cho f(x)=5x^3-7x^2+x+7
g(x)=7x^3 +2x+5
h(x)=2x^3 +4x+1
tính k(x)=f(x) -g(x) +h(x)
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KN
22 tháng 6 2019
a) \(f\left(x\right)=5x^3-7x^2+2x+5\)
\(\Rightarrow f\left(1\right)=5.1^3-7.1^2+2.1+5\)
\(\Rightarrow f\left(1\right)=5.1-7.1+2+5\)
\(\Rightarrow f\left(1\right)=5-7+7\)
\(\Rightarrow f\left(1\right)=5\)
Vậy f(1) = 5.
\(g\left(x\right)=7x^3-7x^2+2x+5\)
\(\Rightarrow g\left(\frac{1}{2}\right)=7.\left(\frac{1}{2}\right)^3-7.\left(\frac{1}{2}\right)^2+2.\frac{1}{2}+5\)
\(\Leftrightarrow g\left(\frac{1}{2}\right)=7.\frac{1}{8}-7.\frac{1}{4}+1+5\)
\(\Leftrightarrow g\left(\frac{1}{2}\right)=\frac{7}{8}-\frac{14}{8}+6\)
\(\Leftrightarrow g\left(\frac{1}{2}\right)=\frac{-7}{8}+\frac{48}{8}\)
\(\Leftrightarrow g\left(\frac{1}{2}\right)=\frac{41}{8}\)
Vậy \(g\left(\frac{1}{2}\right)=\frac{41}{8}\)
KN
22 tháng 6 2019
\(h\left(x\right)=2x^3+4x+1\)
\(\Rightarrow h\left(0\right)=2.0^3+4.0+1\)
\(\Rightarrow h\left(0\right)=0+0+1\)
\(\Rightarrow h\left(0\right)=1\)
Vậy \(h\left(0\right)=1\)
7 tháng 8 2019
a) \(\) Ta có : \(F\left(x\right)=5x^3-7x^2+x+7\)
\(\Rightarrow F\left(-1\right)=5.\left(-1\right)^3-7.\left(-1\right)^2+\left(-1\right)+7\)
\(=\left(-5\right)-7-1+7\)
\(=-6\)
Vậy : \(F\left(-1\right)=-6\)
b) Ta có : \(K\left(x\right)=F\left(x\right)-G\left(x\right)+H\left(x\right)\)
\(\Leftrightarrow K\left(x\right)=5x^3-7x^2+x+7-\left(7x^3-7x^2+2x+5\right)+\left(2x^3+4x+1\right)\)
\(\Leftrightarrow K\left(x\right)=\left(5x^3-7x^3+2x^3\right)+\left(-7x^2+7x^2\right)+\left(x-2x+4x\right)+\left(7-5+1\right)\)
\(\Leftrightarrow K\left(x\right)=3x+3\)
Vậy : \(K\left(x\right)=3x+3\)
c) Ta có : \(K\left(x\right)=3x+3\)
\(\Rightarrow\) Bậc của \(K\left(x\right)\) là 1.
Xét \(K\left(x\right)=0\Leftrightarrow3x+3=0\)
\(\Leftrightarrow3.\left(x+1\right)=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Vậy : nghiệm của đa thức \(K\left(x\right)\) là \(x=-1\)
VM
7 tháng 8 2019
a) \(F\left(x\right)=5x^3-7x^2+x+7\)
=> \(F\left(-1\right)=5.\left(-1\right)^3-7.\left(-1\right)^2+\left(-1\right)+7\)
\(F\left(-1\right)=\left(-5\right)-7+\left(-1\right)+7\)
\(F\left(-1\right)=\left(-13\right)+7\)
\(F\left(-1\right)=-6.\)
Vậy \(F\left(-1\right)=-6.\)
\(G\left(x\right)=7x^3-7x^2+2x+5\)
=> \(G\left(-\frac{1}{2}\right)=7.\left(-\frac{1}{2}\right)^3-7.\left(-\frac{1}{2}\right)^2+2.\left(-\frac{1}{2}\right)+5\)
\(G\left(-\frac{1}{2}\right)=\left(-\frac{7}{8}\right)-\frac{7}{4}+\left(-1\right)+5\)
\(G\left(-\frac{1}{2}\right)=\left(-\frac{29}{8}\right)+5\)
\(G\left(-\frac{1}{2}\right)=\frac{11}{8}.\)
Vậy \(G\left(-\frac{1}{2}\right)=\frac{11}{8}.\)
\(H\left(x\right)=2x^3+4x+1\)
=> \(H\left(0\right)=2.0^3+4.0+1\)
\(H\left(0\right)=0+0+1\)
\(H\left(0\right)=1.\)
Vậy \(H\left(0\right)=1.\)
Chúc bạn học tốt!
