3-4*[(x-1)+(x+1)]=2x+1*1-3x-2x+1+x

   -1x=2x-3x-2x+1+x

    -1x=-3x+1+x

    x+x-x=-3+1+1

    x=-1

  Mk làm luôn ko ghi đề bài 

  Mk ko chắc chắn lắm

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

\(\Leftrightarrow x^3-3x^2+3x-1-2x^2-4x-2=2x-6x^2+1-3x-2x+2x^2\)

\(\Leftrightarrow x^3-5x^2-x-3=-4x^2-3x+1\)

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

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

\(\Leftrightarrow\left(x-1\right)\left(x^2+2x+4\right)=0\)

=>x-1=0

hay x=1

a)để -3/x-1 thuộc Z

=>-3 chia hết x-1

=>x-1\(\in\){1,-1,3,-3}

=>x\(\in\){2,0,4,-2}

b)để -4/2x-1 thuộc Z

=>4 chia hết 2x-1

=>2x-1\(\in\){1,-1,2,-2,4,-4}

=>x\(\in\){1;-3;3;-5;7;-9}

c)\(\frac{3x+7}{x-1}=\frac{3\left(x-1\right)+10}{x-1}=\frac{3\left(x-1\right)}{x-1}+\frac{10}{x-1}\in Z\)

=>10 chia hết x-1

=>x-1\(\in\)Ư(10)

bạn tự làm tiếp nhé

Bài 1: 

a: f(0)=1

f(2)=-3x2+1=-6+1=-5

f(-2)=-3x2+1=-5

f(-1/2)=-3x1/2+1=-3/2+1=-1/2

b: f(x)=-3

=>-3|x|+1=-3

=>-3|x|=-4

=>|x|=4/3

=>x=4/3 hoặc x=-4/3

1/ 

a, (x-3)²+(4+x)(4-x)=10

<=>x²-6x+9+(16-x²)=10

<=>-6x+25=10

<=>-6x=-15

<=>x=5/2

còn lại tương tự a 

2/

a, \(a^2\left(a+1\right)+2a\left(a+1\right)=\left(a^2+2a\right)\left(a+1\right)=a\left(a+1\right)\left(a+2\right)\)

Vì a(a+1)(a+2) là tích 3 nguyên liên tiếp nên a(a+1)(a+2) chia hết cho 2,3

Mà (2,3)=1

=>a(a+1)(a+2) chia hết cho 6 (đpcm)

b, \(x^2+2x+2=\left(x^2+2x+1\right)+1=\left(x+1\right)^2+1\)

Vì \(\left(x+1\right)^2\ge0\Rightarrow\left(x+1\right)^2+1\ge1>0\left(đpcm\right)\)

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

Vì \(\left(x-\frac{1}{2}\right)^2\ge0\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\)(đpcm)

d, \(-x^2+4x-5=-\left(x^2-4x+4\right)-1=-\left(x-2\right)^2-1\)

Vì \(-\left(x-2\right)^2\le0\Rightarrow-\left(x-2\right)^2-1\le-1< 0\) (đpcm)

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

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

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

\(\Leftrightarrow8x=2\)

\(\Leftrightarrow x=\frac{1}{4}\)

bn xem lại đi nha

a) \(2x^2-2x-x^2+6=0\) 

\(\Leftrightarrow x^2-2x+1+5=0\)

\(\Leftrightarrow\left(x-1\right)^2=-5\) ( vô lý)

Vậy không có x thoả mãn \(2x.\left(x-1\right)-x^2+6=0\)

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

\(\Leftrightarrow x^4-6x^2-4x^4+3x^4+3x^2+3=0\)

\(\Leftrightarrow3-3x^2=0\)

\(\Leftrightarrow3x^2=3\Leftrightarrow x^2=1\) \(\Leftrightarrow x\in\left\{-1;1\right\}\)

Vậy \(x\in\left\{-1;1\right\}\)

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

\(\Leftrightarrow x^3+1-2x-x.\left(x^2-4\right)=0\)

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

\(\Leftrightarrow1+2x=0\Leftrightarrow x=\dfrac{-1}{2}\)

