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a)f(x)+g(x)=\(x^5-4x^4-2x^2-7-2x^5+6x^4-2x^2+6.\)
=\(-x^5+2x^4-4x^2-1\)
f(x)-g(x)=\(x^5-4x^4-2x^2-7+2x^5-6x^4+2x^2-6\)
=\(3x^5-10x^4-13\)
b)f(x)+g(x)=\(5x^4+7x^3-6x^2+3x-7-4x^4+2x^3-5x^2+4x+5\)
=\(x^4+9x^3-11x^2+7x-2\)
f(x)-g(x)=\(5x^4+7x^3-6x^2+3x-7+4x^4-2x^3+5x^2-4x-5\)
=\(9x^4+5x^3-x^2-x-12\)
a )
\(f\left(x\right)+g\left(x\right)=x^5-4x^4-2x^2-7+-2x^5+6x^4-2x^2+6\)
\(\Rightarrow f\left(x\right)+g\left(x\right)=\left(x^5-2x^5\right)+\left(6x^4-4x^4\right)-\left(2x^2+2x^2\right)+\left(6-7\right)\)
\(\Rightarrow f\left(x\right)+g\left(x\right)=-x^5+2x^4-4x^2-1\)
\(f\left(x\right)-g\left(x\right)=x^5-4x^4-2x^2-7-\left(-2x^5+6x^4-2x^2+6\right)\)
\(\Rightarrow f\left(x\right)-g\left(x\right)=x^5-4x^4-2x^2-7+2x^5-6x^4+2x^2-6\)
\(\Rightarrow f\left(x\right)-g\left(x\right)=\left(x^5+2x^5\right)-\left(4x^4+6x^4\right)+\left(2x^2-2x^2\right)-\left(6+7\right)\)
\(\Rightarrow f\left(x\right)-g\left(x\right)=3x^5-10x^4-13\)
\(F=-3\left(x-8\right)\left(2x+1\right)-\left(x+5\right)\left(2-3x\right)-4x\left(x-6\right)\)
\(=-3\left(-3-8\right)\left(-6+1\right)-\left(5-3\right)\left(2+9\right)+12\left(-9\right)\)
\(=-3\left(-11\right)\left(-5\right)-\left(-2\right)11-12.9\)
\(=-165+22-108=22-273=-251\)
\(G=\left(5x-4\right)\left(5-2x\right)-7x\left(x^2-4x+3\right)+\left(x^2-4x\right)\left(7x-2\right)\)
\(=\left(5-4\right)\left(5-2\right)-7\left(1-4+3\right)+\left(1-4\right)\left(7-2\right)\)
\(=3-7.0+5.\left(-3\right)=3-15=-12\)
\(H=\left(-3x+5\right)\left(x-6\right)-\left(x-1\right)\left(x^2-2x+3\right)+\left(x+2\right)\left(x^2-3\right)\)
\(=\left(3+5\right)\left(-1-6\right)-\left(-1-1\right)\left(1+2+3\right)+\left(-1+2\right)\left(1-3\right)\)
\(=8\left(-7\right)-\left(-2\right)6+1\left(-2\right)=-56+12-2=-46\)
\(L=5x\left(x-1\right)\left(2x+3\right)-10x\left(x^2-4x+5\right)-\left(x-1\right)\left(x-4\right)\)
\(=-\frac{5}{3}\left(-\frac{4}{3}\right)\left(-\frac{2}{3}+3\right)+\frac{10}{3}\left(\frac{1}{9}+\frac{4}{3}+5\right)-\left(-\frac{4}{3}\right)\left(-\frac{1}{3}-4\right)\)
\(=\frac{20}{9}\left(\frac{7}{3}\right)+\frac{10}{3}\left(\frac{13}{9}+5\right)+\frac{4}{3}\left(-\frac{13}{3}\right)\)
\(=\frac{140}{27}+\frac{10}{3}.\frac{58}{9}-\frac{52}{9}\)
\(=\frac{140}{27}+\frac{580}{27}-\frac{156}{27}=\frac{140+580-156}{27}=\frac{720-156}{27}=\frac{564}{27}\)
\(M=-7x\left(x-5\right)-\left(x-1\right)\left(x^2-x-2\right)+x^2\left(x-3\right)-5x\left(x-8\right)\)
