|\(x+\frac{3}{4}\left|-\frac{1}{2}=0\right|\)
Những câu hỏi liên quan
a)\(\left(\frac{5}{7}x-\frac{1}{4}\right)\left(\frac{-3}{4}x+\frac{1}{2}\right)=0\)
b)\(\left(\frac{4}{5}+x\right).\left(x-\frac{8}{13}\right)=0\)
c)\(\left(2x-\frac{1}{2}\right).\left(x-3\right)=0\)
d)\(x+3\frac{1}{2}x+x=\frac{1}{2}\)
a) \(\left(\frac{5}{7}x-\frac{1}{4}\right)\left(\frac{-3}{4}x+\frac{1}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{5}{7}x-\frac{1}{4}=0\\\frac{-3}{4}x+\frac{1}{2}=0\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{5}{7}x=\frac{1}{4}\\\frac{-3}{4}x=\frac{-1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{20}\\x=\frac{2}{3}\end{cases}}\)
Vậy \(x=\frac{7}{20}\) hoặc x=\(\frac{2}{3}\)
b) \(\left(\frac{4}{5}+x\right)\left(x-\frac{8}{13}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{4}{5}+x=0\\x-\frac{8}{13}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-4}{5}\\x=\frac{8}{13}\end{cases}}\)
Vậy x=-4/5 hoặc x=8/13
c) \(\left(2x-\frac{1}{2}\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{2}=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=3\end{cases}}\)
Vậy x=1/4 hoặc x=3
\(x+\frac{7}{2}x+x=\frac{1}{2}\)
\(2x+\frac{7}{2}x=\frac{1}{2}\)
\(\left(2+\frac{7}{2}\right)x=\frac{1}{2}\)
\(\frac{11}{2}x=\frac{1}{2}\)
\(x=\frac{1}{2}:\frac{11}{2}\)
\(x=\frac{1}{11}\)
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\(\left(4\frac{1}{6}x^2-\frac{2}{3}\right)\left(-0,75x-\frac{21}{32}\right)\left(\frac{5}{6}\left|x\right|-3\frac{1}{3}\right)\)\(\left(4\frac{1}{2}x^4+1\frac{1}{3}x\right)=0\)
\(\left(4\frac{1}{6}x^2-\frac{2}{3}\right)\left(-0,75x-\frac{21}{32}\right)\left(\frac{5}{6}\left|x\right|-3\frac{1}{3}\right)\)\(\left(4\frac{1}{2}x^4+1\frac{1}{3}x\right)=0\)
\(\left(\frac{1}{7}x-\frac{2}{7}\right).\left(-\frac{1}{5}x+\frac{3}{5}\right).\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)0
b) \(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{5}x+1=0\)
a) \(\left(\frac{1}{7}x-\frac{2}{7}\right)\cdot\left(-\frac{1}{5}x+\frac{3}{5}\right)\cdot\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
\(\Rightarrow\)TH1 : \(\frac{1}{7}x-\frac{2}{7}=0\) TH2 : \(-\frac{1}{5}x+\frac{3}{5}=0\) TH3 : \(\frac{1}{3}x+\frac{4}{3}=0\)
\(\frac{1}{7}x=\frac{2}{7}\) \(-\frac{1}{5}x=\frac{3}{5}\) \(\frac{1}{3}x=\frac{4}{3}\)
\(x=\frac{2}{7}\cdot7\) \(x=\frac{3}{5}\cdot-5\) \(x=\frac{4}{3}\cdot3\)
\(x=2\) \(x=-3\) \(x=4\)
Vậy x = 2 hoặc x = -3 hoặc x = 4
b) \(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{5}x+1=0\)
\(x\cdot\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{5}\right)=1\)
\(x\cdot\frac{5+3-24}{30}=1\)
\(x\cdot\frac{-8}{15}=1\)
\(x=1\cdot\frac{-15}{8}=\frac{-15}{8}\)
Vậy x = \(\frac{-15}{8}\)
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\(0=-\frac{\left(x+2\right)^2+12}{\left(x+2\right)^2}+\frac{\left(x+1\right)^2+1}{\left(x+1\right)^2}-\frac{\left(x+3\right)^2+3}{\left(x+3\right)^2}+\frac{\left(x+4\right)^2+4}{\left(x+4\right)^2}\)
1. tinh` giá trị biểu thức ( tính nhanh nếu có thế )
a)frac{-6}{11}.frac{5}{13}+frac{-6}{11}.frac{8}{13}-left(frac{-2}{5}right)^0 b)left(2frac{2}{3}+3frac{1}{2}right);left(4frac{3}{4}-2frac{1}{6}right)+frac{19}{31} c)2,4:left(-2right)^3+left(3-frac{9}{11}right).1frac{3}{8}
d)left(-frac{3}{4}right)^2:frac{-3}{8}+frac{1}{2}-frac{3}{4}-left(frac{-78}{57}right)^0
2. tìm x
a)x+frac{-1}{5}left(-frac{3}{4^{ }}right)^2 b)left|frac{5}{2}x+frac{2}{3}right|-frac{1}{4}0...
