`a)2x^2+3(x-1)(x+1)=5x(x+1)`

`<=>2x^2+3x^2-3=5x^2+5x`

`<=>5x=-3`

`<=>x=-3/5`

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`b)(x-3)^3+3-x=0` nhỉ?

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

`<=>(x-3)(x^2-1)=0`

`<=>[(x=3),(x^2=1<=>x=+-1):}`

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`c)5x(x-2000)-x+2000=0`

`<=>5x(x-2000)-(x-2000)=0`

`<=>(x-2000)(5x-1)=0`

`<=>[(x=2000),(x=1/5):}`

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`d)3(2x-3)+2(2-x)=-3`

`<=>6x-9+4-2x=-3`

`<=>4x=2`

`<=>x=1/2`

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`e)x+6x^2=0`

`<=>x(1+6x)=0`

`<=>[(x=0),(x=-1/6):}`

\(a,\left(x-3\right)\left(x-1\right)=\left(x-3\right)^2\\ \Leftrightarrow\left(x-3\right)\left(x-1-x+3\right)=0\\ \Leftrightarrow2\left(x-3\right)=0\\ \Leftrightarrow x=3\)

\(b,4x^2-9=0\\ \Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

\(c,x^2+6x+9=0\\ \Leftrightarrow\left(x+3\right)^2=0\\ \Leftrightarrow x+3=0\\ \Leftrightarrow x=-3\)

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

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

\(\Leftrightarrow2\left(x-3\right)=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

b: Ta có: \(4x^2-9=0\)

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

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

c: Ta có: \(x^2+6x+9=0\)

\(\Leftrightarrow x+3=0\)

hay x=-3

Bài 1:

a: \(3x-6y=3\cdot x-3\cdot2y=3\left(x-2y\right)\)

b: \(14x^2y-21xy^2+28x^2y^2\)

\(=7xy\cdot2x-7xy\cdot3y+7xy\cdot4xy\)

\(=7xy\left(2x-3y+4xy\right)\)

c: \(10x\left(x-y\right)-8y\cdot\left(y-x\right)\)

\(=10x\left(x-y\right)+8y\left(x-y\right)\)

\(=\left(x-y\right)\left(10x+8y\right)\)

\(=\left(2\cdot5x+2\cdot4y\right)\left(x-y\right)\)

\(=2\left(5x+4y\right)\left(x-y\right)\)

bài 2:

a: Đề thiếu vế phải rồi bạn

b: \(x^3-13x=0\)

=>\(x\left(x^2-13\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\x^2-13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=13\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=0\\x=\pm\sqrt{13}\end{matrix}\right.\)

Bài 1:

a, $3x-6y$

$=3(x-2y)$

b, $14x^2y-21xy^2+28x^2y^2$

$=7xy(2x-3y+4xy)$

c, $10x(x-y)-8y(y-x)$

$=10x(x-y)-8y[-(x-y)]$

$=10x(x-y)+8y(x-y)$

$=(x-y)(10x+8y)$

$=2(x-y)(5x+4y)$

Bài 2:

a, Đề thiếu rồi bạn nhé.

b, \(x^3-13x=0\)

\(\Rightarrow x\left(x^2-13\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-13=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x^2=13\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{13}\\x=-\sqrt{13}\end{matrix}\right.\)

\(5x\left(x-3\right)=x-3\)

\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right)\left(5x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\5x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{1}{5}\end{cases}}}\)

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

\(\Leftrightarrow2x-3-x+5=x+2-x+1\)

\(\Leftrightarrow x+2=3\)

hay x=1

Vậy: x=1

b) Ta có: \(2\left(x-1\right)-5\left(x+2\right)=-10\)

\(\Leftrightarrow2x-2-5x-10=-10\)

\(\Leftrightarrow-3x=-10+10+2=2\)

hay \(x=-\dfrac{2}{3}\)

Vậy: \(x=-\dfrac{2}{3}\)

a, (2x - 3) - (x - 5) = (x + 2) - (x - 1)

 2x - 3 - x + 5 = x + 2 - x + 1

(2x - x) + (-3 + 5) = (x - x) + (2 + 1)

x + 2 = 3

x = 1

a)\(2x\left(x-2016\right)-2x+4032=0\)

\(\Leftrightarrow2x\left(x-2016\right)-2\left(x-2016\right)=0\)

\(\Leftrightarrow\left(2x-2\right)\left(x-2016\right)=0\)

\(\Leftrightarrow2\left(x-1\right)\left(x-2016\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x-2016=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=2016\end{array}\right.\)

b)\(5x\left(x-3\right)=x-3\)

\(\Leftrightarrow5x\left(x-3\right)-\left(x-3\right)=0\)

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-3=0\\5x-1=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=3\\x=\frac{1}{5}\end{array}\right.\)

