Bài 5 : 

f, bạn xem lại đề hay là tìm x chứa tham số a ? 

g, \(x^2+3x-\left(2x+6\right)=0\Leftrightarrow x\left(x+3\right)-2\left(x+3\right)=0\)

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

h, \(5x+20-x^2-4x=0\Leftrightarrow5\left(x+4\right)-x\left(x+4\right)=0\)

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

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

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-5\right)=0\Leftrightarrow x=\pm1;x=5\)

n, \(x\left(x-3\right)-7x+21=0\Leftrightarrow x\left(x-3\right)-7\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left(x-3\right)=0\Leftrightarrow x=3;x=7\)

x=7 nha

1. \(x^6-2x^3+1=0\Leftrightarrow\left(x^3-1\right)^2=0\Leftrightarrow x=1\)

2. \(x^6+\dfrac{1}{4}x^3+\dfrac{1}{64}=0\Leftrightarrow\left(x^3\right)^2+2.x^3.\dfrac{1}{8}+\left(\dfrac{1}{8}\right)^2=0\Leftrightarrow\left(x+\dfrac{1}{8}\right)^2=0\Leftrightarrow x=-\dfrac{1}{2}\)4. \(x^3-10x^2+25x=0\Leftrightarrow x^3-5x^2-5x^2+25x=0\)

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

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

5. \(\dfrac{1}{4}x^3-3x^2+9x=0\)

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

\(\Leftrightarrow x\left[\left(\dfrac{1}{2}x\right)^2-2.\dfrac{1}{2}x.3+3^2\right]=0\)

\(\Leftrightarrow x\left(\dfrac{1}{2}x-3\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

6. \(x^5-16x=0\Leftrightarrow x\left(x^4-16\right)=0\Leftrightarrow x\left(x^2-4\right)\left(x^2+4\right)=0\)

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

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

7. \(4x^2+4x-3=0\Leftrightarrow4x^2-2x^2-6x-3=0\)

\(\Leftrightarrow2x\left(2x-1\right)-3\left(2x-1\right)=0\)

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

8. \(4x^2+28x+48=0\Leftrightarrow4x^2+12x+14x+48=0\)

\(\Leftrightarrow4x\left(x+3\right)+12\left(x+4\right)=0\)

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

9. \(9x^2-12x+3=0\Leftrightarrow9x^2-9x-3x+3=0\Leftrightarrow9x\left(x-1\right)-3\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(9x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)

|2 - x|² + 6x - 3 = 0

<=> (x - 2)² + 6x - 3 = 0

<=> x² - 4x + 4 + 6x - 3 = 0

<=> x² + 2x + 1 = 0

<=> (x + 1)² = 0

<=> x + 1 = 0

<=> x = -1

Bắt phải thể hiện -_-

bạn có thể check lại đề bài câu a được không ạ

là x^2 = x^5 á bạn

ứaedrfgthyjk

a) 2840 + [(999 - 9x) : 60 ] .24 = 3200

=> [(999-9x) : 60 ] . 24 = 2840 - 3200 = -360

=>  (999 - 9x) = \(\frac{-360}{24}=-15\)

=>  9x = 999 - ( - 15) = 999 + 15 = 1014

=>  x = \(\frac{1014}{9}=\frac{338}{3}\)

b) (3x - 48) . 6 = 3³.2² - 2³.3²

(3x - 48) . 6 = 27 . 4  -  8 . 9

(3x -48).6 = 36

(3x - 48 = 36 : 6 = 6

3x = 54

x = 54 : 3 =18

 t ick cho mik nha

1)

a)    2840 + [(999-9x) :60] . 24=3200

                 [(999-9x): 60] . 24=3200- 2840= 360

                (999-9x) : 60=    360 : 24

                (999-9x) : 60= 15

                999-9x        =15 . 60

                  999-9x       =900

                         9x=    999-900=9

                          x=9:9=1

b)   (3x - 48).6=    3³.2²-2³.3²

      (3x-48).6  =   27.4-8.9

       (3x - 48)6  =  108-72

        (3x - 48)6  = 36

       (3x-48)    =   36:6=6

        3x        = 6 + 48

        3x=54

          x=54:3=18

a, 720:[118-(2x -10) ]=60

           [118-(2x-10)]=720:60

           118-(2x-10)=12

                   2x-10=118-12

                    2x-10=106

                     2x     =106+10

                        2x  =116

                         x=116:2

                          x=58

\(a,\left(-5\right).\left|x\right|=-75\)

