Lữ Nguyễn Duy Đức : bắt chước Nguyễn Khắc Vinh

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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$(x-2)(3x^2+6x+12)-(120^2x^2y^2):(60xy^2)$

$=3(x-2)(x^2+2x+4)-240x$

$=3(x^3-2^3)-240x=3x^3-240x-24$

$=3(x^3-80x-8)$

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) \(3x^3+2x-5\)

\(=3x^3+3x^2+5x-3x^2-3x-5\)

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

\(=\left(x-1\right)\left(3x^2+3x+5\right)\)

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 : 

+) \(9x^2-6xy+y^2-21x+7y\)

\(=\left(3x-y\right)^2-7\left(3x-y\right)\)

\(=\left(3x-y\right)\left(3x-y-7\right)\)

+) \(x^2+2x-35\)

\(=x^2+2x+1-36\)

\(=\left(x+1-6\right)\left(x+1+6\right)\)

\(=\left(x-5\right)\left(x+7\right)\)

+) \(2x^2+9x-5\)

\(=2x^2-x+10x-5\)

\(=x\left(2x-1\right)+5\left(2x-1\right)\)

\(=\left(2x-1\right)\left(x+5\right)\)

+) \(6x^2+23x+15\)

\(=6x^2+18x+5x+15\)

\(=6x\left(x+3\right)+5\left(x+3\right)\)

\(=\left(x+3\right)\left(6x+5\right)\)

\(-5x^2+16x-3=-5x^2+15x+x-3=-5x\left(x-3\right)+\left(x-3\right)=\left(x-3\right)\left(-5x+1\right)\)

Phần 1 ) Đánh lộn dấu " ) " thành số 0 rồi bạn .... 

\(\left(x+1\right).\left(x+2\right).\left(x+3\right).\left(x+4\right)-15\)

\(=\left[\left(x+1\right).\left(x+4\right)\right].\left[\left(x+2\right).\left(x+3\right)\right]-15\)

\(=\left(x^2+5x+4\right).\left(x^2+5x+6\right)-15\)

\(\text{Đặt }k=x^2+5x+4,\text{ta có: }\)

\(\left(x^2+5x+4\right).\left(x^2+5x+6\right)-15=k.\left(k+2\right)-15=k^2+2k-15=k^2-5k+3k-15=k.\left(k-5\right)+3.\left(k-5\right)=\left(k-5\right).\left(k+3\right)\)

tự thay vô nha : )

ko ra được đâu

Ta có:  \(\Delta\)= b^2 - 4ac

                    = (-6)^2  -  4.1.15 

                    =-24

     Vì\(\Delta\)< 0  nên phương trình vô nghiệm => không thể phân tích

nbhb hdbbvha

a.\(6x^2-11x+3\)

=\(6x^2-2x-9x+3\)

=\(2x\left(3x-1\right)-3\left(3x-1\right)\)

=\(\left(2x-3\right)\left(3x-1\right)\)

b.\(x^3+3x^2-16x-48\)

=\(x^2\left(x+3\right)-16\left(x+3\right)\)

=\(\left(x^2-16\right)\left(x+3\right)\)

a,\(6x^2-9x-2x+3=3x\left(2x-3\right)-\left(2x-3\right)=\left(3x-1\right)\left(2x-3\right)\)

b,\(x^3+3x^2-16x-48=x\left(x^2-16\right)+3\left(x^2-16\right)=\left(x+3\right)\left(x-4\right)\left(x+4\right)\)