Bạn cần bổ sung yêu cầu đề bài để được hỗ trợ tốt hơn.

a: \(x^3-2y^2=2^3-2\cdot\left(-2\right)^2=8-2\cdot4=0\)

=>\(C=x\left(x^2-y\right)\left(x^3-2y^2\right)\left(x^4-3y^3\right)\left(x^5-4y^4\right)=0\)

b: x+y+1=0

=>x+y=-1

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

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

\(=-x^2+y^2+x^2-y^2-2+3\)

=1

c: 

Trường hợp 1: x<-3

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

\(\Leftrightarrow-5x=4\)

hay \(x=-\dfrac{4}{5}\left(loại\right)\)

Trường hợp 2: -3<=x<-1

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

hay \(x=\dfrac{2}{3}\left(loại\right)\)

Trường hợp 3: x>=-1

\(\Leftrightarrow2x+4=3x\)

hay x=4(nhận)

ta có :

`x^2 = 4`

`=> x = 2 ;-2`

TH1 :

thay `x=2 ; y = 5` ta có :

`2(3.5 -1) = 2.14 = 28`

TH2 :

thay `x= -2 , y = 5` ta có:

`(-2)(3.5-1) = (-2).14 = -28`

`b)`

ta có : `y^2 =1 `

`=> y = 1 ; -1;`

TH1:

thay `x=5 ; y=1` vào ta có:

`(5-3)(1-4)`

`=2.(-3)`

`=-6`

TH2:

thay `x = 5 ; y = -1` vào ta có :

`(5-3)(-1-4) `

`= 2 . (-5)`

`= -10`

a. \(x^2=4\\ \Leftrightarrow x=\sqrt{4}=2\)

Thay \(x=2;y=5\) vào ta được:

\(2\left(3\cdot5-1\right)\)

\(30-2=28\)

b. \(y^2=1\\ \Leftrightarrow y=\sqrt{1}=1\)

Thay \(x=5;y=1\) vào ta được:

\(\left(5-3\right)\left(1-4\right)\)

\(1\cdot\left(-3\right)=-3\)

\(B=\frac{\left(1,09-0,29\right).1\frac{1}{4}}{\left(18,9-16\frac{13}{20}\right).\frac{8}{9}}=\frac{0,8.\frac{5}{4}}{\left[\left(18+0,9\right)-\left(16+\frac{13}{20}\right)\right].\frac{8}{9}}\)

\(B=\frac{\frac{4}{5}.\frac{5}{4}}{\left(18+\frac{9}{10}-16-\frac{13}{20}\right).\frac{8}{9}}\)

\(B=\frac{1}{\left[\left(18-16\right)+\left(\frac{9}{10}-\frac{13}{20}\right)\right].\frac{8}{9}}\)

\(B=\frac{1}{\left(2+\frac{1}{4}\right).\frac{8}{9}}\)

\(B=\frac{1}{\frac{9}{4}.\frac{8}{9}}=\frac{1}{2}\)

thanks

a,thay x=1,y=-1

=>A=(15.1+2.-1)-[(2.1+3)-(5.1+-1)]=13-[5-4]=12

b,thay=-1/2,y=1/7

=>B=4

thks

\(x+y+1=0\\ \Leftrightarrow x+y=-1\)

Thay x+y=-1 vào C ta có:

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

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

\(\Rightarrow C=-x^2+y^2+x^2-y^2-2+3\)

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

\(\Rightarrow C=0+0+1\)

\(\Rightarrow C=1\)

\(x+y+1=0\) =>\(x+y=-1\)

- Thay \(x+y=-1\) vào C ta được:

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

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

a: Ta có: \(\left(x-2\right)\left(x+1\right)< 0\)

\(\Leftrightarrow-1< x< 2\)

b: Ta có: \(\left(x+\dfrac{1}{3}\right)\left(x-1\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}x\le-\dfrac{1}{3}\\x\ge1\end{matrix}\right.\)