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

\(x^2-6x+9+x^2+4x+4-2x^2=3\)

\(-2x+13=3\)

\(-2x=3-13\)

\(-2x=-10\)

\(x=\frac{-10}{-2}\)

\(x=5\)

Vậy \(x=5\)

\(x^2-2x+2\)

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

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

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^

a: \(P=\left(\dfrac{3}{2\left(x+2\right)}-\dfrac{x}{x-2}+\dfrac{2x^2+3}{\left(x-2\right)\left(x+2\right)}\right)\cdot\dfrac{4\left(x-2\right)}{2x-1}\)

\(=\left(\dfrac{3\left(x-2\right)}{2\left(x+2\right)\left(x-2\right)}-\dfrac{2x\left(x+2\right)}{2\left(x-2\right)\left(x+2\right)}+\dfrac{4x^2+6}{2\left(x-2\right)\left(x+2\right)}\right)\cdot\dfrac{4\left(x-2\right)}{2x-1}\)

\(=\dfrac{3x-6-2x^2-4x+4x^2+6}{2\left(x+2\right)\left(x-2\right)}\cdot\dfrac{4\left(x-2\right)}{2x-1}\)

\(=\dfrac{2x^2-x}{x+2}\cdot\dfrac{2}{2x-1}=\dfrac{2x}{x+2}\)

b: Khi 4x²-1=0 thì (2x-1)(2x+1)=0

=>x=1/2(loại) và x=-1/2(nhận)

Khi x=-1/2 thì \(P=\left(2\cdot\dfrac{-1}{2}\right):\left(-\dfrac{1}{2}+2\right)=-1:\dfrac{3}{2}=-\dfrac{2}{3}\)

Với `x \ne +-2,x \ne 1/2,x \ne0`. Ta có:

`(3/[2x+4]+x/[2-x]+[2x^2+3]/[x^2-4]):[2x-1]/[4x-8]`

`=(3/[2(x+2)]-x/[x-2]+[2x^2+3]/[(x-2)(x+2)]).[4(x-2)]/[2x-1]`

`=[3(x-2)-2x(x+2)+2(2x^2+3)]/[x(x-2)(x+2)].[4(x-2)]/[2x-1]`

`=[3x-6-2x^2-4x+4x^2+6]/[x(x+2)]. 4/[2x-1]`

`=[2x^2-x]/[x(x+2)]. 4/[2x-1]`

`=[x(2x-1)]/[x(x+2)] . 4/[2x-1]`

`=4/[x+2]`

a: |2x-3|=1

=>2x-3=1 hoặc 2x-3=-1

=>x=1(nhận) hoặc x=2(loại)

KHi x=1 thì \(A=\dfrac{1+1^2}{2-1}=2\)

b: ĐKXĐ: x<>-1; x<>2

\(B=\dfrac{2x^2-4x+3x+3-2x^2-1}{\left(x-2\right)\left(x+1\right)}=\dfrac{-x+2}{\left(x-2\right)\left(x+1\right)}=\dfrac{-1}{x+1}\)

( 2x - 3 )( x + 1 ) - 2x² + 6x = 0

<=> 2x² - x - 3 - 2x² + 6x = 0

<=> 5x - 3 = 0

<=> 5x = 3

<=> x = 3/5

( x² - x + 1 )( x - 3 ) - x³ + 4x² = 0

<=> x³ - 4x² + 4x - 3 - x³ + 4x² = 0

<=> 4x - 3 = 0

<=> 4x = 3

<=> x = 3/4

( x² - 2 )( x² + 2 ) - x⁴ - 2x + 5 = 0

<=> ( x² )² - 4 - x⁴ - 2x + 5 = 0

<=> x⁴ + 1 - x⁴ - 2x = 0

<=> 1 - 2x = 0

<=> 2x = 1

<=> x = 1/2

( x - 3 )( x² - 3x + 2 ) - ( x² - 2x - 7 )( x - 2 ) + 2x² - 2x = 0

<=> x³ - 6x² + 11x - 6 - ( x³ - 4x² - 3x + 14 ) + 2x² - 2x = 0

<=> x³ - 6x² + 11x - 6 - x³ + 4x² + 3x - 14 + 2x² - 2x = 0

<=> 12x - 20 = 0

<=> 12x = 20

<=> x = 20/12 = 5/3

a, \(\left(2x-3\right)\left(x+1\right)-2x^2+6x=0\)

