a: \(B=1-\sqrt{\left(x-1\right)^2+1}\)

(x-1)^2+1>=1

=>\(\sqrt{\left(x-1\right)^2+1}>=1\)

=>\(B< =0\)

Dấu = xảy ra khi x=1

b: 

ĐKXĐ: -(x+2)^2+2>=0

=>-(x+2)^2>=2

=>(x+2)^2<=2

=>\(-\sqrt{2}-2< =x< =\sqrt{2}-2\)

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

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

=>\(0< =\sqrt{4x-x^2-2}< =\sqrt{2}\)

=>1<=C<=căn 2+1

\(C_{max}=\sqrt{2}+1\Leftrightarrow x=2\)

Bạn xem lại đề nhé.

a) \(A=x^2+5y^2+2xy-4x-8y+2015\)

\(A=x^2-4x+4-2y\left(x-2\right)+y^2+2011+4y^2\)

\(A=\left(x-2\right)^2-2y\left(x-2\right)+y^2+2011+4y^2\)

\(A=\left(x-2-y\right)^2+4y^2+2011\)

Vì \(\left(x-y-2\right)^2\ge0;4y^2\ge0\)

\(\Rightarrow A_{min}=2011\)

Dấu bằng xảy ra : \(\Leftrightarrow\left\{{}\begin{matrix}x-y-2=0\\4y^2=0\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)

b) \(B=\left(x-2012\right)^2+\left(x+2013\right)^2\)

\(B=x^2-4024x+2012^2+x^2+4026x+2013^2\)

\(B=2x^2+2x+2012^2+2013^2\)

\(B=2\left(x^2+x+\dfrac{1}{4}\right)+2012^2+2013^2-\dfrac{1}{2}\)

\(B=2\left(x+\dfrac{1}{2}\right)^2+2012^2+2013^2-\dfrac{1}{2}\)

\(\Rightarrow B_{min}=2012^2+2013^2-\dfrac{1}{2}\)

Dấu bằng xảy ra : \(\Leftrightarrow x=-\dfrac{1}{2}\)

Bài 1:

a) \(A=-\left(2x-5\right)^2+6\left|2x-5\right|+4=-\left[\left(2x-5\right)^2-6\left|2x-5\right|+9\right]+13=-\left(\left|2x-5\right|-3\right)^2+13\le13\)

\(maxA=13\Leftrightarrow\) \(\left[{}\begin{matrix}2x-5=3\\2x-5=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\end{matrix}\right.\)

b) \(B=-x^2-y^2+2x-6y+9=-\left(x^2-2x+1\right)-\left(y^2+6y+9\right)+19=-\left(x-1\right)^2-\left(y+3\right)^2+19\le19\)

\(maxC=19\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)

Bài 2:

\(A=2\left(x^3-y^3\right)-3\left(x+y\right)^2=2\left(x-y\right)\left(x^2+xy+y^2\right)-3\left(x^2+2xy+y^2\right)=4\left(x^2+xy+y^2\right)-3\left(x^2+2xy+y^2\right)=x^2-2xy+y^2=\left(x-y\right)^2=2^2=4\)

bài 2
\(A=2\left(x-y\right)\left(x^2+xy+y^2\right)-3\left(x^2+2xy+y^2\right)\)
\(A=2.2\left(x^2+xy+y^2\right)-3\left(x^2+2xy+y^2\right)\)
\(A=\left(4x^2+4xy+4y^2\right)+\left(-3x^2-6xy-3y^2\right)\)
\(A=x^2-2xy+y^2=\left(x-y\right)^2=2^2=4\)

Bài 1: Tìm x: (2x-6)^3 + (5-x)^3 + (1-x)^3 = 0

​Bài 2: Tìm GTNN :​

A= x^2 -2x -4

B= x^2 -x +5

C= 4x^2 +2x -9

D= 2x^2 -4x +7

Giúp tớ với, tớ đang cần gấp

a) Ta có: \(Q=-x^2-y^2+4x-4y+2=-\left(x^2+y^2-4x+4y-2\right)\)

\(=-\left(x^2-4x+4+y^2+4y+4\right)+10\)

\(=-\left[\left(x-2\right)^2+\left(y+2\right)^2\right]+10\le10\forall x,y\)

Vậy Max_Q=10 khi x=2, y=-2

b) +Ta có: \(A=-x^2-6x+5=-\left(x^2+6x-5\right)=-\left(x^2+6x+9-14\right)\)

\(=-\left(x^2+6x+9\right)+14=-\left(x+3\right)^2+14\le14\forall x\)

