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\(9\left(x-3\right)^2-4\left(x+1\right)^2\)
\(=\left[3\left(x-3\right)-2\left(x+1\right)\right].\left[3\left(x-3\right)+2\left(x+1\right)\right]\)
\(=\left(3x-9-2x-2\right)\left(3x-9+2x+2\right)\)
\(=\left(x-11\right)\left(5x-7\right)\)
\(9\left(x-3\right)^2-4\left(x+1\right)^2\)
\(=3^2\left(x-3\right)^2-2^2\left(x+1\right)^2\)
\(=\left[3\left(x-3\right)\right]^2-\left[2\left(x+1\right)\right]^2\)
\(=\left[3x-9\right]^2-\left[2x+2\right]^2\)
\(=\left[3x-9-2x-2\right].\left[3x-9+2x+2\right]\)
\(=\left(x-11\right)\left(5x-7\right)\)
Chúc bạn học tốt.
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Phân tích đa thức thành nhân tử
27y2-9(x+y)2=\(9\left(3y^2-\left(x+y\right)^2\right)\)
=\(9\left(\sqrt{3}y+x+y\right)\left(\sqrt{3}y-x-y\right)\)
Rút gọn biểu thức
(2x4-x3+3x2): (-1/3x)
=\(\frac{2x^4-x^3+3x^2}{-\frac{1}{3x}}=3x^3\left(-2x^2+x-3\right)\)
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Ta có:\(3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)^2=3x^4+3x^2+3-x^4-x^2-1-2x^3-2x-2x^2\)
\(=2x^4-2x^3-2x+2=2x^3\left(x-1\right)-2\left(x-1\right)=2\left(x^3-1\right)\left(x-1\right)\)
\(=2\left(x-1\right)^2\left(x^2+x+1\right)\)
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3(x4+x+1)-(x2+x+1)2
=3(x2+x+1)(x2-x+1)-(x2+x+1)2
=(x2+x+1)[3(x2-x+1)-(x2-x+1)
=(x2+x+1)(3x2-3x+3-x2+x-1)
=(x2+x+1)(2x2-2x+2)
=(x2+x+1)2(x2-x+1)
bạn vu cong thien làm sai rồi.
\(x^4+x^2+1=\left(x^2+x+1\right)\left(x^2-x+1\right)\)
chứ không phải là:
\(x^4+x+1=\left(x^2+x+1\right)\left(x^2-x+1\right)\)đâu!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)^2\)
\(=3\left(x^4+x^2+1\right)-\left(x^4+x^2+1+2x^3+2x^2+2x\right)\)
\(=\left(x^4+x^2+1\right)\left(3-2x^3-2x^2-2x\right)\)
\(3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)^2\)
\(=3\left(x^2+x+1\right)\left(x+x^2+x+1\right)\)
\(=3\left(x^2+x+1\right)\left(x^2+2x+1\right)\)
\(=3\left(x^2+x+1\right)\left(x+1\right)^2\)
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\(x^8+3x^4+4\)
\(=\left(x^8-x^6+2x^4\right)+\left(x^6-x^4+2x^2\right)+\left(2x^4-2x^2+4\right)\)
\(=x^4\left(x^4-x^2+2\right)+x^2\left(x^4-x^2+2\right)+2\left(x^4-x^2+2\right)\)
\(=\left(x^4+x^2+2\right)\left(x^4-x^2+2\right)\)
\(4x^4+4x^3+5x^2+2x+1\)
\(=\left(4x^4+2x^3+2x^2\right)+\left(2x^3+x^2+x\right)+\left(2x^2+x+1\right)\)
\(=2x^2\left(2x^2+x+1\right)+x\left(2x^2+x+1\right)+\left(2x^2+x+1\right)\)
\(=\left(2x^2+x+1\right)^2\)
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a) \(4x^2-8x+4-9\left(x-y\right)^2\)
\(=4\left(x^2-2x+1\right)-9\left(x-y\right)^2\)
\(=\left[2\left(x-1\right)\right]^2-\left[3\left(x-y\right)\right]^2\)
\(=\left(2x-2+3x-3y\right)\left(2x-2-3x+3y\right)\)
\(=\left(5x-3y-2\right)\left(3y-x-2\right)\)
b) \(x^3-4x^2+12x-27\)
\(=\left(x^3-27\right)-\left(4x^2-12x\right)\)
\(=\left(x-3\right)\left(x^2+3x+9\right)-4x\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-x+9\right)\)
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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 ;)
\(9\left(x-3\right)^2-4\left(x-1\right)^2=\left(3x-9\right)^2-\left(2x-2\right)^2\)
\(=\left[\left(3x-9\right)-\left(2x-2\right)\right]\left[\left(3x-9\right)+\left(2x-2\right)\right]\)
\(=\left(3x-9-2x+2\right)\left(3x-9+2x-2\right)\)
\(=\left(x-7\right)\left(5x-11\right)\)