a) Thể tích của hình hộp chữ nhật là:

x × (x+1) (x-1)

=x × (x^2-1)

= x^3-x

b) Thể tích của hình hộp chữ nhật tại x = 4 là :

4^3 - 4= 60

a) Thể tích của hình hộp chữ nhật là:

x × (x+1) (x-1)

=x × (x^2-1)

= x^3-x

b) Thể tích của hình hộp chữ nhật tại x = 4 là :

4^3 - 4= 60

5x × 4x^2 + 5x × (-2x) × 10x^2 + (-2x) × (-5x) + (-2x) × 2 = -36

20x^3 + (-10x^2) + 5x + (-20x^3) + 10x^2 + (-4x) = -36

Vậy x = -36

a) x^4 - 5x^3 + 4x - 5 - x^4 + 3x^2 + 2x + 1

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

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

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

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

Ta có f(x)=100^x/100^x+10

=>f(a)=100^a/100^a+10

f(b)=100^b/100^b+10

=> f(a)+f(b)=100^a/100^a+10+100^b/100^b+10

=100^a(100^b+10)+10(100^a+10)/100^b(100^a+10)+10(100^a+10)

=100^a × 100^b+100^a × 10 + 100^b,100^b × 100^a/100^b × 100^a + 100^b × 10 + 100^a × 10 + 100

= 100^a + b + 100^a × 10 + 100^b + a + 100^b × 10 × 100^b + a + 100^b × 10 +100^a × 10 +100

Thế a+b=1

=>100+100^a×10+100+100^b×10/100+100^b+10+100^a×10+1000

<=> f(a) + f(b)= 1

Ta có f(x)=100^x/100^x+10

=>f(a)=100^a/100^a+10

f(b)=100^b/100^b+10

=> f(a)+f(b)=100^a/100^a+10+100^b/100^b+10

=100^a(100^b+10)+10(100^a+10)/100^b(100^a+10)+10(100^a+10)

=100^a × 100^b+100^a × 10 + 100^b,100^b × 100^a/100^b × 100^a + 100^b × 10 + 100^a × 10 + 100

= 100^a + b + 100^a × 10 + 100^b + a + 100^b × 10 × 100^b + a + 100^b × 10 +100^a × 10 +100

Thế a+b=1

=>100+100^a×10+100+100^b×10/100+100^b+10+100^a×10+1000

<=> f(a) + f(b)= 1

Ta có f(x)=100^x/100^x+10

=>f(a)=100^a/100^a+10

f(b)=100^b/100^b+10

=> f(a)+f(b)=100^a/100^a+10+100^b/100^b+10

=100^a(100^b+10)+10(100^a+10)/100^b(100^a+10)+10(100^a+10)

=100^a × 100^b+100^a × 10 + 100^b,100^b × 100^a/100^b × 100^a + 100^b × 10 + 100^a × 10 + 100

= 100^a + b + 100^a × 10 + 100^b + a + 100^b × 10 × 100^b + a + 100^b × 10 +100^a × 10 +100

Thế a+b=1

=>100+100^a×10+100+100^b×10/100+100^b+10+100^a×10+1000

<=> f(a) + f(b)= 1

Ta có f(x)=100^x/100^x+10

=>f(a)=100^a/100^a+10

f(b)=100^b/100^b+10

=> f(a)+f(b)=100^a/100^a+10+100^b/100^b+10

=100^a(100^b+10)+10(100^a+10)/100^b(100^a+10)+10(100^a+10)

=100^a × 100^b+100^a × 10 + 100^b,100^b × 100^a/100^b × 100^a + 100^b × 10 + 100^a × 10 + 100

= 100^a + b + 100^a × 10 + 100^b + a + 100^b × 10 × 100^b + a + 100^b × 10 +100^a × 10 +100

Thế a+b=1

=>100+100^a×10+100+100^b×10/100+100^b+10+100^a×10+1000

<=> f(a) + f(b)= 1

Ta có f(x)=100^x/100^x+10

=>f(a)=100^a/100^a+10

f(b)=100^b/100^b+10

=> f(a)+f(b)=100^a/100^a+10+100^b/100^b+10

=100^a(100^b+10)+10(100^a+10)/100^b(100^a+10)+10(100^a+10)

=100^a × 100^b+100^a × 10 + 100^b,100^b × 100^a/100^b × 100^a + 100^b × 10 + 100^a × 10 + 100

= 100^a + b + 100^a × 10 + 100^b + a + 100^b × 10 × 100^b + a + 100^b × 10 +100^a × 10 +100

Thế a+b=1

=>100+100^a×10+100+100^b×10/100+100^b+10+100^a×10+1000

<=> f(a) + f(b)= 1