\(\left(a-b\right)^2\ge0\Rightarrow a^2+b^2\ge\frac{\left(a+b\right)^2}{2}>\frac{1}{2}\)

\(\Rightarrow a^4+b^4\ge\frac{\left(a^2+b^2\right)^2}{2}>\frac{1}{8}\)( đpcm )

Đẳng thức xảy ra <=> a = b = 1/2 

Ta có : a + b > 1 > 0 (1)

Bình phương hai vế : (a + b)² > 1 => a² + 2ab + b² > 1 (2)

Mặt khác (a - b)² \(\ge\)0 => a² - 2ab + b² \(\ge\)0       (3)

Cộng từng vế của (2) hoặc (3) : \(2\left(a^2+b^2\right)>1\)=> a² + b² \(\ge\frac{1}{2}\)(4)

Bình phương hai vế của (4) : \(a^4+2a^2b^2+b^4>\frac{1}{4}\)(5)

Mặt khác \(\left(a^2-b^2\right)^2\ge0\)=> a⁴ + 2a²b² + b⁴ \(\ge\)0 (6)

Cộng từng vế (5) và (6) : \(2\left(a^4+b^4\right)>\frac{1}{4}\)=> \(a^4+b^4>\frac{1}{8}\)

\(3\left(x+3\right)-x^2+9=0\)

\(< =>3x+18-x^2=0\)

Ta có : \(\Delta=3^2-4.\left(-1\right).18=81\)

\(< =>\orbr{\begin{cases}x_1=\frac{-3+9}{-2}=-3\\x_2=\frac{-3-9}{-2}=6\end{cases}}\)

Vậy 

3( x + 3 ) - x² + 9 = 0

<=> 3x + 9 - x² + 9 = 0

<=> -x² + 3x + 18 = 0

<=> -x² - 3x + 6x + 18 = 0

<=> -x( x + 3 ) 6( x + 3 ) = 0

<=> ( -x + 6 )( x + 3 ) = 0

<=> \(\orbr{\begin{cases}-x+6=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=-3\end{cases}}\) 

( 2x - 1 )² + ( x + 3 )² - 5( x + 7 )( x - 7 ) = 0

<=> ( 2x - 1 )² + ( x + 3 )² - 5( x² - 7² ) = 0

<=> 4x² - 4x + 1 + x² + 6x + 9 - 5x² + 245 = 0

<=> 2x + 255 = 0

<=> 2x = -255

<=> x = -255/2

\(pt< =>4x^2-4x+1+x^2+6x+9-5x^2+5.49=0\)

\(< =>2x+255=0< =>x=-\frac{255}{2}\)

Đề là \(x+y=a+b\) hả ?

Vì \(x+y=a+b\Rightarrow\left(x+y\right)^2=\left(a+b\right)^2\)

\(\Rightarrow x^2+2xy+y^2=a^2+2ab+b^2\)

Mà \(x^2+y^2=a^2+b^2\Rightarrow2xy=2ab\Rightarrow xy=ab\)

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

\(\Rightarrow x^3+y^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)

\(\Rightarrow x^3+y^3=a^3+b^3\left(đpcm\right)\)

Bài làm:

Ta có: \(x+y=a+b\Leftrightarrow x-a=b-y\left(1\right)\)

Thay (1) vào ta có: \(x^2+y^2=a^2+b^2\Leftrightarrow x^2-a^2=b^2-y^2\)

\(\Leftrightarrow\left(x-a\right)\left(x+a\right)=\left(b-y\right)\left(b+y\right)\)

\(\Rightarrow x+a=b+y\left(2\right)\)

Cộng vế (1) và (2) lại: \(2x=2b\Rightarrow x=b\)

\(\Rightarrow y=a\)

\(\Rightarrow\hept{\begin{cases}x^3=b^3\\y^3=a^3\end{cases}}\)

\(\Rightarrow x^3+y^3=a^3+b^3\)

Nếu \(x-a=b-y=0\Rightarrow\hept{\begin{cases}x=a\\b=y\end{cases}}\Leftrightarrow\hept{\begin{cases}x^3=a^3\\b^3=y^3\end{cases}}\)

\(\Rightarrow x^3+y^3=a^3+b^3\)

=> đpcm

-4x⁵(x³ - 4x² + 7x - 3)

= -4x⁸ + 16x⁷ - 28x⁶ + 12x⁵

Ta có : \(-4x^5\left(x^3-4x^2+7x-3\right)\)

\(=-4x^8+4x^7-28x^6+12x^5\)

@Hoc tot@

\(a,b,c>0;ab+ac+bc=abc\)

<=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\)

Đặt: \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z>0\)=> x + y + z = 1

Ta có:\(P=\frac{1}{bc\left(1+\frac{1}{a}\right)}+\frac{1}{ac\left(1+\frac{1}{b}\right)}+\frac{1}{ab\left(1+\frac{1}{c}\right)}\)

Viết lại  \(P=\frac{yz}{1+x}+\frac{xz}{1+y}+\frac{xy}{1+z}\)

\(=\frac{yz}{\left(x+z\right)+\left(x+y\right)}+\frac{xz}{\left(x+y\right)+\left(z+y\right)}+\frac{xy}{\left(x+z\right)+\left(y+z\right)}\)

\(\le\frac{1}{4}\left(\frac{yz}{x+z}+\frac{yz}{x+y}\right)+\frac{1}{4}\left(\frac{xz}{x+y}+\frac{xz}{y+z}\right)+\frac{1}{4}\left(\frac{xy}{x+z}+\frac{xy}{y+z}\right)\)

\(\le\frac{1}{4}\left(\frac{yz+xy}{x+z}+\frac{yz+xz}{x+y}+\frac{xz+xy}{y+z}\right)=\frac{1}{4}\left(x+y+z\right)=\frac{1}{4}\)

Dấu "=" xảy ra <=> x = y = z = 1/3 <=> a= b = c = 3

max P = 1/4 tại a = b = c = 3

Đặt ⎧⎪⎨⎪⎩a+b−c=xb+c−a=yc+a−b=z(x,y,z>0){a+b−c=xb+c−a=yc+a−b=z(x,y,z>0)

⇒⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩a=z+x2b=x+y2c=y+z2⇒{a=z+x2b=x+y2c=y+z2

⇒√a(1b+c−a−1√bc)=√2(z+x)2(1y−2√(x+y)(y+z))≥√x+√z2(1y−2√xy+√yz)=√x+√z2y−1√y⇒a(1b+c−a−1bc)=2(z+x)2(1y−2(x+y)(y+z))≥x+z2(1y−2xy+yz)=x+z2y−1y
Tương tự

⇒∑√a(1b+c−a−1√bc)≥∑√x+√z2y−∑1√y⇒∑a(1b+c−a−1bc)≥∑x+z2y−∑1y

⇒VT≥∑[x√x(y+z)]2xyz−∑√xy√xyz≥2√xyz(x+y+z)2xyz−x+y+z√xyz≐x+y+z√xyz−x+y+z√xyz=0⇒VT≥∑[xx(y+z)]2xyz−∑xyxyz≥2xyz(x+y+z)2xyz−x+y+zxyz≐x+y+zxyz−x+y+zxyz=0

(∑√xy≤x+y+z,x√x(y+z)≥2x√xyz)(∑xy≤x+y+z,xx(y+z)≥2xxyz)

dấu = ⇔x=y=z⇔a=b=c

Mai Anh ! cậu giỏi quá, cậu nè :33