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\(\Leftrightarrow\left(3x-1\right)^2-4^2=0\)
\(\Leftrightarrow\left(3x-1-4\right)\left(3x-1+4\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(3x+3\right)=0\Leftrightarrow\orbr{\begin{cases}3x-5=0\\3x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}}\)
Ta có
a, x2-x-y2-y
=x2-y2-(x+y)
=(x-y)(x+y) - (x+y)
=(x+y)(x-y-1)
b, x2-2xy+y2-z2
=(x-y)2-z2
=(x-y-z)(x-y+z)
\(9\left(x+1\right)^2-\left(3x-2\right)^2\)
\(=9\left(x^2+2x+1\right)-\left(9x^2-12x+4\right)\)
\(=9x^2+18x+9-9x^2+12x-4\)
\(=30x+5\)
\(=5\left(6x+1\right)\)
\(9\left(x+1\right)^2-\left(3x-2\right)^2\)
\(=\left[3\left(x+1\right)+3x-2\right]\left[3\left(x+1\right)-3x+2\right]\)
\(=\left(3x+3+3x-2\right)\left(3x+3-3x+2\right)\)
\(=5\left(6x+1\right)\)
PTĐTTNT ??? :)) bn phân tích rồi đấy, đề là tìm x thôi
Giải ( suỵt :), đừng ai nhìn thấy ... :v
\(\left(2x-10\right)\left(x+10\right)\left(x+\sqrt{3}\right)=0\)
TH1 : \(2x-10=0\Leftrightarrow x=5\)
TH2 : \(x+10=0\Leftrightarrow x=-10\)
TH3 : \(x+\sqrt{3}=0\Leftrightarrow x=-\sqrt{3}\)( vô lí )
Vậy x = {5;-10}
\(3x^2-3y^2-2\left(x-y\right)^2\)
\(=3\left(x^2-y^2\right)-2\left(x-y\right)^2\)
\(=3\left(x-y\right)\left(x+y\right)-2\left(x-y\right)^2\)
\(=\left(x-y\right)\left[3\left(x+y\right)-2\left(x-y\right)\right]\)
\(=\left(x-y\right)\left(3x+3y-2x+2y\right)\)
\(=\left(x-y\right)\left(x+5y\right)\)
\(=x^4-16x^2+100=x^4+20x^2+100-36x^2\)
\(=\left(x^2+10\right)^2-\left(6x\right)^2=\left(x^2-6x+10\right)\left(x^2+6x+10\right)\)
\(\left(x^2-8\right)^2+36\)
\(=x^4-16x^2+64+36\)
\(=x^4-16x^2+100\)
\(=x^4+20x^2+100-36x^2\)
\(=\left(x^2+10\right)^2-\left(6x\right)^2\)
\(=\left(x^2-6x+10\right)\left(x^2+6x+10\right)\)
\(3x^2-3y^2-2\left(x-y\right)^2\)
\(=3\left(x^2-y^2\right)-2\left(x-y\right)^2\)
\(=3\left(x-y\right)\left(x+y\right)-2\left(x-y\right)^2\)
\(=\left(x-y\right)\left[3\left(x+y\right)-2\left(x-y\right)\right]\)
\(=\left(x-y\right)\left(3x+3y-2x+2y\right)\)
\(=\left(x-y\right)\left(x+5y\right)\)
Chúc bạn học tốt!!!
\(=\left(2x+1\right)^2+2\left(2x+1\right)=\left(2x+1\right)\left(2x+1+2\right)=\left(2x+1\right)\left(2x+3\right)\)
\(\)
a) 16x2-(x2+4)2= (4x)2-(x2+4)2
= (4x-x2-4)(4x+x2+4)
\(\text{b) 27x^3-54x^2+36x-8=[(3x)^3-3.(3x)^2.2+3.3x.2^2-2^3}]\)
= (3x-2)3
\(\text{c) (x+y)^3 - (x-y)^3= (x+y-x+y)[(x+y)^2+(x+y)(x-y)+(x-y)^2]}\)
=2y(x2+2xy+y2+x2-y2+x2-2xy+y2)
= 2y(3x2+y2)
( 3x - 1 )2 - 16
= ( 3x - 1 )2 - 42
= ( 3x - 1 - 4 )( 3x - 1 + 4 )
= ( 3x - 5 )( 3x + 3 )
= 3( 3x - 5 )( x + 1 )