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dễ mà bro 

a)x+5= 9

x=9-5

x=4

b) -3x=24

x=24:-3

x=-8

c) [x]-9=15

[x]=15+9

[x]=24

x=24

d)[x-9]=-2+17

[x-9]=15

[x]=15+9

[x]=24

x=24

a) \(x+5=-2+11\)                                b)\(-3x=5+29\)

\(\Rightarrow x=-2+11-5\)                                   \(\Rightarrow-3x=24\)

\(\Rightarrow x=4\)                                                       \(\Rightarrow-8\)        

c)\(\left|x\right|-9=-2+17\)                         d)\(\left|x-9\right|=-2+17\)

     \(\left|x\right|=15+9\)                                    \(\left|x-9\right|=15\)

      \(\left|x\right|=24\)                                          \(x-9=15\) hoặc \(x-9=15\)

x = 24 hoặc - 24                                    \(x=24\) hoặc \(x=-6\)

Trường hợp 1: x<2/3

Pt sẽ là 2-3x+11-3x=9

=>-6x+13=9

=>-6x=-4

hay x=2/3(loại)

Trường hợp 2: 2/3<=x<11/3

Pt sẽ là 3x-2+11-3x=9

=>9=9(luôn đúng)

Trường hợp 3: x>=11/3

Pt sẽ là 3x-2+3x-11=9

=>6x-13=9

=>6x=22

hay x=11/3(nhận)

\(x^{11}+3x^{10}+x^9+3x^8+x^7-3x^6-17x^5+3x^4+x^3+3x^2+x+3=0\)

\(\Leftrightarrow\left(x^{11}+2x^{10}+4x^9+6x^8+9x^7+6x^6+4x^5+2x^4+x^3\right)+\left(x^{10}+2x^9+4x^8+6x^7+9x^6+6x^5+4x^4+2x^3+x^2\right)-\left(5x^9+10x^8+20x^7+30x^6+45x^5+30x^4+20x^3+10x^2+5x\right)+\left(3x^8+6x^7+12x^6+18x^5+27x^4+18x^3+12x^2+6x+3\right)=0\)

\(\Leftrightarrow x^3\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)+x^2\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)-5\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)+3\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)=0\)

\(\Leftrightarrow\left(x^3+x^2-5x+3\right)\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)=0\)

\(\Leftrightarrow\left(x^2-2x+1\right)\left(x+3\right)\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2\left(x+3\right)\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)=0\)

Dễ thấy: \(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1>0\forall x\)

Nên \(\left[{}\begin{matrix}\left(x-1\right)^2=0\\x+3=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)

đex ~ vừa thấy trên face lướt qua luôn

a) x = 4

b) x = -8

c) | x | - 9 = -2 + 17

| x | = 15 + 9

| x | = 24

x = 24 hoặc x = -24

d) |x – 9| = -2 + 17

|x – 9| = 15

x – 9 = 15 hoặc x – 9 = -15

x = 24 hoặc x = -6

Lời giải:

a. $x(3x+1)+(x-1)^2-(2x+1)(2x-1)=0$

$\Leftrightarrow (3x^2+x)+(x^2-2x+1)-(4x^2-1)=0$

$\Leftrightarrow 3x^2+x+x^2-2x+1-4x^2+1=0$

$\Leftrightarrow (3x^2+x^2-4x^2)+(x-2x)+(1+1)=0$

$\Leftrightarrow -x+2=0$

$\Leftrightarrow x=2$

b.

$(x+1)^3+(2-x)^3-9(x-3)(x+3)=0$

$\Leftrightarrow [(x+1)+(2-x)][(x+1)^2-(x+1)(2-x)+(2-x)^2]-9(x-3)(x+3)=0$

$\Leftrightarrow 3[x^2+2x+1-(x-x^2+2)+(x^2-4x+4)]-9(x-3)(x+3)=0$

$\Leftrightarrow 3(3x^2-3x+3)-9(x^2-9)=0$

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

$\Leftrightarrow 9(x^2-x+1-x^2+9)=0$
$\Leftrightarrow 9(-x+10)=0$

$\Leftrightarrow -x+10=0\Leftrightarrow x=10$

c.

$(x-1)^3-(x+3)(x^2-3x+9)+3x^2=25$

$\Leftrightarrow (x^3-3x^2+3x-1)-(x^3+3^3)+3x^2=25$

$\Leftrightarrow x^3-3x^2+3x-1-x^3-27+3x^2=25$
$\Leftrightarrow (x^3-x^3)+(-3x^2+3x^2)+3x-28=25$

$\Leftrightarrow 3x-28=25$

$\Leftrightarrow x=\frac{53}{3}$

d.

$(x+2)^3-(x+1)(x^2-x+1)-6(x-1)^2=23$
$\Leftrightarrow (x^3+6x^2+12x+8)-(x^3+1)-6(x^2-2x+1)=23$

$\Leftrightarrow x^3+6x^2+12x+8-x^3-1-6x^2+12x-6=23$

$\Leftrightarrow (x^3-x^3)+(6x^2-6x^2)+(12x+12x)+(8-1-6)=23$
$\Leftrightarrow 24x+1=23$

$\Leftrgihtarrow 24x=22$

$\Leftrightarrow x=\frac{11}{12}$

e.

$(x+3)(x^2-3x+9)-x(x-2)(x+2)+11=0$

$\Leftrightarrow x^3+3^3-x(x^2-4)+11=0$

$\Leftrightarrow x^3+27-x^3+4x+11=0$

$\Leftrightarrow (x^3-x^3)+4x+(27+11)=0$

$\Leftrightarrow 4x+38=0$

$\Leftrightarrow x=\frac{-19}{2}$

f.

$x(x-3)-x+3=0$

$\Leftrightarrow x(x-3)-(x-3)=0$

$\Leftrightarrow (x-3)(x-1)=0$

$\Leftrightarrow x-3=0$ hoặc $x-1=0$

$\Leftrightarrow x=3$ hoặc $x=1$

\(\dfrac{2}{3}\)\(x\) + (\(\dfrac{4}{5}\)\(x\) - \(\dfrac{11}{15}\)) = \(\dfrac{5}{9}\)

\(\dfrac{2}{3}\)\(x\) + \(\dfrac{4}{5}\)\(x\) - \(\dfrac{11}{15}\) = \(\dfrac{5}{9}\)

(\(\dfrac{2}{3}+\dfrac{4}{5}\))\(x\) = \(\dfrac{5}{9}\) + \(\dfrac{11}{5}\)

\(\dfrac{22}{15}\)\(x\)        = \(\dfrac{58}{45}\)

     \(x\)        = \(\dfrac{58}{45}\): \(\dfrac{22}{15}\)

     \(x\)       = \(\dfrac{29}{33}\)