a,sửa đề :  \(\left(\frac{1}{x^2+4x+4}-\frac{1}{x^2-4x+4}\right):\left(\frac{1}{x+2}+\frac{1}{x^2-4}\right)\)

\(=\left(\frac{1}{\left(x+2\right)^2}-\frac{1}{\left(x-2\right)^2}\right):\left(\frac{x-2+1}{\left(x+2\right)\left(x-2\right)}\right)\)

\(=\left(\frac{x^2-4x+4-x^2-4x-4}{\left(x+2\right)^2\left(x-2\right)^2}\right):\left(\frac{x-1}{\left(x+2\right)\left(x-2\right)}\right)\)

\(=\frac{-8x\left(x+2\right)\left(x-2\right)}{\left(x+2\right)^2\left(x-2\right)^2\left(x-1\right)}=\frac{-8x}{\left(x-1\right)\left(x^2-4\right)}\)

b, \(\left(\frac{2x}{2x-y}-\frac{4x^2}{4x^2+4xy+y^2}\right):\left(\frac{2x}{4x^2-y^2}+\frac{1}{y-2x}\right)\)

\(=\left(\frac{2x}{2x-y}-\frac{4x^2}{\left(2x+y\right)^2}\right):\left(\frac{2x}{\left(2x-y\right)\left(2x+y\right)}-\frac{1}{2x-y}\right)\)

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

\(=\left(\frac{8x^3+8x^2y+2xy^2-8x^3+4x^2y}{\left(2x-y\right)\left(2x+y\right)^2}\right):\left(\frac{-y}{\left(2x-y\right)\left(2x+y\right)}\right)\)

\(=-\left(\frac{12x^2y+xy^2}{2x+y}\right)=\frac{-12x^2y-xy^2}{2x+y}\)

p) \(\left(9-x\right)\left(x^2+2x-3\right)\)

\(=9\left(x^2+2x-3\right)-x\left(x^2+2x-3\right)\)

\(=9x^2+18x-27-x^3-2x^2+3x\)

\(=-x^3+7x^2+21x-27\)

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

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

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

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

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

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

\(=-6x^3+24x^2-12x+\dfrac{1}{2}x^2-2x+1\)

\(=-6x^3+\dfrac{49}{2}x^2-14x+1\)

q) \(\left(6x+1\right)\left(x^2-2x-3\right)\)

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

\(=6x^3-12x^2-18x+x^2-2x-3\)

\(=6x^3-11x^2-20x-3\)

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

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

\(=-2x^3-6x^2+2x-x^2-3x+1\)

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

u) \(\left(2x-3\right)\left(-x^2+x+6\right)\)

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

\(=-2x^3+2x^2+12x+3x^2-3x-18\)

\(=-2x^3+5x^2+9x-18\)

s) \(\left(-4x+5\right)\left(x^2+3x-2\right)\)

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

\(=-4x^3-12x^2+8x+5x^2+15x-10\)

\(=-4x^3-7x^2+23x-10\)

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

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

\(=-x^2-3+2x^4+6x+18-12x^3\)

\(=2x^4-12x^3-x^2+6x+15\)

p: (-x+9)(x^2+2x-3)

=-x^3-2x^2+3x+9x^2+18x-27

=-x^3+7x^2+21x-27

n: (-x+3)(x^2+x+1)

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

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

o: (-6x+1/2)(x^2-4x+2)

=-6x^3+24x^2-12x+1/2x^2-2x+1

=-64x^3+49/2x^2-14x+1

q: (6x+1)(x^2-2x-3)

=6x^3-12x^2-18x+x^2-2x-3

=6x^3-11x^2-20x-3

r: (2x+1)(-x^2-3x+1)

