Question
Mathematics
Perform the nnamerome
1. $\left(12{y}^{2}+17y-4\right)÷\left(9{y}^{2}-13y+3\right)=$(12y^(2)+17 y-4)-:(9y^(2)-13 y+3)=
$2\left(2{x}^{3}+7x2+x\right)+\left(2{x}^{2}-4x-12\right)=$2(2x^(3)+7x2+x)+(2x^(2)-4x-12)=
2. $\left(-3{m}^{2}+m\right)+\left(4{m}^{2}+6m\right)=$(-3m^(2)+m)+(4m^(2)+6m)=
3. $\left(7{z}^{3}+4z-1\right)+\left(2{z}^{2}-6z+2\right)=$(7z^(3)+4z-1)+(2z^(2)-6z+2)=
4. $\left(3{a}^{2}+2a-2\right)-\left({a}^{2}-3a+7\right)=$(3a^(2)+2a-2)-(a^(2)-3a+7)=
Solve problem with AI
1. $\left(12{y}^{2}+17y-4\right)÷\left(9{y}^{2}-13y+3\right)=\frac{4}{3}$(12y^(2)+17 y-4)-:(9y^(2)-13 y+3)=(4)/(3)
Solution:
We can solve this problem using long division method or synthetic division method. Using synthetic division method, we get:

Therefore, $\left(12{y}^{2}+17y-4\right)÷\left(9{y}^{2}-13y+3\right)=\frac{4}{3}$(12y^(2)+17 y-4)-:(9y^(2)-13 y+3)=(4)/(3)
2. $2\left(2{x}^{3}+7x2+x\right)+\left(2{x}^{2}-4x-12\right)=4{x}^{3}+18{x}^{2}-5x-12$2(2x^(3)+7x2+x)+(2x^(2)-4x-12)=4x^(3)+18x^(2)-5x-12
Solution:
We can simplify the given expression as follows:
$2\left(2{x}^{3}+7x2+x\right)+\left(2{x}^{2}-4x-12\right)=4{x}^{3}+14{x}^{2}+2x+2{x}^{2}-4x-12$2(2x^(3)+7x2+x)+(2x^(2)-4x-12)=4x^(3)+14x^(2)+2x+2x^(2)-4x-12
$=4{x}^{3}+18{x}^{2}-5x-12$=4x^(3)+18x^(2)-5x-12
Therefore, $2\left(2{x}^{3}+7x2+x\right)+\left(2{x}^{2}-4x-12\right)=4{x}^{3}+18{x}^{2}-5x-12$2(2x^(3)+7x2+x)+(2x^(2)-4x-12)=4x^(3)+18x^(2)-5x-12
3. $\left(-3{m}^{2}+m\right)+\left(4{m}^{2}+6m\right)={m}^{2}+7m$(-3m^(2)+m)+(4m^(2)+6m)=m^(2)+7m
Solution:
We can simplify the given expression as follows:
$\left(-3{m}^{2}+m\right)+\left(4{m}^{2}+6m\right)=-3{m}^{2}+4{m}^{2}+m+6m$(-3m^(2)+m)+(4m^(2)+6m)=-3m^(2)+4m^(2)+m+6m
$={m}^{2}+7m$=m^(2)+7m
Therefore, $\left(-3{m}^{2}+m\right)+\left(4{m}^{2}+6m\right)={m}^{2}+7m$(-3m^(2)+m)+(4m^(2)+6m)=m^(2)+7m
4. $\left(7{z}^{3}+4z-1\right)+\left(2{z}^{2}-6z+2\right)=7{z}^{3}+2{z}^{2}-2z+1$(7z^(3)+4z-1)+(2z^(2)-6z+2)=7z^(3)+2z^(2)-2z+1
Solution:
We can simplify the given expression as follows:
$\left(7{z}^{3}+4z-1\right)+\left(2{z}^{2}-6z+2\right)=7{z}^{3}+2{z}^{2}+\left(-6z+4z\right)-1+2$(7z^(3)+4z-1)+(2z^(2)-6z+2)=7z^(3)+2z^(2)+(-6z+4z)-1+2
$=7{z}^{3}+2{z}^{2}-2z+1$=7z^(3)+2z^(2)-2z+1
Therefore, $\left(7{z}^{3}+4z-1\right)+\left(2{z}^{2}-6z+2\right)=7{z}^{3}+2{z}^{2}-2z+1$(7z^(3)+4z-1)+(2z^(2)-6z+2)=7z^(3)+2z^(2)-2z+1
5. $\left(3{a}^{2}+2a-2\right)-\left({a}^{2}-3a+7\right)=2{a}^{2}-a-9$(3a^(2)+2a-2)-(a^(2)-3a+7)=2a^(2)-a-9
Solution:
We can simplify the given expression as follows:
$\left(3{a}^{2}+2a-2\right)-\left({a}^{2}-3a+7\right)=3{a}^{2}+2a-2-{a}^{2}+3a-7$(3a^(2)+2a-2)-(a^(2)-3a+7)=3a^(2)+2a-2-a^(2)+3a-7
$=2{a}^{2}-a-9$=2a^(2)-a-9
Therefore, $\left(3{a}^{2}+2a-2\right)-\left({a}^{2}-3a+7\right)=2{a}^{2}-a-9$(3a^(2)+2a-2)-(a^(2)-3a+7)=2a^(2)-a-9
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