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 \(g\left(\frac{1}{2}\right)=\frac{41}{8}\)

\(h\left(x\right)=2x^3+4x+1\)

\(\Rightarrow h\left(0\right)=2.0^3+4.0+1\)

\(\Rightarrow h\left(0\right)=0+0+1\)

\(\Rightarrow h\left(0\right)=1\)

Vậy \(h\left(0\right)=1\)

b)\(f\left(x\right)-g\left(x\right)+h\left(x\right)\)

\(=5x^3-7x^2+2x+5-2x^3-4x-1+7x^3-7x^2+2x+5\)

Rút gọn rồi tìm k(x)

Tìm M(x) tương tự

c) Bậc của k(x) là đơn thức có bậc cao nhất là 3

Nghiệm của k(x) là khi k(x) = 0 . Như câu a)

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 \(x=-1\)

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!

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 ) 

\(f\left(x\right)+g\left(x\right)=x^5-4x^4-2x^2-7+-2x^5+6x^4-2x^2+6\)

\(\Rightarrow f\left(x\right)+g\left(x\right)=\left(x^5-2x^5\right)+\left(6x^4-4x^4\right)-\left(2x^2+2x^2\right)+\left(6-7\right)\)

\(\Rightarrow f\left(x\right)+g\left(x\right)=-x^5+2x^4-4x^2-1\)

\(f\left(x\right)-g\left(x\right)=x^5-4x^4-2x^2-7-\left(-2x^5+6x^4-2x^2+6\right)\)

\(\Rightarrow f\left(x\right)-g\left(x\right)=x^5-4x^4-2x^2-7+2x^5-6x^4+2x^2-6\)

\(\Rightarrow f\left(x\right)-g\left(x\right)=\left(x^5+2x^5\right)-\left(4x^4+6x^4\right)+\left(2x^2-2x^2\right)-\left(6+7\right)\)

\(\Rightarrow f\left(x\right)-g\left(x\right)=3x^5-10x^4-13\)

b ) Làm tương tự 

a) \(f\left(x\right)=5x^3-7x^2+x+7+4x^5\)

\(f\left(-1\right)=5.\left(-1\right)^3-7.\left(-1\right)^2+\left(-1\right)+7+4.\left(-1\right)^5\)

\(f\left(-1\right)=\left(-5\right)-7+\left(-1\right)+7+\left(-4\right)\)

\(f\left(-1\right)=-10\)

\(\Rightarrow f\left(x\right)=-10\)

\(g\left(x\right)=4x^5-3x^3-7x^2+2x+5\)

\(g\left(0\right)=4.0^5-3.0^3-7.0^2+2.0+5\)

\(g\left(0\right)=5\)

\(\Rightarrow g\left(x\right)=0\)

\(h\left(x\right)=x^2-4x-5\)

\(h\left(-\frac{1}{2}\right)=\left(-\frac{1}{2}\right)^2-4.\left(-\frac{1}{2}\right)-5\)

\(h\left(-\frac{1}{2}\right)=\frac{1}{4}-\left(-2\right)-5\)

\(h\left(-\frac{1}{2}\right)=-\frac{11}{4}\)

\(\Rightarrow h\left(x\right)=-\frac{11}{4}\)

\(f\left(-1\right)=5\left(-1\right)^3-7\left(-1\right)^2+\left(-1\right)+7+4\left(-1\right)^5\)

\(f\left(-1\right)=-5-7-1+7-4\)

\(f\left(-1\right)=-10\)

\(g\left(0\right)=4.0^5-3.0^3-7.0^2+2.0+5\)

\(g\left(0\right)=0-0-0+0+5\)

\(g\left(0\right)=5\)

\(h\left(-\frac{1}{2}\right)=\left(-\frac{1}{2}\right)^2-4\left(-\frac{1}{2}\right)-5\)

\(h\left(-\frac{1}{2}\right)=\frac{1}{4}-\left(-2\right)-5\)

\(h\left(-\frac{1}{2}\right)=\frac{1}{4}+2-5\)

\(h\left(-\frac{1}{2}\right)=-\frac{11}{4}\)

f(x) = 6x⁷ - 5x³ + 1

g(x) = -3 + 2x - 4x⁷

h(x) = -2x⁷ - x⁵ + 7x² + x⁶

\(f\left(x\right)+g\left(x\right)+h\left(x\right)=x^6-x^5-5x^3-7x^2+2x-2\)

a) Thu gọn, sắp xếp các đa thức theo lũy thừa tăng của biến

$f\left(x\right)=x^2+2x^3-7x^5-9-6x^7+x^3+x^2+x^5-4x^2+3x^7$

= -9 - 2x² + 3x³ - 6x⁵ - 3x⁷

$g\left(x\right)=x^5+2x^3-5x^8-x^7+x^3+4x^2-5x^7+x^4-4x^2-x^6-12$

$\mathrm{=\; -12\; +\; 3x3\; +\; x4\; +\; x5\; -\; x6\; -\; 6x7\; -\; 5x8}$

$h\left(x\right)=x+4x^5-5x^6-x^7+4x^3+x^2-2x^7+x^6-4x^2-7x^7+x$

$\mathrm{=\; 2x\; -\; 3x2\; +\; 4x3\; +4x5\; -4x6\; -\; 10x7}$

b) Tính $f\left(x\right)+g\left(x\right)-h(x)\; =\; ($-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

a, 4x^3 +3x^2+7x

b, = 0