\(\frac{7x+6}{2x\left(x+7\right)}-\frac{3x+6}{2x^2+14x}\)

\(=\frac{7x+6}{2x^2+14x}-\frac{3x+6}{2x^2+14x}=\frac{7x+6-3x-6}{2x^2+14x}\)

\(=\frac{4x}{2x^2+14x}=\frac{2\cdot2x}{2\left(x^2+7x\right)}=\frac{2x}{x\left(x+7\right)}=\frac{2}{x+7}\)

\(\frac{7x+6}{2x\left(x+7\right)}-\frac{3x+6}{2x^2+14x}\)( ĐKXĐ : \(x\ne0;x\ne-7\))

\(=\frac{7x+6}{2x\left(x+7\right)}-\frac{3x+6}{2x\left(x+7\right)}\)

\(=\frac{7x+6-\left(3x+6\right)}{2x\left(x+7\right)}\)

\(=\frac{7x+6-3x-6}{2x\left(x+7\right)}\)

\(=\frac{4x}{2x\left(x+7\right)}=\frac{2x\cdot2}{2x\left(x+7\right)}=\frac{2}{x+7}\)

\(=\frac{6x-3}{x}.\left(\frac{3x^2}{4x^2-1}\right)\)

\(=\frac{9x}{\left(2x-1\right)\left(2x+1\right)}.\left(\frac{2x-1}{1}\right)\)

\(=\frac{9x}{2x+1}\)

Bài làm 

\(\frac{6x-3}{x}:\frac{4x^2-1}{3x^2}=\frac{6x-3}{x}.\frac{3x^2}{4x^2-1}\)

\(=\frac{18x^3-9x^2}{4x^3-x}\)

Bài làm 

\(\frac{x^2-9}{2x+6}:\frac{3-x}{2}=\frac{x^2-9}{2x+6}.\frac{2}{3-x}\)

\(=\frac{2x^2-18}{6x-2x^2+18-6x}=\frac{2x^2-18}{-2x^2+19}=1\)

\(\frac{x^2-9}{2x+6}:\frac{3-x}{2}\)

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

\(=-1\)

\(\frac{4xy+2\left(x+y\right)\left(x-y\right)}{2\left(x+y\right)\left(x-y\right)}\)

\(=\frac{4xy+2\left(x^2-y^2\right)}{2\left(x^2-y^2\right)}\)

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

\(4x^2-28x+49=\left(2x\right)^2-2\cdot2x\cdot7+7^2=\left(2x-7\right)^2\)

thay x=4 vào ta được \(\left(2\cdot4-7\right)^2=\left(8-7\right)^2=1^2=1\)

vậy \(4x^2-28x+49=1\)khi x=4

\(9x^2+42x+49=\left(3x\right)^2+2\cdot3x\cdot7+7^2=\left(3x+7\right)^2\)

thay x=1 và ta được \(\left(3\cdot1+7\right)^2=10^2=100\)

vậy \(9x^2+42x+49=100\)đạt được khi x=1

\(25x^2-2xy+\frac{1}{25y^2}=\left(5x\right)^2-2\cdot5x\cdot\frac{1}{5y}+\left(\frac{1}{5y}\right)^2=\left(5x-\frac{1}{5y}\right)^2\)

thay x=\(\frac{-1}{5}\)và y=-5 vào ta được \(\left[5\cdot\left(\frac{-1}{5}\right)-\frac{1}{5\cdot\left(-5\right)}\right]^2=\left(1-\frac{1}{-25}\right)^2=\left(\frac{26}{25}\right)^2=...\)

vậy \(25x^2-2xy+\frac{1}{25y^2}=\left(\frac{26}{25}\right)^2\)khi x=\(\frac{-1}{5}\)và y=-5

4x² - 28x + 49 = ( 2x )² - 2.2x.7 + 7² = ( 2x - 7 )²

Thế x = 4 ta được : ( 2 . 4 - 7 )² = 1² = 1

9x² + 42x + 49 = ( 3x )² + 2.3x.7 + 7² = ( 3x + 7 )²

Thế x = 1 ta được : ( 3.1 + 7 )² = 10² = 100

25x² - 2xy + 1/25y² = ( 5x )² - 2.5x.1/5y + ( 1/5y )² = ( 5x - 1/5y )²

Thế x = -1/5 , y = -5 ta được : \(\left[5\cdot\left(-\frac{1}{5}\right)-\frac{1}{5}\cdot\left(-5\right)\right]^2=\left[-1+1\right]^2=0\)

a) 4xy + 2(x + y)(x - y)

= 4xy + 2[x(x - y) + y(x - y)]

= 4xy + 2[x² - xy + xy - y² ]

= 4xy + 2.(x² - y²)

= 4xy + 2x² - 2y²

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

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

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

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

a) \(x^2+1\ge1\)

\(\Rightarrow x^2+1< 1\)( Vô lí )

=> BPT vô nghiệm 

b) \(x^2+2x< 2x\)

\(\Leftrightarrow x^2+2x-2x< 0\)

\(\Leftrightarrow x^2< 0\)( vô lí )

Vậy BPT vô nghiệm

c) \(x^2-2x+3< -2x+3\)

\(\Leftrightarrow x^2-2x+3+2x-3< 0\)

\(\Leftrightarrow x^2< 0\)

Vậy,,,,,,,,,,,,,,,,,,,

a, \(x^2+1< 1\)(*)

Ta có : \(x^2\ge0< =>x^2+1\ge1\)

Nên không thể bé hơn 1 

Nên (*) vô lí 

b, \(x^2+2x< 2x\)(**)

Ta có : \(x^2\ge0< =>x^2+2x\ge2x\)

Nên không thể bé hơn 2x

Nên (**) vô lí 

c, \(x^2-2x+3< -2x+3\)

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

\(< =>x^2< 0\)( vô lí )

a) \(-x^2+3x+4>0\)

\(\Leftrightarrow-\left(x^2-3x-4\right)>0\)

\(\Leftrightarrow x^2-3x-4< 0\)

\(\Leftrightarrow x^2-2.x.\frac{3}{2}+\frac{9}{4}-\frac{25}{4}< 0\)

\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2-\frac{25}{4}< 0\)

\(\Leftrightarrow\left(x-\frac{3}{2}-\frac{5}{2}\right)\left(x-\frac{3}{2}+\frac{5}{2}\right)< 0\)

\(\Leftrightarrow\left(x-4\right)\left(x-1\right)< 0\)

\(\Leftrightarrow1< x< 4\)

b) \(x^2-6x+5\ge0\)

\(\Leftrightarrow x^2-2.3x+9-4\ge0\)

\(\Leftrightarrow\left(x-3\right)^2-4\ge0\)

\(\Leftrightarrow\left(x-3-2\right)\left(x-3+3\right)\ge0\)

\(\Leftrightarrow x\left(x-5\right)\ge0\)

Còn lại tự làm

\(-A=2x^2-0,5x+8\)

=> \(-2A=4x^2-x+16\)

=> \(-2A=\left(2x-\frac{1}{4}\right)^2+\frac{255}{16}\)

Có: \(\left(2x-\frac{1}{4}\right)^2\ge0\)

=> \(-2A\ge\frac{255}{16}\)

=> \(A\le-\frac{255}{16}:2\)

=>  \(A\le-\frac{255}{32}\)

DẤU "=" XẢY RA <=> \(\left(2x-\frac{1}{4}\right)^2=0\)

<=> \(x=\frac{1}{8}\)