\(=\frac{-7}{2}\left(\frac{1}{2}-5\right)+\frac{\left(\frac{1}{4}-\frac{1}{2}-2\right)}{2}+\frac{1}{4}\left(\frac{1}{2}-3\right)-\frac{5}{2}\left(\frac{1}{2}-8\right)\)
\(=\frac{7}{2}.\frac{9}{2}-\frac{9}{8}-\frac{1}{4}.\frac{5}{2}+\frac{5}{2}.\frac{15}{2}\)
\(=\frac{63}{4}-\frac{9}{8}-\frac{5}{8}+\frac{75}{4}=\frac{138}{4}-\frac{7}{4}=\frac{131}{4}\)
\(f\left(x\right)=x^5-4x^4-2x^2-7\)
\(g\left(x\right)=-2x^5+6x^4-2x^2+6\)
\(f\left(x\right)+g\left(x\right)=-x^5+2x^4-4x^2-1\)
\(f\left(x\right)-g\left(x\right)=3x^5-10x^4-13\)
a) Ta có:
\(f\left(x\right)+g\left(x\right)=\left(2x^3-x^2+5\right)+\left(x^2+2x-2x^3-1\right)\)
\(f\left(x\right)+g\left(x\right)=2x^3-x^2+5+x^2+2x-2x^3-1\)
\(f\left(x\right)+g\left(x\right)=2x-4\)
\(f\left(x\right)+g\left(x\right)=2\left(x-2\right)\)
Ta có:
\(f\left(x\right)-g\left(x\right)=\left(2x^3-x^2+5\right)-\left(x^2+2x-2x^3-1\right)\)
\(f\left(x\right)-g\left(x\right)=2x^3-x^2+5-x^2-2x+2x^3+1\)
\(f\left(x\right)-g\left(x\right)=4x^3-2x+6\)
b)
\(f\left(0\right)=2.0^3-0^2+5\)
\(f\left(0\right)=5\)
\(f\left(\dfrac{1}{2}\right)=2.\left(\dfrac{1}{2}\right)^3-\left(\dfrac{1}{2}\right)^2+5\)
\(f\left(\dfrac{1}{2}\right)=2.\dfrac{1}{8}-\dfrac{1}{4}+5\)
\(f\left(\dfrac{1}{2}\right)=\dfrac{1}{4}-\dfrac{1}{4}+5\)
\(f\left(\dfrac{1}{2}\right)=5\)
\(f\left(-5\right)=2.\left(-5\right)^3-\left(-5\right)^2+5\)
\(f\left(-5\right)=2.\left(-125\right)-25+5\)
\(f\left(-5\right)=-250-25+5\)
\(f\left(-5\right)=-270\)
c) Ta có:
\(f\left(x\right)+g\left(x\right)=0\)
\(\Leftrightarrow2\left(x-2\right)=0\)
\(\Rightarrow x-2=0\)
\(\Rightarrow x=2\)
Vậy nghiệm cùa f(x) + g(x) là 2
4x^3-3x^2 +1 x^2+2x-1 4x 4x^3+8x^2-4x - -11x^2+4x+1 -11 -11x^2-22x+11 - 26x-10
OLM chỉ có phần chụp ảnh cho CTV
Lưu ý bạn cố phải viết thẳng hàng vì OLM ko viết đc
c) thay x=1 vào đa thức f(x) ta có: f(1)=4.1^3-1^2+2.1-5
=4-2+2-5
=- 1
vậy 1 k phải là nghiệm của đa thức f(x)
MÌNH CHỈ LÀM ĐƯỢC C THÔI HOK TỐT
làm sai nha chỗ nào là 1 thì thay bằng -1 nha kq sẽ ra nha
\(f\left(x\right)+g\left(x\right)=\left[x\left(1-2x\right)+\left(2x^2-x+4\right)\right]+\left[x\left(x-5\right)-x\left(x+2\right)+7x\right]\)
\(=x-2x^2+2x^2-x+4+x^2-5x-x^2-2x+7x\)
\(=4\)
\(f\left(x\right)-g\left(x\right)=\left[x\left(1-2x\right)+\left(2x^2-x+4\right)\right]-\left[x\left(x-5\right)-x\left(x+2\right)+7x\right]\)
\(=x-2x^2+2x^2-x+4-x^2+5x+x^2+2x-7x\)
\(=4\)