Bài 2:

Vì \(a+b=1\)\(\Rightarrow b=1-a\)

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

\(=a^2-ab+b^2+ab=a^2+b^2\)

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

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

\(=2.\left(a-\frac{1}{2}\right)^2+\frac{1}{2}\)

Vì \(\left(a-\frac{1}{2}\right)^2\ge0\forall a\)\(\Rightarrow2\left(a-\frac{1}{2}\right)^2\ge0\forall a\)

\(\Rightarrow2\left(a-\frac{1}{2}\right)^2+\frac{1}{2}\ge\frac{1}{2}\forall a\)

hay \(a^3+b^3+ab\ge\frac{1}{2}\)

Dấu " = " xảy ra \(\Leftrightarrow a-\frac{1}{2}=0\)\(\Leftrightarrow a=\frac{1}{2}\)

\(\Rightarrow b=1-\frac{1}{2}=\frac{1}{2}\)

1. x² - y² + 2x - 4y - 10 = 0

<=> ( x² + 2x + 1 ) - ( y² + 4y + 4 ) - 7 = 0

<=> ( x + 1 )² + ( y + 2 )² = 7

<=> ( x + 1 + y + 2 ) ( x + 1 - y - 2 ) = 7

<=> ( x + y + 3 ) ( x - y - 1 ) = 7

Vì x ; y nguyên dương nên : ( x + y + 3 ) ( x - y - 1 ) = 7 . 1

=>\(\orbr{\begin{cases}x+y=4\\x-y=2\end{cases}}\)=>\(\orbr{\begin{cases}x=3\\y=1\end{cases}}\)

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

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

\(=2x^2-2xy+4y^2\)

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

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

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

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

\(=2x^2-2xy-4y^2\)

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

\(7x.\left(x-2\right)-5.\left(x-1\right)=7x^2+3\)

\(7x^2-14x-5x+5=7x^2+3\)

\(7x^2-19x+5=7x^2+3\)

\(7x^2-19x+5-\left(7x^2+3\right)=0\)

\(7x^2-19x+5-7x^2-3=0\)

\(-19x+2=0\)

\(-19x=0-2\)

\(-19x=-2\)

\(x=\frac{2}{19}\)

a) Xét hiệu ta có:

\(a^2+b^2+c^2-ab-bc-ca\)

\(=\frac{1}{2}.\left(2a^2+2b^2+2c^2-2ab-2bc-2ca\right)\)

\(=\frac{1}{2}.\left[\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)\right]\)

\(=\frac{1}{2}.\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\right]\)

Vì \(\left(a-b\right)^2\ge0\forall a,b\); \(\left(b-c\right)^2\ge0\forall b,c\); \(\left(a-c\right)^2\ge0\forall a,c\)

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

\(\Rightarrow\frac{1}{2}.\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\right]\ge0\forall a,b,c\)

\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ca\)

Dấu " = " xảy ra \(\Leftrightarrow a=b=c\)

a,Ta có:\(a^2+b^2\ge2ab\)

\(b^2+c^2\ge2bc\)

\(a^2+c^2\ge2ca\)

Cộng theo từng vế ba bđt trên,ta được:

\(2\left(a^2+b^2+c^2\right)\ge2\left(ab+ac+bc\right)\)

\(\Leftrightarrow a^2+b^2+c^2\ge ab+ac+bc\)

Dấu "="xảy ra khi a=b=c

b,\(a^3+b^3\ge ab\left(a+b\right)\)(chia cả 2 vế cho a+b)

\(\Leftrightarrow a^2-ab+b^2\ge ab\)

\(\Leftrightarrow a^2-ab+b^2-ab\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\)đúng với mọi a,b

Dấu"=" xảy ra khi a=b

c,\(a^2+b^2+c^2\ge a\left(b+c\right)\)

\(\Leftrightarrow a^2+b^2+c^2\ge ab+ac\)

\(\Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2ac\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac\ge0\)

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

\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+b^2+c^2\ge0\)đúng với mọi a,b,c

Dấu"=" xảy ra khi a=b=c=0

Ta có\(\frac{1}{x}+\frac{1}{y}=\frac{1}{3}\)(x;y > 0)

=> \(\frac{x+y}{xy}=\frac{1}{3}\)

=> 3(x + y) = xy

=> 3x + 3y = xy

=> xy - 3x - 3y = 0

=> x(y - 3) - 3y + 9 = 9

=> x(y - 3) - 3(y - 3) = 9

=> (x - 3)(y - 3) = 9

Vì x;y > 0

=> x - 3 > -3 ; y - 3 > -3 (1)

mà 9 = 1.9 = (-1).(-9) = 3.3 = (-3).(-3) (2)

Từ (1)(2) 

=> x - 3 = 1 ; y - 3 = 9 

=> x = 4 ; y = 12

hoặc x = 12 ; y = 4

Vậy các cặp (x ; y) thỏa mãn là (4;12);(12;4)

Ta có \(\frac{1}{x}+\frac{1}{y}=\frac{1}{3}\)

\(\Leftrightarrow\frac{y}{xy}+\frac{x}{xy}=\frac{1}{3}\)

\(\Leftrightarrow\frac{x+y}{xy}=\frac{1}{3}\)

