1​) x = -1\(\frac{1}{2}\)

_​​2) x = \(\pm2\); x = -1

3) x = -1 ; x = 1 ; x = -2i ; x = 2i

4) x \(\approx-1,59385382059801\) ; x \(\approx0,405545399146862\); x \(\approx6,18830832902225\)

Kế​t quả​ rú​t gọ​n: Unknown!!!

_​​

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

d) \(\frac{4x^2-12x+9}{9-4x^2}=-\frac{\left(2x+3\right)^2}{\left(2x-3\right)\left(2x+3\right)}=\frac{2x+3}{2x-3}\)

a: A=x+3+|x-3|

=x+3+3-x(x<=3)

=6

b:\(B=\sqrt{x^2+4x+4}-\sqrt{x^2}\)

\(=\left|x+2\right|-\left|x\right|\)

=x+2-x=2

c: \(C=\dfrac{\sqrt{x^2-2x+1}}{x-1}\)

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

????  

xin lỗi nha ! 

mình mới học lớp 3 

mà bài này khó nắm 

ko bt thì ko nhắn nha

a.A=\(\sqrt{12}-\sqrt{27}+\sqrt{4+2\sqrt{3}}\)\(=2\sqrt{3}-3\sqrt{3}+\sqrt{\left(\sqrt{3}+1\right)^2}\)               \(=-\sqrt{3}+\sqrt{3}+1\)                                                                                  =1                                                                                                        b. \(x^2-2x-4=0\)                                                                                   Δ= \(\left(-2\right)^2-4\times1\times-4=20>0\)                                             \(\Rightarrow\)   phương trình có 2 nghiệm pb                                                \(x1=\dfrac{2+\sqrt{20}}{2}=1+\sqrt{5}\)     \(x2=\dfrac{2-\sqrt{20}}{2}=1-\sqrt{5}\)                      c. \(\left\{{}\begin{matrix}2x-y=5\\x+3y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-y=5\\2x+6y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-7y=7\\2x-y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\2x+1=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=2\end{matrix}\right.\)

yggucbsgfuyvfbsudy

????????

Câu 3: 

=>|2x-1|=x+10

TH1: x>=1/2

=>2x-1=x+10

=>x=11(nhận)

TH2: x<1/2

=>1-2x=x+10

=>-3x=9

=>x=-3(nhận)

Bài 2:

(1 + x)³ + (1 - x)³ - 6x(x + 1) = 6

<=> x³ + 3x² + 3x + 1 - x³ + 3x² - 3x + 1 - 6x² - 6x = 6

<=> -6x + 2 = 6

<=> -6x = 6 - 2

<=> -6x = 4

<=> x = -4/6 = -2/3

Bài 3: 

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

<=> 10x³ + 25x² - 15x = 0

<=> 5x(2x - 1)(x + 3) = 0

<=> 5x = 0 hoặc 2x - 1 = 0 hoặc x + 3 = 0

<=> x = 0 hoặc x = 1/2 hoặc x = -3

b) (4x - 1)(x - 3) - (x - 3)(5x + 2) = 0

<=> 4x² - 13x + 3 - 5x² + 13x + 6 = 0

<=> -x² + 9 = 0

<=> -x² = -9

<=> x² = 9

<=> x = +-3

c) (x + 4)(5x + 9) - x² + 16 = 0

<=> 5x² + 9x + 20x + 36 - x² + 16 = 0

<=> 4x² + 29x + 52 = 0

<=> 4x² + 13x + 16x + 52 = 0

<=> 4x(x + 4) + 13(x + 4) = 0

<=> (4x + 13)(x + 4) = 0

<=> 4x + 13 = 0 hoặc x + 4 = 0

<=> x = -13/4 hoặc x = -4

Lê Nhật Hằng cảm ơn bạn nha