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Phân tích đa thức thành nhân tử:(em làm luôn đấy,ko ghi lại đề)
\(\left(x^3+y^3\right)-\left(x+y\right)+3xy\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)-\left(x+y\right)+3xy\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)\(=\left(x+y\right)\left[\left(x+y\right)^2-1^2\right]\)
\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
\(8x^3+12x^2+6x+1=0.\)
\(\Leftrightarrow\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1^2+1^3=0\)
\(\Leftrightarrow\left(2x+1\right)^3=0\)
\(\Leftrightarrow2x+1=0\)
\(\Leftrightarrow x=-\frac{1}{2}\)
\(2x^2+5x-3=0\Leftrightarrow\left(2x^2+6x\right)+\left(-x-3\right)=0\)
\(\Leftrightarrow2x\left(x+3\right)-\left(x+3\right)=0\Leftrightarrow\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\x+3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{2}\\x=-3\end{cases}}\)
\(x^2-2x-3=0\Leftrightarrow\left(x^2-3x\right)+\left(x-3\right)=0\)
\(\Leftrightarrow x\left(x-3\right)+\left(x-3\right)=0\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}.}\)
\(\left(5x-1\right)+2\left(1-5x\right)\left(4+5x\right)+\left(5x+4\right)^2\)
\(=5x-1+2\left(4+5x-20x-25x^2\right)+25x^2+40x+16\)
\(=25x^2+45x+15+8+10x-40x-50x^2\)
\(=-25x^2+15x+23\)
\(\left(x-y\right)^3+\left(y+x\right)^3+\left(y-x\right)^3-3xy\left(x+y\right)\)
\(=\left(x-y\right)^3-\left(x-y\right)^3+\left(x+y\right)^3-3x^2y-3xy^2\)
\(=\left(x+y\right)^3-3x^2y-3xy^2\)
\(=x^3+3x^2y+3xy^2+y^3-3xy^2-3x^2y\)
\(=x^3+y^3\)
x3-x+3x2y+3xy2+y3-y
=x2(x-1)+3(x2y+xy2)+y2(y-1)
=x2(x-1)+3(x2.y+y2.x)+y2(y-1)
=x2(x-1)+3{[x(x+1)+y(y+1)]}+y2(y-1)
=x2(x-1)+3.x(x+1)+3.y(y+1)+y2(y-1)
=x2(x-1)+2x2+3.x(x+1)+3.y(y+1)+y2(y-1)+2y2-2x2-2y2
=x2(x+1)+3.x(x+1)+3.y(y+1)+y2(y+1)-2x2-2y2
=(x2+3)(x+1)+(y2+3)(y+1)-2(x2+y2)
a, \(x^3-x+3x^2y+3xy^2+y^3-y\)
= \((x^3+3x^2y+3xy^2+y^3)-x-y\)
= \(\left(x+y\right)^3-\left(x+y\right)\)
= \(\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
= \(\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
\(1.x^3+2x+x^2=x\left(x^2+x+2\right)\)
\(2.2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)^2\)
\(3.-3x^3-5x^2+8x=-3x^3+3x^2-8x^2+8x\)
\(=-3x^2\left(x-1\right)-8x\left(x-1\right)=\left(3x^2+8x\right)\left(1-x\right)\)
\(=x\left(3x+8\right)\left(1-x\right)\)
\(4.x^2+4x-5=x^2-x+5x-5=\left(x-1\right)\left(x+5\right)\)
\(5.6x^2-3x-3=6x^2-6x+3x-3=3\left(x-1\right)\left(2x+1\right)\)
\(6.3x^2-2x-5=3x^2+3x-5x-5=\left(x+1\right)\left(3x-5\right)\)
\(8.x^2-2x-4y^2-4y=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)\(=\left(x+2y\right)\left(x-y-2\right)\)
\(9.x^3+2x^2y+xy^2-9x=x\left(x^2+2xy+y^2-9\right)\)
\(=x\left(x+y-3\right)\left(x+y+3\right)\)
\(10.x^2-y^2+6x+9=\left(x+3-y\right)\left(x+3+y\right)\)
Trả lời tội ghê đó bạn nhưng mk gửi một bài mà sao bạn trả lời một câu vậy bạn nhưng dù sao vẫn cảm on nha
1.\(x^3+6x^2+12xy+8=x^3+3.2x^2+3.2^2x+2^3=\left(x+2\right)^3\)
3.\(x^4+2x^3+x^2-y^2=\left(x^2\right)^2+2x^2.x+x^2-y^2\)\(=\left(x^2+x\right)^2-y^2=\left(x^2+x-y\right)\left(x^2+x+y\right)\)
k mình nha bn !!!!!!! cái 2 bn xem lại đề đi, rồi mình giải cho
\(1.5x\left(x^2+2x-1\right)-3x^2\left(x-2\right)=5x^3+10x^2-5x-3x^3+6x^2\)
\(=2x^3+16x^2-5x\)
\(=\left(2x^3-x\right)+\left(16x^2-4x\right)\)
\(=x\left(2x^2-1\right)+4x\left(4x-1\right)\left(ĐCCM\right)\)
1. \(x\left(x^2+2x^2y+y^2-4\right)\)
2. = \(\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left(\left(x+y\right)^2-1\right)=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
3. tớ chưa ra đến :">>
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tks bạn nhaz. còn câu 3 :(((