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a: \(A\left(x\right)=2x^4-x^3+3x^2+9x-2\)
\(B\left(x\right)=2x^4-5x^3-x+9\)
\(C\left(x\right)=x^4+4x^2+5\)
A(x): bậc 4; hệ số cao nhất là 2; hệ số tự do là -2
B(x): bậc 4; hệ số cao nhất là 4; hệ số tự do là 9
b: M(x)=A(x)+B(x)=4x^4-6x^3+3x^2+8x+7
N(x)=B(x)-A(x)=-4x^3-3x^2-10x+11
c: Q(x)=-N(x)=4x^3+3x^2+10x-11
20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)
Vậy...
a) Ta có: \(x^2+4x+3\)
\(=x^2+x+3x+3\)
\(=x\left(x+1\right)+3\left(x+1\right)\)
\(=\left(x+1\right)\left(x+3\right)\)
b) Ta có: \(16x-5x^2-3\)
\(=-5x^2+16x-3\)
\(=-5x^2+15x+x-3\)
\(=-5x\left(x-3\right)+\left(x-3\right)\)
\(=\left(x-3\right)\left(-5x+1\right)\)
c) Ta có: \(2x^2+7x+5\)
\(=2x^2+2x+5x+5\)
\(=2x\left(x+1\right)+5\left(x+1\right)\)
\(=\left(x+1\right)\left(2x+5\right)\)
d) Ta có: \(2x^2+3x-5\)
\(=2x^2+5x-2x-5\)
\(=x\left(2x+5\right)-\left(2x+5\right)\)
\(=\left(2x+5\right)\left(x-1\right)\)
e) Ta có: \(x^3-3x^2+1-3x\)
\(=\left(x+1\right)\cdot\left(x^2-x+1\right)-3x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1-3x\right)\)
\(=\left(x+1\right)\left(x^2-4x+1\right)\)
f) Ta có: \(x^2-4x-5\)
\(=x^2-4x+4-9\)
\(=\left(x-2\right)^2-3^2\)
\(=\left(x-2-3\right)\left(x-2+3\right)\)
\(=\left(x-5\right)\left(x+1\right)\)
g) Ta có: \(\left(a^2+1\right)^2-4a^2\)
\(=\left(a^2+1\right)^2-\left(2a\right)^2\)
\(=\left(a^2+1-2a\right)\left(a^2+1+2a\right)\)
\(=\left(a-1\right)^2\cdot\left(a+1\right)^2\)
h) Ta có: \(x^3-3x^2-4x+12\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-4\right)\)
\(=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
i) Ta có: \(x^4+x^3+x+1\)
\(=x^3\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3+1\right)\)
\(=\left(x+1\right)^2\cdot\left(x^2-x+1\right)\)
k) Ta có: \(x^4-x^3-x^2+1\)
\(=x^3\left(x-1\right)-\left(x^2-1\right)\)
\(=x^3\left(x-1\right)-\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x^3-x-1\right)\)
l) Ta có: \(\left(2x+1\right)^2-\left(x-1\right)^2\)
\(=\left(2x+1-x+1\right)\left(2x+1+x-1\right)\)
\(=3x\left(x+2\right)\)
m) Ta có: \(x^4+4x^2-5\)
\(=x^4-x^2+5x^2-5\)
\(=x^2\left(x^2-1\right)+5\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2+5\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)
a) Ta có: \(4x^2-6x\)
\(=2x\left(2x-3\right)\)
b) Ta có: \(9x^4y^3+3x^2y^4\)
\(=3x^2y^3\left(3x^2+y\right)\)
c) Ta có: 3(x-y)-5x(y-x)
=3(x-y)+5x(x-y)
=(x-y)(3+5x)
d) Ta có: \(x^3-2x^2+5x\)
\(=x\left(x^2-2x+5\right)\)
e) Ta có: \(5\left(x+3y\right)-15x\left(x+3y\right)\)
\(=\left(x+3y\right)\left(5-15x\right)\)
\(=5\left(x+3y\right)\cdot\left(1-3x\right)\)
f) Ta có: \(2x^2\left(x+1\right)+4\left(x+1\right)\)
\(=\left(x+1\right)\left(2x^2+4\right)\)
\(=2\left(x+1\right)\left(x^2+2\right)\)
Ta có: \(P\left(x\right)=x^2+5x^4-3x^3+x^2+4x^4+3x^3-x+5\)
\(=9x^4+2x^2-x+5\)
Ta có: \(Q\left(x\right)=x-5x^3-x^2-x^4+4x^3-x^2-3x-1\)
\(=-x^4-x^3-2x^2-2x-1\)
Ta có: P(x)+Q(x)
\(=9x^4+2x^2-x+5-x^4-x^3-2x^2-2x-1\)
\(=8x^4-x^3-3x+4\)
Ta có: P(x)-Q(x)
\(=9x^4+2x^2-x+5+x^4+x^3+2x^2+2x+1\)
\(=10x^4+x^3+4x^2+x+6\)