Vậy x=\(\dfrac{-1}{2}\)

d) \(\left(x+3\right).\left(x^2-3x+9\right)-x.\left(x-2\right).\left(x+2\right)=15\)

\(\Leftrightarrow x^3+27-x.\left(x^2-4\right)-15=0\)

\(\Leftrightarrow x^3-27-x^3+4x-15=0\)

\(\Leftrightarrow4x-42=0\)

\(\Leftrightarrow x=10,5\)

Vậy x=10,5

10 + (2x - 1) ² : 3 = 13

=>    (2x - 1) ² : 3 = 13 - 10

=>    (2x - 1) ² : 3 = 3

=>    (2x - 1) ²      =  3 . 3 

=>    (2x - 1) ²      =  3 ²  

=>              2x - 1 = 3 

=>                   2x = 3 + 1 

=>                   2x = 4

=>                      x = 2

10 + (2x - 1)² : 3 = 13 

=> (2x - 1)² : 3 = 13 - 10

=> (2x - 1 )² : 3 = 3

=>  (2x - 1)²      = 9

=>  (2x - 1)²      = 3²

=>  2x  - 1         = 3

 => 2x                = 4

 => x    = 2

Vậy x = 2

\(10+\left(2x-1\right)^2:3=13\)

\(\left(2x-1\right)^2:3=13-10\)

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

\(\left(2x-1\right)^2=3.3\)

\(\left(2x-1\right)^2=9\)

\(\left(2x-1\right)^2=\left(\pm3\right)^2\)

\(2x-1=\pm3\)

\(TH1:2x-1=3\)                                                                    \(TH2:2x-1=-3\)

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

\(2x=4\)                                                                                                \(2x=-2\)

\(x=4:2\)                                                                                                \(x=-2:2\)

\(x=2\)                                                                                                    \(x=-1\)

                                  Vậy \(x\in\left\{2,-1\right\}\)

1) Ta có: \(4x^2-1=\left(2x+1\right).\left(3x-5\right)\)

\(\Leftrightarrow\left(2x+1\right).\left(2x-1\right)-\left(2x+1\right).\left(3x-5\right)=0\)

\(\Leftrightarrow\left(2x+1\right).\left[\left(2x-1\right)-\left(3x-5\right)\right]=0\)

\(\Leftrightarrow\left(2x+1\right).\left(2x-1-3x+5\right)=0\)

\(\Leftrightarrow\left(2x+1\right).\left(4-x\right)=0\)

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

Vậy \(x=-\frac{1}{2}\) hoặc \(x=4\)

2) Ta có: \(\left(x+1\right)^2=4.\left(x^2-2x+1\right)\)

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

\(\Leftrightarrow\left[\left(x+1\right)+2.\left(x-1\right)\right].\left[\left(x+1\right)-2.\left(x-1\right)\right]=0\)

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

\(\Leftrightarrow\left(3x-1\right).\left(3-x\right)=0\)

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

Vậy \(x=\frac{1}{3}\) hoặc \(x=3\)

3) Ta có: \(2x^3+5x^2-3x=0\)

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

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

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

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

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

Vậy \(x=0\) hoặc \(x=-3\) hoặc \(x=-\frac{1}{2}\)

4) Ta có: \(2x=3x-2\)

\(\Leftrightarrow2x-3x=-2\)

\(\Leftrightarrow-x=-2\)

\(\Leftrightarrow x=2\left(TM\right)\)

Vậy \(x=2\)

5) Ta có: \(x+15=3x-1\)

\(\Leftrightarrow x-3x=-1-15\)

\(\Leftrightarrow-2x=-16\)

\(\Leftrightarrow x=8\left(TM\right)\)

Vậy \(x=8\)

6) Ta có: \(2-x=0,5x-4\)

\(\Leftrightarrow-x-0,5x=-4-2\)

\(\Leftrightarrow-1,5x=-6\)

\(\Leftrightarrow x=4\left(TM\right)\)

Vậy \(x=4\)

1) 4x²-1=(2x+1)(3x-5)

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

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

<=> (2x+1)(4-x)=0

<=>\([^{2x+1=0}_{4-x=0}< =>[^{2x=-1}_{x=4}< =>[^{x=\frac{-1}{2}}_{x=4}\)

