\(5\frac{4}{7}\): [ x : 1,3 + 8,4 . \(\frac{6}{7}\). ( 6 - \(\frac{\left(2,3+5\div6,25\right)\times7}{8\times0,-125+6,9}\)) ] = \(1\frac{1}{14}\)

\(\frac{39}{7}\): [ x : 1,3 + \(\frac{36}{5}\). ( 6 - \(\frac{\left(2,3+0,8\right).7}{0,1+6,9}\)) ] = \(\frac{15}{14}\)

\(\frac{39}{7}\): [ x : 1,3 + \(\frac{36}{5}\). ( 6 - \(\frac{3,1.7}{7}\)) ] = \(\frac{15}{14}\)

\(\frac{39}{7}\): [ x : 1,3 + \(\frac{36}{5}\). ( 6 - 3,1 ) ] = \(\frac{15}{14}\)

x : 1,3 + \(\frac{36}{5}\). 2,9 = \(\frac{39}{7}\): \(\frac{15}{14}\)

x : 1,3 + 20,88 = 5,2

x : 1,3 = - 15,68

x = - 15,68 . 1,3

x = - 20,384

ta có 

\(5\frac{4}{7}:\left\{x:1,3+8,4.\frac{6}{7}.\left[6-\frac{\left(2,3+5:6,25\right).7}{8.0,0125+6,9}\right]\right\}=1\frac{1}{14}\)

\(\Leftrightarrow\frac{39}{7}:\left\{x:1,3+7,2.\left[6-\frac{\left(2,3+0,8\right).7}{0,1+6,9}\right]\right\}=\frac{15}{14}\)

\(\Leftrightarrow\frac{39}{7}:\left\{x:1,3+7,2.\left[6-\frac{3,1.7}{7}\right]\right\}=\frac{15}{14}\)

\(\Leftrightarrow\frac{39}{7}:\left\{x:1,3+7,2.2,9\right\}=\frac{15}{14}\Leftrightarrow\left\{x:1,3+7,2.2,9\right\}=\frac{39}{7}:\frac{15}{14}\)

\(\Leftrightarrow x:1,3+20,88=5,2\Leftrightarrow x:1,3=-15,68\Leftrightarrow x=-20,384\)

a) X = 15

b) X = 4

c ) X= 23

d) X= 11

( Chỉ là ý kiến riêng thôi nhé, nhận gạch đá )

a) \(\frac{6+x}{33}=\frac{7}{11}\)

=> (6 + x). 11 = 33.7

=> 66 + 11x = 231

=> 11x = 231 - 66

=> 11x = 165

=> x = 165 : 11

=> x = 15

b) 15/26 + x/13 = 46/52

=> x/13 = 23/26 - 15/26

=> x/13 = 4/13

=> x = 4

c) 121/27 x 54/11 < x < 100/21 : 25/126

=> 22 < x < 24

=> x = 23 (vì x là số tự nhiên)

d) 1 < 11/x < 12

=> 11/x \(\in\){2; 3; 4 ; ...; 11}

=> x \(\in\) {11/2; 11/3; ...; 1}

Vì x là số tự nhiên => x = 1

a)\(x^{2016}=x^{2017}\)

 \(\Leftrightarrow x^{2017}-x^{2016}=0\)

\(\Leftrightarrow x^{2016}.\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^{2016}=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)

Vay ...

b) \(2y.\left(x+1\right)-x-7=0\)

\(\Leftrightarrow2y.\left(x+1\right)-\left(x+1\right)=6\)

\(\Leftrightarrow\left(x+1\right).\left(2y+1\right)=6\)

Đến chỗ này bạn tự tìm các cặp x,y nha

N-NHOK TÌ làm đúng rùi đo kkk

\(2y\left(x+1\right)-x-7=0\)

\(\Rightarrow2y\left(x+1\right)-x-1-6=0\)

\(\Rightarrow2y\left(x+1\right)-\left(x+1\right)=6\)

\(\Rightarrow\left(x+1\right)\left(2y-1\right)=6\)

..........

Chia ra các trường hợp em nhé

em cảm ơn chị ạ .

\(2y\left(x+1\right)-x-7=0\)

<=> \(2xy+2y-x-7=0\)

<=> \(x\left(2y-1\right)+\left(2y-1\right)-6=0\)

<=> \(\left(x+1\right)\left(2y-1\right)=6\)

bn tự giải nốt nhé

a)33:11=3

7x3=21,21-6=15

x=15

a, \(\frac{6+x}{33}=\frac{7}{11}\)

\(\Leftrightarrow\left(6+x\right).11=7.33\)

\(\Leftrightarrow66+11x=231\)

\(\Leftrightarrow11x=231-66\)

\(\Leftrightarrow x=\frac{165}{11}=15\)

Vậy x = 15.

b,\(\frac{12+x}{43-x}=\frac{2}{3}\)

\(\Leftrightarrow3.\left(12+x\right)=2\left(43-x\right)\)

\(\Leftrightarrow36+3x=86-2x\)

\(\Leftrightarrow3x+2x=-36+86\)

\(\Leftrightarrow5x=50\)

\(\Leftrightarrow x=10\)

Vây x = 10.

tích đúng mình làm cho

A/ \(\frac{x}{17}=\frac{60}{204}\)

\(204x=60.17\)

\(204x=1020\)

\(x=\frac{1020}{204}\)

\(x=\frac{17}{4}=4,25\)

vậy x= 4,25

B/ \(\left(6+x\right).11=33.7\)

\(66+11x=231\)

\(11x=231-66=165\)

\(x=\frac{165}{11}\)

\(x=15\)

vậy x = 15

C/ \(\left(12+x\right).3=\left(43-x\right).2\)

\(36+3x=86-2x\)

\(3x+2x=86-36\)

\(5x=50\)

\(x=\frac{50}{5}=10\)

vậy x=10

mk nhầm 

phần a là \(\frac{1020}{204}\)= 5 nha

mk nhầm, xin lỗi

a)\(\frac{x}{17}=\frac{60}{204}=\frac{5}{17}\Rightarrow x=5\)

b)\(\frac{6+x}{33}=\frac{7}{11}\Rightarrow11\left(6+x\right)=7.33\Rightarrow11.6+11x=231\Rightarrow66+11x=231\)

\(\Rightarrow11x=231-66\Rightarrow11x=165\Rightarrow x=\frac{165}{11}=15\)

c)\(\frac{12+x}{43-x}=\frac{2}{3}\Rightarrow2\left(43-x\right)=3\left(12+x\right)\Rightarrow2.43-2x=3.12+3x\)

\(86-2x=36+3x\Rightarrow86-36=3x+2x\Rightarrow50=5x\Rightarrow x=\frac{50}{5}=10\)

\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)

\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)

\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\)

Dễ thấy: \(\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\ne0\Rightarrow x+2004=0\Leftrightarrow x=-2014\)

x = -2014

ti-ck nha

.........

\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

\(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)\left(x+1\right)=0\)

\(\Rightarrow x+1=0\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\right)\)

\(x=-1\)

Vậy \(x=-1\)