\(a,\Leftrightarrow\left(2-x\right)\left(x^2+4\right)>0\Leftrightarrow2-x>0\Leftrightarrow x< 2\\ b,\Leftrightarrow x+3>0\Leftrightarrow x>-3\\ c,\Leftrightarrow\left[{}\begin{matrix}x< -3\\x>4\end{matrix}\right.\)

b: \(\Leftrightarrow x+3>0\)

hay x>-3

A, số nguyên âm lớn nhất là -1

=> 10-x=-1

          X= 10 - (-1)

          X= 11

B, |x-1| = |-4|

TH1: x-1=-4

         X  = -4 + 1

         X = -3

TH2: x-1 = 4

         X   = 4+1

         X   = 5

Mình làm vật thôi ^_^ chúc bạn học tốt 

Đức™

🤟🏿🤟🏿🤟🏿🤟🏿🤟🏿🌹

a/ 10-x là số nguyên âm lớn nhất => 10-x=-1

=>x=11

b/  I x-1 I =I-4I 

=> Ix-1I=4 =>x-1=4 hoặc x-1=-4

=> x= 5 hoặc x= -3

c/(4-61-19x).(-4)²  =0 => -57-19x=0

=> 19x=-57 => x=-3

d/ 5x/-12=1/4+3/-2

=> 5x/-12=5/-12

=> 5x=5 => x=1

a; \(\dfrac{2}{3}\)\(x\) - \(\dfrac{3}{2}\)\(x\) = \(\dfrac{5}{12}\)

    (\(\dfrac{2}{3}\) - \(\dfrac{3}{2}\))\(x\) = \(\dfrac{5}{12}\)

     - \(\dfrac{5}{6}\)\(x\) = \(\dfrac{5}{12}\)

     \(x\)   = \(\dfrac{5}{12}\) : (- \(\dfrac{5}{6}\))

     \(x=\) - \(\dfrac{1}{2}\)

Vậy \(x=-\dfrac{1}{2}\) 

b; \(\dfrac{2}{5}\) + \(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\)

           \(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\) - \(\dfrac{2}{5}\)

          \(\dfrac{3}{5}\).(3\(x\) - 3,7) = - \(\dfrac{57}{10}\)

         3\(x\) - 3,7 = - \(\dfrac{57}{10}\) : \(\dfrac{3}{5}\)

         3\(x\)   - 3,7 = - \(\dfrac{19}{2}\)

         3\(x\)         = - \(\dfrac{19}{2}\) + 3,7

          3\(x\)        = - \(\dfrac{29}{5}\)

           \(x\)         = - \(\dfrac{29}{5}\) : 3

           \(x\)        = - \(\dfrac{29}{15}\)

Vậy \(x\) \(\in\) - \(\dfrac{29}{15}\) 

c; \(\dfrac{7}{9}\) : (2 + \(\dfrac{3}{4}\)\(x\)) + \(\dfrac{5}{9}\) = \(\dfrac{23}{27}\)

    \(\dfrac{7}{9}\): (2 + \(\dfrac{3}{4}\)\(x\)) = \(\dfrac{23}{27}\) - \(\dfrac{5}{9}\)

     \(\dfrac{7}{9}\):(2 + \(\dfrac{3}{4}\)\(x\)) = \(\dfrac{8}{27}\)

      2 + \(\dfrac{3}{4}\)\(x\) = \(\dfrac{7}{9}\) : \(\dfrac{8}{27}\)

      2 + \(\dfrac{3}{4}\)\(x\) =  \(\dfrac{21}{8}\)

             \(\dfrac{3}{4}x\) = \(\dfrac{21}{8}\) - 2

             \(\dfrac{3}{4}\)\(x\) =  \(\dfrac{5}{8}\)

               \(x\) = \(\dfrac{5}{8}\) : \(\dfrac{3}{4}\)

              \(x\) =  \(\dfrac{5}{6}\)

