\(x,y\inℤ\)phải không?

Ta có:

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

\(\Rightarrow x^2y^2+4x^2+2y^2-4-x^2y^2-5x^2-y^2+3=0\) (bỏ ngoặc đổi dấu)

\(\Rightarrow\left(x^2y^2-x^2y^2\right)+\left(4x^2-5x^2\right)+\left(2y^2-y^2\right)+\left(-4+3\right)=0\)

\(\Rightarrow0-x^2+y^2-1=0\)

\(\Rightarrow y^2-x^2=1\)

\(\Rightarrow\left(y-x\right)\left(y+x\right)=1\)

Vậy ta có

\(\left(y-x\right)=1;\left(y+x\right)=1\)\(\Rightarrow y=1;x=0\)

Hoặc \(\left(y-x\right)=-1;\left(y+x\right)=-1\)\(\Rightarrow y=-1;x=0\)

Vậy ...

(Không biết đúng không nữa, nếu thấy đúng thì t***k mik nhé!)

A=2018x-2017x-2016x-x

A=(2018-2017-2016-...-1)x

A=[(2018-2017)-(2017-2016)-....-(2-1)].x

A=(1-1-....-1)x

A=[1-(1+1+...+1)]x

A=(1-1008)x

A=-1007x

Thay x=2017 vào A ta có

A=-1007.2017= -2031119

Vậy A=-2031119

tổng số lần bắn súng là bao nhiêu vậy bạn???

tổng số điểm bắn là

10*25+9*10+8n+7*4+6*5=398+8n

và 8,92(25+10+n+4+5)=392,48+8,92n

như vậy ta có pt:

398+8n=392,48=8,92n

5,52n=0,92

n=6

89 độ nha !

https://olm.vn/hoi-dap/detail/212899860100.html , tham gia có thưởng

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xin lỗi mk ko bt giải vì chưa ok 

ai như vậy thì k mk nha

a) \(P+\left(4x^2-5xy-y^2\right)=5x^2+10xy-2y^2\)

\(P=5x^2+10xy-2y^2-4x^2+5xy+y^2\)

\(P=x^2+15xy-y^2\)

Vậy....

b) \(\left(2xy+y^2\right)-P=3x^2-6xy+y^2\)

\(P=2xy+y^2-3x^2+6xy-y^2\)

\(P=-3x^2+8xy\)

Vậy....

a)    P + ( 4x² - 5xy - y² ) = 5x² + 10xy - 2y²

<=> P = 5x² + 10xy - 2y² - ( 4x² - 5xy - y² )

           = 5x² + 10xy - 2y² - 4x² + 5xy + y²

           = x² + 15xy - y²

b)    ( 2xy + y² ) - P = 3x² -6xy + y²

<=> P = ( 2xy + y²) - ( 3x² - 6xy + y² )

           = 2xy + y² - 3x² + 6xy -y²

           = 8xy - 3x²