\(\left|x\right|=\frac{-75}{-5}=15\)

\(\Rightarrow\orbr{\begin{cases}x=15\\x=-15\end{cases}}\)

Vậy....

\(b,\left(-6\right)^3.x^2=-1944\)

\(-216.x^2=-1944\)

\(x^2=9\)

\(\Rightarrow x=\pm3\)

Vậy....

\(d,\left|9-x\right|=-7+64\)

\(\left|9-x\right|=57\)

\(\Rightarrow\orbr{\begin{cases}9-x=57\\9-x=-57\end{cases}\Rightarrow\orbr{\begin{cases}x=-48\\x=66\end{cases}}}\)

Vậy...

\(e,\left|x+101\right|-\left(-16\right)=\left(-43\right).\left(-5\right)\)

\(\left|x+101\right|+16=215\)

\(\left|x+101\right|=199\)

\(\Rightarrow\orbr{\begin{cases}x+101=199\\x+101=-199\end{cases}\Rightarrow\orbr{\begin{cases}x=98\\x=-300\end{cases}}}\)

Vậy..

hok tốt!!

a,\(\left(-5\right).\left|x\right|=-75\)

\(=>\left|x\right|=-75:\left(-5\right)=15\)

\(=>\orbr{\begin{cases}x=15\\x=-15\end{cases}}\)

b,\(\left(-6\right)^3.x^2=-1944\)

\(=>\frac{1944}{216}=x^2\)

\(=>x=\sqrt{\frac{1944}{216}}=3\)

CẢM ƠN MỌI NGƯỜI NHIỀU Ạ 

\(a.x-143=57\)

\(x=200\)

\(b.\left(8x-12\right):4=3^3\)

\(8x-12=27.4\)

\(8x-12=108\)

\(8x=120\)

\(x=15\)

\(d.10+2x=4^2\)

\(2x=16-10\)

\(2x=6\)

\(x=3\)

a) 716-(x-143)=659

<=>x-143=57

<=>x=57+143

<=>x=200

b) [(8x-12):4].3³=3⁶

<=>(8x-12):4=3³

<=>8x-12=27.4

<=>8x-12=108

<=>8x=120

<=>x=5

c) -2<|x|≤1, x thuộc Z

Vì -2<|x|≤1

=>|x| thuộc {-1;0;1}

+)Với |x|=-1=>Vô lí

+)Với |x|=0=>x=0(tm)

+)Với |x|=1=>x=-1;1(tm) 

Vậy x thuộc {-1;1;0}

d) 10+2x=4⁵:4³

<=>10+2x=16

<=>2x=6

<=>x=3

e) 4^x+1+4⁰=65

<=>4^x+1+1=65

<=>4^x+1=64

<=>4^x+1=4³

<=>x+1=3

<=>x=2

g) 96-2(x+1)=-42

<=>2(x+1)=96+42

<=>2(x+1)=138

<=>x+1=69

<=>x=68

h) 4x-20=2⁵:2²

<=>4x-20=2³

<=>4x=8+20

<=>4x=28

<=>x=7

k) 8x-75=5x+21

<=>8x-5x=75+21

<=>3x=96

<=>x=32

i) [(8x-14):2-2].31=341

<=>(8x-14):2-2=11

<=>(8x-14):2=13

<=>8x-14=26

<=>8x=40

<=>x=5