Tìm x biết (x -\(\frac{1}{3}\)) 2 - \(\frac{1}{4}\)= 0
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d,
\(|x-\frac{1}{3}|=\frac{5}{6}\Rightarrow \left[\begin{matrix} x-\frac{1}{3}=\frac{5}{6}\\ x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{7}{6}\\ x=\frac{-1}{2}\end{matrix}\right.\)
e,
\(\frac{3}{4}-2|2x-\frac{2}{3}|=2\)
\(\Leftrightarrow 2|2x-\frac{2}{3}|=\frac{3}{4}-2=\frac{-5}{4}\)
\(\Leftrightarrow |2x-\frac{2}{3}|=-\frac{5}{8}<0\) (vô lý vì trị tuyệt đối của 1 số luôn không âm)
Vậy không tồn tại $x$ thỏa mãn đề bài.
f,
\(\frac{2x-1}{2}=\frac{5+3x}{3}\Leftrightarrow 3(2x-1)=2(5+3x)\)
\(\Leftrightarrow 6x-3=10+6x\)
\(\Leftrightarrow 13=0\) (vô lý)
Vậy không tồn tại $x$ thỏa mãn đề bài.
a,
$0-|x+1|=5$
$|x+1|=0-5=-5<0$ (vô lý do trị tuyệt đối của một số luôn không âm)
Do đó không tồn tại $x$ thỏa mãn điều kiện đề.
b,
\(2-|\frac{3}{4}-x|=\frac{7}{12}\)
\(|\frac{3}{4}-x|=2-\frac{7}{12}=\frac{17}{12}\)
\(\Rightarrow \left[\begin{matrix} \frac{3}{4}-x=\frac{17}{12}\\ \frac{3}{4}-x=\frac{-17}{12}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-2}{3}\\ x=\frac{13}{6}\end{matrix}\right.\)
c,
\(2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}\)
\(2|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{4}\)
\(|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{8}\)
\(\Rightarrow \left[\begin{matrix} \frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\ \frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{29}{12}\\ x=\frac{-13}{12}\end{matrix}\right.\)
\(\frac{x-1}{99}-\frac{x+1}{101}+\frac{x-2}{98}-\frac{x+2}{102}+\frac{x-3}{97}-\frac{x+3}{103}+\frac{x-4}{96}-\frac{x+4}{104}=0\)
\(\Rightarrow\frac{x-1}{99}-1-\frac{x+1}{101}+1+\frac{x-2}{98}-1-\frac{x+2}{102}+1+\frac{x-3}{97}-1-\frac{x+3}{103}+1+\frac{x-4}{96}-1-\frac{x+4}{104}+1=0\)
\(\Rightarrow\frac{x-100}{99}-\frac{x-100}{101}+\frac{x-100}{98}-\frac{x-100}{102}+\frac{x-100}{97}-\frac{x-100}{103}+\frac{x-100}{96}-\frac{x-100}{104}=0\)
\(\Rightarrow\left(x-100\right).\left(\frac{1}{99}-\frac{1}{101}+\frac{1}{98}-\frac{1}{102}+\frac{1}{97}-\frac{1}{103}+\frac{1}{96}-\frac{1}{104}\right)=0\)
Vì \(\frac{1}{99}>\frac{1}{101};\frac{1}{98}>\frac{1}{102};\frac{1}{97}>\frac{1}{103};\frac{1}{96}>\frac{1}{104}\)
\(\Rightarrow\frac{1}{99}-\frac{1}{101}+\frac{1}{98}-\frac{1}{102}+\frac{1}{97}-\frac{1}{103}+\frac{1}{96}-\frac{1}{104}\ne0\)
\(\Rightarrow x-100=0\)
\(\Rightarrow x=100\)
Vậy \(x=100\)
\(a,\left(\frac{1}{7}x-\frac{2}{7}\right)\left(-\frac{1}{5}x+\frac{3}{5}\right)\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
TH1 : \(\frac{1}{7}x-\frac{2}{7}=0\Rightarrow\frac{x-2}{7}=0\Rightarrow x-2=0\Leftrightarrow x=2\)
