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20 tháng 9 2022

`7x.(4x-2) - (x-3).(x-1)+16`

`=28x^2 -14x - x^2 +4x-3 +16`

`=27x^2 -10x +13`

20 tháng 9 2022

`7x(4x-2)-(x-3)(x-1)+16`

`=28x^2-14x-x^2+x+3x-3+16`

`=27x^2-10x+13`

8 tháng 5 2022

Giúp tui vs mn

 

4 tháng 2 2022

lớp 8 có pt bậc 2 ak??

4 tháng 2 2022

Có nhưng giải bằng PT tích nhé

a:

ĐKXĐ: x<>-1/2

Để \(\dfrac{2x^3+x^2+2x+2}{2x+1}\in Z\) thì

\(2x^3+x^2+2x+1+1⋮2x+1\)

=>\(2x+1\inƯ\left(1\right)\)

=>2x+1 thuộc {1;-1}

=>x thuộc {0;-1}

b:

ĐKXĐ: x<>1/3

 \(\dfrac{3x^3-7x^2+11x-1}{3x-1}\in Z\)

=>3x^3-x^2-6x^2+2x+9x-3+2 chia hết cho 3x-1

=>2 chia hết cho 3x-1

=>3x-1 thuộc {1;-1;2;-2}

=>x thuộc {2/3;0;1;-1/3}

mà x nguyên

nên x thuộc {0;1}

c: 

ĐKXĐ: x<>2

\(\dfrac{x^4-16}{x^4-4x^3+8x^2-16x+16}\in Z\)

=>\(\left(x^2-4\right)\left(x^2+4\right)⋮\left(x-2\right)^2\left(x^2+4\right)\)

=>\(x+2⋮x-2\)

=>x-2+4 chia hết cho x-2

=>4 chia hết cho x-2

=>x-2 thuộc {1;-1;2;-2;4;-4}

=>x thuộc {3;1;4;0;6;-2}

 

a: Để A là số nguyên thì

x^3-2x^2+4 chia hết cho x-2

=>\(x-2\in\left\{1;-1;2;-2;4;-4\right\}\)

=>\(x\in\left\{3;1;4;0;6;-2\right\}\)

b: Để B là số nguyên thì

\(3x^3-x^2-6x^2+2x+9x-3+2⋮3x-1\)

=>\(3x-1\in\left\{1;-1;2;-2\right\}\)

=>\(x\in\left\{\dfrac{2}{3};0;1;-\dfrac{1}{3}\right\}\)

 

28 tháng 8 2019

a) = (x+3).(x-3)^2-(x-3)(x+3)^2

=(x^2-9)(x-3)-(x^2-9)(x+3)

=(x^2-9)(x-3-x-3)

=-6(x^2-9)

các câu còn lại tương tự

28 tháng 8 2019

\(a,\left(x+3\right)\left(x^2-3x+9\right)-\left(x-3\right)\left(x^2+3x+9\right)\)

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

\(=x^3+3-x^3+3\)

\(=6\)

\(b,\left(x-5\right)\left(x^2+5x+25\right)-\left(x+5\right)\left(x^2-5x+25\right)\)

\(=x^3-5^3-x^3-5^3\)

\(=-125-125\)

\(=-250\)

2 tháng 1 2023

\(a,đk:x\ne0;4;1\)

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

\(đk:x\ne-2;1\)

\(\dfrac{x}{x+2}+\dfrac{7x-16}{\left(x+2\right)\left(7x-7\right)}\\ =\dfrac{x\left(7x-7\right)}{\left(x+2\right)\left(7x-7\right)}+\dfrac{7x-16}{\left(x+2\right)\left(7x-7\right)}\\ =\dfrac{7x^2-7x+7x-16}{\left(x+2\right)\left(7x-7\right)}\\ =\dfrac{7x^2-16}{\left(x+2\right)\left(7x-7\right)}\)

 

2 tháng 1 2023

a)

\(\dfrac{x-1}{x^2-5x+4}-\dfrac{4}{x^2-4x}\) \(ĐKXĐ:x\ne0;x\ne4;x\ne1\)

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

\(=\dfrac{x-1}{x\left(x-4\right)-\left(x-4\right)}-\dfrac{4}{x\left(x-4\right)}\)

\(=\dfrac{x-1}{\left(x-1\right)\left(x-4\right)}-\dfrac{4}{x\left(x-4\right)}\)

\(=\dfrac{x^2-x}{x\left(x-1\right)\left(x-4\right)}-\dfrac{4\left(x-1\right)}{x\left(x-1\right)\left(x-4\right)}\)

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

\(=\dfrac{x\left(x-1\right)-4\left(x-1\right)}{x\left(x-1\right)\left(x-4\right)}\)

\(=\dfrac{\left(x-1\right)\left(x-4\right)}{x\left(x-1\right)\left(x-4\right)}\\ =\dfrac{1}{x}\)

b)

\(\dfrac{x}{x+2}+\dfrac{7x-16}{\left(x+2\right)\left(7x-7\right)}\)  \(ĐKXĐ:x\ne-2;x\ne1\)

