Lời giải:

a. $\frac{2-x}{4}=\frac{3x-1}{3}$

$\Rightarrow 3(2-x)=4(3x-1)$

$\Rightarrow 6-3x=12x-4$

$\Rightarrow 6+4=12x+3x$

$\Rightarrow 10=15x$

$\Rightarrow x=\frac{10}{15}=\frac{2}{3}$

b.

$\frac{x}{7}=\frac{x+16}{35}$

$\Rightarrow \frac{5x}{35}=\frac{x+16}{35}$

$\Rightarrow 5x=x+16$

$\Rightarrow 4x=16$

$\Rightarrow x=4$

c.

$\sqrt{x^2+1}=3$

$\Rightarrow x^2+1=9$

$\Rightarrow x^2=8\Rightarrow x=\pm \sqrt{8}=\pm 2\sqrt{2}$

Lời giải:

$3^6-M=3^0+3^1+3^2+3^3+3^4+3^5$

$3(3^6-M)=3^1+3^2+3^3+3^4+3^5+3^6$

$\Rightarrow 3(3^6-M)-(3^6-M)=3^6-3^0$

$\Rightarrow 2(3^6-M)=3^6-1$

$\Rightarrow 2M = 2.3^6-(3^6-1)=3^6+1$

$\Rightarrow M=\frac{3^6+1}{2}$

M=36-(35+34+...+31+30)

   Đặt A=35+34+...+31+30

        3A=3⁶+3⁵+...+3²+31

        3A-A=3⁶+3⁵+...+3²+31-35-34-...-31-30

        2A=3⁶-30=>A=\(\dfrac{3^6-3^0}{2}\)

 Thay A vào M ta có:

M=3⁶-\(\dfrac{3^6-3^0}{2}\)

M=\(\dfrac{2.3^6}{2}\)-\(\dfrac{3^6-3^0}{2}\)

M=\(\dfrac{3^6.\left(2-1\right)-1}{2}\)

M=\(\dfrac{3^6.1-1}{2}\)

M=\(\dfrac{3^6-1}{2}\)

M=364

Lời giải:

$A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{99}}$

$3A=1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{98}}$

$\Rightarrow 3A-A=1-\frac{1}{3^{99}}< 1$

$\Rightarrow 2A< 1\Rightarrow A< \frac{1}{2}$

       A =  \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\) + ... + \(\dfrac{1}{3^{99}}\) 

     3A =  1  + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + ... + \(\dfrac{1}{3^{98}}\)

3A - A = ( 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + ... + \(\dfrac{1}{3^{98}}\)) - (\(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + ... + \(\dfrac{1}{3^{98}}\) - \(\dfrac{1}{3^{99}}\))

2A     = 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + ... + \(\dfrac{1}{3^{98}}\)  - \(\dfrac{1}{3}\) - \(\dfrac{1}{3^2}\) - ... - \(\dfrac{1}{3^{98}}\) - \(\dfrac{1}{3^{99}}\)

2A = ( \(\dfrac{1}{3}\) - \(\dfrac{1}{3}\)) + (\(\dfrac{1}{3^2}\)  - \(\dfrac{1}{3^2}\)) + ... + (1 - \(\dfrac{1}{3^{99}}\))

2A   = 0 + 0 + ... + 0 + 1 - \(\dfrac{1}{3^{99}}\)

2A = (1 - \(\dfrac{1}{3^{99}}\))

 A = \(\dfrac{1}{2}\) - \(\dfrac{1}{2.3^{99}}\) < \(\dfrac{1}{2}\)

Giá tiền mỗi bó hoa sau khi giảm:

\(85000-85000.15\%=72250\) (đồng)

Giá tiền 9 bó hoa đầu tiên:

\(72250.9=650250\) (đồng)

Giá tiền 36 bó hoa còn lại:

\(36.72250.80\%=2080800\) (đồng)

Tổng số tiền công ty phải trả:

\(2080800+650250=2731050\) (đồng)

Nếu không có thêm điều kiện gì thì biểu thức này không có giá trị lớn nhất bạn nhé.

a) \(4.\left(-\dfrac{1}{2}\right)^3-2.\left(-\dfrac{1}{2}\right)^2+3.\left(-\dfrac{1}{2}\right)+1\)

\(=4.\left(-\dfrac{1}{8}\right)-2.\dfrac{1}{4}+3.\left(-\dfrac{1}{2}\right)+1\)

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

\(=-\dfrac{3}{2}\)

b) \(8.\sqrt{9}-\sqrt{64}\)

\(=8.3-8\)

\(=24-8\)

\(=16\)

c) \(\sqrt{\dfrac{9}{16}}+\dfrac{25}{46}:\dfrac{5}{23}-\dfrac{7}{4}\)

\(=\dfrac{3}{4}+\dfrac{5}{2}-\dfrac{7}{4}\)

\(=-1+\dfrac{5}{2}\)

\(=\dfrac{3}{2}\)

56:54=

a) 2/3 + 3/4 . (-4/9)

= 2/3 - 1/3

= 1/3

b) -5/7 . 31/33 + (-5/7) : 33/2

= -5/7 . 31/33 - 5/7 . 2/33

= -5/7 . (31/33 + 2/33)

= -5/7 . 1

= -5/7

c) -3/5 . 13/11 - (-3/5) . 2/11

= -3/5 . (13/11 - 2/11)

= -3/5 . 1

= -3/5

\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}\)

\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{c}+\dfrac{1}{a}\)

\(\Rightarrow\dfrac{1}{a}=\dfrac{1}{b}=\dfrac{1}{c}=\dfrac{1+1+1}{a+b+c}=\dfrac{3}{a+b+c}=\dfrac{3}{1}=3\)

\(\Rightarrow a=b=c=\dfrac{1}{3}\)

\(\Rightarrow A=\dfrac{a^3\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}=a^3=\left(\dfrac{1}{3}\right)^3=\dfrac{1}{27}\)

\(A=1^3+2^3+3^3+...+n^3\)

Ta chứng minh

\(A=1^3+2^3+3^3+...+n^3=\left(1+2+3+...+n\right)^2\) (1)

+ Với \(n=3\)

\(1^3+2^3+3^3=36\)

\(\left(1+2+3\right)^2=36\)

=> (1) đúng

+ Giả sử (1) đúng với \(n=k\)

\(\Rightarrow1^3+2^3+3^3+...+k^3=\left(1+2+3+...+k\right)^2\)

+ Ta cần chứng minh (1) đúng với \(n=k+1\) Khi đó

\(VT=1^3+2^3+3^3+...+k^3+\left(k+1\right)^3=\)

\(=\left(1+2+3+...+k\right)^2+\left(k+1\right)^3=\)

\(=\left[\dfrac{k\left(k+1\right)}{2}\right]^2+\left(k+1\right)^3=\)

\(=\dfrac{k^2\left(k+1\right)^2+4\left(k+1\right)^3}{4}=\dfrac{\left(k+1\right)^2\left(k^2+4k+4\right)}{4}\)

\(VP=\left[1+2+3+...+k+\left(k+1\right)\right]^2=\)

\(=\left[\dfrac{\left(k+1\right)\left(k+1+1\right)}{2}\right]^2=\)

\(\dfrac{\left(k+1\right)^2\left(k+2\right)^2}{4}=\dfrac{\left(k+1\right)^2\left(k^2+4k+4\right)}{4}\)

Như vậy VT=VP nên (1) đúng  với \(n=k+1\)

Theo nguyên tắc của phương pháp quy nạp => (1) đúng