a) ta có : \(\left(1-2x\right)\left(x-1\right)-5=x-1-2x^2+2x-5\)

\(=-2x^2+3x-6=-\left(2x^2-3x+6\right)=-\left(\left(\sqrt{2}x\right)^2-2.\sqrt{2}.\dfrac{3}{2\sqrt{2}}x+\left(\dfrac{3}{2\sqrt{2}}\right)^2+\dfrac{39}{8}\right)\)

\(=-\left(\left(\sqrt{2}x-\dfrac{3}{2\sqrt{2}}\right)^2+\dfrac{39}{8}\right)=-\left(\sqrt{2}x-\dfrac{3}{2\sqrt{2}}\right)^2-\dfrac{39}{8}\)

ta có : \(\left(\sqrt{2}x-\dfrac{3}{2\sqrt{2}}\right)^2\ge0\) với mọi \(x\) \(\Rightarrow-\left(\sqrt{2}x-\dfrac{3}{2\sqrt{2}}\right)^2\le0\) với mọi \(x\)

\(-\left(\sqrt{2}x-\dfrac{3}{2\sqrt{2}}\right)^2-\dfrac{39}{8}\le\dfrac{-39}{8}< 0\) với mọi \(x\)

vậy \(\left(1-2x\right)\left(x-1\right)-5< 0\) (đpcm)

b) ta có : \(-x^2-y^2+2x+2y-3\)

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

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

ta có : \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge\forall x\\\left(y-1\right)^2\ge\forall y\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}-\left(x-1\right)^2\le0\forall x\\-\left(y-1\right)^2\le0\forall y\end{matrix}\right.\)

\(\Rightarrow-\left(x-1\right)^2-\left(y-1\right)^2\le0\) với mọi \(x;y\)

\(\Leftrightarrow-\left(x-1\right)^2-\left(y-1\right)^2-1\le-1< 0\) với mọi \(x;y\)

vậy \(-x^2-y^2+2x+2y-3< 0\) (đpcm)

\(a,A=\left(1-2x\right)\left(x-1\right)-5\)

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

\(=-2x^2+3x-6\)

\(=-\left(2x^2-3x+\dfrac{9}{8}\right)-\dfrac{39}{8}\)

\(=-\left[\left(\sqrt{2}x\right)^2-2.\sqrt{2}x.\dfrac{3}{2\sqrt{2}}+\left(\dfrac{3}{2\sqrt{2}}\right)^2\right]-\dfrac{39}{8}\)

\(=-\left(\sqrt{2}x-\dfrac{3}{2\sqrt{2}}\right)^2-\dfrac{39}{8}\)

Ta có :

\(-\left(\sqrt{2}x-\dfrac{3}{2\sqrt{2}}\right)^2\le0\) \(\Rightarrow-\left(\sqrt{2}x-\dfrac{3}{2\sqrt{2}}\right)^2-\dfrac{39}{8}\le-\dfrac{39}{8}\)

Hay A \(\le-\dfrac{39}{8}\)

Dấu = xảy ra \(\Leftrightarrow\left(\sqrt{2}x-\dfrac{3}{2\sqrt{2}}\right)^2=0\)

\(\Leftrightarrow\sqrt{2}x-\dfrac{3}{2\sqrt{2}}=0\) \(\Leftrightarrow\sqrt{2}x=\dfrac{3}{2\sqrt{2}}\Leftrightarrow x=\dfrac{3}{2\sqrt{2}}:\sqrt{2}\)

\(\Leftrightarrow x=\dfrac{3}{4}\)

Vậy \(Min_A=-\dfrac{39}{8}\Leftrightarrow x=\dfrac{3}{4}\)

a) ( 1 - 2x ).( x - 1) -5

= x - 1 - 2x² + 2x -5

= x - 1 - x² - x² + 2x -\(\dfrac{1}{4}+\dfrac{19}{4}\)

= - ( x² - 2x +1) - [x² - 2.\(\dfrac{1}{2}\)x + ( \(\dfrac{1}{2}\))² ] + \(\dfrac{19}{4}\)

