\(P=\dfrac{x+3}{x-2}+\dfrac{x-3}{x+2}-\dfrac{2x^2+3x+6}{x^2-4}\left(x\ne\pm2\right)\)

\(=\dfrac{\left(x+3\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-3\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x^2+3x+6}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{x^2+5x+6}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-5x+6}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x^2+3x+6}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{\left(x^2+5x+6\right)+\left(x^2-5x+6\right)-\left(2x^2+3x+6\right)}{\left(x-2\right)\left(x+2\right)}\)\

\(=\dfrac{x^2+5x+6+x^2-5x+6-2x^2-3x-6}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{-3x+6}{\left(x-2\right)\left(x+2\right)}=\dfrac{-3\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=-\dfrac{3}{x+2}\)

Với x khác cộng trừ 2

`3(x-2)^2+(x-1)^3-x^3=7`

`<=>3x^2-12x+12+x^3-3x^2+3x-1-x^3=7`

`<=>-9x=-4`

`<=>x=4/9`

Vậy `S={4/9}`

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

\(\Leftrightarrow3\left(x^2-4x+4\right)+x^3-3x^2+3x-1-x^3=7\)

\(\Leftrightarrow3x^2-12x+12+\left(x^3-x^3\right)-3x^2+3x-1=7\)

\(\Leftrightarrow\left(3x^2-3x^2\right)+\left(-12x+3x\right)+\left(12-1\right)=7\)

\(\Leftrightarrow-9x+11=7\)

\(\Leftrightarrow9x=4\)

\(\Leftrightarrow x=\dfrac{4}{9}\)

Ta có:

(x+a)(x+b)(x+c) = x^3 + (a+b+c)x^2 +(ab+bc+ca)x + abc

VT = (x^2+ax+bx+ab)(x+c)

= x^3 + ax^2 + bx^2 + abx + cx^2 + cax + bcx + abc (1)

VP = x^3 + (a+b+c)x^2 +(ab+bc+ca)x + abc

= x^3 + ax^2 + bx^2 + abx + cx^2 + cax + bcx + abc (2)

Từ (1) và (2), suy ra:

(x+a)(x+b)(x+c) = x^3 + (a+b+c)x^2 +(ab+bc+ca)x + abc

Mình chỉ làm đc phần a thui, thông cảm nha :)))

H=x²+y²-xy-x+y+1

=>4H=4x²+4y²-4xy -4x +4y +4

4H= (4x²+y²+1-4xy-4x+2y)+3y²+2y +3

4H=(2x-y-1)²+3(y²+\(\dfrac{2}{3}y\) +\(\dfrac{1}{9}\))+ \(\dfrac{8}{3}\)

4H = (2x-y-1)²+3(y+\(\dfrac{1}{3}\) )² \(+\dfrac{8}{3}\) \(\ge\)\(\dfrac{8}{3}\)(vì \(\left(2x-y-1\right)^2\ge0\forall x,y;3\left(y+\dfrac{1}{3}\right)^2\ge0\forall y\)

=> H \(\ge\dfrac{2}{3}\) 

Dấu ''=' xảy ra khi \(\left\{{}\begin{matrix}y=-\dfrac{1}{3}\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy Hmin= 2/3khi x=1/3;y=-1/3

\(H=x^2+y^2-xy-x+y+1\)

\(\Rightarrow4H=4x^2+4y^2-4xy-4x+4y+4\)

\(\Leftrightarrow4A=4x^2+y^2+1-4xy+2y-4x+3y^2+2y+3\)

\(\Rightarrow3.4A=3\left(2x-y-1\right)^2+9y^2+6y+9\)

\(\Leftrightarrow12A=3\left(2x-y-1\right)^2+\left(9y^2+6y+1\right)+8\)

\(\Leftrightarrow12A=3\left(2x-y-1\right)^2+\left(3y+1\right)^2+8\)

Mà \(3\left(2x-y-1\right)^2+\left(3y+1\right)^2+8\ge8\forall x,y\)

\(\Rightarrow12A\ge8\)

\(\Rightarrow A\ge\dfrac{8}{12}=\dfrac{2}{3}\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}2x-y-1=0\\3y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=-\dfrac{1}{3}\end{matrix}\right.\)

Như này thì tìm Dmin nhỉ ?

D=x²-xy+x+y²-2y+2015

=>4D=4x²-4xy+4x-8y +8060

4D = (4x²+y²+1-4xy+4x-2y) +3y²-6y +8059

4D = (2x-y+1)² +3(y²-2y+1) +8056

4D=(2x-y+1)²+3(y-1)²+8056

Vì (2x-y+1)² \(\ge0\forall xy\)    3(y-1)² \(\ge0\forall y\)

=> 4D\(\ge\) 8056

=> D\(\ge\) 2014

   Dấu ''='' xảy ra khi \(\left\{{}\begin{matrix}\left(2x-y+1\right)^2=0\\3\left(y-1\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x-y+1=0\\y-1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x=y-1\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\)

Vậy Dmin =2014 khi x=0 ;y=1

Đề yêu cầu gì bạn?

Gọi x(m) (x>0) là chiều dài mảnh đất hình chữ nhật

Chiều rộng mảnh đất hình chữ nhật là: x-6(m)

Ta có pt : x(x-6)=91

       <=>x²-6x-91=0

\(\Leftrightarrow\left\{{}\begin{matrix}x_1=13\\x_2=-7\left(loại\right)\left(x>0\right)\end{matrix}\right.\)

Vậy chiều dài là 13m 

Chiều rộng là 13-6=7 (m)

Chu vi vườn hoa là :

2(13+7)=40(m)

\(\left(\sqrt{3}-\sqrt{2}\right)\sqrt{5+2\sqrt{6}}\)

\(=\left(\sqrt{3}-\sqrt{2}\right)\sqrt{3+2+2.\sqrt{3}.\sqrt{2}}\)

\(=\left(\sqrt{3}-\sqrt{2}\right)\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2}\)

\(=\left(\sqrt{3}-\sqrt{2}\right)\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)

\(=\left(\sqrt{3}-\sqrt{2}\right)\left|\sqrt{3}+\sqrt{2}\right|\)

\(=\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)\)

\(=1\)

\(P=\left(\dfrac{2}{x-3}+\dfrac{x^2+3}{9-x^2}+\dfrac{x-1}{x+3}\right):\left(\dfrac{2x-1}{2x+1}-1\right)\)

\(=\left(\dfrac{2}{x-3}-\dfrac{x^2+3}{x^2-9}+\dfrac{x-1}{x+3}\right):\left(\dfrac{2x-1}{2x+1}-\dfrac{2x+1}{2x+1}\right)\)

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

\(=\dfrac{2x+6-x^2-3+x^2-4x+3}{\left(x-3\right)\left(x+3\right)}:\dfrac{-2}{2x+1}\)

\(=\dfrac{-2x+6}{\left(x-3\right)\left(x+3\right)}:\dfrac{-2}{2x+1}\)

\(=\dfrac{-2}{\left(x+3\right)}\times\dfrac{2x+1}{-2}\)

\(=\dfrac{2x+1}{x+3}\)