bạn tự quy đồng lên mà giải

Hình như sai ở đâu thì phải 

lay x phần 7 - 1 phần 4 khác ra

\(a/\frac{7}{9}-\frac{x}{3}=\frac{1}{9}\)

\(\Rightarrow\frac{x}{3}=\frac{7}{9}-\frac{1}{9}\)

\(\Rightarrow\frac{x}{3}=\frac{2}{3}\)

\(\Rightarrow x=2\)

Vậy \(x=2\)

\(b/\frac{1}{x}-\frac{-2}{15}=\frac{7}{15}\)

\(\Rightarrow\frac{1}{x}=\frac{7}{15}+\frac{-2}{15}\)

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

\(\Rightarrow x=3\)

Vậy \(x=3\)

\(c/\frac{-11}{14}-\frac{-4}{x}=\frac{-3}{14}\)

\(\Rightarrow\frac{-4}{x}=\frac{-11}{14}-\frac{-3}{14}\)

\(\Rightarrow\frac{-4}{x}=\frac{-4}{7}\)

\(\Rightarrow x=7\)

Vậy \(x=7\)

\(d/\frac{x}{21}-\frac{2}{3}=\frac{5}{21}\)

\(\Rightarrow\frac{x}{21}=\frac{5}{21}+\frac{2}{3}\)

\(\Rightarrow\frac{x}{21}=\frac{19}{21}\)

\(\Rightarrow x=19\)

Vậy \(x=19\)

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CẢM ƠN VƯƠNG MẠT MẠT NHA!

\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{19\cdot21}-\frac{x}{14}=\frac{2}{-7}\)

\(\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{19\cdot21}\right)-\frac{x}{14}=\frac{2}{-7}\)

\(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)-\frac{x}{14}=\frac{2}{-7}\)

\(\frac{1}{2}\left(1-\frac{1}{21}\right)-\frac{x}{14}=\frac{2}{-7}\)

\(\frac{1}{2}\cdot\frac{20}{21}-\frac{x}{14}=\frac{2}{-7}\)

\(\frac{10}{21}-\frac{x}{14}=\frac{2}{-7}\)

\(\frac{x}{14}=\frac{10}{21}-\frac{2}{-7}\)

\(\frac{x}{14}=\frac{16}{21}\)

\(\Rightarrow x\cdot=21=14\cdot16\)

\(\Rightarrow x\cdot21=224\)

\(\Rightarrow x=\frac{224}{21}\)

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

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

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

Vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne=\)

Nên x + 1 = 0 => x = -1

b) \(\frac{x+1}{14}+\frac{x+2}{13}=\frac{x+3}{12}+\frac{x+4}{11}\)

\(\Leftrightarrow\frac{x+1}{14}+1+\frac{x+2}{13}+1=\frac{x+3}{12}+1+\frac{x+4}{11}+1\)

\(\Leftrightarrow\frac{x+15}{14}+\frac{x+15}{13}=\frac{x+15}{12}+\frac{x+15}{11}\)

\(\Leftrightarrow\frac{x+15}{14}+\frac{x+15}{13}-\frac{x+15}{12}-\frac{x+15}{11}=0\)

\(\Leftrightarrow\left(x+15\right)\left(\frac{1}{14}+\frac{1}{13}-\frac{1}{12}-\frac{1}{11}\right)=0\)

Vì \(\frac{1}{14}+\frac{1}{13}-\frac{1}{12}-\frac{1}{11}\ne0\)

Nên x  +15 = 0 => x = -15

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

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

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

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

Vì \(\frac{1}{10}>\frac{1}{13};\frac{1}{11}>\frac{1}{14}\Rightarrow\frac{1}{10}+\frac{1}{11}>\frac{1}{13}+\frac{1}{14}\Rightarrow\frac{1}{10}+\frac{1}{11}+\frac{1}{12}>\frac{1}{13}+\frac{1}{14}\)

\(\Rightarrow\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}>0\)

\(\Rightarrow x+1=0\Rightarrow x=-1\)

b, Bạn cộng thêm 1 vào \(\frac{x+1}{14};\frac{x+1}{13};\frac{x+1}{12};\frac{x+1}{11}\)Mội bên phân số 1 đơn vị rồi áp dụng như bài 1

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

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

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

Vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\)

Nên \(x+1=0\)

\(\Rightarrow\)\(x=-1\)

Vậy \(x=-1\)

Chúc bạn học tốt ~ 

\(1\frac{1}{7}-\frac{5}{7}< \frac{x}{7}< 2\frac{1}{14}-1\frac{3}{14}\)

=> \(\frac{8}{7}-\frac{5}{7}< \frac{x}{7}< \frac{29}{14}-\frac{17}{14}\)

=> \(\frac{3}{7}< \frac{x}{7}< \frac{12}{14}\)

=> \(\frac{3}{7}< \frac{x}{7}< \frac{6}{7}\)

=> 3 < x < 6

=> x thuộc { 3 ; 4 ; 5 ; 6 }

Đặt \(\frac{7}{x+y}=a,\frac{1}{x-y}=b\)

Khi đó ta có:

\(\hept{\begin{cases}2a+3b=5\\a-2b=-1\end{cases}}\Rightarrow\hept{\begin{cases}2a+3b=5\left(1\right)\\2a-4b=-2\left(2\right)\end{cases}}\)

Trứ vế với vế của (1) và (2), ta được:

       \(2a+3b-\left(2a-4b\right)=5-\left(-2\right)\)

\(\Rightarrow7b=7\Rightarrow b=1.\)

Thay b = 1 vào (1): \(2a+3=5\Rightarrow a=1.\)

\(a=1\Rightarrow\frac{7}{x+y}=1\Rightarrow x+y=7\)

\(b=1\Rightarrow\frac{1}{x-y}=1\Rightarrow x-y=1\)

Từ đó tính được \(x=4,y=3\)

Chúc bạn học tốt.

\(\begin{array}{l}a)x + 0,25 = \frac{1}{2}\\x = \frac{1}{2} - 0,25\\x = \frac{1}{2} - \frac{1}{4}\\x = \frac{2}{4} - \frac{1}{4}\\x = \frac{1}{4}\end{array}\)

Vậy \(x = \frac{1}{4}\)

\(\begin{array}{l}b)x - \left( { - \frac{5}{7}} \right) = \frac{9}{{14}}\\x = \frac{9}{{14}} + \left( { - \frac{5}{7}} \right)\\x = \frac{9}{{14}} + \left( { - \frac{{10}}{{14}}} \right)\\x = \frac{{ - 1}}{{14}}\end{array}\)

Vậy \(x = \frac{{ - 1}}{{14}}\)