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Bài 1 :
a, \(A=x\left(x-6\right)+10\)
=x^2 - 6x + 10
=x^2 - 2.3x+9+1
=(x-3)^2 +1 >0 Với mọi x dương
+) \(ax-a+bx-b+x-1=a\left(x-1\right)+b\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(a+b+1\right)\)
+) Xem lại đề
ax - a + bx - b + x - 1
= a( x - 1 ) + b( x - 1 ) + 1( x - 1 )
= ( x - 1 )( a + b + 1 )
x3 - 2x2 - 2x + 4 ( sửa -4 thành +4 )
= x2( x - 2 ) - 2( x - 2 )
= ( x - 2 )( x2 - 2 )
Bonus = ( x - 2 )[ x2 - ( √2 )2 ]
= ( x - 2 )( x - √2 )( x + √2 )
Ta có : 6x2 - 11x + 3
= 6x2 - 2x - 9x + 3
= (6x2 - 2x) - (9x - 3)
= 2x(3x - 1) - 3(3x - 1)
= (2x - 3)(3x - 1)
Ta có:
\(x^3+x^2-4x=4\)
\(\Rightarrow x^3+x^2-4x-4=0\)
\(\Rightarrow\left(x^3+x^2\right)-\left(4x+4\right)=0\)
\(\Rightarrow x^2\left(x+1\right)-4\left(x+1\right)=0\)
\(\Rightarrow\left(x^2-4\right)\left(x+1\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x+2\right)\left(x+1\right)=0\)
\(\Rightarrow x-2=0;x+2=0;x+1=0\)
\(\Rightarrow x\in\left\{2;-2;-1\right\}\)
a)\(2.\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right).\left(2-x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\2-x=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)
b)\(3x^3-48x=0\)
\(\Leftrightarrow3x\left(x^2-16\right)=0\)
\(\Leftrightarrow3x.\left(x-4\right).\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\frac{x=4}{\frac{x=0}{x=-4}}}\)
c)\(x^3+x^2-4x=4\)
\(\Leftrightarrow x^2\left(x+1\right)-4\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)\left(x+2\right)\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{x=0}{x=2}\\\overline{x=-2}\end{cases}}\)
ax - bx + ab - x2
= ( ax + ab ) - ( x2 + bx )
= a ( x + b ) - x ( x + b )
= ( a - x ) ( x + b )
ax - bx + ab - x2
= ( ax + ab ) - ( x2 + bx )
= a( x + b ) - x( x + b )
= ( x + b )( a - x )