phân tích đa thức thành nhân tử (x^2-6x+15)(x^2-16x+60)-21x^2
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Lời giải:
$(4x-2)^2-16x+16x=(4x-2)^2$
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$(3x-4)(3x+4)-(20x^2y-15xy^2):(5xy)$
$=(3x-4)(3x+4)-(4x-3y)$ không phân tích được thành nhân tử.
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$(x-2)(3x^2+6x+12)-(120^2x^2y^2):(60xy^2)$
$=3(x-2)(x^2+2x+4)-240x$
$=3(x^3-2^3)-240x=3x^3-240x-24$
$=3(x^3-80x-8)$
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x^3+4x^2-21x\)
\(=x\left(x^2+4x-21\right)\)
\(=x\left(x^2-3x+7x-21\right)\)
\(=x\left[x\left(x-3\right)+7\left(x-3\right)\right]\)
\(=x\left(x-3\right)\left(x+7\right)\)
b) \(5x^3+6x^2+x\)
\(=x\left(5x^2+6x+1\right)\)
\(=x\left(5x^2+5x+x+1\right)\)
\(=x\left[5x\left(x+1\right)+\left(x+1\right)\right]\)
\(=x\left(x+1\right)\left(5x+1\right)\)
c) \(x^3-7x+6\)
\(=x^3+2x^2-3x-2x^2-4x+6\)
\(=x\left(x^2+2x-3\right)-2\left(x^2+2x-3\right)\)
\(=\left(x-2\right)\left(x^2+2x-3\right)\)
\(=\left(x-2\right)\left(x-1\right)\left(x+3\right)\)
d) \(3x^3+2x-5\)
\(=3x^3+3x^2+5x-3x^2-3x-5\)
\(=x\left(3x^2+3x+5\right)-\left(3x^2+3x+5\right)\)
\(=\left(x-1\right)\left(3x^2+3x+5\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1 :
\(49\left(x-2\right)^2-25\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left[7\left(x-2\right)-5\left(2x+1\right)\right]\left[7\left(x-2\right)+5\left(2x+1\right)\right]=0\)
\(\Leftrightarrow\left(7x-14-10x-5\right)\left(7x-14+10x+5\right)=0\)
\(\Leftrightarrow\left(-3x-19\right)\left(17x-9\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-3x=19\\17x=9\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-19}{3}\\x=\frac{9}{17}\end{cases}}}\)
Bài 2 :
+) \(9x^2-6xy+y^2-21x+7y\)
\(=\left(3x-y\right)^2-7\left(3x-y\right)\)
\(=\left(3x-y\right)\left(3x-y-7\right)\)
+) \(x^2+2x-35\)
\(=x^2+2x+1-36\)
\(=\left(x+1-6\right)\left(x+1+6\right)\)
\(=\left(x-5\right)\left(x+7\right)\)
+) \(2x^2+9x-5\)
\(=2x^2-x+10x-5\)
\(=x\left(2x-1\right)+5\left(2x-1\right)\)
\(=\left(2x-1\right)\left(x+5\right)\)
+) \(6x^2+23x+15\)
\(=6x^2+18x+5x+15\)
\(=6x\left(x+3\right)+5\left(x+3\right)\)
\(=\left(x+3\right)\left(6x+5\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(-5x^2+16x-3=-5x^2+15x+x-3=-5x\left(x-3\right)+\left(x-3\right)=\left(x-3\right)\left(-5x+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\Delta\)= b^2 - 4ac
= (-6)^2 - 4.1.15
=-24
Vì\(\Delta\)< 0 nên phương trình vô nghiệm => không thể phân tích
![](https://rs.olm.vn/images/avt/0.png?1311)