= -( x - 1)² -( x - \(\dfrac{1}{2}\))² + \(\dfrac{19}{4}\)

Do : -( x - 1)² ₌< 0 ; -( x - \(\dfrac{1}{2}\))² ₌< 0

--> ( 1 - 2x ).( x - 1) -5 ₌< -5 < 0 ( ĐPCM)

b) - x² - y² + 2x + 2y -3

= - x² + 2x - 1 - y² + 2y -1 -1

= - ( x² - 2x +1) -( y² - 2y + 1) -1

= -( x - 1)² - ( y - 1)² - 1

Do : -( x - 1)² nhỏ hơn hoặc bằng 0

- ( y - 1)² nhỏ hơn hoặc bằng 0

--> - x² - y² + 2x + 2y -3 ₌< -5 < 0 ( đpcm)

P/s : Chỗ ₌< là nhỏ hơn hoặc bằng nhé

Bài 2: 

a: \(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

a. \(\left(3x-5\right)^2-\left(x+1\right)^2=0\Leftrightarrow\left(3x-5+x+1\right)\left(3x-5-x-1\right)=0\Leftrightarrow\left(4x-4\right)\left(2x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}4x-4=0\\2x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

Vậy ...

b. \(\left(5x-4\right)^2-49x^2=0\Leftrightarrow\left(5x-4\right)^2-\left(7x\right)^2=0\Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\Leftrightarrow\left(-2x-4\right)\left(12x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}-2x-4=0\\12x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy ...

c. \(4x^3-36x=0\Leftrightarrow4x\left(x^2-9\right)=0\Leftrightarrow4x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}4x=0\\x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)

Vậy ...

d. \(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\Leftrightarrow\left(2x+3\right)\left(x-1\right)-\left(2x-3\right)\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(2x+3-2x+3\right)=0\Leftrightarrow6\left(x-1\right)=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)

Vậy ...

Rối mắt , loạn thần kinh toàn là x không

Nhiều quá bạn ơi 

a) 2x(x - 5) - x(3 + 2x) = 26

    2x² - 10x - 3x - 2x² = 26

    -10x - 3x = 26

    -13x = 26 => x = -2

b) 5x(x - 1) = x - 1

    5x(x - 1) - (x - 1) = 0

    (x - 1)(5x - 1) = 0

    x - 1 = 0 => x = 1

    5x - 1 = 0 => x = \(\frac{1}{5}\)

Vậy x = 0; x = \(\frac{1}{5}\)

d) (x - 2)^2 = 1 

= x = 2 + 1 = 3  

c) (x^2 + 1). (x + 2011) = 0

Tim x:

a) x^2 + 2x = 0 

= \(x^2+2x=0\)

= \(x^2=0:2=0\)

b) (x - 3) + 2x^2 - 6x = 0

Rút gọn thừa số chung : 

\(2x^2-5x-3=0\)

x = \(\frac{-1}{2}\)x = 3

=\(x^2=0\)

=> x = 0 

Bạn Nguyễn Phương Trung làm đúng rồi, các bn tk cho bn ấy nha !

a. \(\left(2x-3\right)\left(x+1\right)+\left(2x-3\right)\left(3x-7\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(x+1+3x-7\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(4x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\4x-6=0\end{matrix}\right.\)\(\Leftrightarrow x=\dfrac{3}{2}\)

b. \(\left(x-4\right)\left(3x-2\right)+x^2-16=0\)

\(\Leftrightarrow\left(x-4\right)\left(3x-2\right)+\left(x-4\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(3x-2+x+4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(4x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\4x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{2}\end{matrix}\right.\)

(2x-3)(x+1)+(2x+3)(3x-7)=0

<=> (2x-3)(x+1)-(2x-3)(3x-7)=0

<=> (2x-3)(x+1-3x+7)=0

<=> (2x-3)(-2x+8)=0

<=> 2x-3=0 => x=3/2

Hoặc -2x+8=0 => x= 4

Vậy x thuộc{3/2;4}