Vậy \(x=\dfrac{5}{6}\)

a, (2x-3)^2=(x+5)^2

2x-3=x+5

2x-3-x-5=0

x-8=0

x=8

b, x^2(x-1)-4x^2+8x-4=0

x^2(x-1)-(4x^2-8x+4)=0

x^2(x-1)-4(x^2-2x+1)=0

x^2(x-1)-4(x-1)^2=0

(x-1)(x^2-4)(x-1)=0

(x-1)(x-2)(x+2)(x-1)=0

=>x-1=0=>x=1

=>x-2=0=>x=2

=>x+2=0=>x=-2

=>x-1=0=>x=1

Vậy : x=1 ;x=2 và x=-2

c, (x-4)^2-36=0

(x-4)^2-6^2=0

(x-4-6)(x-4+6)=0

(x-10)(x+2)=0

=>x-10=0=>x=10

=>x+2=0=>x=-2

Vậy : x=10 và x=-2

k đúng cho mình nhé bạn !

a. \(8x\left(x-2007\right)-2x+4034=0\)

\(\Rightarrow\left(x-2017\right)\left(4x-1\right)\)

\(\Rightarrow\left[{}\begin{matrix}x-2017=0\\4x-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2017\\4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)

Vậy x=2017 hoặc x=1/4

b.\(\dfrac{x}{2}+\dfrac{x^2}{8}=0\)

\(\Rightarrow\dfrac{x}{2}\left(1+\dfrac{x}{4}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{2}=0\\1+\dfrac{x}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\\dfrac{x}{4}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)

Vậy x=0 hoặc x=-4

c.\(4-x=2\left(x-4\right)^2\)

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

\(\Rightarrow\left(4-x\right)\left(2x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4-x=0\\2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{7}{2}\end{matrix}\right.\)

Vậy x=4 hoặc x=7/2

d.\(\left(x^2+1\right)\left(x-2\right)+2x=4\)

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

Nxet: (x²+3)>0 với mọi x

=> x-2=0 <=>x=2

Vậy x=2

a, 8\(x\).(\(x-2007\)) - 2\(x\) + 4034 = 0

     4\(x\)(\(x\) - 2007) - \(x\) + 2017 = 0

     4\(x^2\) - 8028\(x\) - \(x\) + 2017 = 0

     4\(x^2\) - 8029\(x\) + 2017 = 0

     4(\(x^2\) - 2. \(\dfrac{8029}{8}\) \(x\) +( \(\dfrac{8029}{8}\))²) - (\(\dfrac{8029}{4}\))²  + 2017 = 0

    4.(\(x\) + \(\dfrac{8029}{8}\))² = (\(\dfrac{8029}{4}\))² - 2017

       \(\left[{}\begin{matrix}x=-\dfrac{8029}{8}+\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\\x=-\dfrac{8029}{8}-\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\end{matrix}\right.\) 

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

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

ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)

Ta có : `(x-1)/x -1/(x+1) =(2x-1)/(x(x+1))`

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

`=> x^2 +x -x-1 -x-2x+1=0`

`<=> x^2 -3x =0`

`<=> x(x-3)=0`

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

__

`(x+2)(5-3x)=0`

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

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

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

__

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

\(\Leftrightarrow\dfrac{20\left(1-2x\right)}{12}+\dfrac{6x}{12}=\dfrac{9\left(x-5\right)}{12}-\dfrac{24}{12}\)

