ta có:

                  \(\left(7\frac{1}{2}.8\frac{3}{70}+8\frac{3}{70}.\frac{9}{4}+\frac{19}{4}.8\frac{3}{70}+5\frac{1}{2}.8\frac{3}{70}\right):x=1126\)

                                                  \(8\frac{3}{70}\left(7\frac{1}{2}+\frac{9}{4}+\frac{19}{4}+5\frac{1}{2}\right):x=1126\)

                                                                \(\frac{563}{70}.\left(\frac{15}{2}+7+\frac{11}{2}\right):x=1126\)

                                                                                             \(\frac{563}{70}.20:x=1126\)

                                                                                                 \(\frac{1126}{70}:x=1126\)

                                                                                                   \(=>x=\frac{1126}{7}:1126\)

                                                                                                    \(=>x=\frac{1}{7}\)

            cho mình nha các bạn.

cho mình nha.

i) (x - 1)(5x + 3) = (3x - 8)(x - 1)

<=> 5x² + 3x - 5x - 3 = 3x² - 3x - 8x + 8

<=> 5x² - 2x - 3 = 3x² - 11x + 8

<=> 5x² - 2x - 3 - 3x² + 11x - 8 = 0

<=> 2x² + 9x - 11 = 0

<=> 2x² + 11x - 2x - 11 = 0

<=> x(2x + 11) - (2x + 11) = 0

<=> (x - 1)(2x + 11) = 0

<=> x - 1 = 0 hoặc 2x + 11 = 0

<=> x = 0 hoặc x = -11/2

m) 2x(x - 1) = x² - 1

<=> 2x² - 2x = x² - 1

<=> 2x² - 2x - x² + 1 = 0

<=> x² - 2x + 1 = 0

<=> (x - 1)² = 0

<=> x - 1 = 0

<=> x = 1

n) (2 - 3x)(x + 11) = (3x - 2)(2 - 5x)

<=> 2x + 22 - 3x² - 33x = 6x - 15x² - 4 + 10x

<=> -31x + 22 - 3x² = 16x - 15x² - 4

<=> 31x - 22 + 3x² + 16x - 15x² - 4 = 0

<=> 47x - 18 - 12x² = 0

<=> -12x² + 47x - 26 = 0

<=> 12x² - 47x + 26 = 0

<=> 12x² - 8x - 39x + 26 = 0

<=> 4x(3x - 2) - 13(3x - 2) = 0

<=> (4x - 13)(3x - 2) = 0

<=> 4x - 13 = 0 hoặc 3x - 2 = 0

<=> x = 13/4 hoặc x = 2/3

i) (x - 1)(5x + 3) = (3x - 8)(x - 1)

<=> 5x² + 3x - 5x - 3 = 3x² - 3x - 8x + 8

<=> 5x² - 2x - 3 = 3x² - 11x + 8

<=> 5x² - 2x - 3 - 3x² + 11x - 8 = 0

<=> 2x² + 9x - 11 = 0

<=> 2x² + 11x - 2x - 11 = 0

<=> x(2x + 11) - (2x + 11) = 0

<=> (x - 1)(2x + 11) = 0

<=> x - 1 = 0 hoặc 2x + 11 = 0

<=> x = 0 hoặc x = -11/2

m) 2x(x - 1) = x² - 1

<=> 2x² - 2x = x² - 1

<=> 2x² - 2x - x² + 1 = 0

<=> x² - 2x + 1 = 0

<=> (x - 1)² = 0

<=> x - 1 = 0

<=> x = 1

n) (2 - 3x)(x + 11) = (3x - 2)(2 - 5x)

<=> 2x + 22 - 3x² - 33x = 6x - 15x² - 4 + 10x

<=> -31x + 22 - 3x² = 16x - 15x² - 4

<=> 31x - 22 + 3x² + 16x - 15x² - 4 = 0

<=> 47x - 18 - 12x² = 0

<=> -12x² + 47x - 26 = 0

<=> 12x² - 47x + 26 = 0

<=> 12x² - 8x - 39x + 26 = 0

<=> 4x(3x - 2) - 13(3x - 2) = 0

<=> (4x - 13)(3x - 2) = 0

<=> 4x - 13 = 0 hoặc 3x - 2 = 0

<=> x = 13/4 hoặc x = 2/3

wifi chán thế?