TH2 : \(-\frac{1}{5}x+\frac{3}{5}=0\Rightarrow\frac{-x+3}{5}=0\Rightarrow-x+3=0\Leftrightarrow x=3\)
TH3 : \(\frac{1}{3}x+\frac{4}{3}=0\Rightarrow\frac{x+4}{3}=0\Rightarrow x+4=0\Leftrightarrow x=-4\)
\(\Rightarrow x\in\left\{2;3;-4\right\}\)
\(b,\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)
\(\Rightarrow\frac{5}{30}x+\frac{3}{30}x-\frac{8}{30}x+1=0\)
\(\Rightarrow\frac{5x+3x-8x}{30}+1=0\)
\(\Rightarrow1=0\)( vô lý )\(\Rightarrow x\in\varnothing\)
b, \(x\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)+1=0\)
\(0+1=0\)
=> x thuoc rong
a, \(\frac{1}{6}x+\frac{1}{10}-\frac{4}{15}x+1=0\)
\(\Leftrightarrow-\frac{1}{10}x=-\frac{11}{10}\)
\(\Leftrightarrow x=11\)
b,\(\left(\frac{1}{7}x-\frac{2}{7}\right)\left(-\frac{1}{5}x+\frac{3}{5}\right)\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
\(\Leftrightarrow\frac{1}{7}x-\frac{2}{7}=0\)hoặc \(-\frac{1}{5}x+\frac{3}{5}=0\)hoặc \(\frac{1}{3}x+\frac{4}{3}=0\)
+) \(\frac{1}{7}x-\frac{2}{7}=0\Leftrightarrow\frac{1}{7}x=\frac{2}{7}\Leftrightarrow x=2\)
+)\(-\frac{1}{5}x+\frac{3}{5}=0\Leftrightarrow-\frac{1}{5}x=-\frac{3}{5}\Leftrightarrow x=3\)
+)\(\frac{1}{3}x+\frac{4}{3}=0\Leftrightarrow\frac{1}{3}x=-\frac{4}{3}\Leftrightarrow x=-4\)
c, \(\frac{1}{2}x-\frac{11}{15}:\frac{33}{35}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{7}{9}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{2}x=\frac{4}{9}\)
\(\Leftrightarrow x=\frac{8}{9}\)
a/ \(\frac{1}{6}x+\frac{1}{10}-\frac{4}{15}x+1=0\)
\(\Rightarrow-\frac{1}{10}x=-\frac{11}{10}\)
\(\Rightarrow x=11\)
b/ \(\left(\frac{1}{7}x-\frac{2}{7}\right)\left(-\frac{1}{5}x+\frac{3}{5}\right)\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
\(\Rightarrow\frac{1}{7}x-\frac{2}{7}=0\Rightarrow\frac{1}{7}x=\frac{2}{7}\Rightarrow x=2\)
hoặc \(-\frac{1}{5}x+\frac{3}{5}=0\Rightarrow-\frac{1}{5}x=-\frac{3}{5}\Rightarrow x=3\)
hoặc \(\frac{1}{3}x+\frac{4}{3}=0\Rightarrow\frac{1}{3}x=-\frac{4}{3}\Rightarrow x=-4\)
Vậy x = 2, x = 3, x = -4
c/ \(\frac{1}{2}x-\frac{11}{15}:\frac{33}{35}=-\frac{1}{3}\)
\(\Rightarrow\frac{1}{2}x-\frac{7}{9}=-\frac{1}{3}\)
\(\Rightarrow\frac{1}{2}x=\frac{4}{9}\Rightarrow x=\frac{8}{9}\)
Vậy x = 8/9
\(a,x\cdot\frac{1}{2}\cdot\frac{2}{3}=4\)
\(\Rightarrow x\cdot\frac{1}{3}=4\)
\(\Rightarrow x=12\)
\(b,-\frac{2}{7}\cdot\frac{5}{7}\cdot x=\frac{7}{21}\)
\(\Rightarrow-\frac{10}{49}x=\frac{7}{21}\)
\(\Rightarrow x=-\frac{49}{30}\)
k đi làm tiếp cho
\(\left(x-\frac{1}{3}\right)^2-\frac{1}{4}=0\)
\(\left(x-\frac{1}{3}\right)^2=0+\frac{1}{4}\)
\(\left(x-\frac{1}{3}\right)^2=\frac{1}{4}\)
\(x-\frac{1}{3}=\sqrt{\frac{1}{4}}\)
\(x-\frac{1}{3}=\frac{1}{2}\)
\(x=\frac{1}{2}+\frac{1}{3}\)
\(x=\frac{5}{6}\)
=>(x-1/3)^2=1/4
=>(x-1/3)^2=(1/2)^2
=>x-1/3=1/2
=>x=1/2+1/3
=>x=5/6