\(P\left(x\right)-Q\left(x\right)=\left(-2x+\frac{1}{2}x^2+3x^4-3x^2-3\right)-\left(3x^4+x^3-4x^2+1,5x^3-3x^4+2x+1\right)\\ P\left(x\right)-Q\left(x\right)=-2x+\frac{1}{2}x^2+3x^4-3x^2-3-3x^4-x^3+4x^2-1,5x^3+3x^4-2x-1\\ P\left(x\right)-Q\left(x\right)=\left(-2x-2x\right)+\left(\frac{1}{2}x^2-3x^2+4x^2\right)+\left(3x^4-3x^4+3x^4\right)+\left(-3-1\right)+\left(-x^3-1,5x^3\right)\\ P\left(x\right)-Q\left(x\right)=-4x+\frac{3}{2}x^2+3x^4-4-\frac{5}{2}x^3\)

\(R\left(x\right)+P\left(x\right)-Q\left(x\right)+x^2=2x^3-\frac{3}{2}x+1\\ \Rightarrow R\left(x\right)+\left(P\left(x\right)-Q\left(x\right)\right)+x^2=2x^3-\frac{3}{2}x+1\\ \Rightarrow R\left(x\right)-4x+\frac{3}{2}x^2+3x^4-4-\frac{5}{2}x^3+x^2=2x^3-\frac{3}{2}x+1\\ \Rightarrow R\left(x\right)-4x+\left(\frac{3}{2}x+x^2\right)+3x^4-4-\frac{5}{2}x^3=2x^3-\frac{3}{2}x+1\\ \Rightarrow R\left(x\right)-4x+\frac{5}{2}x^2+3x^4-4-\frac{5}{2}x^3=2x^3-\frac{3}{2}x+1\\ \Rightarrow R\left(x\right)=2x^3-\frac{3}{2}x+1+4x-\frac{5}{2}x^2-3x^4+4+\frac{5}{2}x^3\\ \Rightarrow R\left(x\right)=\left(2x^3+\frac{5}{2}x^3\right)+\left(\frac{-3}{2}x+4x\right)+\left(1+4\right)-\frac{5}{2}x^2-3x^4\\ \Rightarrow R\left(x\right)=\frac{9}{2}x^3+\frac{5}{2}x+5-\frac{5}{2}x^2-3x^4\)

Rút gọn:

\(P\left(x\right)=2x^2+4x\)

\(Q\left(x\right)=-x^3+2x^2-x+2\)

Để \(R\left(x\right)-P\left(x\right)-Q\left(x\right)=0\)

<=> \(R\left(x\right)=P\left(x\right)+Q\left(x\right)\)

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

= \(-x^3+4x^2+3x+2\)

KL: \(R\left(x\right)=-x^3+4x^2+3x+2\)

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

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

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

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

\(+\left(x-x-2x\right)+\left(-1+2-1\right)\)

\(=2x^4+4x^3-2x\)

`@` `\text {Ans}`

`\downarrow`

`P(x)+Q(x)-R(x)`

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

`= 5x^2 + 5x - 4 + 2x^2 - 3x + 1 - 4x^2 + x - 3`

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

`= 3x^2 + 3x - 6`

Thay `x=-1/2`

`3*(-1/2)^2 + 3*(-1/2) - 6`

`= 3*1/4 - 3/2 - 6`

`= 3/4 - 3/2 - 6`

`= -3/4 - 6 = -27/4`

Vậy, khi `x=-1/2` thì GTr của đa thức là `-27/4`

P(x)+Q(x)-R(x)

=5x^2+5x-4+2x^2-3x+1-4x^2+x-3

=2x^2+3x-6(1)

Khi x=-1/2 thì (1) sẽ là 2*1/4+3*(-1/2)-6=1/2-3/2-6=-7

\(\dfrac{1}{R\left(x\right)}=\dfrac{1}{x\left(x+2\right)}=\dfrac{1}{2}\left(\dfrac{1}{x}-\dfrac{1}{x+2}\right)\)

\(\Rightarrow S=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2022}-\dfrac{1}{2024}+\dfrac{1}{2023}-\dfrac{1}{2025}\right)+\dfrac{1}{2.2023}\)

\(=\dfrac{1}{2}\left(\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{2024}-\dfrac{1}{2025}\right)+\dfrac{1}{2.2023}\)

Một kết quả rất xấu

a)\(\left(2x+1\right)^2+\left(2x-1\right)^2-2\left(1+2x\right)\left(2x-1\right)\)

\(=\left(2x+1-2x+1\right)^2\)

\(=2^2=4\)

b)\(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)+3\left(x-1\right)\left(x+1\right)\)

\(=x^3-3x^2+3x-1-x^3+8+3x^2-3\)

\(=3x+4\)

a) Sắp xếp các hạng tử của mỗi đa thức theo lũy thừa tăng của biến.

Thu gọn: P(x) = 3x² - 5 + x⁴ - 3x³ - x⁶ - 2x² - x³

= x² - 5 + x⁴ - 4x³ - x⁶

Sắp xếp: P(x) = -5 + x² - 4x³ + x⁴ - x⁶

Thu gọn: Q(x) = x³ + 2x⁵ - x⁴ + x² - 2x³ + x - 1= -x³ +2x⁵ - x⁴ + x² + x - 1

Sắp xếp: Q(x) = -1 + x + x² - x³ - x⁴ + 2x⁵

b) Ta có:

.

a) Sắp xếp các hạng tử của mỗi đa thức theo lũy thừa tăng của biến.

Thu gọn: P(x) = 3x² - 5 + x⁴ - 3x³ - x⁶ - 2x² - x³

= x² - 5 + x⁴ - 4x³ - x⁶

Sắp xếp: P(x) = -5 + x² - 4x³ + x⁴ - x⁶

Thu gọn: Q(x) = x³ + 2x⁵ - x⁴ + x² - 2x³ + x - 1= -x³ +2x⁵ - x⁴ + x² + x - 1

Sắp xếp: Q(x) = -1 + x + x² - x³ - x⁴ + 2x⁵

b) Ta có:

\(M\left(x\right)+N\left(x\right)\)

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

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

\(M\left(x\right)-N\left(x\right)\)

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

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

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

\(M\left(x\right)+P\left(x\right)=N\left(x\right)\)

\(\Rightarrow P\left(x\right)=N\left(x\right)-M\left(x\right)\)

\(\Rightarrow P\left(x\right)=2x^4-2x^2+2x+1-\left(5x^3-x^2-4\right)\)

\(\Rightarrow P\left(x\right)=2x^4-2x^2+2x+1-5x^3+x^2+4\)

\(\Rightarrow P\left(x\right)=2x^4-5x^3-x^2+2x+5\)

khó vị

e mới hc lớp 7 thoi hehe......

bạn tham khảo nha

chúc bạn học tốt nha.

P(x)= 2x²(x−1)−5(x+2)−2x(x−2)
       =2x³-2x²-5x-10-2x²+4x
       =(-2x²-2x²)-(5x-4x) +2x³-10
       =2x³-4x²-x-10
Q(x)=x²(2x−3)−x(x+1)−(3x−2)
       =2x³-3x²-x²-x-3x+2
       =2x³-(3x²+x²)-(3x+x)+2
       =2x³-4x²-4x+2

a) B(x)=\(4x^5\) -\(2x^4\) +\(3x^3\) -\(2x^2\) +\(4x\) +\(\dfrac{-1}{2}\)

b) C(x)=\(2x^4-x^3+\dfrac{1}{2}+4x\)