Vậy Max_A=14 khi x=-3

+Ta có: \(B=-4x^2-9y^2-4x+6y+3=-\left(4x^2+9y^2+4x-6y-3\right)\)

\(=-\left(4x^2+4x+1+9y^2-6y+1-5\right)\)

\(=-\left[\left(2x+1\right)^2+\left(3y-1\right)^2\right]+5\le5\forall x,y\)

Vậy Max_B=5 khi x=-1/2, y=1/3

c) Ta có: \(P=x^2+y^2-2x+6y+12=x^2-2x+1+y^2+6y+9+2\)

\(=\left(x-1\right)^2+\left(y+3\right)^2+2\ge2\forall x,y\)

Vậy Min_P=2 khi x=1, y=-3

b) x² - 2x + 1 = 25x²

<=> (x - 1)² - 25x² = 0

<=> (x - 1 - 5x)(x - 1 + 5x)  = 0

<=> (-4x - 1)(6x - 1) = 0

<=> \(\orbr{\begin{cases}-4x-1=0\\6x-1=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-\frac{1}{4}\\x=\frac{1}{6}\end{cases}}\)

c) 4x² - 4x = 24

<=> x² - x - 6 = 0

<=> x² - 3x + 2x - 6 = 0

<=> x(x - 3) + 2(x - 3) = 0

<=> (x + 2)(x - 3) = 0

<=> \(\orbr{\begin{cases}x+2=0\\x-3=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)

a) (3 - x)² - x(x - 4) = 2x - 5

<=> x² - 6x + 9-  x² + 4x = 2x - 5

<=> -2x + 9 = 2x - 5

<=> 2x + 2x = 9 + 5

<=> 4x = 14

<=> x = 7/2

Bài 1.

a)

\((x-2)(2x-1)-(2x-3)(x-1)-2\\=2x^2-x-4x+2-(2x^2-2x-3x+3)-2\\=2x^2-5x+2-(2x^2-5x+3)-2\\=2x^2-5x+2-2x^2+5x-3-2\\=(2x^2-2x^2)+(-5x+5x)+(2-3-2)\\=-3\)

b)

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

Bài 2.

a)

\((14x^3+12x^2-14x):2x=(x+2)(3x-4)\\\Leftrightarrow 14x^3:2x+12x^2:2x-14x:2x=3x^2-4x+6x-8\\ \Leftrightarrow 7x^2+6x-7=3x^2+2x-8\\\Leftrightarrow (7x^2-3x^2)+(6x-2x)+(-7+8)=0\\\Leftrightarrow 4x^2+4x+1=0\\\Leftrightarrow (2x)^2+2\cdot 2x\cdot 1+1^2=0\\\Leftrightarrow (2x+1)^2=0\\\Leftrightarrow 2x+1=0\\\Leftrightarrow 2x=-1\\\Leftrightarrow x=\frac{-1}2\)

b)

\((4x-5)(6x+1)-(8x+3)(3x-4)=15\\\Leftrightarrow 24x^2+4x-30x-5-(24x^2-32x+9x-12)=15\\\Leftrightarrow 24x^2-26x-5-(24x^2-23x-12)=15\\\Leftrightarrow 24x^2-26x-5-24x^2+23x+12=15\\\Leftrightarrow -3x+7=15\\\Leftrightarrow -3x=8\\\Leftrightarrow x=\frac{-8}3\\Toru\)

\(A=4-6x-x^2=-\left(x^2+6x-4\right)=-\left(x^2+6x+9-13\right)\)

\(=-\left[\left(x+3\right)^2-13\right]=-\left(x+3\right)^2+13\le13\)

Vậy \(A_{max}=13\Leftrightarrow x+3=0\Leftrightarrow x=-3\)

\(B=3x^2-6x+1=\left(\sqrt{3}x\right)^2-2.\sqrt{3}x.\sqrt{3}+3-2\)

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

Vậy \(B_{min}=-2\Leftrightarrow\sqrt{3}x-\sqrt{3}=0\Leftrightarrow x=1\)

\(C=5x^2-2x-3=\left(\sqrt{5}x\right)^2-2.\sqrt{5}x.\frac{1}{\sqrt{5}}+\frac{1}{5}-\frac{16}{5}\)

\(=\left(\sqrt{5}x-\frac{1}{\sqrt{5}}\right)^2-\frac{16}{5}\ge-\frac{16}{5}\)

Vậy \(C_{min}=-\frac{16}{5}\Leftrightarrow\sqrt{5}x-\frac{1}{\sqrt{5}}=0\Leftrightarrow\sqrt{5}x=\frac{1}{\sqrt{5}}\Leftrightarrow x=\frac{1}{5}\)