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g) \(x^5-3x^4+3x^3-x^2=x^2\left(x^3-3x^2+3x-1\right)=x^2\left(x-1\right)^3\)
f) \(x^2-25-2xy+y^2=\left(x^2-2xy+y^2\right)-25=\left(x-y\right)^2-5^2=\left(x-y-5\right)\left(x-y+5\right)\)
e) \(16x^3+54y^3=2\left(8x^3+27y^3\right)=2\left[\left(2x\right)^3+\left(3y\right)^3\right]=2\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
d) \(3y^2-3z^2+3x^2+6xy=3\left(x^2+2xy+y^2-z^2\right)=3\left[\left(x+y\right)^2-z^2\right]=3\left(x+y+z\right)\left(x+y-z\right)\)
Answer:
\(5x^2-10xy+5y^2-20z^2\)
\(=5.\left(x^2-2xy+y^2-4z^2\right)\)
\(=5.[\left(x+y\right)^2-\left(2z\right)^2]\)
\(=5.\left(x+y-2z\right).\left(x+y+2z\right)\)
\(16x-5x^2-3\)
\(=\left(-5x^2+15x\right)+\left(x-3\right)\)
\(=-5x.\left(x-3\right)+\left(x-3\right)\)
\(=\left(1-5x\right).\left(x-3\right)\)
\(x^2-5x+5y-y^2\)
\(=(x-y).(x+y)-5.(x-y)\)
\(=(x-y).(x+y-5)\)
\(3x^2-6xy+3y^2-12z^2\)
\(=3.(x^2-2xy+y^2-4z^2)\)
\(=3[\left(x-y\right)^2-\left(2z\right)^2]\)
\(=3.(x-y-2z).(x-y+2z)\)
\(x^2+4x+3\)
\(=(x^2+x)+(3x+3)\)
\(=x.(x+1)+3.(x+1)\)
\(=(x+1).(x+3)\)
\((x^2+1)^2-4x^2\)
\(=(x^2-2x+1).(x^2+2x+1)\)
\(=(x-1)^2.(x+1)^2\)
\(x^2-4x-5\)
\(=(x^2+x)-(5x+5)\)
\(=x.(x+1)-5.(x+1)\)
\(=(x-5).(x+1)\)
\(4x^2-25+\left(2x+7\right).\left(5-2x\right)\)
\(=\left(2x+5\right).\left(2x-5\right)-\left(2x+7\right).\left(2x-5\right)\)
\(=\left(2x+5-2x-7\right).\left(2x-5\right)\)
\(=-2.\left(2x-5\right)\)
\(a^2x^2-a^2x^2-b^2x^2+b^2y^2\)
\(=a^2.\left(x^2-y^2\right)-b^2.\left(x^2-y^2\right)\)
\(=\left(a^2-b^2\right).\left(x^2-y^2\right)\)
\(=\left(a-b\right).\left(a+b\right).\left(x-y\right).\left(x+y\right)\)
\(x^2-y^2+12y-36\)
\(=x^2-\left(y^2-12y+36\right)\)
\(=x^2-\left(y-6\right)^2\)
\(=\left(x-y+6\right).\left(x+y-6\right)\)
\(\left(x+2\right)^2-x^2+2x-1\)
\(=\left(x+2\right)^2-\left(x^2-2x+1\right)\)
\(=\left(x+2\right)^2-\left(x-1\right)^2\)
\(=[x+2-\left(x-1\right)].[x+2+\left(x-1\right)]\)
\(=\left(x+2-x+1\right).\left(x+2+x-1\right)\)
\(=3.\left(2x+1\right)\)
\(16x^2-y^2=\left(4x\right)^2-y^2=\left(4x-y\right).\left(4x+y\right)\)
\(1+27x^3=1^3+\left(3x\right)^3=\left(1+3x\right).\left(1-3x+9x^2\right)\)
a) \(\left(3x^2+10x-8\right)^2=\left(5x^2-2x+10\right)^2\)
\(3x^2+10x-8=5x^2-2x+10\)
\(3x^2-5x^2+10x+2x-8-10=0\)
\(-2x^2+12x-18=0\)
\(x^2-6x+9=0\)
\(\left(x-3\right)^2=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
b) \(\frac{x^2-x-6}{x-3}=0\)
\(\Rightarrow x^2-x-6=0\)
\(\Rightarrow x^2-2x.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}-6=0\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2-\frac{25}{4}=0\)
\(\Rightarrow\left(x-\frac{1}{2}-\frac{5}{2}\right)\left(x-\frac{1}{2}+\frac{5}{2}\right)=0\)
\(\Rightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
c1 thử vào là xong nhé
c2 x^2 -3x+7=1+2x
<=> x^2 -3x-2x+6=0
<=> x^2 -5x+6=0
<=> (x-2)(x-3)=0
<=> x=2
hoặc a+3
=> đpcm
a) = 5( x2 - 9y2 - 6y - 1 ) = 5[ x2 - ( 9y2 + 6y + 1 ) ] = 5[ x2 - ( 3y + 1 )2 ] = 5( x - 3y - 1 )( x + 3y + 1 )
b) = 125x3 - 25x2 + 15x2 - 3x + 5x - 1 = 25x2( 5x - 1 ) + 3x( 5x - 1 ) + ( 5x - 1 ) = ( 5x - 1 )( 25x2 + 3x + 1 )
c) = 5( x - 7 ) + a( x - 7 ) = ( x - 7 )( a + 5 )
d) = ( a - b )2 + ( a - b ) = ( a - b )( a - b + 1 )
e) = ax2 + a - a2x - x = ax( a - x ) + ( a - x ) = ( a - x )( ax + 1 )
f) = ( 10x )2 - ( x2 + 25 )2 = ( 10x - x2 - 25 )( 10x + x2 + 25 ) = -( x - 5 )2( x + 5 )2
\(a,x^2-3x+7=1+2x\\ \Leftrightarrow x^2-3x-2x+7-1=0\\ \Leftrightarrow x^2-5x+6=0\\ \Leftrightarrow x^2-3x-2x+6=0\\ \Leftrightarrow x\left(x-3\right)-2\left(x-3\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)