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

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

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

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}4x+1=0\\2x-3=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{4}\\x=\frac{3}{2}\end{array}\right.\)

a) 1\(\dfrac{2}{3}\).           b)\(\dfrac{1}{7}\).             c) 1               d )0

a: =>x+5>0

hay x>-5

b: =>2x+1<0

hay x<-1/2

c: =>(x-1)(x-4)>0

=>x>4 hoặc x<1

a) x>-5 ĐKXĐ x\(\ne\)-5

b)x<\(-\dfrac{1}{2}\)

c)x>4 hoặc x<1

d)ĐKXĐ x\(\ne\)1, ko tìm đc x

a) \(\left(\frac{1}{7}x-\frac{2}{3}\right)\left(-\frac{1}{5}x+\frac{3}{5}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\frac{1}{7}x-\frac{2}{3}=0\\-\frac{1}{5}x+\frac{3}{5}=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}\frac{1}{7}x=\frac{2}{3}\\-\frac{1}{5}x=-\frac{3}{5}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{14}{3}\\x=3\end{cases}}\)

b)\(\frac{1}{10}x-\frac{4}{5}x+1=0\)

\(\Leftrightarrow x.\left(\frac{1}{10}-\frac{4}{5}\right)+1=0\)

\(\Rightarrow-\frac{7}{10}x=-1\)

\(\Rightarrow x=\frac{10}{7}\)

c)\(\left(2x-\frac{1}{3}\right).\left(5x+\frac{2}{7}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{3}=0\\5x+\frac{2}{7}=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=\frac{1}{3}\\5x=-\frac{2}{7}\end{cases}}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{6}\\x=-\frac{2}{35}\end{cases}}\)

a, (1/7 . x - 2/3) . (-1/5 . x + 3/5) = 0

Suy ra : 1/7 .x -2/3 = 0 hoặc -1/5 .x + 3/5 =0

Vậy : 1/7 .x = 2/3 hoặc -1/5 .x = 3/5

         x =2/3 : 1/7 hoặc x = 3/5 : (-1/5)

        x = 14/3 hoặc x = -3

b, 1/10 .x - 4/5 .x + 1 =0

   x . (1/10 - 4/5) + 1 = 0

   x . (-7/10) + 1 = 0

   x . -7/10 =0 +1 = 1

   x = 1 : (-7/10)

   x = -10/7

c, (2x - 1/3 ) . (5x +2/7) = 0

Suy ra : 2x - 1/3 = 0 hoặc 5x + 2/7 = 0

Vậy : 2x = 1/3 hoặc 5x = 2/7

         x = 1/3 : 2 hoặc x = 2/7 : 5

         x = 1/6 hoặc x = 2/35

cái chỗ x = -3 bạn sửa lại là 3 nha , mk nhấn nhầm

a, 2\(xy\) - 2\(x\) + 3\(y\) = -9

(2\(xy\) - 2\(x\)) + 3\(y\) - 3 = -12

2\(x\)(\(y-1\)) + 3(\(y-1\)) = -12

(\(y-1\))(2\(x\) + 3) = -12

Ư(12) = {-12; -6; -4; -3; -2; -1; 1; 2; 3; 4; 6; 12}

Lập bảng ta có:

Theo bảng trên ta có: Các cặp \(x\);\(y\) nguyên thỏa mãn đề bài là:

(\(x;y\)) = (-1; -11); (0; -3); (-3; 5); ( -2; 13)

b, (\(x+1\))²(\(y\) - 3) = -4 

    Ư(4) = {-4; -2; -1; 1; 2; 4}

Lập bảng ta có: 

Theo bảng trên ta có: các cặp \(x;y\) nguyên thỏa mãn đề bài là: 

(\(x;y\)) = (0; -1); (-3; 2); (1; 2)

c) \(\left(x+3\right)^2+\left(2y-1\right)^2< 44\)

\(\Leftrightarrow\left(x+3\right)^2< 44-\left(2y-1\right)^2< 44\) (do \(-\left(2y-1\right)^2\le0\)) (1) 

mà (x + 3)² là số chính phương 

Kết hợp (1) ta được \(\left(x+3\right)^2\le36\)

\(\Leftrightarrow\left(x+3\right)^2\le6^2\Leftrightarrow\left(x+3\right)^2\in\left\{0;1;4;9;25;36\right\}\)

Với (x + 3)² \(\in\left\{0;1;4\right\}\) ta được (2y - 1)² \(\in\left\{0;1;4;9;25;36\right\}\) 

Với (x + 3)² \(\in\left\{9;16\right\}\) ta được (2y - 1)² \(\in\left\{0;1;4;9;25\right\}\) 

Với (x + 3)² = 25 ta được (2y - 1)² \(\in\left\{0;1;4;9;16\right\}\)

Với (x + 3)² = 36 ta được (2y - 1)² \(\in\left\{0;1;4;9\right\}\)