\(\Leftrightarrow3\left(x+y\right)=xy\)

\(\Leftrightarrow3x+3y-xy=0\)

\(\Leftrightarrow\left(x-3\right)\left(y-3\right)=9\)

\(\Leftrightarrow\left(x-3\right)\left(y-3\right)=9=3.3=\left(-3\right).\left(-3\right)=1.9=9.1=\left(-1\right)\left(-9\right)=\left(-9\right)\left(-1\right)\)

\(th1\hept{\begin{cases}x-3=3\Leftrightarrow x=6\\y-3=3\Leftrightarrow y=6\end{cases}}\left(tm\right)\)

\(th2\hept{\begin{cases}x-3=-3\Leftrightarrow x=0\\y-3=-3\Leftrightarrow y=0\end{cases}}\left(ktm\right)\)

\(th3\hept{\begin{cases}x-3=1\Leftrightarrow x=4\\y-3=9\Leftrightarrow y=12\end{cases}}\left(tm\right)\)

\(th4\hept{\begin{cases}x-3=9\Leftrightarrow x=12\\y-3=1\Leftrightarrow y=4\end{cases}}\left(tm\right)\)

thử các cặp còn lại rồi kl

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

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

\(\Leftrightarrow y^2+2x^2+2x+2=2x+2\)

\(\Leftrightarrow y^2+2x^2=0\)

Vì \(x^2\ge0;y^2\ge0\)

\(\Rightarrow y^2+2x^2\ge0\)

Mà \(y^2+2x^2=0\)

Nên \(\hept{\begin{cases}y^2=0\\2x^2=0\end{cases}}\)

Hay x = y = 0

Ta có (a + b + c)² \(\ge0\forall a;b;c\inℝ\)

=> a² + b² + c² + 2ab + 2bc + 2ca \(\ge\)0

=> a² + b² + c² \(\ge\)0 - (2ab + 2bc + 2ca)

=> a² + b² + c² \(\le\)2ab + 2bc + 2ca

=> a² + b² + c² \(\le\)2(ab + bc + ca) 

Dấu "=" xảy ra <=> a + b + c = 0

Xí bài 2 ý a) trước :>

4x² + 2y² + 2z² - 4xy - 4xz + 2yz - 6y - 10z + 34 = 0

<=> ( 4x² - 4xy + y² - 4xz + 2yz + z² ) + ( y² - 6y + 9 ) + ( z² - 10z + 25 ) = 0

<=> [ ( 4x² - 4xy + y² ) - 2( 2x - y )z + z² ] + ( y - 3 )² + ( z - 5 )² = 0

<=> [ ( 2x - y )² - 2( 2x - y )z + z² ] + ( y - 3 )² + ( z - 5 )² = 0

<=> ( 2x - y - z )² + ( y - 3 )² + ( z - 5 )² = 0

Ta có : \(\hept{\begin{cases}\left(2x-y-z\right)^2\\\left(y-3\right)^2\\\left(z-5\right)^2\end{cases}}\ge0\forall x,y,z\Rightarrow\left(2x-y-z\right)^2+\left(y-3\right)^2+\left(z-5\right)^2\ge0\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}2x-y-z=0\\y-3=0\\z-5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=4\\y=3\\z=5\end{cases}}\)

Thế vào T ta được : 

\(T=\left(4-4\right)^{2014}+\left(3-4\right)^{2014}+\left(5-4\right)^{2014}\)

\(T=0+1+1=2\)

(x² + x)² - 2(x² + x) - 15

= [(x² + x)² - 2(x² + x) + 1] - 16

= (x² + x + 1)² - 4²

= (x² + x + 5)(x² + x - 3)

( x² + x )² - 2 ( x² + x ) - 15

Đặt t = x² + x , đa thức trở thành

t² - 2t - 15

= ( t² + 3t ) - ( 5t + 15 )

= t ( t + 3 ) - 5 ( t + 3 )

= ( t - 5 ) ( t + 3 )

= ( x² + x - 5 ) ( x² + x + 3 )

A=(12x³y⁴-15x²y³-9x³y²):3x²y

  =(12x³y⁴:3x²y)-(15x²y³:3x²y)- (9x³y²:3x²y)

  =4x²y³-5y²-3x²

B, sai đề bạn ơi

Bài 2 : 

a, \(x^2-4x=0\Leftrightarrow x\left(x-4\right)=0\Leftrightarrow x=0;4\)

b, \(5x\left(x-2020\right)-x+2020=0\)

\(\Leftrightarrow5x\left(x-2020\right)-\left(x-2020\right)=0\Leftrightarrow\left(5x-1\right)\left(x-2020\right)=0\)

\(\Leftrightarrow x=\frac{1}{5};2020\)

c, \(\left(4x+5\right)^2-\left(2x-1\right)^2=0\)

\(\Leftrightarrow16x^2+40x+25-\left(4x^2-4x+1\right)=0\)

\(\Leftrightarrow12x^2+44x+24=0\Leftrightarrow4\left(x+3\right)\left(3x+2\right)=0\)

\(\Leftrightarrow x=-3;-\frac{2}{3}\)

a,x²-4x=0

= x.(x-4)=0

=> x=0 hoặc x-4=0

=>x=0 hoặc x=4