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a/ \(x^3-5x^2+6x+3=\left(x-2\right)\left(x^2-3x\right)+3.\)( Dùng phép chia đa thức)
Để A chia hết cho x-2 thì 3 phải chia hết cho x-2 => x-2 là ước của 3
=> x-2={3-; -1; 1; 3} => x={-1; 1; 3; 5}
b/ Chia F(x) cho x-1
\(f\left(x\right)=\left(x-1\right)\left(x^2-5x+6\right)\)
Giải phương trình bậc 2 \(x^2-5x+6=0\) để tìm nghiệm còn lại
a. x3+x2+2x2+2x
= (x3+x2)+(2x2+2x)
= x2(x+1)+2x(x+1)
= (x2+2x)(x+1)
= x(x+2)(x+1)
c: \(x^4+x^3-4x^2+x+1\)
\(=x^4-x^3+2x^3-2x^2-2x^2+2x-x+1\)
\(=\left(x-1\right)\left(x^3+2x^2-2x-1\right)\)
\(=\left(x-1\right)\left[\left(x-1\right)\left(x^2+x+1\right)+2x\left(x-1\right)\right]\)
\(=\left(x-1\right)^2\cdot\left(x^2+3x+1\right)\)
a) ( x 2 – 4x + 1)( x 2 – 2x + 3).
b) ( x 2 + 5x – 1)( x 2 + x – 1).
\(A=444....444=4.111.....111=4.\frac{10^{2n}-1}{9}\)
\(B=888.....888=8.111.....111=8.\frac{10^n-1}{9}\)
\(\Rightarrow A+2B+4=\frac{4.10^{2n}-4+16.10^n-16+36}{9}=\frac{4.10^{2n}+16.10^n+16}{9}=\left(\frac{2.10^n+4}{3}\right)^2\)
là số hính phương (đpcm)
2) Ta có :
\(x^4+6x^2+25=x^4+10x^2+25-4x^2=\left(x^2+5\right)^2-\left(2x\right)^2\)
\(=\left(x^2-2x+5\right)\left(x^2+2x+5\right)\)(1)
\(3x^4+4x^2+28x+5=\left(3x^4+6x^3+x^2\right)+\left(-6x^3-12x^2-2x\right)+\left(15x^2+30x+5\right)\)
\(=x^2\left(3x^2+6x+1\right)-2x\left(3x^2+6x+1\right)+5\left(3x^2+6x+1\right)\)
\(=\left(x^2-2x+5\right)\left(3x^2+6x+1\right)\)(2)
Từ (1) ; (2) \(\Rightarrow f\left(x\right)=x^2-2x+5\Rightarrow f\left(2011\right)=2011^2-2.2011+5=4040104\)
x đầu ở đa thức A là x^3 chăng?
a/ \(A=x^3-5x^2+8x-4\)
\(=\left(x^3-x^2\right)+\left(-4x^2+4\right)+\left(8x-8\right)\)
\(=x^2\left(x-1\right)-4\left(x-1\right)\left(x+1\right)+8\)
\(=\left(x-1\right)\left(x^2-4x-4\right)=\left(x-1\right)\left(x-2\right)^2\)
b/ \(B=\dfrac{x^5}{30}-\dfrac{x^3}{6}+\dfrac{2x}{15}\)
\(=\dfrac{x^5}{30}-\dfrac{5x^3}{30}+\dfrac{4x}{30}\)
\(=\dfrac{x\left(x^4-5x^2+4\right)}{30}\)
\(=\dfrac{x\left(x^4-x^2-4x^2+4\right)}{30}\)
\(=\dfrac{x\left(x+2\right)\left(x-1\right)\left(x+1\right)\left(x-2\right)}{30}\)
f(x) = x4 + 6x3 +11x2 + 6x
\(=x^4+x^3+5x^3+5x^2+6x^2+6x\)
\(=\left(x^4+x^3\right)+\left(5x^3+5x^2\right)+\left(6x^2+6x\right)\)
\(=x^3\left(x+1\right)+5x^2\left(x+1\right)+6x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3+5x^2+6x\right)\)
\(=x\left(x+1\right)\left(x^2+5x+6\right)\)
\(=x\left(x+1\right)\left[x^2+2x+3x+6\right]\)
\(=x\left(x+1\right)\left[\left(x^2+2x\right)+\left(3x+6\right)\right]\)
\(=x\left(x+1\right)\left[x\left(x+2\right)+3\left(x+2\right)\right]\)
\(=x\left(x+1\right)\left(x+2\right)\left(x+3\right)\)
b)Ta có
\(f\left(x\right)+1=x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)
\(=\left[x\left(x+3\right)\right].\left[\left(x+1\right)\left(x+2\right)\right]+1\)
\(=\left(x^2+3x\right).\left(x^2 +3x+2\right)+1\)
\(=\left(x^2+3x+1-1\right).\left(x^2+3x+1+1\right)+1\)
\(=\left[\left(x^2+3x+1\right)-1\right].\left[\left(x^2+3x+1\right)+1\right]+1\)
\(=\left(x^2+3x+1\right)^2-1+1=\left(x^2+3x+1\right)^2\)
Vậy với mọi x nguyên thì f(x) + 1 luôn có giá trị là 1 số chính phương