Chú bé hồng sinh ra trong gia đình nghiện mẹ theo giai bố chết sớm.Càng ngày ra sản cạn kiệt chú bé phải đi trộm chó và đến nhà lão Hạc bắt cậu Vàng đi vì mất ậu Vàng nên Lão Hạc phải ăn bả cho để chết.
:P

Hay !

\(\frac{7}{8}=\frac{14}{16}\)

\(\Rightarrow\frac{14}{16}=\frac{8}{16}+\frac{2}{16}+\frac{4}{16}\)

\(\Rightarrow\frac{14}{16}=\frac{1}{2}+\frac{1}{8}+\frac{1}{4}\)

hay

\(\frac{7}{8}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\)

\(3< \left|3+2x\right|< 4\)

\(\Leftrightarrow\orbr{\begin{cases}3< 3+2x< 4\\-3>3+2x>-4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}0< 2x< 1\\-6>2x>-7\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}0< x< \frac{1}{2}\\-3>x>\frac{-7}{2}\end{cases}}\)

897316542x978226546

=8.7777886e+17

ti ck nha

nè nè

8777788588573743e+17

          [ Giải: ]

( x + 3/4 ) × 5/7 = 10/9

x + 3/4 = 10/9 : 5/7

x + 3/4 = 10/9 . 7/5

x + 3/4 = 14/9

x = 14/9 - 3/4

x = 56/36 - 27/36

x = 29/36

          [ Hoq chắc ]

(x + 3/4)=10/9 : 5/7

x+3/4=14/9

x=14/9 - 3/4

x= 29/36

\(\frac{2}{x-3}+\frac{x-5}{x-1}=1\)

\(ĐKXĐ:x\ne1;x\ne3\)

\(pt\Leftrightarrow\frac{2x-2}{x^2-4x+3}+\frac{x^2-8x+15}{x^2-4x+3}=1\)

\(\Leftrightarrow\frac{x^2-6x+13}{x^2-4x+3}=1\)

\(\Leftrightarrow x^2-6x+13=x^2-4x+3\)

\(\Leftrightarrow-2x+10=0\Leftrightarrow x=-5\left(t/mđkxđ\right)\)

Vậy pt có 1 nghiệm là -5

2/x - 3 + x - 5/x - 1 = 1

2(x - 1) + (x - 5)(x - 3) = (x - 3)(x - 1)

-6x + 13 + x^2 = x^2 - 4x + 3

-6x + 13 = -4x + 3

13 = -4x + 3 + 6x

13 = 2x + 3

13 - 3 = 3x

10 = 2x

5 = x

=> x = 5

\(\frac{14}{27}:\frac{7}{18}\cdot\frac{6}{5}+\frac{3}{10}\)

\(=\frac{4}{3}\cdot\frac{6}{5}+\frac{3}{10}\)

\(=\frac{8}{5}+\frac{3}{10}\)

\(=\frac{19}{10}\)

. là nhân nha

(3x - 5)(5 - 3x) + 9(x + 1)^2 = 30

=> 15x - 9x^2 - 25 + 15x + 9(x^2 + 2x + 1) = 30

=> 30x - 9x^2 - 25 + 9x^2 + 18x + 9 = 30

=> 38x - 16 = 30

=> 38x = 46

=> x = 23/19

\(\left(3x-5\right)\left(5-3x\right)+9\left(x+1\right)^2=30.\)

\(15x-9x^2-25+15x+9\left(x^2+2x+1\right)=30\)

\(15x-9x^2-25+15x+9x^2+18x+9=30\)

\(48x-16=30\)

\(48x=30+16=46\)

\(x=\frac{46}{48}=\frac{23}{24}\)

  \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)

\(x^2+8x+16-\left(x^2-1\right)=16\)

\(x^2+8x+16-x^2+1=16\)

\(8x+17=16\)

\(8x=16-17=-1\)

\(x=-\frac{1}{8}\)

\(M=2x^2-x+1=\left(\sqrt{2}x\right)^2-2.\sqrt{2}x.\frac{1}{\sqrt{8}}+\frac{1}{8}+\frac{7}{8}\)

\(=\left(\sqrt{2}x-\frac{1}{\sqrt{8}}\right)^2+\frac{7}{8}\ge\frac{7}{8}\)

Vậy \(M_{min}=\frac{7}{8}\Leftrightarrow\sqrt{2}x-\frac{1}{\sqrt{8}}=0\)

\(\Leftrightarrow\sqrt{2}x=\frac{1}{\sqrt{8}}\Leftrightarrow x=\frac{1}{16}\)

\(x=\frac{1}{\sqrt{16}}=\frac{1}{4}\)nha. Gõ nhầm

\(B=\left(1-\frac{3}{2.4}\right)\left(1-\frac{3}{3.5}\right)\left(1-\frac{3}{4.6}\right)...\left(1-\frac{3}{n\left(n+2\right)}\right)\)

\(=\frac{1.5}{2.4}.\frac{2.6}{3.5}.\frac{3.7}{4.6}...\frac{\left(n-1\right)\left(n+3\right)}{n\left(n+2\right)}\)

\(=\frac{\left[1.2.3...\left(n-1\right)\right]\left[5.6.7...\left(n+3\right)\right]}{\left(2.3.4...n\right)\left[4.5.6...\left(n+2\right)\right]}\)

\(=\frac{n+3}{4n}< 2\left(đpcm\right)\)