Tìm x biết :
\(9,5-\frac{3}{4}.\left|X-\frac{1}{3}\right|=6\frac{1}{3}-\frac{1}{3}.\left|\frac{1}{3}-X\right|\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
30 tháng 1 2022
\(\Leftrightarrow3\left(x^2-4x+4\right)-\dfrac{5}{4}\left(9x^2+6x+1\right)=\dfrac{4}{3}\left(-x^2+4x-3\right)-\dfrac{7}{6}x\left(x-3\right)\)
\(\Leftrightarrow3x^2-12x+12-\dfrac{45}{4}x^2-\dfrac{15}{2}x-\dfrac{5}{4}=-\dfrac{4}{3}x^2+\dfrac{16}{3}x-4-\dfrac{7}{6}x^2+\dfrac{7}{2}x\)
\(\Leftrightarrow x^2\cdot\dfrac{-33}{4}-\dfrac{39}{2}x+\dfrac{43}{4}+\dfrac{5}{2}x^2-\dfrac{53}{6}x+4=0\)
\(\Leftrightarrow x^2\cdot\dfrac{-23}{4}-\dfrac{85}{3}x+\dfrac{59}{4}=0\)
\(\Leftrightarrow12\left(\dfrac{-23}{4}x^2-\dfrac{85}{3}x+\dfrac{59}{4}\right)=0\)
\(\Leftrightarrow-69x^2-340x+177=0\)
\(\Leftrightarrow69x^2+340x-177=0\)
\(\text{Δ}=340^2-4\cdot69\cdot\left(-177\right)=164452\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-170-\sqrt{41113}}{69}\\x_2=\dfrac{-170+\sqrt{41113}}{69}\end{matrix}\right.\)
10 tháng 5 2016
c)\(\frac{1}{2}x+\frac{1}{8}x=\frac{3}{4}\)
\(\Rightarrow x.\left(\frac{1}{2}-\frac{1}{8}\right)=\frac{3}{4}\)
\(\Rightarrow x.\frac{3}{8}=\frac{3}{4}\)
=>x\(=\frac{3}{4}:\frac{3}{8}\)
=>x=\(2\)
10 tháng 5 2016
a)\(x+\frac{1}{6}=\frac{-3}{8}\)
=>\(x=\frac{-3}{8}-\frac{1}{6}\)
=>\(x=\frac{-9}{24}-\frac{4}{24}\)
=>\(x=\frac{-13}{24}\)
b)\(2-\left|\frac{3}{4}-x\right|=\frac{7}{12}\)
=>\(\left|\frac{3}{4}-x\right|=2-\frac{7}{12}\)
=>\(\left|\frac{3}{4}-x\right|=\frac{24}{12}-\frac{7}{12}\)
\(\Rightarrow\left|\frac{3}{4}-x\right|=\frac{17}{12}\)
TH1: \(\frac{3}{4}-x=\frac{17}{12}\)
=>x=\(\frac{3}{4}-\frac{17}{12}\)
=>x=\(x=-\frac{2}{3}\)
TH2:\(\frac{3}{4}-x=-\frac{17}{12}\)
=>\(x=\frac{3}{4}-\left(-\frac{17}{12}\right)\)
=>x=\(x=\frac{13}{6}\)
Dzồi nhìu phết
TD
0
11 tháng 9 2016
a)\(\left(2x-3\right)\left(x+1\right)< 0\)
\(\Leftrightarrow\begin{cases}2x-3>0\\x+1< 0\end{cases}\) hoặc \(\begin{cases}2x-3< 0\\x+1>0\end{cases}\)
\(\Leftrightarrow\begin{cases}x>\frac{3}{2}\\x< -1\end{cases}\) (loại) hoặc \(\begin{cases}x< \frac{3}{2}\\x>-1\end{cases}\)
\(\Leftrightarrow-1< x< \frac{3}{2}\)
b) \(\left(x-\frac{1}{2}\right)\left(x+3\right)>0\)
\(\Leftrightarrow\begin{cases}x-\frac{1}{2}>0\\x+3>0\end{cases}\) hoặc \(\begin{cases}x-\frac{1}{2}< 0\\x+3< 0\end{cases}\)
\(\Leftrightarrow\begin{cases}x>\frac{1}{2}\\x>-3\end{cases}\) hoặc \(\begin{cases}x< \frac{1}{2}\\x< -3\end{cases}\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x>\frac{1}{2}\\x< -3\end{array}\right.\)
c) Sai đề phải là \(\frac{x}{\left(x+3\right)\left(x+7\right)}\)
Có: \(\frac{3}{\left(x+3\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+3\right)\left(x+17\right)}\)
\(\Leftrightarrow\)\(\frac{1}{x+3}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+7}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)
\(\Leftrightarrow\)\(\frac{1}{x+3}-\frac{1}{x+7}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)
\(\Leftrightarrow\)\(\frac{4}{\left(x+3\right)\left(x+7\right)}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)
\(\Leftrightarrow x=4\)
8 tháng 1 2020
1.
\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
\(MC:12\)
Quy đồng :
\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)
\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)
\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)
\(\Leftrightarrow6x+9-3x=-4-9+16\)
\(\Leftrightarrow-7x=3\)
\(\Leftrightarrow x=\frac{-3}{7}\)
2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
\(MC:20\)
Quy đồng :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
\(9,5-\frac{3}{4}\left|X-\frac{1}{3}\right|=6\frac{1}{3}-\frac{1}{3}\left|\frac{1}{3}-X\right|\)
\(\frac{19}{2}-\frac{3}{4}\left|X-\frac{1}{3}\right|=\frac{19}{3}-\frac{1}{3}\left|X-\frac{1}{3}\right|\)
\(\frac{19}{2}-\frac{3}{4}\left|X-\frac{1}{3}\right|+\frac{1}{3}\left|X-\frac{1}{3}\right|=\frac{19}{3}\)
\(\frac{19}{2}-\left(\frac{3}{4}\left|X-\frac{1}{3}\right|-\frac{1}{3}\left|X-\frac{1}{3}\right|=\frac{19}{3}\right)\)
\(\left|X-\frac{1}{3}\right|\left(\frac{3}{4}-\frac{1}{3}\right)=\frac{19}{2}-\frac{19}{3}\)
\(\frac{5}{12}\left|X-\frac{1}{3}\right|=\frac{19}{6}\)
\(\left|X-\frac{1}{3}\right|=\frac{19}{6}\div\frac{5}{12}\)
\(\left|X-\frac{1}{3}\right|=\frac{38}{5}\)
\(\Rightarrow\orbr{\begin{cases}X-\frac{1}{3}=\frac{38}{5}\\X-\frac{1}{3}=\frac{-38}{5}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{119}{15}\\x=\frac{-109}{15}\end{cases}}\)
Vậy.....................
P/s: sai thì bỏ qua nha!