1,\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\left(7-\frac{1}{6}\right)+\frac{1}{3}\)

  \(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\frac{41}{6}+\frac{1}{3}\)

\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{41}{14}+\frac{1}{3}\)

 \(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{137}{42}\)

\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{137}{42}-\frac{1}{2}\)

\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{58}{21}\)

 \(\left(x-\frac{9}{4}\right)=\frac{5}{2}:\frac{2}{9}\)

\(\left(x-\frac{9}{4}\right)=\frac{45}{4}\)

\(x=\frac{45}{4}+\frac{9}{4}\)

\(x=\frac{27}{2}\)

Bước cưối 58/21 minh man viết nhầm nên sai 

\(\left(x-\frac{9}{4}\right)=\frac{58}{21}:\frac{2}{9}\)

\(\left(x-\frac{9}{4}\right)=\frac{87}{7}\)

\(x=\frac{87}{7}+\frac{9}{4}\)

\(x=\frac{411}{28}\)

\(\frac{1}{3}x:\frac{2}{3}=1\frac{3}{4}:\frac{2}{5}\)

\(\Rightarrow\frac{1}{3}x:\frac{2}{3}=\frac{35}{8}\)

\(\Rightarrow\frac{1}{3}x=\frac{35}{12}\)

\(\Rightarrow x=\frac{35}{4}\)

a) \(\left(\frac{1}{3}\cdot x\right):\frac{2}{3}=1\frac{3}{4}:\frac{2}{5}\)

\(=\left(\frac{1}{3}\cdot x\right):\frac{2}{3}=\frac{35}{8}\)

\(\Rightarrow\frac{1}{3}\cdot x=\frac{35}{8}\cdot\frac{2}{3}\)

\(\Rightarrow x=\frac{35}{12}:\frac{1}{3}=\frac{35}{4}\)

b) \(4,5:0,3=2,5:\left(0,1\cdot x\right)\)

\(=15=2,5:\left(0,1\cdot x\right)\)

\(\Rightarrow0,1\cdot x=2,25:15\)

\(\Rightarrow x=\frac{3}{20}:0,1=\frac{3}{2}=1,5\)

c) \(8:\left(\frac{1}{4}\cdot x\right)=2:0,02\)

\(8:\left(\frac{1}{4}\cdot x\right)=100\)

\(\Rightarrow\frac{1}{4}\cdot x=8:100\)

\(\Rightarrow x=\frac{2}{25}:\frac{1}{4}=\frac{8}{25}=0,32\)

d) \(3:2\frac{1}{4}=\frac{3}{4}:\left(6\cdot x\right)\)

\(=\frac{4}{3}=\frac{3}{4}:\left(6:x\right)\)

\(\Rightarrow6\cdot x=\frac{3}{4}:\frac{4}{3}\)

\(\Rightarrow x=\frac{9}{16}:6=\frac{3}{32}=0,09375\)

a)\(11\frac{1}{4}-\left(2\frac{5}{7}+5\frac{1}{4}\right)\)

\(=\frac{45}{4}-\left(\frac{19}{7}+\frac{21}{4}\right)\)

\(=\frac{45}{4}-\left(\frac{76}{28}+\frac{147}{28}\right)\)

\(=\frac{45}{4}-\frac{223}{28}\)

\(=\frac{315}{28}-\frac{223}{28}\)

\(=\frac{23}{7}\)

b) \(\left(8\frac{5}{11}+3\frac{5}{8}\right)-3\frac{5}{11}\)

   \(=\left(\frac{93}{11}+\frac{29}{8}\right)-\frac{38}{11}\)

   \(=\left(\frac{744}{88}+\frac{319}{88}\right)-\frac{38}{11}\)

   \(=\frac{1063}{88}-\frac{38}{11}=\frac{1063}{88}-\frac{304}{88}\)

    \(=\frac{69}{8}\)

1a) \(11\frac{1}{4}-\left(2\frac{5}{7}+5\frac{1}{4}\right)\)

\(=11\frac{1}{4}-2\frac{5}{7}-5\frac{1}{4}\)

\(=11\frac{1}{4}-5\frac{1}{4}-2\frac{5}{7}\)

\(=6-2\frac{5}{7}=3\frac{2}{7}=\frac{23}{7}\)

2a) \(x\div8,25=\left(-2,2\right)\)

\(\Leftrightarrow x=-18,15\)

\(\left(-\frac{2}{3}+\frac{3}{7}\right):\frac{4}{5}+\left(-\frac{1}{3}+\frac{4}{7}\right):\frac{4}{5}\)

= \(\left(-\frac{14}{21}+\frac{9}{21}\right).\frac{5}{4}+\left(-\frac{7}{21}+\frac{12}{21}\right).\frac{5}{4}\)

= \(-\frac{5}{21}.\frac{5}{4}+\frac{5}{21}.\frac{5}{4}\)

= \(\frac{5}{4}.\left(-\frac{5}{21}+\frac{5}{21}\right)\)

= \(\frac{5}{4}.0\)

= \(0\)

Chúc bạn học tốt !!!

