\(2xy+6x^2-3x-y=11\)

=>\(2x\left(y+3x\right)-\left(y+3x\right)=11\)

=>(2x-1)(3x+y)=11

=>\(\left(2x-1\right)\left(3x+y\right)=1\cdot11=11\cdot1=\left(-1\right)\cdot\left(-11\right)=\left(-11\right)\cdot\left(-1\right)\)

=>\(\left(2x-1;3x+y\right)\in\left\{\left(1;11\right);\left(11;1\right);\left(-1;-11\right);\left(-11;-1\right)\right\}\)

=>\(\left(x;3x+y\right)\in\left\{\left(1;11\right);\left(6;1\right);\left(0;-11\right);\left(-5;-1\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(1;8\right);\left(6;-17\right);\left(0;-11\right);\left(-5;14\right)\right\}\)

Số không không phải là số âm, cũng không phải là số dương.

a) \(P\left(x\right)+Q\left(x\right)=x^4-2x^2+1\)

\(\Rightarrow Q\left(x\right)=x^4-2x^2+1-P\left(x\right)\)

\(=x^4-2x^2+1-\left(x^3-2x^2+x-\dfrac{1}{2}\right)\)

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

\(=x^4-x^3+\left(-2x^2+2x^2\right)-x+\left(1+\dfrac{1}{2}\right)\)

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

b) \(P\left(x\right)-H\left(x\right)=x^3+x^2+2\)

\(\Rightarrow H\left(x\right)=P\left(x\right)-\left(x^3+x^2+2\right)\)

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

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

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

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

là phiên âm đó bạn 

do là j

a; A = (1.2 + 2.3 + 3.4+ ... + 19.20)(13 - 3).(13 - 4)...(13 - 20)

 A =(1.2 + 2.3 + 3.4 + ... + 19.20).(13 - 3).(13 - 4)...(13 - 13)...(13 - 20)

A = (1.2 + 2.3 + 3.4 + ... + 19.20).(13 - 3).(13 - 4)...0...(13 - 20)

A = 0

a; A = (1.2 + 2.3 + 3.4+ ... + 19.20)(13 - 3).(13 - 4)...(13 - 20)

 A =(1.2 + 2.3 + 3.4 + ... + 19.20).(13 - 3).(13 - 4)...(13 - 13)...(13 - 20)

A = (1.2 + 2.3 + 3.4 + ... + 19.20).(13 - 3).(13 - 4)...0...(13 - 20)

A = 0

Chúc bạn học tốt !!!!

a-b=7 nên a=b+7

\(\dfrac{3a-b}{2a+7}+\dfrac{3b-a}{2b-7}\)

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

\(=\dfrac{2b+21}{2b+21}+\dfrac{2b-7}{2b-7}=1+1=2\)