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* là nhân à ???????????
\(x^2-x-2015\times2016=\left(x^2-2016x\right)+\left(2015x-2015\times2016\right)=x.\left(x-2016\right)+2015\left(x-2016\right)=\left(x-2016\right)\left(x+2015\right)\)k nha
1.x2-9
= (x-3)(x+3)
2. -2x2+2x+12
= -2x2+6x-4x+12
= -2x(x+2)+6(x+2)
= (x+2)(-2x+6)
4. -2x2+2x+24
= -2x2+8x-6x+24
= -2x(x+3)+8(x+3)
= (x+3)(-2x+8)
6. x2-5x+4
= x2-4x-x+4
= x(x-1) -4(x-1)
= (x-1)(x-4)
8. x2-7x+6
= x2-6x-x+6
= x(x-1)-6(x-1)
= (x-1)(x-6)
9. x2+5x+4
= x2+4x+x+4
= x(x+1)+4(x+1)
=(x+1)(x+4)
10. x2+7x+6
= x2 +x+6x+6
= x(x+1)+6(x+1)
= (x+6)(x+1)
K nhé
Bài 3 :
\(\frac{x-1}{2016}+\frac{x-2}{2015}=\frac{x-3}{2014}+\frac{x-4}{2013}\)
\(\Leftrightarrow\)\(\left(\frac{x-1}{2016}-1\right)+\left(\frac{x-2}{2015}-1\right)=\left(\frac{x-3}{2014}-1\right)+\left(\frac{x-4}{2013}-1\right)\)
\(\Leftrightarrow\)\(\frac{x-1-2016}{2016}+\frac{x-2-2015}{2015}=\frac{x-3-2014}{2014}+\frac{x-4-2013}{2013}\)
\(\Leftrightarrow\)\(\frac{x-2017}{2016}+\frac{x-2017}{2015}=\frac{x-2017}{2014}+\frac{x-2017}{2013}\)
\(\Leftrightarrow\)\(\frac{x-2017}{2016}+\frac{x-2017}{2015}-\frac{x-2017}{2014}-\frac{x-2017}{2013}=0\)
\(\Leftrightarrow\)\(\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\right)=0\)
Vì \(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\ne0\)
Nên \(x-2017=0\)
\(\Rightarrow\)\(x=2017\)
Vậy \(x=2017\)
Chúc bạn học tốt ~
Bài 1 :
\(\left(8x-5\right)\left(x^2+2014\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}8x-5=0\\x^2+2014=0\end{cases}\Leftrightarrow\orbr{\begin{cases}8x=0+5\\x^2=0-2014\end{cases}}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}8x=5\\x^2=-2014\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{8}\\x=\sqrt{-2014}\left(loai\right)\end{cases}}}\)
Vậy \(x=\frac{5}{8}\)
Chúc bạn học tốt ~
a) \([(x-y)3 + (y-z)3]+ (z-x)3\)=\(\left(x-y+y-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2\right]-\left(x-z\right)^3\)
\(=\left(x-z\right)\left[\left(\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2-\left(x-z\right)^2\right)\right]\)
\(=\left(x-z\right)\left[\left(x-y\right)\left(x-y-y+z\right)+\left(y-z-x+z\right)\left(y-z+x-z\right)\right]=\left(x-z\right)\left[\left(x-2y+z\right)\left(x+z\right)-\left(x-y\right)\left(x+y-2z\right)\right]\)
\(=\left(x-z\right)\left(x-y\right)\left(x-2y+z-x-y+2z\right)=\left(x-z\right)\left(x-y\right)\left(z-y\right)3\)
b) \(=y^2\left(x^2y-x^3+z^3-z^2y\right)-z^2x^2\left(z-x\right)=y^2\left[-y\left(z^2-x^2\right)-\left(z^3-x^3\right)\right]-z^2x^2\left(z-x\right)\)
\(=y^2\left(z-x\right)\left(-yz-xy-z^2-zx-x^2\right)-z^2x^2\left(z-x\right)=\left(z-x\right)\left(-y^3z-xy^2-z^2y^2-xyz-x^2y^2-z^2x^2\right)\)
đến đây coi như là thành nhân tử rồi nha. em muốn gọn thì ráng ngồi nghĩ rồi tách nha. chỉ cần nhóm mấy cái có ngoặc giống nhau là đc. k khó đâu. chịu khó nghĩ để rèn luyện nha
c) \(x^8+2x^4+1-x^4=\left(x^4+1\right)^2-x^4=\left(x^4+1-x^2\right)\left(x^4+1+x^2\right)\)
\(\left(9a^3-6a^2\right)+\left(6a^2-4a\right)+\left(-9a+6\right)=3a^2\left(3a-2\right)+2a\left(3a-2\right)-3\left(3a-2\right)=\left(3a-2\right)\left(3a^2+2a-3\right)\)
d) em sửa đề đi. đề sai rồi. đồng nhất hệ số phải có dấu bằng nha.
có gì liên hệ chị. đúng nha ;)
\(\left(x-y\right)z^3+\left(y-z\right)x^3+\left(z-x\right)y^3\)
\(=\left(x-y\right)z^3-\left[\left(x-y\right)+\left(z-x\right)\right]x^3+\left(z-x\right)y^3\)
\(=\left(x-y\right)z^3-\left(x-y\right)x^3-\left(z-x\right)x^3+\left(z-x\right)y^3\)
\(=\left(x-y\right)\left(z^3-x^3\right)-\left(z-x\right)\left(x^3-y^3\right)\)
\(=\left(x-y\right)\left(z-x\right)\left(z^2+zx+x^2\right)-\left(z-x\right)\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=\left(x-y\right)\left(z-x\right)\left(z^2+zx+x^2-x^2-xy-y^2\right)\)
\(=\left(x-y\right)\left(z-x\right)\left[\left(x^2-x^2\right)+\left(zx-xy\right)+\left(z^2-y^2\right)\right]\)
\(=\left(x-y\right)\left(z-x\right)\left[x\left(z-y\right)+\left(z-y\right)\left(y+z\right)\right]\)
\(=\left(x-y\right)\left(z-x\right)\left(z-y\right)\left(x+y+z\right)\)
\(=-\left(x-y\right)\left(y-z\right)\left(z-x\right)\left(x+y+z\right)\)