a: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)

\(=4x^2-4x+1+4-2\left(4x^2-12x+9\right)\)

\(=4x^2-4x+5-8x^2+24x-18\)

\(=-4x^2+20x-13\)

e: \(\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)=8x^3+27y^3\)

Lời giải:

Áp dụng BĐT Bunhiacopxky:

$P^2\leq (x+y)[(29x+3y)+(29y+3x)]=32(x+y)^2\leq 32.(x^2+y^2)(1+1)=64(x^2+y^2)\leq 64.2=128$

$\Rightarrow P\leq 8\sqrt{2}$
Vậy $P_{\max}=8\sqrt{2}$

a: Ta có: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)

\(=4x^2-4x+1-2\left(4x^2-12x+9\right)+4\)

\(=4x^2-4x+5-8x^2+24x-18\)

\(=-4x^2+20x-13\)

b: \(\left(3x+2\right)^2+2\left(3x+2\right)\left(1-2y\right)+\left(1-2y\right)^2\)

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

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

Lời giải:

a. $=(x-y)(x+y)=[(-1)-(-3)][(-1)+(-3)]=2(-4)=-8$
b. $=3x^4-2xy^3+x^3y^2+3x^2y+12xy+15y-12xy-12$

$=3x^4-2xy^3+x^3y^2+3x^2y+15y-12$
=3-2.1(-2)^3+1^3.(-2)^2+3.1^2(-2)+15(-2)-12$
$=-25$
c.

$=2x^4+3x^3y-4x^3y-12xy+12xy=2x^4-x^3y$

$=x^3(2x-y)=(-1)^3[2(-1)-2]=-1.(-4)=4$

d. 

$=2x^2y+4x^2-5xy^2-10x+3xy^2-3x^2y$

$=(2x^2y-3x^2y)+4x^2+(-5xy^2+3xy^2)-10x$

$=-x^2y+4x^2-2xy^2-10x$

$=-3^2.(-2)+4.3^2-2.3(-2)^2-10.3=0$

Áp dụng bất đẳng thức Cosi ta có:

1 32 32 x 29 x + 3 y  ≤  1 4 2 32 x + 29 x + 3 y 2 = 1 8 2 61 x + 3 y

Tương tự

1 32 32 y 29 y + 3 x  ≤  1 8 2 61 y + 3 x

=> P ≤  4 2 x + y  ≤  4 2 x 2 + 1 2 + y 2 + 1 2 = 8 2

Vậy P min =  8 2 <=> x = y = 1

a: \(\dfrac{\left(x+1\right)}{x^2+2x-3}=\dfrac{\left(x+1\right)}{\left(x+3\right)\cdot\left(x-1\right)}=\dfrac{\left(x+1\right)\left(x+2\right)\left(x+5\right)}{\left(x+3\right)\left(x-1\right)\left(x+2\right)\left(x+5\right)}\)

\(\dfrac{-2x}{x^2+7x+10}=\dfrac{-2x}{\left(x+2\right)\left(x+5\right)}=\dfrac{-2x\left(x+3\right)\left(x-1\right)}{\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x-1\right)}\)

b: \(\dfrac{x-y}{x^2+xy}=\dfrac{x-y}{x\left(x+y\right)}=\dfrac{y^2\left(x-y\right)}{xy^2\left(x+y\right)}\)

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

c: \(\dfrac{x-2y}{2}=\dfrac{\left(x-2y\right)\left(x-xy\right)}{2\left(x-xy\right)}\)

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

Ta có: \(\hept{\begin{cases}\left(x-1\right)^2\ge0\left(2\right)\\\left(y-1\right)^2\ge0\left(3\right)\end{cases}\Leftrightarrow\hept{\begin{cases}x^2+1\ge2x\\y^2+1\ge2y\end{cases}\left(\forall x;y\inℝ\right)}}\)

\(\Rightarrow VT_{\left(1\right)}\ge\left(2x+2y+2\right)\left(2x+2y+2\right)\left(x;y\ge0\right)\)

\(\Leftrightarrow VT_{\left(1\right)}\ge4\left(x+y+1\right)^2\)(4)

Đặt \(3x+y+2=a;3y+x+b\Rightarrow a+b=4\left(x+y+1\right)\)

Lại có: \(\left(a-b\right)^2\ge0\left(\forall a;b\inℝ\right)\left(5\right)\)

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

\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)

\(\Leftrightarrow\frac{\left(a+b\right)^2}{4}\ge ab\)

\(\Leftrightarrow\frac{16\left(x+y+1\right)^2}{4}\ge\left(3x+y+2\right)\left(3y+x+2\right)\)

\(\Leftrightarrow4\left(x+y+1\right)^2\ge\left(3x+y+2\right)\left(3y+x+2\right)=VP_{\left(1\right)}\left(6\right)\)

Từ (4) và (6) => \(VT_{\left(1\right)}\ge VP_{\left(1\right)}\)

\(\Rightarrow VT_{\left(1\right)}=VP_{\left(1\right)}\)

Dấu '=' xảy ra đồng thời ở (2), (3), (5) 

\(\Leftrightarrow\hept{\begin{cases}x=1\\y=1\\3x+y+2=3y+x+2\end{cases}}\Leftrightarrow x=y=1\)