\(\frac{14}{5}\) -x =\(\frac65\) + \(\frac34\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.


\(x^2-x-\frac{3}{4}=0\)
\(\Rightarrow x^2+\frac{1}{2}x-\frac{3}{2}x-\frac{3}{4}=0\)
\(\Rightarrow x\left(x+\frac{1}{2}\right)-\frac{3}{2}\left(x+\frac{1}{2}\right)=0\)
\(\Rightarrow\left(x-\frac{3}{2}\right)\left(x+\frac{1}{2}\right)=0\)
\(\Rightarrow\hept{\begin{cases}x-\frac{3}{2}=0\\x+\frac{1}{2}=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{2}\end{cases}}}\)
Chúc bạn học tốt.


nhưng mà mình chỉ biết thế thôi thiếu gì thì bổ sung, còn làm dc thì làm luôn nha đung mình k

3/14 × 5/17 + 11/14 × 5/17 + 12/17 × 5/16 + 12/17 × 11/16
= 5/17 × ( 3/14 + 11/14) + 12/17 × ( 5/16 + 11/16)
= 5/17 × 1 + 12/17 × 1
= 5/17 + 12/17
= 1
3/14 × 5/17 + 11/14 × 5/17 + 12/17 × 5/16 + 12/17 × 11/16
= 5/17 × ( 3/14 + 11/14) + 12/17 × ( 5/16 + 11/16)
= 5/17 × 1 + 12/17 × 1
= 5/17 + 12/17
= 1

Mình lỡ đánh nhầm 2 lần \(\frac{5}{14}\)nha :)) chỉ 1 lần thôi

\(x^2-x-\frac{3}{4}=0\)
\(\Rightarrow\frac{x^2-x}{1}-\frac{3}{4}=0\)
\(\Rightarrow\frac{\left(x^2-x\right).4}{4}-\frac{3}{4}=0\)
\(\Rightarrow\frac{4x^2-4x-3}{4}=0\)
\(\Rightarrow4x^2-4x-3=0\)
\(\Rightarrow4x^2-6x+2x-3=0\)
\(\Rightarrow2x\left(2x-3\right)+\left(2x-3\right)=0\)
\(\Rightarrow\left(2x+1\right)\left(2x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+1=0\\2x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{3}{2}\end{cases}}}\)

\(a,\left(\frac{5}{7}+\frac{9}{7}\right)x\frac{21}{28}\)
\(C1:=\frac{14}{7}x\frac{21}{28}=\frac{3}{2}\)
\(C2:=\frac{5}{7}x\frac{21}{28}+\frac{9}{7}x\frac{21}{28}=\frac{15}{28}+\frac{27}{28}=\frac{3}{2}\)
\(b,\frac{4}{5}x\frac{13}{14}+\frac{13}{14}x\frac{1}{5}\)
\(C1:=\frac{26}{35}+\frac{13}{70}=\frac{13}{14}\)
\(C2:=\frac{13}{14}x\left(\frac{4}{5}+\frac{1}{5}\right)=\frac{13}{14}x1=\frac{13}{14}\)
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 }

a.\(\frac{11}{14}-\frac{3}{x-1}=\frac{5}{14}\)
\(\frac{3}{x-1}=\frac{11}{14}-\frac{5}{14}\)
\(\frac{3}{x-1}=\frac{3}{7}\)
\(\Rightarrow x-1=7\)
\(x=7+1\)
Vậy \(x=8\)
b.\(\frac{42}{25}:\frac{2x+1}{5}=\frac{6}{5}\)
\(\frac{2x+1}{5}=\frac{42}{25}:\frac{6}{5}\)
\(\frac{2x+1}{5}=\frac{7}{5}\)
\(\Rightarrow2x+1=7\)
\(2x=7-1\)
\(2x=6\)
\(x=6:2\)
\(x=3\)
a.\(\frac{11}{14}-\frac{3}{x-1}=\frac{5}{14}\)
\(\frac{3}{x-1}=\frac{11}{14}-\frac{5}{14}=\frac{3}{7}\)
\(\Rightarrow x-1=7\)
\(x=7+1=8\)
VẬY, \(x=8\)
b. \(\frac{42}{25}:\frac{2x+1}{5}=\frac{6}{5}\)
\(\frac{2x+1}{5}=\frac{42}{25}:\frac{6}{5}=\frac{7}{5}\)
\(\Rightarrow2x+1=7\)
\(2x=7-1=6\)
\(x=6:2=3\)
VẬY, \(x=3\)

\(\frac{2}{x-14}-\frac{5}{x-13}=\frac{2}{x-9}-\frac{5}{x-11}.\) \(\left(ĐKXĐ:x\ne14;13;11;9\right)\)
\(\Leftrightarrow\frac{2}{x-14}-\frac{2}{x-9}=\frac{5}{x-13}-\frac{5}{x-11}\)
\(\Leftrightarrow\frac{10}{\left(x-14\right)\left(x-9\right)}=\frac{10}{\left(x-13\right)\left(x-11\right)}\)
\(\Rightarrow\left(x-14\right)\left(x-9\right)=\left(x-13\right)\left(x-11\right)\)
\(\Leftrightarrow x^2-23x+126=x^2-24x+143\)
\(\Leftrightarrow x=17\left(tm\right)\)
Vậy phương trình có tập nhiệm \(S=\left\{17\right\}\)
`14/5 - x = 6/5 + 3/4 `
`14/5 - x = 24/20 + 15/20`
`14/5 - x = 39/20`
` x = 14/5 - 39/20`
` x = 56/20 - 39/20`
` x =17/20`
Vậy ...
14/5- x= 6/5 + 3/4
14/5- x= 39/20
x= 39/20- 14
x= -17/20