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a) Kết quả N = (x + 1)(x + 2);
b) Kết quả N = 2(x + 3)(x - 3).
1. \(\left(x+1\right)^3-125\)
\(=\left(x+1\right)^3-5^3\)
\(=\left(x+1-5\right).\left[\left(x+1\right)^2+\left(x+1\right).5+5^2\right]\)
2. \(\left(x+4\right)^3-64\)
\(=\left(x+4\right)^3-4^3\)
\(=\left(x+4-4\right).\left[\left(x+4\right)^2+\left(x+4\right).4+4^2\right]\)
3. \(x^3-\left(y-1\right)^3\)
\(=(x^3-y+1).\left[\left(x^2\right)+x.\left(y+1\right)+\left(y+1\right)^2\right]\)
\(\)4. \(\left(a+b\right)^3-c^3\)
\(=\left[\left(a+b\right)-c\right].\left[\left(a+b\right)^2+\left(a+b\right).c+c^2\right]\)
5. \(125-\left(x+2\right)^3\)
\(=5^3-\left(x+2\right)^3\)
\(=\left(5-x-2\right).\left[5^2+5.\left(x+2\right)+\left(x+2\right)^2\right]\)
6. \(\left(x+1\right)^3+\left(x-2\right)^3\)
\(=\left[\left(x+1\right)+\left(x-2\right)\right].\left[\left(x+1\right)^2-\left(x+1\right).\left(x-2\right)+\left(x-2\right)^2\right]\)
a) Kết quả M = (x + l)(2x - 3);
b) Kết quả M = (2x - 1)(x - 2).
a: \(\Leftrightarrow\dfrac{5x^2-13x+6}{A}=\dfrac{5x-3}{2x+5}\)
\(\Leftrightarrow\dfrac{5x^2-10x-3x+6}{A}=\dfrac{5x-3}{2x+5}\)
\(\Leftrightarrow A=\dfrac{\left(2x+5\right)\left(x-2\right)\left(5x-3\right)}{\left(5x-3\right)}=\left(2x+5\right)\left(x-2\right)\)
b: \(\Leftrightarrow\dfrac{x\left(x+4\right)}{A}=\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(2x-1\right)}=\dfrac{x}{\left(2x-1\right)}\)
=>A=(x+4)(2x-1)
a: |4x-1|=1
=>\(\left[\begin{array}{l}4x-1=1\\ 4x-1=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}4x=2\\ 4x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac12\\ x=0\end{array}\right.\)
Thay x=1/2 vào A(x), ta được:
\(A\left(\frac12\right)=\left(\frac12\right)^4-4\cdot\left(\frac12\right)^3+2\cdot\left(\frac12\right)^2-5\cdot\frac12+6\)
\(=\frac{1}{16}-4\cdot\frac18+2\cdot\frac14-\frac52+6=\frac{1}{16}-\frac12+\frac12-\frac52+6\)
\(=\frac{1}{16}-\frac{40}{16}+\frac{96}{16}=\frac{97-40}{16}=\frac{57}{16}\)
Thay x=0 vào A(x), ta được:
\(A\left(0\right)=0^4-4\cdot0^3+2\cdot0^2-5\cdot0+6=6\)
b: \(A\left(x\right)-B\left(x\right)=3x^2-x-3x^3-x^2+x^4-2x^2+6\)
=>A(x)-B(x)=\(x^4-3x^3+\left(3x^2-x^2-2x^2\right)-x+6\)
=>A(x)-B(x)=\(x^4-3x^3-x+6\)
=>\(B\left(x\right)=A\left(x\right)-\left(x^4-3x^3-x+6\right)\)
=>\(B\left(x\right)=x^4-4x^3+2x^2-5x+6-x^4+3x^3+x-6=-x^3+2x^2-4x\)
c: Đặt B(x)=0
=>\(-x^3+2x^2-4x=0\)
=>\(x^3-2x^2+4x=0\)
=>\(x\left(x^2-2x+4\right)=0\)
mà \(x^2-2x+4=x^2-2x+1+3=\left(x-1\right)^2+3>0\forall x\)
nên x=0