\(=\dfrac{x\left(7x-7\right)}{\left(x+2\right)\left(7x-7\right)}+\dfrac{7x-16}{\left(x+2\right)\left(7x-7\right)}\)

\(=\dfrac{7x^2-7x+7x-16}{\left(x+2\right)\left(7x-7\right)}\)

\(=\dfrac{7x^2-16}{\left(x+2\right)\left(7x-7\right)}\)

1 tháng 1 2018

a)\(\dfrac{27-x^3}{5x+5}:\dfrac{2x-6}{3x+3}\)

\(=\dfrac{\left(3-x\right)\left(9+3x+x^2\right)}{5\left(x+1\right)}:\dfrac{2\left(x-3\right)}{3\left(x+1\right)}\)

\(=\dfrac{\left(3-x\right)\left(9+3x+x^2\right)3\left(x+1\right)}{5\left(x+1\right)2\left(x-3\right)}\)

\(=\dfrac{-\left(x-3\right)\left(9+3x+x^2\right)3\left(x+1\right)}{5\left(x+1\right)2\left(x-3\right)}\)

\(=\dfrac{-\left(9+3x+x^2\right)3}{10}\)

b)\(4x^2-16:\dfrac{3x+6}{7x-2}\)

\(=4\left(x^2-4\right):\dfrac{3\left(x+2\right)}{7x-2}\)

\(=4\left(x-2\right)\left(x+2\right)\cdot\dfrac{7x-2}{3\left(x+2\right)}\)

\(=\dfrac{4\left(x-2\right)\left(x+2\right)\left(7x-2\right)}{3\left(x+2\right)}\)

\(=\dfrac{4\left(x-2\right)\left(7x-2\right)}{3}\)

c)\(\dfrac{3x^3+3}{x-1}:x^2-x+1\)

\(=\dfrac{3\left(x^3+1\right)}{x-1}:x^2-x+1\)

\(=\dfrac{3\left(x+1\right)\left(x^2-x+1\right)}{x-1}\cdot\dfrac{1}{x^2-x+1}\)

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

d)\(\dfrac{4x+6y}{x-1}:\dfrac{4x^2+12xy+9y^2}{1-x^3}\)

\(=\dfrac{2\left(2x+3y\right)}{x-1}\cdot\dfrac{\left(1-x\right)\left(1+x+x^2\right)}{\left(2x+3y\right)^2}\)

\(=\dfrac{2\left(2x+3y\right)}{x-1}\cdot\dfrac{-\left(x-1\right)\left(1+x+x^2\right)}{\left(2x+3y\right)^2}\)

\(=\dfrac{-2\left(1+x+x^2\right)}{2x+3y}\)

ngoamthanghoa

1 tháng 1 2018

a) \(\dfrac{27-x^3}{5x+5}:\dfrac{2x-6}{3x+3}\)

\(=\dfrac{27-x^3}{5x+5}.\dfrac{3x+3}{2x-6}\)

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

\(=-\dfrac{3\left(x-3\right)\left(x^2+3x+9\right)\left(x+1\right)}{10\left(x+1\right)\left(x-3\right)}\)

\(=-\dfrac{3\left(x^2+3x+9\right)}{10}\)

b) \(4x^2-16:\dfrac{3x+6}{7x-2}\)

\(=4x^2-16.\dfrac{7x-2}{3x+6}\)

\(=\dfrac{4\left(x^2-4\right)\left(7x-2\right)}{3\left(x+2\right)}\)

\(=\dfrac{4\left(x-2\right)\left(x+2\right)\left(7x-2\right)}{3\left(x+2\right)}\)

\(=\dfrac{4\left(x-2\right)\left(7x-2\right)}{3}\)

c) \(\dfrac{3x^3+3}{x-1}:x^2-x+1\)

\(=\dfrac{3x^3+3}{x-1}.\dfrac{1}{x^2-x+1}\)

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

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

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

d) \(\dfrac{4x+6y}{x-1}:\dfrac{4x^2+12xy+9y^2}{1-x^3}\)

\(=\dfrac{4x+6y}{x-1}.\dfrac{1-x^3}{4x^2+12xy+9y^2}\)

\(=\dfrac{2\left(2x+3y\right)\left(1-x\right)\left(1+x+x^2\right)}{\left(x-1\right)\left(2x+3y\right)^2}\)

\(=-\dfrac{2\left(2x+3y\right)\left(x-1\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(2x+3y\right)^2}\)

\(=-\dfrac{2\left(x^2+x+1\right)}{2x+3y}\)

a: \(\Leftrightarrow\left(3x+2\right)\left(5-x\right)=-9x^2+4\)

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

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

=>x=-2/3 hoặc x=-3/2

b: \(\Leftrightarrow4x\left(x+5\right)+x^2-25=0\)

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

=>x=-5 hoặc x=1

c: \(\Leftrightarrow3x\left(x-1\right)=\left(x-1\right)^2\)

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

=>x=1 hoặc x=-1/2