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

b/ mình không chắc nữa

bài 3

a/ \(9x^2-49=0 \Leftrightarrow x^2=\frac{49}{9} \Leftrightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=-\frac{7}{3}\end{cases}}\)

b/ \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x+1\right)\left(x-1\right)-27=0 \Leftrightarrow x^3+27-x\left(x^2-1\right)-27=0\)

\(\Leftrightarrow x^3-x^3+x=0\Leftrightarrow x=0\)

c/\(\left(x-1\right)\left(x+2\right)-x-2=0 \Leftrightarrow \left(x-1\right)\left(x+2\right)-\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-1\right)^2=0\Leftrightarrow\orbr{\begin{cases}x+2=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}}}\)

d/ \(x\left(3x+2\right)+\left(x+1\right)^2-\left(2x-5\right)\left(2x+5\right)=0\)

\(\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=0\)

\(\Leftrightarrow4x+25=0 \Leftrightarrow x=\frac{-25}{4}\)

e/ mình lười qá ko viết đề đâu 

\(\Leftrightarrow4x^2-7x-2-4x^2+4x+3=7\)

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

có gì sai bn sửa lại nha 

Bài 1:

\(a,=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+2y^2}{2\left(x-y\right)\left(x+y\right)}=\dfrac{2y\left(x+y\right)}{2\left(x-y\right)\left(x+y\right)}=\dfrac{y}{x-y}\\ b,Sửa:\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\\ =\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}:\dfrac{3x-9-x^2}{3x\left(x+3\right)}=\dfrac{x^2+3x+9}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{-3x\left(x+3\right)}{x^2-3x+9}\\ =\dfrac{-3}{x-3}\)

Bài  2:

\(a,\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\\ b,\Leftrightarrow x^3+x^2+x+a=\left(x+1\right)\cdot a\left(x\right)\\ \text{Thay }x=-1\Leftrightarrow-1+1-1+a=0\Leftrightarrow a=1\)

5: =>4x^2-1/9=0

=>(2x-1/3)(2x+1/3)=0

=>x=1/6 hoặc x=-1/6

6: =>x-1=2

=>x=3

7:=>(2x-1)^3=-27

=>2x-1=-3

=>2x=-2

=>x=-1

8: =>1/8(x-1)^3=-125

=>(x-1)^3=-1000

=>x-1=-10

=>x=-9

3: =>(5x-5)^2-4=0

=>(5x-7)(5x-3)=0

=>x=3/5 hoặc x=7/5

4: =>(5x-1)^2=0

=>5x-1=0

=>x=1/5

1: =>(3x-1)(2x-1)=0

=>x=1/3 hoặc x=1/2

2: =>x^2(2x-3)-4(2x-3)=0

=>(2x-3)(x^2-4)=0

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

=>x=3/2;x=2;x=-2

`@` `\text {Answer}`

`\downarrow`

`1,`

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

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

`<=> 2x(3x - 1) - (3x - 1) = 0`

`<=> (2x - 1)(3x-1) = 0`

`<=>`\(\left[{}\begin{matrix}2x-1=0\\3x-1=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}2x=1\\3x=1\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy,  `S = {1/2; 1/3}`

`2,`

\(x^2\left(2x-3\right)+12-8x=0\)

`<=> x^2(2x - 3) - 8x + 12 =0`

`<=> x^2(2x - 3) - (8x - 12) = 0`

`<=> x^2(2x - 3) - 4(2x - 3) = 0`

`<=> (x^2 - 4)(2x - 3) = 0`

`<=>`\(\left[{}\begin{matrix}x^2-4=0\\2x-3=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x^2=4\\2x=3\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x^2=\left(\pm2\right)^2\\x=\dfrac{3}{2}\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=\pm2\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy, `S = {+-2; 3/2}`

`3,`

\(25\left(x-1\right)^2-4=0\)

`<=> 25(x-1)(x-1) - 4 = 0`

`<=> 25(x^2 - 2x + 1) - 4 = 0`

`<=> 25x^2 - 50x + 25 - 4 = 0`

`<=> 25x^2 - 15x - 35x + 21 = 0`

`<=> (25x^2 - 15x) - (35x - 21) = 0`

`<=> 5x(5x - 3) - 7(5x - 3) = 0`

`<=> (5x - 7)(5x - 3) = 0`

`<=>`\(\left[{}\begin{matrix}5x-7=0\\5x-3=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}5x=7\\5x=3\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=\dfrac{7}{5}\\x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy, `S = {7/5; 3/5}`

`4,`

\(25x^2-10x+1=0\)

`<=> 25x^2 - 5x - 5x + 1 = 0`

`<=> (25x^2 - 5x) - (5x - 1) = 0`

`<=> 5x(5x - 1) - (5x - 1) = 0`

`<=> (5x - 1)(5x-1)=0`

`<=> (5x-1)^2 = 0`

`<=> 5x - 1 = 0`

`<=> 5x = 1`

`<=> x = 1/5`

Vậy,` S = {1/5}.`

1) \(2x\left(x-3\right)+5x-15=0\)

\(2x\left(x-3\right)+5\left(x-3\right)=0\)

\(\left(x-3\right)\left(2x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{-5}{2}\end{matrix}\right.\)

2) \(x\left(2x-7\right)-4x+14=0\)

\(x\left(2x-7\right)-2\left(2x-7\right)=0\)

\(\left(2x-7\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x-7=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\end{matrix}\right.\)

