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a)\(P=x^3+6x^2+12x+8+x^3-6x^2+12x-8-2x^3-24x=0\)
Vậy g/t P không phụ thuộc vào biến.
b)\(Q=x^3-3x^2+3x-1-\left(x^3+3x^2+3x+1\right)+6\left(x^2-1\right)=-6x^2-2+6x^2-6=-8\)
Vậy g/t Q không phụ thuộc vào biến.
b) Ta có: \(Q=\left(x-1\right)^3-\left(x+1\right)^3+6\left(x+1\right)\left(x-1\right)\)
\(=\left(x-1-x-1\right)\left[\left(x-1\right)^2+\left(x-1\right)\left(x+1\right)+\left(x+1\right)^2\right]+6\left(x^2-1\right)\)
\(=-2\left(x^2-2x+1+x^2-1+x^2+2x+1\right)+6\left(x^2-1\right)\)
\(=-2\left(3x^2+1\right)+6\left(x^2-1\right)\)
\(=-6x^2-2+6x^2-6\)
=-8
Ta có:
P(x)=x2+(x+2)2+(x+3)2+...+(x+98)2−[(x+1)2+(x+3)2+...+(x+99)2]
=[x2-(x+1)2]+[(x+2)2-(x+3)2]+[(x+3)2-(x+4)2]+...+[(x+98)2-(x+99)2]
=(x-x-1)(x+x+1)+(x+2-x-3)(x+2+x+3)+...+(x+98-x-99)(x+98+x+99)
=-(2x+1)-(2x+5)-....-(2x+197)
=(-2x-2x-...-2x)+(-1-5-...-197)
Vì đa thức trên có \(\dfrac{197-1}{4}+1=50\text{ số hạng => -2x có 50 hạng tử}\)
Nên ta có:
=(-2x*50)+\(\left(\dfrac{\left(-197-1\right)\cdot50}{2}\right)\)
=-100x-4950
Mà P(x)=ax+b =>{a=-100; b=-4950}
Vậy a-b= -100-(-4950)= 4850 (Hihi! Mình tự làm nên ko biết đúng hay ko?)
a) \(\left(1+x\right)^2+\left(1-x\right)^2\)
\(=1+2x+x^2+1-2x+x^2\)
\(=2x^2+2\)
b) \(\left(x+2\right)^2+\left(1+x\right)\left(1-x\right)\)
\(=x^2+4x+4+1-x^2\)
\(=4x+5\)
c) \(\left(x-3\right)^2+3\left(x+1\right)^2\)
\(=x^2-6x+9+3x^2+6x+3\)
\(=4x^2+12\)
d)\(\left(2+3x\right)\left(3x-2\right)-\left(3x+1\right)^2\)
\(=9x^2-4-9x^2-6x-1\)
\(=-6x-5\)
e) \(\left(x+5\right)\left(x-2\right)-\left(x+2\right)^2\)
\(=x^2-2x+5x-10-x^2-4x-4\)
\(=-x-14\)
f) \(\left(x+3\right)\left(2x-5\right)-2\left(1+x\right)^2\)
\(=2x^2-5x+6x-15-2-4x-2x^2\)
\(=-3x-17\)
g) \(\left(4x-1\right)\left(4x+1\right)-4\left(1-2x\right)^2\)
\(=16x^2-1-4+16x-16x^2\)
\(=16x-5\)
#Học tốt!
Chắc là \(q\left(x\right)=x^2-4????\)
\(f\left(2\right)=2^5+2^2+1=37\) ; \(f\left(-2\right)=-27\)
Do \(f\left(x\right)\) có 5 nghiệm nên f(x) có dạng:
\(f\left(x\right)=\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)\left(x-x_4\right)\left(x-x_5\right)\)
\(\Rightarrow f\left(2\right)=\left(2-x_1\right)\left(2-x_2\right)\left(2-x_3\right)\left(2-x_4\right)\left(2-x_5\right)=37\)
\(f\left(-2\right)=\left(-2-x_1\right)\left(-2-x_2\right)\left(-2-x_3\right)\left(-2-x_4\right)\left(-2-x_5\right)=-27\)
\(\Rightarrow\left(2+x_1\right)\left(2+x_2\right)\left(2+x_3\right)\left(2+x_4\right)\left(2+x_5\right)=27\)
\(A=\left(x_1^2-4\right)\left(x^2_2-4\right)\left(x_3^2-4\right)\left(x_4^2-4\right)\left(x^2_5-4\right)\)
\(A=-\left(2-x_1\right)\left(2-x_2\right)\left(2-x_3\right)\left(2-x_4\right)\left(2-x_5\right)\left(2+x_1\right)\left(2+x_2\right)\left(2+x_3\right)\left(2+x_4\right)\left(2+x_5\right)\)
\(A=-37.27=-999\)
a, \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)
\(\Leftrightarrow x^2+8x+16-\left(x^2-x+x-1\right)=16\)
\(\Leftrightarrow8x+1=0\Leftrightarrow x=-\frac{1}{8}\)
b, \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)
\(\Leftrightarrow2x+255=0\Leftrightarrow x=-\frac{225}{2}\)
c, \(\left(x+2\right)\left(x-2\right)-x^3-2x=15\)
\(\Leftrightarrow x^2-4-x^3-2x=15\)( vô nghiệm )
d, \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)
\(\Leftrightarrow x^3+9x^2+27x+27-9x^3+6x^2-x+8x^3+1=28\)
\(\Leftrightarrow15x^2+26=0\Leftrightarrow x^2\ne-\frac{26}{15}\)( vô nghiệm )
Tính nhẩm hết á, sai bỏ quá nhá, sắp đi hc ... nên chất lượng hơi kém xíu ~~~
\(Q\left(x\right).\left(x-2\right)+28=\left(x^2+x+1\right)\left(x+2\right)\)
\(\Rightarrow Q\left(x\right).\left(x-2\right)+28=x^3+2x^2+x^2+2x+x+2\)
\(\Rightarrow Q\left(x\right).\left(x-2\right)=x^3+3x^2+3x-26\)
\(\Rightarrow Q\left(x\right)=\frac{x^3-2x^2+5x^2-10x+13x-26}{x-2}\)
\(=\frac{x^2.\left(x-2\right)+5x.\left(x-2\right)+13.\left(x-2\right)}{x-2}\)
\(=\frac{\left(x-2\right).\left(x^2+5x+13\right)}{x-2}\)
\(=x^2+5x+13\)