`<=> 2x- 40x + 6x = 9x - 45 -24`

`<=>  2x- 40x + 6x-9x + 45 +24=0`

`<=>-41x+69=0`

`<=>-41x=-69`

`<=> x=69/41`

Cậu tách 2 câu 1 lượt mn trl nhanh hơn đó ạ

a:=>x^2-1-x=2x-1

=>x^2-x-1=2x-1

=>x^2-3x=0

=>x=0(loại) hoặc x=3(nhận)

b:=>x+2=0 hoặc 5-3x=0

=>x=-2 hoặc x=5/3

c:=>20(1-2x)+6x=9(x-5)-24

=>20-40x+6x=9x-45-24

=>-34x+20=9x-69

=>-43x=-89

=>x=89/43

d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3

=>2x^2+4x-19=-2x+7

=>2x^2+6x-26=0

=>x^2+3x-13=0

=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)

e: =>(2x-3)(2x-3-x-1)=0

=>(2x-3)(x-4)=0

=>x=4 hoặc x=3/2

a) x = 1; x = - 1 3                 b) x = 2.

c) x = 3; x = -2.                 d) x = -3; x = 0; x = 2.

a:=>x^2-1-x=2x-1

=>x^2-x-1=2x-1

=>x^2-3x=0

=>x=0(loại) hoặc x=3(nhận)

b:=>x+2=0 hoặc 5-3x=0

=>x=-2 hoặc x=5/3

c:=>20(1-2x)+6x=9(x-5)-24

=>20-40x+6x=9x-45-24

=>-34x+20=9x-69

=>-43x=-89

=>x=89/43

d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3

=>2x^2+4x-19=-2x+7

=>2x^2+6x-26=0

=>x^2+3x-13=0

=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)

e: =>(2x-3)(2x-3-x-1)=0

=>(2x-3)(x-4)=0

=>x=4 hoặc x=3/2

a)\(x\left(x+7\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x=0\\x+7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-7\end{cases}}}\)

b)\(\left(x+12\right)\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+12=0\\x-3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-12\\x=3\end{cases}}}\)

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

\(\Leftrightarrow\hept{\begin{cases}-x+5=0\\x-3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=5\\x=3\end{cases}}}\)

d)\(x\left(2+x\right)\left(7-x\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x=0\\2+x=0\\7-x=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=-2\\x=7\end{cases}}}\)

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

\(\Leftrightarrow\hept{\begin{cases}x-1=0\\x+2=0\\-x-3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\x=-2\\x=-3\end{cases}}}\)

#H

Mấy phép tính này bạn áp dụng công thức \(a.b=0\Rightarrow\hept{\begin{cases}a=0\\b=0\end{cases}}\)để làm nên mấy phần đầu bạn tự làm

d)\(x.\left(2+x\right).\left(7+x\right)=0\)

\(\Rightarrow\hept{\begin{cases}x=0\\2+x=0\\7+x=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x=-2\\x=-7\end{cases}}}\)

Vậy \(x\in\left\{0;2;7\right\}\)

e)\(\left(x-1\right).\left(x+2\right).\left(-x-3\right)=0\)

\(\Rightarrow\hept{\begin{cases}x-1=0\\x+2=0\\-x-3=0\end{cases}\Rightarrow\hept{\begin{cases}x=1\\x=-2\\x=-3\end{cases}}}\)

Vậy \(x\in\left\{1;-2;-3\right\}\)

Chúc bạn học tốt

Trả lời:

a, x.(x + 7) = 0

=> x = 0 hoặc x + 7 = 0

=>                      x = -7

Vậy x = 0; x = -7

b, (x + 12).(x - 3) = 0

=> x + 12 = 0 hoặc x - 3 = 0

=>      x = -12             x = 3

Vậy x = -12; x = 3

c, (-x + 5).(x - 3) = 0

=> -x + 5 = 0 hoặc x - 3 = 0

=> -x = -5                  x = 3

=>  x = 5

Vậy x = 5; x = 3

d, x.(2 + x).(7 - x) = 0

=> x = 0 hoặc 2 + x = 0 hoặc 7 - x = 0

=>                      x = -2                x = 7

Vậy x = 0; x = -2; x = 7

e, (x - 1).(x + 2).(-x - 3) = 0

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

=>      x = 1             x = -2               -x = 3

=>                                                    x = -3

Vậy x = 1; x = -2; x = -3