\(\Leftrightarrow2x^2+2x-3x-3-2x^2+6x=0\Leftrightarrow5x-3=0\Leftrightarrow x=\frac{3}{5}\)

b, \(\left(x^2-x+1\right)\left(x-3\right)-x^3+4x^2=0\)

\(\Leftrightarrow x^3-3x^2-x^2+3x+x-3-x^3+4x^2=0\Leftrightarrow4x-3=0\Leftrightarrow x=\frac{3}{4}\)

c ; d tương tự nhé ! 

a)\(\left(3x+2\right)\left(2x-3\right)=6x^2-5x-6\)

b) viết lại đề nhin (dùng f(x) viết mới rõ ra dduocj) ko phải dùng {[(...)]} cho chuẩn vào

c) \(\left(x-2\right)^3-x^2.\left(x-6\right)=x^3-3.x.2\left(x-2\right)-8-x^3+6x^2\)

\(=x^3-6x^2+12x-8-x^3+6x^2=12x-8=4\Rightarrow x=1\)

a: 

ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

b: \(A=\left(\dfrac{x-2}{2x-2}+\dfrac{3}{2x-2}-\dfrac{x+3}{2x+2}\right):\left(1-\dfrac{x-3}{x+1}\right)\)

\(=\left(\dfrac{x-2}{2\left(x-1\right)}+\dfrac{3}{2\left(x-1\right)}-\dfrac{x+3}{2\left(x+1\right)}\right):\dfrac{x+1-x+3}{x+1}\)

\(=\dfrac{\left(x-2\right)\left(x+1\right)+3\left(x+1\right)-\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{2}\)

\(=\dfrac{x^2-x-2+3x+3-x^2-2x+3}{2\left(x-1\right)}\cdot\dfrac{1}{2}\)

\(=\dfrac{-2}{4\left(x-1\right)}=\dfrac{-1}{2\left(x-1\right)}\)

Khi x=2005 thì \(A=\dfrac{-1}{2\cdot\left(2005-1\right)}=-\dfrac{1}{4008}\)

Vì x=1 không thỏa mãn ĐKXĐ

nên khi x=1 thì A không có giá trị

c: Để A=-1002 thì \(\dfrac{-1}{2\left(x-1\right)}=-1002\)

=>\(2\left(x-1\right)=\dfrac{1}{1002}\)

=>\(x-1=\dfrac{1}{2004}\)

=>\(x=\dfrac{1}{2004}+1=\dfrac{2005}{2004}\left(nhận\right)\)

1) `2x(3x-1)-(2x+1)(x-3)`

`=6x^2-2x-2x^2+6x-x+3`

`=4x^2+3x+3`

2) `3(x^2-3x)-(4x+2)(x-1)`

`=3x^2-9x-4x^2+4x-2x+2`

`=-x^2-7x+2`

3) `3x(x-5)-(x-2)^2-(2x+3)(2x-3)`

`=3x^2-15x-(x^2-4x+4)-(4x^2-9)`

`=3x^2-15x-x^2+4x-4-4x^2+9`

`=-2x^2-11x+5`

4) `(2x-3)^2+(2x-1)(x+4)`

`=4x^2-12x+9+2x^2+8x-x-4`

`=6x^2-5x+5`

a: A(x)=3x^3+3x-1

B(x)=-2x^3+x^2+4x-3

b: A(x)+B(x)

=3x^3+3x-1-2x^3+x^2+4x-3

=x^3+x^2+7x-4

B(x)-A(x)

=-2x^3+x^2+4x-3-3x^3-3x+1

=-5x^3+x^2+x-2

c; M(x)=x^3+x^2+7x-4

M(-3)=-27+9-21-4=-31-21+9=-43