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a, Ta có \(P\left(x\right)=8x^3+2x^2-3x-3x^3+10-x-2x^2-3\)
\(=5x^3-4x-7\)
\(Q\left(x\right)=9x^3-4x^2+2x-3+2x+3x^2+4x^3-2\)
\(=13x^3-x^2+4x-5\)
b, Ta có : \(P\left(-\frac{1}{2}\right)=5.\left(-\frac{1}{2}\right)^3-4.\left(-\frac{1}{2}\right)-7=-\frac{45}{8}\)
c , \(M\left(x\right)=P\left(x\right)+Q\left(x\right)\)
\(5x^3-4x-7+13x^3-x^2+4x-5=18x^3-x^2-12\)
\(N\left(x\right)=P\left(x\right)-Q\left(x\right)\)
\(5x^3-4x-7-13x^3+x^2-4x+5=-8x^3-8x-2+x^2\)
d, Đặt \(5x^3-4x-7=0\)( vô nghiệm )
a) Điều kiện: \(x\ne\pm1\)
\(B=\frac{x-1}{x+1}-\frac{x+1}{x-1}-\frac{4}{1-x^2}\)
\(B=\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}-\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}-\frac{-4}{\left(x-1\right).\left(x+1\right)}\)
\(B=\frac{x^2-x-x+1-x^2-x-x-1+4}{\left(x-1\right).\left(x+1\right)}\)
\(B=\frac{-4x+4}{\left(x-1\right).\left(x+1\right)}=\frac{-4.\left(x-1\right)}{\left(x-1\right).\left(x+1\right)}=\frac{-4}{x+1}\)
b) \(x^2-x=0\Leftrightarrow x.\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
Khi \(x=0\Leftrightarrow\frac{-4}{0-1}=\frac{-4}{-1}=4\)
Khi \(x=1\Leftrightarrow\frac{-4}{1-1}=0\)
c) \(\frac{-4}{x+1}=-3\Leftrightarrow-3.\left(x+1\right)=-4\Leftrightarrow x+1=\frac{4}{3}\Leftrightarrow x=\frac{1}{3}\)
a) x2 - 5x - y2 -5y
= ( x2 - y2 ) + ( -5x - 5y)
= ( x - y ) ( x + y) - 5( x + y )
= ( x + y ) ( x - y -5)
b) x3 + 2x2 - 4x - 8
= x2 ( x + 2 ) - 4 ( x + 2 )
= ( x +2 ) ( x2 -4 )
= ( x+2)2 ( x-2)
Bai 2 :
a, \(A=\left(x+3\right)^2+\left(x-2\right)^2-2\left(x+3\right)\left(x-2\right)\)
\(=x^2+6x+9+x^2-4x+4-2\left(x^2-2x+3x-6\right)\)
\(=2x^2+2x+13-2x^2-2x+12=25\)
b, \(B=\left(x-2\right)^2-x\left(x-1\right)\left(x-3\right)+3x^2-9x+8\)
\(=x^2-4x+4-x\left(x^2-3x-x+3\right)+3x^2-9x+8\)
\(=4x^2-13x+12-x^3+4x^2-3x=-16x+12-x^3\)