2>4

Dấu '=' xảy ra khi x-y+6=0 hoặc y-1=0

ta có: y-1=0=>y=1

thay y=1 vào x-y+6=0

=>x-1+6=0

=>x+5=0=>x=0-5=>x=-5

vậy Amin=4 khi x=-5;y=1

sửa phần dấu "=" xảy ra khi x-y+6=0 hay 5(y-1)=0

Ta có: 5(y-1)=0=>y-1=0=>y=1

\(4a^3-4a^2+a\)

\(=a.\left(4a^2-4a+1\right)\)

\(=a.\left(2a-1\right)^2\)

\(x^3-x^2y+25.\left(y-x\right)\)

\(=x^2.\left(x-y\right)-25.\left(x-y\right)\)

\(=\left(x-y\right).\left(x^2-25\right)\)

\(=\left(x-y\right).\left(x-5\right).\left(x+5\right)\)

Ta có: 72⁷⁵ = 72^{4 x 18 + 3} = ( 72 ⁴ )¹⁸ x 72³ = (....6)¹⁸ x 72³ = (...6) x (...8) = (...8)

128¹²⁹ = 128^{4 x 32 + 1} = (128⁴)³² x 128 = (...6)³² x 128 = (...6) x 128 = (...8)

(234⁵)⁴² = 234^{5 x 42} = 234²¹⁰ = 234^{4 x 52 + 2} = (234⁴)⁵² x 234² = (...6)⁵² x 234² = (...6) x (...6) = (...6)

(579⁶)³⁵ = 579^{6 x 35} = 579²¹⁰= 579^{4 x 5 + 2} = (579⁴)⁵² x 579² = (...1)⁵² x (...8 ) = (...1) x (...8) = (...8)

HT~

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

`<=>3x^2+2x=0`

`<=>x(3x+2)=0`

`<=>x=0,3x+2=0`

`<=>x=0,x=(-2)/3`

Vậy `x=0,x=(-2)/3`

help me

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

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

xin tiick

`(x^2-2xy+y^2)-9`

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

`=(x-y-3)(x-y+3)`