a) \(7xy^3-x^3y\)

\(=xy\left(7y^2-x^2\right)\)

\(=xy\left(y\sqrt{7}-x\right)\left(y\sqrt{7}+x\right)\)

b) \(121x^2y^2-1\)

\(=\left(11xy\right)^2-1^2\)

\(=\left(11xy-1\right)\left(11xy+1\right)\)

\(25n^2+\left(-20n\right)+4+\left(-4n^2\right)+20n-25⋮21\)

\(\left(25n^2-4n^2\right)+\left(-20n+20n\right)+\left(4-25\right)⋮21\)

\(=21n^2-21⋮21\)

 \(\left(5n-2\right)^2-\left(2n-5\right)^2=\left(5n-2+2n-5\right)\left(5n-2-2n+5\right)\)

\(=\left(7n-7\right)\left(3n+3\right)\)     

\(=21\left(n-1\right)\left(n+1\right)⋮21\forall n\in Z\)     

=> đpcm

ko ghi đề

\(3x^3=12\)

\(x^3=4\)

x từ đây sẽ ra số lẻ

\(\text{x=1.587401}\)

câu b)

\(x\in\left\{\frac{5}{3};-1\right\}\)

mik gợi ý đáp án câu b còn lại bạn tự làm cho chắc kiến thức

 TL:

\(x^2y-xy+x-1\)

\(=xy\left(x-1\right)+\left(x-1\right)\)

\(=\left(xy+1\right)\left(x-1\right)\)

câu b ko ghi lại đề bài

\(\left(x-1\right)\left(x+2\right)\)

\(\Leftrightarrow1.\left[\left(x-1\right)+\left(x+2\right)\right]\)

\(=1.\left(2.x+1\right)\)

\(=2x+1\)

\(a,A=4x^2-20x+27=\left(2x\right)^2-2.2x.5+5^2+2\)\(=\left(2x-5\right)^2+2\)

Mà \(\left(2x-5\right)^2\ge0\Rightarrow\left(2x-5\right)^2+2>0\Rightarrow A>0\)

\(b,B=x^2+x+1=x^2+2.x.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+1\)\(=\left(x-\frac{1}{4}\right)^2+\frac{3}{4}\)

Mà \(\left(x-\frac{1}{4}\right)^2\ge0\Rightarrow\left(x-\frac{1}{4}\right)^2+\frac{3}{4}>0\Rightarrow B>0\)

\(c,C=x^2+4x+y^2-6y+15=x^2+4x+4+y^2-6y+9+2\)

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

Mà \(\left(x+2\right)^2+\left(y-3\right)^2\ge0\Rightarrow\left(x+2\right)^2+\left(y-3\right)^2+2>0\Rightarrow C>0\)

 TL:

\(a,1-2m+m^2-x^2-4x-4\)

\(=\left(m-1\right)^2-\left(x-2\right)^2\)

\(=\left(m+x-3\right)\left(m-x+1\right)\)

6x^3+3x^2+4x+2 3x^2+2 2x+1 6x^3 +4x - 3x^2+2 3x^2+2 - 0

x^5+4x^3+3x^2-5x+15 x^3-x+3 x^2 x^5-x^3+3x^2 - 5x^3-5x+15 +5 5x^3-5x+15 - 0

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

\(\Rightarrow ab-a-b+1=0\Leftrightarrow a\left(b-1\right)-\left(b-1\right)=0\Leftrightarrow\left(a-1\right)\left(b-1\right)=0\Leftrightarrow\orbr{\begin{cases}a=1\\b=1\end{cases}}\) 

\(+,a=1\Rightarrow b=0\Rightarrow P=1\) 

\(+,b=1\Rightarrow a=0\Rightarrow P=1\)