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A = \(4\left(x-5\right)-x^2\left(x+1\right)-x^3\left(x-3\right)-\left(x-4+x^2\right)\)
A = \(4x-20-x^3-x^2-x^4+3x^3-x+4-x^2\)
A = \(-x^3-3x^3-x^2+x^2-x^4+4x-x-20+4\)
A = \(-4x^3-x^4+4x-16\)
B = \(-3\left(x^2-x+1\right)-2\left(4-x^2\right)-6\left(x+1\right)-x^4-x^3\)
B = \(-3x^2+3x-3-8+2x^2-6x-6-x^4-x^3\)
B = \(-x^4-x^3-3x^2-2x^2+3x-6x-3-8-6\)
B = \(-x^4-x^3-5x^2-3x-17\)
C = \(-\left(x^4+3x^2-2\right)-x^2\left(5-x\right)+3\left(x-1\right)\)
C = \(-x^4-3x^2+2-5x^2+x^3+3x-3\)
C = \(-x^4+x^3-3x^2+5x^2+3x+2-3\)
C = \(-x^4+x^3-2x^2+3x-1\)
#Yiin
\(A=4x-20-x^3-x-x^4+3x^3-x+4-x^2\)
\(=-x^4+2x^3-x^2+2x-16\)
\(B=-3x^2+3x-3-8+2x^2-6x-6-x^4-x^3\)
\(=-x^4-x^3-x^2-3x-17\)
\(C=-x^4-3x^2+2-5x^2+x^3+3x-3\)
\(=-x^4+x^3-8x^2+3x-1\)
Từ đó có:
\(A-B=-x^4+2x^3-x^2+2x-16-\left(-x^4-x^3-x^2-3x-17\right)\)
\(=-x^4+2x^3-x^2+2x-16+x^4+x^3+x^2+3x+17\)\(=3x^3+5x+1\)
\(B-C=-x^4-x^3-x^2-3x-17-\left(-x^4+x^3-8x^2+3x-1\right)\)
\(=-x^4-x^3-x^2-3x-17+x^4-x^3+8x^2-3x+1\)
\(=-2x^3+7x^2-6x-16\)
\(C-A=-x^4+x^3-8x^2+3x-1-\left(-x^4+2x^3-x^2-2x-16\right)\)
\(=-x^4+x^3-8x^2+3x-1+x^4-2x^3+x^2+2x+16\)
\(=-x^3-7x^2+5x+15\)
a ) M(x) + N(x) + P(x) = (\(3x^3+x^2+4x^4-x-3x^3+5x^4+x^2-6\)) + (\(-x^2-x^4+4x^3-x^2-5x^3+3x+1+x\)) + (\(1+2x^5-3x^2+x^5+3x^3-x^4-2x\))
= \(3x^3+x^2+4x^4-x-3x^3+5x^4+x^2-6\) \(-x^2-x^4+4x^3-x^2-5x^3+3x+1+x\)\(1+2x^5-3x^2+x^5+3x^3-x^4-2x\)
= ( \(3x^3-3x^3+4x^3-5x^3+3x^3\) ) + ( \(x^2+x^2-x^2-x^2-3x^2\) ) + (\(4x^4+5x^4-x^4-x^4\) ) + ( \(-x+3x+x-2x\) ) + ( \(-6+1+1\) ) + (\(2x^5+x^5\) )
= \(2x^3-3x^2+7x^4+x-4+3x^5\)
Căng, sự thật là nó rất căng
Nhg dù sao thì.....
1) \(A\left(x\right)=\left(x-4\right)^2-\left(2x+1\right)^2\)
Xét \(A\left(x\right)=0\)
\(\Rightarrow\left(x-4\right)^2-\left(2x+1\right)^2=0\)
\(\Rightarrow x^2-8x+16-4x^2-4x-1=0\)
\(\Rightarrow-3x^2-12x+15=0\)
\(\Rightarrow-3x^2+3x-15x+15=0\)
\(\Rightarrow-3x\left(x-1\right)-15\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(-3x-15\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\-3x-15=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
2)(Sửa đề nha, sai cmnr) \(B\left(x\right)=x^3+x^2-4x-4\)
Xét \(B\left(x\right)=0\)
\(\Rightarrow x^3+x^2-4x-4=0\)
\(\Rightarrow x^2\left(x+1\right)-4\left(x+1\right)=0\)
\(\Rightarrow\left(x^2-4\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2-4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\pm2\\x=-1\end{matrix}\right.\)
Đó là những j mình biết