b)

S2=6/2x5+6/5x8+6/8x11+...+6/29x32

=2.(3/2.5+3/5.8+...+3/29.32)

=2.(1/2-1/5+1/5-1/8+...+1/29-1/32)

=2.(1/2-1/32)

=2.15/32

=15/16

a)

Ta có:

S1=2/3x5+2/5x7+2/7x9+...+2/97x99

=1/3-1/5+1/5-1/7+...+1/97-1/99

=1/3-1/99

=32/99

b)

S2=6/2x5+6/5x8+6/8x11+...+6/29x32

=2.(3/2.5+3/5.8+...+3/29.32)

=2.(1/2-1/5+1/5-1/8+...+1/29-1/32)

=2.(1/2-1/32)

=2.15/32

=15/16

 Đúng 1 

a)

Ta có:

S1=2/3x5+2/5x7+2/7x9+...+2/97x99

=1/3-1/5+1/5-1/7+...+1/97-1/99

=1/3-1/99

=32/99

a)3/2 x (1/6+2/6+3/6)

=3/2 x 1

=3/2

mình chắc chắn 100%

\(\frac{3}{2}\cdot\frac{1}{6}+\frac{3}{2}\cdot\frac{2}{6}+\frac{3}{2}\cdot\frac{3}{6}\)

\(=\frac{3}{2}\cdot\left(\frac{1}{6}+\frac{2}{6}+\frac{3}{6}\right)\)

\(=\frac{3}{2}\cdot\frac{6}{6}\)

\(=\frac{3}{2}\cdot1=\frac{3}{2}\)

**** !

3\2

a) \(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}-\dfrac{6}{\sqrt{6}}=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{6}-\dfrac{6}{\sqrt{6}}\)

\(=\dfrac{1}{6\sqrt{6}}-\dfrac{6}{\sqrt{6}}=-\dfrac{35}{6\sqrt{6}}\)

b)\(\left(\sqrt{6}+\sqrt{5}\right)^2+\left(\sqrt{6}-\sqrt{5}\right)^2\)

\(=6+2\sqrt{30}+5+6-2\sqrt{30}+5=22\)

làm họ mình đi

huhu co to chua day

Phương pháp giải:

Thực hiện phép tính lần lượt từ trái sang phải rồi trình bày theo mẫu.

Lời giải chi tiết:

a)

2 × 6 : 3 = 12 : 3 = 4

5 × 4 : 2 = 20 : 2 = 10

b)

6 : 2 × 4 = 3 × 4 = 12

10 : 5 × 7 = 2 × 7 = 14

\(A=1+6+6^2+6^3+...+\)6^2010+6^2011

6A=6+6^2+6^3+6^4...+6^2011+6^2012

6A - A= 6^2012 - 1

5A= 6^2012 - 1

A=  (6^2012 - 1) / 5

Vậy A=(6^2012-1)/5

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#HỌC TỐT NHAAAA

a: \(A=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right)\cdot\dfrac{x+2}{6}\)

\(=\dfrac{x-2x-4+x-2}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{x+2}{6}=\dfrac{-6}{6}\cdot\dfrac{1}{x-2}=\dfrac{-1}{x-2}\)

b: x=2 ko thỏa mãn ĐKXĐ

=>Loại

Khi x=3 thì A=-1/(3-2)=-1

c: A=2

=>x-2=-1/2

=>x=3/2

Có (3x - \(\dfrac{1}{6}\))² ≥ 0 ∀ x; |2y-6| ≥ 0 ∀ y

=> (3x - \(\dfrac{1}{6}\))² + |2y-6| ≥ 0 ∀x,y

Mà (3x - \(\dfrac{1}{6}\))² + |2y-6| ≤ 0

=> (3x - \(\dfrac{1}{6}\))² = 0; |2y - 6| = 0

=> x = \(\dfrac{1}{18}\); y = 3;

=> A = \(\left(\dfrac{1}{18}\right)^2\) + 3² = \(9\dfrac{1}{324}\)

1. a) 40

    b) 3

    c) 18

    d) 3

2. a) 291

    b) 37

    c) 61

    d) 284