Đọc tiếp
1. tinh` giá trị biểu thức ( tính nhanh nếu có thế )
\(a)\frac{-6}{11}.\frac{5}{13}+\frac{-6}{11}.\frac{8}{13}-\left(\frac{-2}{5}\right)^0\) \(b)\left(2\frac{2}{3}+3\frac{1}{2}\right);\left(4\frac{3}{4}-2\frac{1}{6}\right)+\frac{19}{31}\) \(c)2,4:\left(-2\right)^3+\left(3-\frac{9}{11}\right).1\frac{3}{8}\)
\(d)\left(-\frac{3}{4}\right)^2:\frac{-3}{8}+\frac{1}{2}-\frac{3}{4}-\left(\frac{-78}{57}\right)^0\)
2. tìm x
\(a)x+\frac{-1}{5}=\left(-\frac{3}{4^{ }}\right)^2\) \(b)\left|\frac{5}{2}x+\frac{2}{3}\right|-\frac{1}{4}=0\) \(c)\frac{2}{3}x-\frac{1}{2}=\frac{5}{12}+\frac{1}{2}x\) \(d)\left(x-\frac{1}{4}\right)^4=\frac{1}{81}\)
\(e)4x+3\frac{1}{4}=x-\frac{1}{4}\) \(g)\left(x-\frac{1}{3}\right)^3=\frac{1}{27}\)
Tìm x biết
a)\(\frac{x+1}{x-4}>0\)
b)\(\left|x+\frac{3}{4}\right|+\left|y-\frac{1}{5}\right|+\left|x+y+z\right|=0\)
c)\(\left(x+2\right)\left(x-3\right)< 0\)
d)\(\left|x+\frac{3}{4}\right|+\left|y-\frac{2}{5}\right|+\left|z+\frac{1}{2}\right|\le0\)
Ta có : \(\frac{x+1}{x-4}>0\)
Thì sảy ra 2 trường hợp
Th1 : x + 1 > 0 và x - 4 > 0 => x > -1 ; x > 4
Vậy x > 4
Th2 : x + 1 < 0 và x - 4 < 0 => x < -1 ; x < 4
Vậy x < (-1) .
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Ta có : \(\left(x+2\right)\left(x-3\right)< 0\)
Th1 : \(\hept{\begin{cases}x+2< 0\\x-3>0\end{cases}\Rightarrow\hept{\begin{cases}x< -2\\x>3\end{cases}}\left(\text{Vô lý }\right)}\)
Th2 : \(\hept{\begin{cases}x+2>0\\x-3< 0\end{cases}\Rightarrow\hept{\begin{cases}x>-2\\x< 3\end{cases}\Rightarrow}-2< x< 3}\)
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\(\Rightarrow\frac{x-4}{x-4}+\frac{5}{x-4}>0\)
\(\Rightarrow1+\frac{5}{x-4}>0\)
\(\Rightarrow\frac{5}{x-4}>-1\)
\(\Rightarrow\frac{-5}{-x+4}>-\frac{5}{5}\)
\(\Rightarrow-x+4< -5\)
\(\Rightarrow-x< -9\)
\(\Rightarrow x>9\)
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a,\(8x^3-12x^2+6x-5=0\Leftrightarrow8\left(x^3-\frac{3}{2}x^2+\frac{3}{4}x-\frac{1}{8}\right)-4=0\)
\(\Leftrightarrow8\left(x-\frac{1}{2}\right)^3=4\Leftrightarrow\left(x-\frac{1}{2}\right)^3=\frac{1}{2}\Leftrightarrow x=\frac{1}{\sqrt[3]{2}}+\frac{1}{2}\)
\(1.\frac{1}{x^2-2x+2}+\frac{2}{x^2-2x+3}=\frac{6}{x^2-2x+4}
\)
2.\(\frac{2x^4}{\left(x+1\right)^2}-\frac{5x^2}{x+1}+2=0\)
3.\(\left(x+\frac{1}{x}\right)^2-6\left(x+\frac{1}{x}\right)+8=0\)
4.\(\left(x^2+\frac{1}{x^2}\right)-4\left(x+\frac{1}{x}\right)+6=0\)
5.\(\frac{2x}{3x^2-x+2}-\frac{7x}{3x^2+5x+2}=1\)
\(\frac{-3x.\left(5x+3\right)}{1+3x}>=0\)\(\frac{-2x^2+5x-3}{-x.\left(3x+7\right)}>0\)\(\frac{1}{x-2}-\frac{4}{x^2-4}< \frac{1}{3}\)\(x^2-20x+51>0\)\(\left(x-3\right).\left(2x+1\right)\left(1-5x\right)< 0\)\(\left(x-2\right)\left(x+3\right)=< 0\)