Đề như thế này đúng không bạn ? :)

(x + 5)(4 - 3x) - (3x + 2)² + (2x + 1)³ = (2x - 1)(4x² + 2x + 1)

=> x(4 - 3x) + 5(4 - 3x) - [(3x)² + 2.3x.2 + 2² ] + [(2x)³ + 3.(2x)².1 + 3.2x.1² + 1³ ] = (2x - 1)[(2x)² + 2.x.1 + 1² ]

=> 4x - 3x² + 20 - 15x - (9x² + 12x + 4) + (8x³ + 12x² + 6x + 1) = (2x)³ - 1³

=> 4x - 3x² + 20 - 15x - 9x² - 12x - 4 + 8x³ + 12x² + 6x + 1 = 8x³ - 1

=> 4x - 3x² + 20 - 15x - 9x² - 12x - 4 + 8x³ + 12x² + 6x + 1 - 8x³ + 1 = 0

=> (4x - 15x - 12x + 6x) + (-3x² - 9x² + 12x²) + (20 - 4 + 1 + 1) + (8x³ - 8x³) = 0

=> -17x + 18 = 0

=> -17x = -18

=> 17x = 18

=> x = 18/17

Vậy x = 18/17 

Đk: x \(\ge\)2013; y \(\ge\)2013

Ta có: A = \(\frac{\sqrt{x-2013}}{x}+\frac{\sqrt{y-2012}}{y+1}\ge0\forall x;y\)

(vì \(\sqrt{x-2013}\ge0\); x > 0; \(\sqrt{y-2012}\ge0\); y + 1 > 0)

Dấu "=" xảy ra <=> x - 2013 = 0 và y - 2012 = 0 <=> x = 2013 và y = 2012

Vậy MinA = 0 khi x =2013 và y = 2012

Ta lại có: A = \(\frac{\sqrt{x-2013}}{x}+\frac{\sqrt{y-2012}}{y+1}\le\frac{\frac{x-2013+1}{2}}{x}+\frac{\frac{y-2012+1}{2}}{y+1}\)(bđt cosi)

<=> A \(\le\frac{x-2012}{2x}+\frac{y-2011}{2\left(y+1\right)}=\frac{1}{2}-\frac{1006}{x}+\frac{y+1-2012}{2\left(y+1\right)}\)

<= > A \(\le\frac{1}{2}-\frac{1006}{x}+\frac{1}{2}-\frac{1006}{y+1}=1-1006\left(\frac{1}{x}+\frac{1}{y+1}\right)\)

Áp dụng bđt \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) (1)

CM bđt đúng: Từ (1) => (a + b)² > = 4ab <=> (a - b)² > = 0 (luôn đúng với mọi a,b > 0)

Khi đó: A \(\le1-\frac{1006.4}{x+y+1}=1-\frac{4024}{x+y+1}\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}\sqrt{x-2013}=1\\\sqrt{y-2012}=1\\x=y+1\end{cases}}\) <=> \(\hept{\begin{cases}x=2014\\y=2013\end{cases}}\)

Vậy MaxA = \(1-\frac{4024}{2014+2013+1}=1-\frac{1006}{1007}=\frac{1}{1007}\) <=> x = 2014 và y = 2013

PT đa thức thành nhân tử ?

a) \(x\left(x+2\right)+x\left(x-5\right)-5\left(x+2\right)\)

\(=\left[x\left(x+2\right)-5\left(x+2\right)\right]+x\left(x-5\right)\)

\(=\left(x-5\right)\left(x+2\right)+x\left(x-5\right)\)

\(=\left(x-5\right)\left(2x+2\right)\)

\(=2\left(x+1\right)\left(x-5\right)\)

b) \(x^2y^2+y^3+zx^2+yz\)

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

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

c) \(1-2x+2yz+x^2-y^2-z^2\)

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

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

\(=\left(x-y+z-1\right)\left(x+y-z-1\right)\)

d) \(ab\left(x^2+y^2\right)+xy\left(a^2+b^2\right)\)

\(=abx^2+aby^2+a^2xy+b^2xy\)

\(=\left(abx^2+b^2xy\right)+\left(aby^2+a^2xy\right)\)

\(=bx\left(ax+by\right)+ay\left(by+ax\right)\)

\(=\left(ay+bx\right)\left(ax+by\right)\)

(60+30):2=45(km/giờ)

       van to trung binh tu A den B la :

           (  60+30 ) : 2 = 45  km\h 

                Dap so : 45 km\h

3² . 5³ - 20{300 - [540 - 2³ (7⁸ : 7⁶ + 7⁰)]}

= 9 . 125 - 20{300 - [540 - 8(7² + 7⁰)]}

= 9 . 125 - 20{300 - [540 - 8(49 + 1)]

= 9 . 125 - 20{300 - [540 - 8.50]

= 9 . 125 - 20{300 - [540 - 400]}

= 9 . 125 - 20{300 - 140}

= 9 . 125 - 20. 160

= 1125 - 3200 = -2075

\(=9\cdot125-20\cdot\left\{300-\left[540-8\cdot\left(7^2+1\right)\right]\right\}\)   

\(=1125-20\cdot\left\{300-\left[540-8\cdot\left(49+1\right)\right]\right\}\)   

\(=1125-20\cdot\left[300-\left(540-8\cdot50\right)\right]\)   

\(=1125-20\cdot\left[300-\left(540-400\right)\right]\)   

\(=1125-20\cdot\left(300-140\right)\)   

\(=1125-20\cdot160\)    

\(=1125-3200\)    

\(=-2075\)

ghykujk47586

đk: \(x>0;x\ne9\)

a) \(P=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)

b) Với x=0,25 ta có: \(P=\frac{\left(\sqrt{0,25}-1\right)^2}{\sqrt{0,25}}=0,5\)

c) \(P=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}=\sqrt{x}+\frac{1}{\sqrt{x}}-2\ge2\sqrt{\sqrt{x}.\frac{1}{\sqrt{x}}}-2=2-2=0\)

Dấu '=' xảy ra khi x=1 (tmdk). Vậy Min p =0 khi và chỉ khi x=1