=-2x^3-6x^2+2x-x^2-3x+1

=-2x^3-7x^2-x+1

u: =-2x^3+2x^2+12x+3x^2-3x-18

=-2x^3+5x^2+9x-18

s: =-4x^3-12x^2+8x+5x^2+15x-10

=-4x^3-7x^2+23x-10

`@` `\text {Ans}`

`\downarrow`

`a)`

`3x(4x-1) - 2x(6x-3) = 30`

`=> 12x^2 - 3x - 12x^2 + 6x = 30`

`=> 3x = 30`

`=> x = 30 \div 3`

`=> x=10`

Vậy, `x=10`

`b)`

`2x(3-2x) + 2x(2x-1) = 15`

`=> 6x- 4x^2 + 4x^2 - 2x = 15`

`=> 4x = 15`

`=> x = 15/4`

Vậy, `x=15/4`

`c)`

`(5x-2)(4x-1) + (10x+3)(2x-1) = 1`

`=> 5x(4x-1) - 2(4x-1) + 10x(2x-1) + 3(2x-1)=1`

`=> 20x^2-5x - 8x + 2 + 20x^2 - 10x +6x - 3 =1`

`=> 40x^2 -17x - 1 = 1`

`d)`

`(x+2)(x+2)-(x-3)(x+1)=9`

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

`=> 6x + 7 =9`

`=> 6x = 2`

`=> x=2/6 =1/3`

Vậy, `x=1/3`

`e)`

`(4x+1)(6x-3) = 7 + (3x-2)(8x+9)`

`=> 24x^2 - 12x + 6x - 3 = 7 + (3x-2)(8x+9)`

`=> 24x^2 - 12x + 6x - 3 = 7 + 24x^2 +11x - 18`

`=> 24x^2 - 6x - 3 = 24x^2 + 18x -11`

`=> 24x^2 - 6x - 3 - 24x^2 + 18x + 11 = 0`

`=> 12x +8 = 0`

`=> 12x = -8`

`=> x= -8/12 = -2/3`

Vậy, `x=-2/3`

`g)`

`(10x+2)(4x- 1)- (8x -3)(5x+2) =14`

`=> 40x^2 - 10x + 8x - 2 - 40x^2 - 16x + 15x + 6 = 14`

`=> -3x + 4 =14`

`=> -3x = 10`

`=> x= - 10/3`

Vậy, `x=-10/3`

Hello các bạn còn đó ko?

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

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

\(=-x^2-3x+2c^3x+6x+18-12c^3\)

\(=-x^2+3x+2c^3x+18-12c^3\)

f) \(\left(2x-5\right)\left(x^2-x+3\right)\)

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

\(=2x^3-2x^2+6x-5x^2+5x-15\)

\(=2x^3-7x^2+11x-15\)

w) \(\left(3x+1\right)\left(x^2-2x-5\right)\)

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

\(=3x^3-6x^2-15x+x^2-2x-5\)

\(=3x^3-5x^2-17x-5\)

x) \(\left(6x-3\right)\left(x^2+x-1\right)\)

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

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

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

y) \(\left(5x-2\right)\left(3x+1-x^2\right)\)

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

\(=15x^2+5x-5x^3-6x-2+2x^2\)

\(=-5x^3+17x^2-x-2\)

z) \(\left(\dfrac{3}{4}x+1\right)\left(4x^2+4x+4\right)\)

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

\(=3x^3+3x^2+3x+4x^2+4x+4\)

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

f: =2x^3-2x^2+6x-5x^2+5x-15

=2x^3-7x^2+11x-15

w: =3x^3-6x^2-15x+x^2-2x-5

=3x^3-5x^2-17x-5

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

=6x^3+3x^2-9x+3

y: =(5x-2)(-x^2+3x+1)

=-5x^3+15x^2+5x+2x^2-6x-2

=-5x^3+17x^2-x-2

z: =3x^3+3x^2+3x+4x^2+4x+4

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

1.

\(x^4-6x^2-12x-8=0\)

\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)

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

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

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

\(\Leftrightarrow x=1\pm\sqrt{5}\)

3.

ĐK: \(x\ge-9\)

\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)

\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)

Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)

\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)

\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)

\(\Leftrightarrow...\)

2.

ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)

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

\(\Leftrightarrow\dfrac{1}{2x+\dfrac{1}{x}+4}+\dfrac{1}{2x+\dfrac{1}{x}-4}=\dfrac{3}{5}\)

Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)

\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)

Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:

\(\left(1\right)\Leftrightarrow\dfrac{1}{b+8}+\dfrac{1}{b}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)

\(\Leftrightarrow10b+40=3\left(b+8\right)b\)

\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)

TH1: \(b=2\Leftrightarrow...\)

TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)

\(a,\Leftrightarrow\left(3x-7\right)\left(3x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{7}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-1-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\\ c,\Leftrightarrow4x^2-7x-2-4x^2+4x+3=7\\ \Leftrightarrow-3x=6\Leftrightarrow x=-2\\ d,\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=0\\ \Leftrightarrow4x=-26\Leftrightarrow x=-\dfrac{13}{2}\\ e,\Leftrightarrow x^3+27-x^3+x-27=0\\ \Leftrightarrow x=0\\ f,\Leftrightarrow\left(4x-3\right)\left(4x-3+3x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)

a) 9x²-49=0

(3x)²-7²=0

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

th1: 3x-7=0

<=>3x=7

<=>x=\(\dfrac{7}{3}\)

th2: 3x+7=0

<=>3x=-7

<=>x=\(-\dfrac{7}{3}\)

a, \(\left(x+3\right)^3-\left(x+2\right)\left(x-2\right)-6x^2-20\)

\(=x^3+9x^2+27x+27-\left(x^2-4\right)-6x^2-20\)

\(=x^3+9x^2+27x+27-x^2+4+6x^2+20\)

\(=x^3+14x^2+27x+51\)

b, \(\left(2x+3\right)\left(4x^2-6x+9\right)-\left(2x-3\right)\left(4x^2+6x+9\right)\)

\(=8x^3-12x^2+18x+12x^2-18x+18-\left(8x^3+12x^2+18x-12x^2-18x-18\right)\)

\(=8x^3+18-8x^3+18=36\)

c, \(\left(2x-1\right)\left(4x^2+2x+1\right)\left(2x+1\right)\left(4x^2-2x+1\right)\)

\(=\left(8x^3+4x^2+2x-4x^2-2x-1\right)\left(8x^3-4x^2+2x+4x^2-2x+1\right)\)

\(=\left(8x^3-1\right)\left(8x^3+1\right)=\left(8x^3\right)^2-1\)

\(=64x^5-1\)

d, \(\left(x+4\right)\left(x^2-4x+16\right)-\left(50+x^2\right)\)

\(=x^3-4x^2+16x+4x^2-16x+64-50-x^2\)

\(=x^3-x^2+14\)

Chúc bạn học tốt!!!

\(a.x=-0,6\)

\(c.x=-11,6\)

Pt nhju ak!!!

a) Ta có: \(\left(x-2\right)\cdot x=2x\cdot\left(x+5\right)\)

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

\(\Leftrightarrow x\cdot\left[x-2-2\left(x+5\right)\right]=0\)

\(\Leftrightarrow x\left(x-2-2x-10\right)=0\)

\(\Leftrightarrow x\left(-x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\-x=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\)

Vậy: S={0;-8}

b) Ta có: \(\left(2x-5\right)\left(x+11\right)=\left(5-2x\right)\left(2x+1\right)\)

\(\Leftrightarrow\left(2x-5\right)\left(x+11\right)-\left(5-2x\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(x+11\right)+\left(2x-5\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(x+11+2x+1\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(3x+12\right)=0\)

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

Vậy: \(S=\left\{\dfrac{5}{2};-4\right\}\)

c) Ta có: \(x^2+6x+9=4x^2\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}-x+3=0\\3x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=-3\\3x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

Vậy: S={3;-1}

d) Ta có: \(\left(x+2\right)\left(5-4x\right)=x^2+4x+4\)

\(\Leftrightarrow\left(x+2\right)\left(5-4x\right)-\left(x^2+4x+4\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(5-4x\right)-\left(x+2\right)^2=0\)

\(\Leftrightarrow\left(x+2\right)\left(5-4x-x-2\right)=0\)

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

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

Vậy: \(S=\left\{-2;\dfrac{3}{5}\right\}\)