3) \(x^2-12x+36=0\)

\(\left(x-6\right)^2=0\)

\(x-6=0\)

\(x=6\)

4) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-1\right)\left(x+1\right)-27=0\)

\(\left(x^3+3^3\right)-x\left(x^2-1\right)-27=0\)

\(x^3+27-x^3+x-27=0\)

\(x=0\)

a)

pt <=>     \(x^2+4x+4+x^2-6x+9=2x^2+14x\)

<=>     \(2x^2-2x+13=2x^2+14x\)

<=>     \(16x=13\)

<=>     \(x=\frac{13}{16}\)

b)

pt <=>     \(x^3+3x^2+3x+1+x^3-3x^2+3x-1=2x^3\)

<=>   \(2x^3+6x=2x^3\)

<=>   \(6x=0\)

<=>   \(x=0\)

c)

pt <=>    \(\left(x^3-3x^2+3x-1\right)-125=0\)

<=>   \(\left(x-1\right)^3=125\)

<=>   \(x-1=5\)

<=>   \(x=6\)

d)

pt <=>   \(\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\)

<=>   \(\left(x-1\right)^2+\left(y+2\right)^2=0\)     (1)

CÓ:   \(\left(x-1\right)^2;\left(y+2\right)^2\ge0\forall x;y\)

=>   \(\left(x-1\right)^2+\left(y+2\right)^2\ge0\)       (2)

TỪ (1) VÀ (2) =>    DÁU "=" XẢY RA <=>   \(\hept{\begin{cases}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{cases}}\)

<=>     \(\hept{\begin{cases}x=1\\y=-2\end{cases}}\)

e)

pt <=>   \(2x^2+8x+8+y^2-2y+1=0\)

<=>   \(2\left(x+2\right)^2+\left(y-1\right)^2=0\)

TA LUÔN CÓ:   \(2\left(x+2\right)^2+\left(y-1\right)^2\ge0\forall x;y\)

=> DẤU "=" XẢY RA <=>   \(\hept{\begin{cases}2\left(x+2\right)^2=0\\\left(y-1\right)^2=0\end{cases}}\) 

<=>     \(\hept{\begin{cases}x=-2\\y=1\end{cases}}\)

a) ( x + 2 )² + ( x - 3 )² = 2x( x + 7 )

<=> x² + 4x + 4 + x² - 6x + 9 = 2x² + 14x

<=> x² + 4x + x² - 6x - 2x² - 14x = -4 - 9

<=> -16x = -13

<=> x = 13/16

b) ( x + 1 )³ + ( x - 1 )³ = 2x³

<=> x³ + 3x² + 3x + 1 + x³ - 3x² + 3x - 1 = 2x³

<=> x³ + 3x² + 3x + x³ - 3x² + 3x - 2x³ = -1 + 1

<=> 6x = 0

<=> x = 0

c) x³ - 3x² + 3x - 126 = 0

<=> ( x³ - 3x² + 3x - 1 ) - 125 = 0

<=> ( x - 1 )³ = 125

<=> ( x - 1 )³ = 5³

<=> x - 1 = 5

<=> x = 6

d) x² + y² - 2x + 4y + 5 = 0

<=> ( x² - 2x + 1 ) + ( y² + 4y + 4 ) = 0

<=> ( x - 1 )² + ( y + 2 )² = 0 (*)

\(\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\\\left(y+2\right)^2\ge0\forall y\end{cases}}\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\forall x,y\)

Đẳng thức xảy ra ( tức (*) ) <=> \(\hept{\begin{cases}x-1=0\\y+2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}\)

e) 2x² + 8x + y² - 2y + 9 = 0

<=> 2( x² + 4x + 4 ) + ( y² - 2y + 1 ) = 0

<=> 2( x + 2 )² + ( y - 1 )² = 0 (*)

\(\hept{\begin{cases}2\left(x+2\right)^2\ge0\forall x\\\left(y-1\right)^2\ge0\forall y\end{cases}}\Rightarrow2\left(x+2\right)^2+\left(y-1\right)^2\ge0\forall x,y\)

Đẳng thức xảy ra ( tức xảy ra (*) ) <=> \(\hept{\begin{cases}x+2=0\\y-1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-2\\y=1\end{cases}}\)

a) ( x - 3 )² - 4 = 0

<=> ( x - 3 )² = 4

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

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

<=> \(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)

Vậy S = { 5 ; 1 }

b) x² - 9 = 0

<=> x² = 9

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

Vậy S = { 3 ; -3 }

c) x( x - 2x ) - x² - 8 = 0

<=> x² - 2x² - x² - 8 = 0

<=> -2x² - 8 = 0

<=> -2x² = 8

<=> x² = -4 ( vô lí )

<=> x = \(\varnothing\)

Vậy S = { \(\varnothing\)}

d) 2x( x - 1 ) - 2x² + x - 5 = 0

<=> 2x² - 2x - 2x² + x - 5 = 0

<=> -x - 5 = 0

<=> -x = 5

<=> x = -5

Vậy S = { -5 }

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

<=> x² - 3x - ( x² - x - 2 ) = 0

<=> x² - 3x - x² + x + 2 = 0

<=> - 2x + 2 = 0

<=> -2x = -2

<=> x = 1

Vậy S = { 1 }

f) x( 3x - 1 ) - 3x² - 7x = 0

<=> 3x² - x - 3x² - 7x = 0

<=> -8x = 0

<=> x = 0

Vậy S = { 0 }