2) (x+1)²= 4(x²-2x+1)

<=> x²+2x+1-4(x²-2x+1)=0

<=> x²+2x+1-4x²+8x-4=0

<=> -3x²+10x-3=0

<=> -3x²+x+9x-3=0

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

<=> (3x-1)(3-x)=0

<=> \([^{3x-1=0}_{3-x=0}< =>[^{3x=1}_{x=3}< =>[^{x=\frac{1}{3}}_{x=3}\)

3) 2x³+5x²-3x=0

<=> 2x(x²+\(\frac{5}{2}x-\frac{3}{2})=0\)

<=> 2x\(\left[x^2+2.\frac{5}{4}x+\frac{25}{16}-\left(\frac{25}{16}+\frac{3}{2}\right)\right]=0\)

<=> 2x\(\left[\left(x+\frac{5}{4}\right)^2-\frac{49}{16}\right]=0\)

<=> 2x\(\left(x+\frac{5}{4}-\frac{7}{4}\right)\left(x+\frac{5}{4}+\frac{7}{4}\right)=0\)

<=> x\(\left(x-\frac{1}{2}\right)\left(x+3\right)=0\)

<=>\(\left[{}\begin{matrix}x=0\\x=\frac{1}{2}\\x=-3\end{matrix}\right.\)

4) 2x=3x-2

<=> 2x-3x=-2

<=> -x=-2

<=> x=2

5) x+15=3x-1

<=> x-3x=1-15

<=> -2x=-14

<=> x=-14:-2

<=> x=7

6) 2-x=0,5x-4

<=> -x-0,5x=-4-2

<=> -1,5x=-6

<=> x= -6: -1,5

<=> x=4

học tốt nghen

1, \(4x^2-1=\left(2x+1\right)\left(3x-5\right)\)

\(\Leftrightarrow4x^2-1=6x^2-10x+3x-5\)

\(\Leftrightarrow4x^2-1=6x^2-7x-5\)

\(\Leftrightarrow6x^2-7x-5-4x^2+1=0\)

\(\Leftrightarrow2x^2-7x-4=0\)

\(\Leftrightarrow\left(2x^2-8x\right)-\left(x-4\right)=0\)

\(\Leftrightarrow2x\left(x-4\right)-\left(x-4\right)=0\)

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

Vậy phương trình có tập nghiệm là \(S=\left\{4;\frac{1}{2}\right\}\)

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

\(\Leftrightarrow x^2+2x+1=4x^2-8x+4\\ \Leftrightarrow4x^2-8x+4-x^2-2x-1=0\\ \Leftrightarrow3x^2-10x+3=0\\ \Leftrightarrow\left(3x^2-x\right)-\left(9x-3\right)=0\\ \Leftrightarrow x\left(3x-1\right)-3\left(3x-1\right)=0\\ \Leftrightarrow\left(3x-1\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=3\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là \(S=\left\{3;\frac{1}{3}\right\}\)

3,

\(2x^3+5x^2-3x=0\\ \Leftrightarrow x\left(2x^2+5x-3\right)=0\\ \Leftrightarrow x\left[\left(2x^2+x\right)-\left(6x+3\right)\right]=0\\ \Leftrightarrow x\left[x\left(2x+1\right)-3\left(2x+1\right)\right]=0\\ \Leftrightarrow x\left(2x+1\right)\left(x-3\right)=0\\ \left[{}\begin{matrix}x=0\\2x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\frac{1}{2}\\x=3\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là \(S=\left\{0;-\frac{1}{2};3\right\}\)

a)   \(\left(x-3\right)\left(6-x\right)>0\)

\(\Rightarrow\)\(\hept{\begin{cases}x-3>0\\6-x>0\end{cases}\Leftrightarrow\hept{\begin{cases}x>3\\x< 6\end{cases}\Leftrightarrow}3< x< 6}\)

hoặc   \(\hept{\begin{cases}x-3< 0\\6-x< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 3\\x>6\end{cases}}}\)(vô lí)

Vậy    \(3< x< 6\)