a) Áp dụng bất đẳng thức Cosi ta có :

\(x^2+1\geq 2x\\ 4y^2+1\geq 4y\\ 9z^2+1\geq 6z\)

Suy ra \(S\leq 6\)

Dấu = xảy ra khi \(x=1;y=\frac{1}{2}; z=\frac{1}{3}\)

Đặt \(P=\dfrac{xy}{xy+1}\Rightarrow\dfrac{1}{P}=\dfrac{xy+1}{xy}=1+\dfrac{1}{xy}\)

Ta có : \(xy\le\dfrac{x^2+y^2}{2}=\dfrac{8}{2}=4\Rightarrow\dfrac{1}{xy}\ge4\)

\(\Rightarrow\dfrac{1}{P}\ge5\Rightarrow P\le\dfrac{1}{5}\)

Dấu "=" xảy ra khi $x=y=2$

Bài 13:

a) \(501^2\)

\(=\left(500+1\right)^2\)

\(=500^2+2\cdot500\cdot1+1^2\)

\(=250000+1000+1\)

\(=251001\)

b) \(88^2+24\cdot88+12^2\)

\(=88^2+2\cdot12\cdot88+12^2\)

\(=\left(88+12\right)^2\)

\(=100^2\)

\(=10000\)

c) \(52\cdot48\)

\(=\left(50+2\right)\left(50-2\right)\)

\(=50^2-2^2\)

\(=2500-4\)

\(=2496\)

Bài 14:

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

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

\(P=8x^3+x^3\)

\(P=9x^3\)

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

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

\(Q=-2y^3+2y^3\)

\(Q=0\)

Bài `14`

`a. P = ( 2x - 1 ) ( 4x^2 + 2x + 1 ) + ( x + 1 ) ( x^2 -x+1)`

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

`=8x^3-1+x^3+1`

`= 9x^3`

__

`b, Q = ( x - y ) ( x^2 + xy + y^2 ) - ( x + y ) ( x^2 - xy + y^2)+2y^3`

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

`=x^3-y^3 -x^3-y^3+2y^3`

`= 0`

1) 

Ta có: x+y=2

nên \(\left(x+y\right)^2=4\)

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

\(\Leftrightarrow2xy=2\)

hay xy=1

Ta có: \(x^3+y^3\)

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

\(=2^3-3\cdot1\cdot2\)

=2

2)\(x^2+y^2=\left(x+y\right)^2-2xy=8^2-2\cdot\left(-20\right)=104\)

\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=8^3-3\cdot\left(-20\right)\cdot8=512+480=992\)

\(x^2+y^2+xy=\left(x+y\right)^2-xy=8^2-\left(-20\right)=64+20=84\)

Lời giải:
a.

$x^3+y^3=(x+y)^3-3xy(x+y)=9^3-3.9.18=243$

$x^4+y^4=(x^2+y^2)^2-2x^2y^2=[(x+y)^2-2xy]^2-2x^2y^2$

$=[9^2-2.18]^2-2.18^2=1377$

Nếu $x\geq y$ thì:

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

$=|x-y|[(x+y)^2-xy]=\sqrt{(x+y)^2-4xy}[(x+y)^2-xy]$

$=\sqrt{9^2-4.18}(9^2-18)=189$

Nếu $x< y$ thì $x^3-y^3=-189$

b.

$A=(x+y)^2-6(x+y)+y-5$

$=(-9)^2-6(-9)+y-5=130+y$

Chưa đủ cơ sở để tính biểu thức.

a) \(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=9^3-3\cdot18\cdot9=243\)

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

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

\(=\left(9^2-2\cdot18\right)^2-2\cdot18^2\)

\(=45^2-2\cdot324\)

=1377

+) Tìm trên mạng thì đề thiếu xy + yz - zx = 7 

+) Nếu bổ sung đề: Tìm x; y ; z nguyên dương thì có thể làm như sau: 

Không mất tính tổng quát: g/s: 

x ≥ y ≥ z

Vì x2 + y2 + z2 = 14 => 

x 2 ≤ 14

⇒ x ≤ √ 14 < 4

  Vì x nguyên dương 

=> x  ∈ { 1; 2; 3}

+)Vớix=3=>\hept{y+z=3y2+z2=5⇒\hept{y+z=y2≤5

\(a,=\dfrac{\left(x+1\right)\left(x+y\right)}{\left(x-y\right)\left(x+1\right)}=\dfrac{x+y}{x-y}\\ b,=\dfrac{\left(x-3\right)^2}{3x\left(x-3\right)}=\dfrac{x-3}{3x}\\ c,=\dfrac{\left(y-x\right)\left(y+x\right)}{xy\left(x-y\right)}=\dfrac{-x-y}{xy}\)

Lời giải:

a.

\(\frac{x^2+xy+x+y}{x^2-xy+x-y}=\frac{x(x+y)+(x+y)}{x(x+1)-y(x+1)}=\frac{(x+y)(x+1)}{(x+1)(x-y)}=\frac{x+y}{x-y}\)

b.

\(\frac{x^2-6x+9}{3x^2-9x}=\frac{(x-3)^2}{3x(x-3)}=\frac{x-3}{3x}\)

c.

\(\frac{y^2-x^2}{x^2y-xy^2}=\frac{(y-x)(y+x)}{-xy(y-x)}=\frac{x+y}{-xy}\)

fnf tha