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8 tháng 9 2016

Ta có : 

\(a^2+b^2+2ab=\left(a+b\right)^2\)( Hẳng đẳng thức )

\(\Leftrightarrow\left(x+3\right)^2+\left(3-5x\right)^2+2\left(x+3\right)\left(3-5x\right)\)

\(=\left[\left(x+3\right)+\left(3-5x\right)\right]^2\)

\(=\left[x+3+3-5x\right]^2\)

\(=\left(6-4x\right)^2\)

\(=4\left(3-2x\right)^2\)

11 tháng 6 2019

Phân tích đa thức thành nhân tử:(em làm luôn đấy,ko ghi lại đề)

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

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

\(=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)\(=\left(x+y\right)\left[\left(x+y\right)^2-1^2\right]\)

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

11 tháng 6 2019

\(8x^3+12x^2+6x+1=0.\)

\(\Leftrightarrow\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1^2+1^3=0\)

\(\Leftrightarrow\left(2x+1\right)^3=0\)

\(\Leftrightarrow2x+1=0\)

\(\Leftrightarrow x=-\frac{1}{2}\)

\(2x^2+5x-3=0\Leftrightarrow\left(2x^2+6x\right)+\left(-x-3\right)=0\)

\(\Leftrightarrow2x\left(x+3\right)-\left(x+3\right)=0\Leftrightarrow\left(x+3\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\x+3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{2}\\x=-3\end{cases}}\)

\(x^2-2x-3=0\Leftrightarrow\left(x^2-3x\right)+\left(x-3\right)=0\)

\(\Leftrightarrow x\left(x-3\right)+\left(x-3\right)=0\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}.}\)

\(\left(5x-1\right)+2\left(1-5x\right)\left(4+5x\right)+\left(5x+4\right)^2\)

\(=5x-1+2\left(4+5x-20x-25x^2\right)+25x^2+40x+16\)

\(=25x^2+45x+15+8+10x-40x-50x^2\)

\(=-25x^2+15x+23\)

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

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

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

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

\(=x^3+y^3\)

1 tháng 8 2018

\(a.x^2-y^2-5x+5y=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\) \(b.5x^3-5x^2y-10x^2+10xy=5x^2\left(x-y\right)-10x\left(x-y\right)=5x\left(x-y\right)\left(x-2\right)\) \(c.x^3-2x^2-4xy^2+x=x\left(x^2-2x+1-4y^2\right)=x\left[\left(x-1\right)^2-4y^2\right]=\left(x-1-2y\right)\left(x-1+2y\right)\) \(d.\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\)

Đặt : \(x^2+7x+11=t\) , ta có :

\(\left(t+1\right)\left(t-1\right)-8=t^2-1-8=\left(t-3\right)\left(t+3\right)=\left(x^2+7x+8\right)\left(x^2+7x+14\right)\)

\(e.2x^2-5x-7=2x^2+2x-7x-7=2x\left(x+1\right)-7\left(x+1\right)=\left(x+1\right)\left(2x-7\right)\) \(f.x^2-12x+36=\left(x-6\right)^2=\left(x-6\right)\left(x-6\right)\)

\(g.x^4-5x^2+4=x^4-x^2-4x^2+4=x^2\left(x^2-1\right)-4\left(x^2-1\right)=\left(x^2-1\right)\left(x^2-4\right)=\left(x+1\right)\left(x-1\right)\left(x+2\right)\left(x-2\right)\) \(g.a^3+b^3+c^3-3abc=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\)

3 tháng 7 2016

q=x^3 - y^3 + y^3 + x^3 + y^3 - x^3 - 3x^2y + 3xy^2

rồi thu gọn

3 tháng 7 2016

a đề sai

a) P= (5x-1)+2(1-5x)(4+5x)+(5x+4)2

P = (5x-1)2 + 2(1 - 5x)(4 + 5x)+ (5x + 4)2 - (5x - 1)2

P = (5x - 1 + 5x + 4)2 - (5x - 1)2

P = (5x - 1 + 5x + 4 - 5x + 1) (5x - 1 + 5x + 4 + 5x - 1)

b) làm tương tự nhé!! kết quả: Q = x3 + y3

t i c k nha!! 5364574674675678568585785789769647568585