\(\left(\frac{-2}{3}+\frac{3}{7}\right):\frac{4}{5}+\left(-\frac{1}{3}+\frac{4}{7}\right):\frac{4}{5}\)

\(=\left(\frac{3}{7}-\frac{2}{3}\right).\frac{4}{5}+\left(\frac{4}{7}-\frac{1}{3}\right).\frac{4}{5}\)

\(=\left(\frac{3}{7}-\frac{2}{3}+\frac{4}{7}-\frac{1}{3}\right).\frac{4}{5}\)

\(=\left[\left(\frac{3}{7}+\frac{4}{7}\right)-\left(\frac{2}{3}+\frac{1}{3}\right)\right].\frac{4}{5}\)

\(=\left(1-1\right).\frac{4}{5}\)

\(=0.\frac{4}{5}\)

\(=0\)

\(\left(\frac{-2}{3}+\frac{3}{7}\right):\frac{4}{5}+\left(\frac{-1}{3}+\frac{4}{7}\right):\frac{4}{5}\)

\(=\left(\frac{-14}{21}+\frac{9}{21}\right):\frac{4}{5}+\left(\frac{-7}{21}+\frac{12}{21}\right):\frac{4}{5}\)

\(=\frac{-5}{21}:\frac{4}{5}+\frac{5}{21}:\frac{4}{5}\)

\(=\frac{-5}{21}.\frac{5}{4}+\frac{5}{21}.\frac{5}{4}\)

\(=\frac{5}{4}.\left(\frac{-5}{21}+\frac{5}{21}\right)\)

\(=\frac{5}{4}.0\)

\(=0\)

a,x=(-1/2)(-2)^3=4

b, x=1/16

anh có thể viết phân số ra như này ko ạ:
\(\frac{3}{4}\)

viết như vậy em nhìn rối mắt lắm ạ!

(0,25)^8 =(0,5)^16

(0,125)^4=(0,5)^12

Bài 3:

a, Đặt \(A=\left|2x-\frac{1}{5}\right|+2017\)

Để A đạt GTNN thì \(\left|2x-\frac{1}{5}\right|\)đạt GTNN

Mà \(\left|2x-\frac{1}{5}\right|\ge0\)

Do đó \(\left|2x-\frac{1}{5}\right|=0\)thì A đạt GTNN tức là A = 0 + 2017 = 2017 khi

\(2x-\frac{1}{5}=0=>2x=0+\frac{1}{5}=\frac{1}{5}=>x=\frac{1}{5}.\frac{1}{2}=\frac{1}{10}\)

b, Đặt \(B=\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{4}\right|\)

Ta thấy \(\frac{1}{2}>\frac{1}{3}>\frac{1}{4}=>x+\frac{1}{2}>x+\frac{1}{3}>x+\frac{1}{4}\)

Do đó để B đạt GTNN thì \(x+\frac{1}{2}\)đạt GTNN

mà \(x+\frac{1}{2}\ge0\)

Từ 2 điều trên => \(x+\frac{1}{2}=0=>x=-\frac{1}{2}\)

Khi đó \(x+\frac{1}{3}=-\frac{1}{2}+\frac{1}{3}=-\frac{1}{6}\)

và \(x+\frac{1}{4}=-\frac{1}{2}+\frac{1}{4}=-\frac{1}{4}\)

Vậy GTNN của \(B=\left|0\right|+\left|-\frac{1}{6}\right|+\left|-\frac{1}{4}\right|=0+\frac{1}{6}+\frac{1}{4}=\frac{10}{24}\)khi x = -1/2

Phần b này thì mình không chắc lắm bạn tự xem lại nhé

Bài 1: 

\(M=\frac{2017}{11-x}\)đạt GTLN <=> 11 - x đạt GTNN và 11 - x > 0 (nếu không thì M đạt giá trị âm (vô lí))

=> 11 - x = 1

=> x = 10

Vậy x = 10 thì M đạt GTLN tức là bằng \(\frac{2017}{1}=2017\)

Bài 2

a, Đặt \(A=-2\left|x-\frac{3}{4}\right|-\left|y+\frac{3}{4}\right|+\frac{5}{6}\)

Để A đạt GTLN <=> \(-2\left|x-\frac{3}{4}\right|\)đạt GTLN và \(\left|y+\frac{3}{4}\right|\)đạt GTNN

mà \(\left|x-\frac{3}{4}\right|\ge0=>-2\left|x-\frac{3}{4}\right|\le0\)

và \(\left|y+\frac{3}{4}\right|\ge0\)

Do đó \(-2\left|x-\frac{3}{4}\right|=0\)và \(\left|y+\frac{3}{4}\right|=0\)

Vậy GTLN của A = 0 - 0 + 5/6 = 5/6 khi

\(\left|x-\frac{3}{4}\right|=0=>x-\frac{3}{4}=0=>x=\frac{3}{4}\)

Và \(\left|y+\frac{3}{4}\right|=0=>y+\frac{3}{4}=0=>y=-\frac{3}{4}\)

b, Đặt \(B=-\left(x+\frac{1}{2}\right)^2+\frac{5}{7}\)

Để B đạt GTLN thì \(-\left(x+\frac{1}{2}\right)^2\)đạt GTLN

Mà \(\left(x+\frac{1}{2}\right)^2\ge0=>-\left(x+\frac{1}{2}\right)^2\le0\)

Do đó để B đạt GTLN thì \(-\left(x+\frac{1}{2}\right)^2=0\)

Khi đó GTLN của B = 0 + 5/7 = 5/7 khi

\(\left(x+\frac{1}{2}\right)^2=0=>x+\frac{1}{2}=0=>x=-\frac{1}{2}\)