Cho x là số nguyên. Cmr B= x4 - 4x3 - 2x2 + 12x +9
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Bài 3:
\(B=x^4-4x^3-2x^2+12x+9=\left(x^4+x^3\right)-\left(5x^3+5x^2\right)+\left(3x^2+3x\right)+\left(9x+9\right)=\left(x^3-5x^2+3x+9\right)\left(x+1\right)=\left[\left(x^3+x^2\right)-\left(6x^2+6x\right)+\left(9x+9\right)\right]\left(x+1\right)=\left(x^2-6x+9\right)\left(x+1\right)^2=\left(x-3\right)^2\left(x+1\right)^2=\left[\left(x-3\right)\left(x+1\right)\right]^2\)
Bài 3:
\(B=x^4-4x^3-2x^2+12x+9\)
\(=x^4-3x^3-x^3+3x^2-5x^2+15x-3x+9\)
\(=\left(x-3\right)\left(x^3-x^2-5x-3\right)\)
\(=\left(x-3\right)\left(x^3-3x^2+2x^2-6x+x-3\right)\)
\(=\left(x-3\right)^2\cdot\left(x+1\right)^2\)
\(=\left(x^2-2x-3\right)^2\)
\(P\left(-1\right)=\left(-1\right)^4+2.\left(-1\right)^2+1=4\\ P\left(1\right)=1^4+2.1^2+1=4\)
\(P\left(-1\right)=\left(-1\right)^4+2\cdot\left(-1\right)^2+1=4\)
\(P\left(1\right)=P\left(-1\right)=4\)
\(Q\left(2\right)=2^4+4\cdot2^3+2\cdot2^2-4\cdot2+1=49\)
\(Q\left(1\right)=1^4+4\cdot1^3+2\cdot1^2-4\cdot1+1=4\)
a, \(\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)-\left(3x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(\left(9x^2-4\right)-\left(\left(3x+2\right)\left(x-1\right)\right)\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-\left(3x^2-x-2\right)\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-3x^2+x+2\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x+1\right)=0;3x^2+x-2=0\)
=> x=-1
với \(3x^2+x-2=0\)
ta sử dụng công thức bậc 2 suy ra : \(x=\dfrac{2}{3};x=-1\)
Vậy ghiệm của pt trên \(S\in\left\{-1;\dfrac{2}{3}\right\}\)
b: \(\Leftrightarrow x^2-2x+1-1+x^2=x+3-x^2-3x\)
\(\Leftrightarrow2x^2-2x=-x^2-2x+3\)
\(\Leftrightarrow3x^2=3\)
hay \(x\in\left\{1;-1\right\}\)
c: \(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x-3\right)-\left(x-1\right)\left(x-2\right)\left(x+2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+1\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-2x-3-x^2-3x+10\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(-5x+7\right)=0\)
hay \(x\in\left\{1;-2;\dfrac{7}{5}\right\}\)
a.
\(x^3-7x+6=0\)
\(\Leftrightarrow x^3-3x^2+2x+3x^2-9x+6=0\)
\(\Leftrightarrow x\left(x^2-3x+2\right)+3\left(x^2-3x+2\right)=0\)
\(\Leftrightarrow\left(x^2-3x+2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x^2-x-2x+2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[x\left(x-1\right)-2\left(x-1\right)\right]\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-3\end{matrix}\right.\)
f.
\(x^4-4x^3+12x-9=0\)
\(\Leftrightarrow x^4-4x^3+3x^2-3x^2+12x-9=0\)
\(\Leftrightarrow x^2\left(x^2-4x+3\right)-3\left(x^2-4x+3\right)=0\)
\(\Leftrightarrow\left(x^2-4x+3\right)\left(x^2-3\right)=0\)
\(\Leftrightarrow\left(x^2-x-3x+3\right)\left(x^2-3\right)=0\)
\(\Leftrightarrow\left[x\left(x-1\right)-3\left(x-1\right)\right]\left(x^2-3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(x^2-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=3\\x=\pm\sqrt{3}\end{matrix}\right.\)
c) Ta có: \(C=4x^2+y^2-4xy+8x-4y+4\)
\(=\left(2x-y\right)^2+2\cdot\left(2x-y\right)\cdot2+2^2\)
\(=\left(2x-y+2\right)^2\)
Cho các đa thức: P(x)=x4+2x2+1 và Q(x)=x4+4x3+2x2-4x+1
Tính P(x)-Q(x) các bn giải giùm mik nhé.......
`Q(-2)=(-2)^4+4*(-2)^3+2*(-2)^2-4*(-2)+1`
`= 16+4*(-8)+2*4+8+1`
`= 16-32+8+8+1`
`= -16+8+8+1`
`= -8+8+1=1`
`Q(1)=1^4+4*1^3+2*1^2-4*1+1`
`= 1+4+2-4+1`
`= 2+2+4-4=4`
Q(-2) = (-2)⁴ + 4.(-2)³ + 2.(-2)² - 4.(-2) + 1
= 16 - 32 + 8 + 8 + 1
= 1
--------------------
Q(1) = 1⁴ + 4.1³ + 2.1² - 4.1 + 1
= 1 + 4 + 2 - 4 + 1
= 4
\(Q\left(x\right)=x^4+4x^3+2x^2-4x+1\)
\(\Rightarrow Q\left(-2\right)=\left(-2\right)^4+4\cdot\left(-2\right)^3+2\cdot\left(-2\right)^2-4\cdot\left(-2\right)+1=1\)
\(Q\left(-1\right)=\left(-1\right)^4+4\cdot\left(-1\right)^3+2\cdot\left(-1\right)^2-4\cdot\left(-1\right)+1=4\)