b) Ta có:

Tại x= 5 thì biểu thức P xác định nên giá trị của biểu thức P tại x = 5 là:

b)

Sửa đề: f(x)=A(x)+B(x)

Ta có: f(x)=A(x)+B(x)

\(=x^5+7x^4-9x^3-2x^2-\dfrac{1}{4}x-x^5+5x^4-2x^3+4x^2-\dfrac{1}{4}\)

\(=12x^4-11x^3+2x^2-\dfrac{1}{4}x-\dfrac{1}{4}\)

a) Ta có: \(A\left(x\right)=x^5-3x^2+7x^4-9x^3+x^2-\dfrac{1}{4}x\)

\(=x^5+7x^4-9x^3+\left(-3x^2+x^2\right)-\dfrac{1}{4}x\)

\(=x^5+7x^4-9x^3-2x^2-\dfrac{1}{4}x\)

Ta có: \(B\left(x\right)=5x^4-x^5+x^2-2x^3+3x^2-\dfrac{1}{4}\)

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

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

b) Ta có: G(x)=A(x)-B(x)

\(=x^5+7x^4-9x^3-2x^2-\dfrac{1}{4}x+x^5-5x^4+2x^3-4x^2+\dfrac{1}{4}\)

\(=2x^5+2x^4-7x^3-6x^2-\dfrac{1}{4}x+\dfrac{1}{4}\)

1: 7+7x2+7x3+7x4=7x1+7x2+7x3+7x4

=7x(1+2+3+4)=7x10

2:a) 54-18x3:2+4=54-54:2+4=54-27+4=31

b)54-18x3:(2+4)=54-54:6=54-9=45

3:Xx7+X=384

Xx(1+7)=384

Xx8=384

X=384:8

X=48.Vậy X =48

a) P(x) = 7x² . (x² – 5x + 2 ) – 5x .(x³ – 7x² + 3x)

= 7x² . x² + 7x² . (-5x) + 7x² . 2 – [5x. x³ + 5x . (-7x²) + 5x . 3x]

= 7. (x² . x²) + [7.(-5)] . (x² . x) + (7.2).x²  – {5. (x.x³) + [5.(-7)]. (x.x²) + (5.3).(x.x)}

= 7x⁴ + (-35). x³ + 14x²  – [ 5x⁴ + (-35)x³ + 15x² ]

= 7x⁴ + (-35). x³ + 14x²   - 5x⁴ + 35x³ - 15x²

= (7x⁴ – 5x⁴) + [(-35). x³ + 35x³ ] + (14x² - 15x² )

= 2x⁴ + 0 - x² 

= 2x⁴ – x²

b) Thay x = \( - \dfrac{1}{2}\) vào P(x), ta được:

P(\( - \dfrac{1}{2}\)) = 2. (\( - \dfrac{1}{2}\))⁴ –  (\( - \dfrac{1}{2}\))² \))

 \(\begin{array}{l} = 2.\dfrac{1}{{16}} - \dfrac{1}{4} \\ = \dfrac{1}{8} - \dfrac{{2}}{8} \\ = \dfrac{-1}{8} \end{array}\)

\(50\%+\dfrac{1}{2}+\dfrac{1}{7}\times\dfrac{2}{5}+\dfrac{1}{7}\times\dfrac{3}{5}\)

\(=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{2}{35}+\dfrac{3}{35}\)

\(=1+\dfrac{5}{35}\)

\(=1+\dfrac{1}{7}\)

\(=\dfrac{8}{7}\)

\(50\%+\dfrac{1}{2}+\dfrac{1}{7}x\dfrac{2}{5}+\dfrac{1}{7}x\dfrac{3}{5}\)

\(=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{7}x\left(\dfrac{2}{5}+\dfrac{3}{5}\right)\)

\(=1+\dfrac{1}{7}x1\)

\(=\dfrac{7}{7}+\dfrac{1}{7}=\dfrac{8}{7}\)

50% + 1/2 + 1/7 × 2/5 + 1/7 × 3/5

= 1/2 + 1/2 + 1/7 × (2/5 + 3/5)

= 1 + 1/7 × 1

= 1 + 1/7

= 7/7 + 1/7

= 8/7

8/7

=7x3³-8=7x27-8=181

\(7.3^3-\left(8.1\right)^2=7.27-8^2=189-64=125\)

Bài 4: 

b: \(=x^2z\left(-1+3-7\right)=-5x^2z=-5\cdot\left(-1\right)^2\cdot\left(-2\right)=10\)

c: \(=xy^2\left(5+0.5-3\right)=2.5xy^2=2.5\cdot2\cdot1^2=5\)

`@` `\text {Ans}`

`\downarrow`

`a)`

Thu gọn:

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

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

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

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

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

`= x^5 - x^4 - x^3 - x^2 + 7x - 2`

`@` Tổng:

`P(x)+Q(x)=`\((-x^5 + 5x^4 + 2x^2 + 2x - 4) + (x^5 - x^4 - x^3 - x^2 + 7x - 2)\)

`= -x^5 + 5x^4 + 2x^2 + 2x - 4 + x^5 - x^4 - x^3 - x^2 + 7x - 2`

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

`= 4x^4 - x^3 + x^2 + 9x - 6`

`@` Hiệu:

`P(x) - Q(x) =`\((-x^5 + 5x^4 + 2x^2 + 2x - 4) - (x^5 - x^4 - x^3 - x^2 + 7x - 2)\)

`= -x^5 + 5x^4 + 2x^2 + 2x - 4 - x^5 + x^4 + x^3 + x^2 - 7x + 2`

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

`= -2x^5 + 6x^4 + x^3 + 3x^2 - 5x - 2`

`b)`

`@` Thu gọn:

\(H (x) = ( 3x^5 - 2x^3 + 8x + 9) - ( 3x^5 - x^4 + 1 - x^2 + 7x)\)

`= 3x^5 - 2x^3 + 8x + 9 - 3x^5 + x^4 - 1 + x^2 - 7x`

`= (3x^5 - 3x^5) + x^4 - 2x^3 - x^2 + (8x + 7x) + (9+1)`

`= x^4 - 2x^3 - x^2 + 15x + 10`

\(R( x) = x^4 + 7x^3 - 4 - 4x ( x^2 + 1) + 6x\)

`= x^4 + 7x^3 - 4 - 4x^3 - 4x + 6x`

`= x^4 + (7x^3 - 4x^3) + (-4x + 6x) - 4`

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

`@` Tổng:

`H(x)+R(x)=` \((x^4 - 2x^3 - x^2 + 15x + 10)+(x^4 + 3x^3 + 2x - 4)\)

`= x^4 - 2x^3 - x^2 + 15x + 10+x^4 + 3x^3 + 2x - 4`

`= (x^4 + x^4) + (-2x^3 + 3x^3) - x^2 + (15x + 2x) + (10-4)`

`= 2x^4 + x^3 - x^2 + 17x + 6`

`@` Hiệu: 

`H(x) - R(x) =`\((x^4 - 2x^3 - x^2 + 15x + 10)-(x^4 + 3x^3 + 2x - 4)\)

`=x^4 - 2x^3 - x^2 + 15x + 10-x^4 - 3x^3 - 2x + 4`

`= (x^4 - x^4) + (-2x^3 - 3x^3) - x^2 + (15x - 2x) + (10+4)`

`= -5x^3 - x^2 + 13x + 14`

`@` `\text {# Kaizuu lv u.}`