7 tháng 7 2018
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)=\(5x^4+7x^3-6x^2+3x-7+4x^4-2x^3+5x^2-4x-5\)
=\(9x^4+5x^3-x^2-x-12\)
AK
7 tháng 7 2018
a )
\(f\left(x\right)+g\left(x\right)=x^5-4x^4-2x^2-7+-2x^5+6x^4-2x^2+6\)
\(\Rightarrow f\left(x\right)+g\left(x\right)=\left(x^5-2x^5\right)+\left(6x^4-4x^4\right)-\left(2x^2+2x^2\right)+\left(6-7\right)\)
\(\Rightarrow f\left(x\right)+g\left(x\right)=-x^5+2x^4-4x^2-1\)
\(f\left(x\right)-g\left(x\right)=x^5-4x^4-2x^2-7-\left(-2x^5+6x^4-2x^2+6\right)\)
\(\Rightarrow f\left(x\right)-g\left(x\right)=x^5-4x^4-2x^2-7+2x^5-6x^4+2x^2-6\)
\(\Rightarrow f\left(x\right)-g\left(x\right)=\left(x^5+2x^5\right)-\left(4x^4+6x^4\right)+\left(2x^2-2x^2\right)-\left(6+7\right)\)
\(\Rightarrow f\left(x\right)-g\left(x\right)=3x^5-10x^4-13\)
7 tháng 4 2019
a) \(f\left(x\right)=5x^3-7x^2+x+7+4x^5\)
\(f\left(-1\right)=5.\left(-1\right)^3-7.\left(-1\right)^2+\left(-1\right)+7+4.\left(-1\right)^5\)
\(f\left(-1\right)=\left(-5\right)-7+\left(-1\right)+7+\left(-4\right)\)
\(f\left(-1\right)=-10\)
\(\Rightarrow f\left(x\right)=-10\)
\(g\left(x\right)=4x^5-3x^3-7x^2+2x+5\)
\(g\left(0\right)=4.0^5-3.0^3-7.0^2+2.0+5\)
\(g\left(0\right)=5\)
\(\Rightarrow g\left(x\right)=0\)
\(h\left(x\right)=x^2-4x-5\)
\(h\left(-\frac{1}{2}\right)=\left(-\frac{1}{2}\right)^2-4.\left(-\frac{1}{2}\right)-5\)
\(h\left(-\frac{1}{2}\right)=\frac{1}{4}-\left(-2\right)-5\)
\(h\left(-\frac{1}{2}\right)=-\frac{11}{4}\)
\(\Rightarrow h\left(x\right)=-\frac{11}{4}\)
7 tháng 4 2019
\(f\left(-1\right)=5\left(-1\right)^3-7\left(-1\right)^2+\left(-1\right)+7+4\left(-1\right)^5\)
\(f\left(-1\right)=-5-7-1+7-4\)
\(f\left(-1\right)=-10\)
\(g\left(0\right)=4.0^5-3.0^3-7.0^2+2.0+5\)
\(g\left(0\right)=0-0-0+0+5\)
\(g\left(0\right)=5\)
\(h\left(-\frac{1}{2}\right)=\left(-\frac{1}{2}\right)^2-4\left(-\frac{1}{2}\right)-5\)
\(h\left(-\frac{1}{2}\right)=\frac{1}{4}-\left(-2\right)-5\)
\(h\left(-\frac{1}{2}\right)=\frac{1}{4}+2-5\)
\(h\left(-\frac{1}{2}\right)=-\frac{11}{4}\)
NH
1
RN
20 tháng 6 2020
f(x) = 6x7 - 5x3 + 1
g(x) = -3 + 2x - 4x7
h(x) = -2x7 - x5 + 7x2 + x6
\(f\left(x\right)+g\left(x\right)+h\left(x\right)=x^6-x^5-5x^3-7x^2+2x-2\)
15 tháng 5 2017
a) Thu gọn, sắp xếp các đa thức theo lũy thừa tăng của biến
= -9 - 2x2 + 3x3 - 6x5 - 3x7
b) Tính -9 - 2x2 + 3x3 - 6x5 - 3x7 ) + (-12 + 3x3 + x4 + x5 - x6 - 6x7 - 5x8 ) - (2x - 3x2 + 4x3 +4x5 -4x6 - 10x7)
= - 9 - 2x2 + 3x3 - 6x5 - 3x7 -12 + 3x3 + x4 + x5 - x6 - 6x7 - 5x8 - 2x + 3x2 - 4x3 - 4x5 + 4x6 + 10x7
= -21 - 2x + x2 + 2x3 + x4 - 9x5 + 3x6 + x7 - 5x8
Có: \(f\left(x\right)=5x^3-7x^2+x+7\)
\(g\left(x\right)=7x^3+2x+5\)
\(h\left(x\right)=2x^3+4x+1\)
\(\Rightarrow k\left(x\right)=f\left(x\right)-g\left(x\right)+h\left(x\right)\\ =5x^3-7x^2+x+7-7x^3-2x-5+2x^3+4x+1\\ =\left(5x^3-7x^3+2x^3\right)+\left(7-5+1\right)+\left(4x-2x+x\right)-7x^2\\ =0+3+3x-7x^2\\ =-7x^2+3x+3\)