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31 tháng 10 2018

Vậy (6x3 – x2 – 14x + 3) : (2x – 3) = 3x2 + 4x – 1

2 tháng 8 2023

\(\dfrac{6}{x^2+4x}+\dfrac{3}{2x+8}\\ =\dfrac{6}{x\left(x+4\right)}+\dfrac{3}{2\left(x+4\right)}\\ =\dfrac{6.2}{2x\left(x+4\right)}+\dfrac{3x}{2x\left(x+4\right)}\\ =\dfrac{12+3x}{2x\left(x+4\right)}\\ =\dfrac{3\left(4+x\right)}{2x\left(x+4\right)}\\ =\dfrac{3}{2x}\)

________

\(\dfrac{x+1}{x-2}+\dfrac{x-2}{x+2}+\dfrac{x-14}{x^2-4}\\ \left(\text{đ}k\text{x}\text{đ}:x\ne\pm2\right)\\ =\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}+\dfrac{x-14}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{x^2+2x+x+2+x^2-4x+4+x-14}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{2x^2-8}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{2\left(x^2-4\right)}{x^2-4}\\ =2\)

 

a: \(=\dfrac{6}{x\left(x+4\right)}+\dfrac{3}{2\left(x+4\right)}\)

\(=\dfrac{12+3x}{2x\left(x+4\right)}=\dfrac{3\left(x+4\right)}{2x\left(x+4\right)}=\dfrac{3}{2x}\)

b: \(=\dfrac{\left(x+1\right)\left(x+2\right)+\left(x-2\right)^2+x-14}{x^2-4}\)

\(=\dfrac{x^2+3x+2+x^2-4x+4+x-14}{x^2-4}=\dfrac{2x^2-8}{x^2-4}=2\)

Bài 1: 

b: \(=\dfrac{x+3-4-x}{x-2}=\dfrac{-1}{x-2}\)

Bài 2: 

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

\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)

d: \(=\dfrac{3}{2x^2y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)

\(=\dfrac{3y^2+10xy+2x^3}{2x^2y^3}\)

e: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2x^2-4xy}{\left(x+2y\right)\cdot\left(x-2y\right)}=\dfrac{2x}{x+2y}\)

24 tháng 2 2023

`!`

`(x^3-3x^2):(x-3)`         

`= x^2(x-3)  :(x-3)` 

`=x^2`

`----`

`(2x^2+2x-4):(x+2)`

`=2(x^2+x-2) :(x+2)`

`= 2 (x^2+2x-x-2) :(x+2)`

\(=2\left[x\left(x+2\right)-\left(x+2\right)\right]:\left(x+2\right)\)

`= 2(x+2)(x-1) :(x+2)`

`=2(x-1)`

`-------`

`(x^4-x-14):(x-2)`

https://hoc24.vn/cau-hoi/x4-x-14-x-2-giup-minh.204306769717

bn tham khảo ở đây nha

`----`

`(x^3-3x^2+x-3):(x-3)`

`= x^2(x-3) +(x-3) :(x-3)`

`=(x-3)(x^2+1):(x-3)`

`=x^2+1`

24 tháng 2 2023

thank

a)

(x*211)+(1+2+...+211)=23632

(x*211)+22366=23632

x*211=23632-22366

x*211=1266

x=1266:211

x=6

Vậy x=6

Bài 3:

3: \(6x\left(x-y\right)-9y^2+9xy\)

\(=6x\left(x-y\right)+9xy-9y^2\)

\(=6x\left(x-y\right)+9y\left(x-y\right)\)

\(=\left(x-y\right)\left(6x+9y\right)\)

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

Bài 4:

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27 tháng 8 2023

Bài 1 :

\(M=\dfrac{30-2^{20}}{2^{18}}=\dfrac{2.15-2^{20}}{2^{18}}=\dfrac{15}{2^{17}}-2^2=\dfrac{15}{2^{17}}-4< 0\left(\dfrac{15}{2^{17}}< 1\right)\)

\(N=\dfrac{3^5}{1^{2021}+2^3}=\dfrac{3^5}{9}=\dfrac{3^5}{3^2}=3^3=27\)

\(\Rightarrow M< N\)

27 tháng 8 2023

Bài 3 :

a) \(t^2+5t-8\) khi \(t=2\)

\(=5^2+2.5-8\)

\(=25+10-8\)

\(=27\)

b) \(\left(a+b\right)^2-\left(b-a\right)^3+2021\left(1\right)\)

\(\left\{{}\begin{matrix}a=5\\b=a+1=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=11\\b-a=1\end{matrix}\right.\)

\(\left(1\right)=11^2-1^3+2021=121-1+2021=2141\)

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

\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)

\(\left(1\right)=1^3=1\)

a: \(\dfrac{x^2}{3x+6}+\dfrac{4x+4}{3x+6}=\dfrac{x^2+4x+4}{3x+6}=\dfrac{x+2}{3}\)

b: \(\dfrac{x+3}{x}+\dfrac{x}{3-x}-\dfrac{9}{3x-x^2}\)

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

=0

6 tháng 8 2018

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

\(=4x\left(x^2-9\right)-x^3+27\)

\(=4x^3-36x-x^3+27\)

\(=3x^3-36x+27\)

6 tháng 8 2018

\(\left(x+6\right)^2-2x.\left(x+6\right)+\left(x-6\right).\left(x+6\right)\)

\(=\left(x+6\right).\left(x+6-2x+x-6\right)\)

\(=\left(x+6\right).0\)

\(=0\)