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a) 5x.(x+3/4) = 0
=> x = 0
x+3/4 = 0 => x = -3/4
b) \(\frac{x+7}{2010}+\frac{x+6}{2011}=\frac{x+5}{2012}+\frac{x+4}{2013}.\)
\(\Rightarrow\frac{x+7}{2010}+\frac{x+6}{2011}-\frac{x+5}{2012}-\frac{x+4}{2013}=0\)
\(\frac{x+7}{2010}+1+\frac{x+6}{2011}+1-\frac{x+5}{2012}-1-\frac{x+4}{2013}-1=0\)
\(\left(\frac{x+7}{2010}+1\right)+\left(\frac{x+6}{2011}+1\right)-\left(\frac{x+5}{2012}+1\right)-\left(\frac{x+4}{2013}+1\right)=0\)
\(\frac{x+2017}{2010}+\frac{x+2017}{2011}-\frac{x+2017}{2012}-\frac{x+2017}{2013}=0\)
\(\left(x+2017\right).\left(\frac{1}{2010}+\frac{1}{2011}-\frac{1}{2012}-\frac{1}{2013}\right)=0\)
=> x + 2017 = 0
x = -2017
a) để 2x - 3 > 0
=> 2x > 3
x > 3/2
b) 13-5x < 0
=> 5x < 13
x < 13/5
c) \(\frac{x+3}{2x-1}>0\)
=> x + 3 > 0
x > -3
d) \(\frac{x+7}{x+3}=\frac{x+3+4}{x+3}=1+\frac{4}{x+3}\)
Để x+7/x+3 < 1
=> 1 + 4/x+3 < 1
=> 4/x+3 < 0
=> không tìm được x thỏa mãn điều kiện
\(A=\left(x-2\right)^2\ge0\forall x\)
Dấu '=' xảy ra khi x=2
\(B=\left(2x-1\right)^2+1\ge1\forall x\)
Dấu '=' xảy ra khi x=1/2
\(D=\left(x^2-9\right)^4+\left|y-2\right|-1\ge-1\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x^2-9=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left(x,y\right)\in\left\{\left(-3;2\right);\left(3;2\right)\right\}\)
Tìm x . biết :
\(a,\frac{2}{5}:\left(-x-\frac{1}{2}\right)=\frac{4}{5}\)
\(\Rightarrow-x-\frac{1}{2}=\frac{2}{5}:\frac{4}{5}\)
\(\Rightarrow-x-\frac{1}{2}=\frac{2}{5}.\frac{5}{4}\)
\(\Rightarrow-x-\frac{1}{2}=\frac{1}{2}\)
\(\Rightarrow-x=\frac{1}{2}+\frac{1}{2}\)
\(\Rightarrow-x=1\)
\(\Rightarrow x=-1\)
Vậy \(x=-1\)
a. \(\frac{2}{5}.\left(-x-\frac{1}{2}\right)=\frac{4}{5}\)
\(\Rightarrow-x-\frac{1}{2}=\frac{2}{5}:\frac{4}{5}\)
\(\Rightarrow-x-\frac{1}{2}=\frac{2}{5}.\frac{5}{4}\)
\(\Rightarrow-x-\frac{1}{2}=\frac{1}{2}\)
\(\Rightarrow-x=\frac{1}{2}+\frac{1}{2}\)
\(\Rightarrow-x=1\)
\(\Rightarrow x=-1\)
\(\left|2x-\frac{1}{2}\right|+1=3x\)
\(\Leftrightarrow\left|2x-\frac{1}{2}\right|=3x-1\)
\(\Leftrightarrow\orbr{\begin{cases}2x-\frac{1}{2}=3x-1\\2x-\frac{1}{2}=1-3x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3x=-1+\frac{1}{2}\\2x+3x=1+\frac{1}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-x=-\frac{1}{2}\\5x=\frac{3}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{3}{10}\end{cases}}\)
x2 + 2x = 0
=> x(x + 2) = 0
=> \(\orbr{\begin{cases}x=0\\x+2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)
(x - 2) + 3.x2 - 6x = 0
=> (x - 2) + 3x2 - 3x . 2 = 0
=> (x - 2) + 3x.(x - 2) = 0
=> (1 + 3x)(x - 2) = 0
=> \(\orbr{\begin{cases}1+3x=0\\x-2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{1}{3}\\x=2\end{cases}}\)
a)\(\frac{x}{2}=\frac{y}{3};\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{x}{10}=\frac{y}{15};\frac{y}{15}=\frac{z}{21}\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{21}\)
áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x+y+z}{10+15+21}=\frac{98}{48}=\frac{49}{23}\)
suy ra :
\(\frac{x}{10}=\frac{49}{23}\Rightarrow x=\frac{490}{23}\)
\(\frac{y}{15}=\frac{49}{23}\Rightarrow y=\frac{735}{23}\)
\(\frac{z}{21}=\frac{49}{23}\Rightarrow z=\frac{1029}{23}\)
bạn xem lại đề ra số hơi xấu
a)\(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}\)
\(\frac{2}{5}+x=\frac{1}{4}\)
\(x=\frac{1}{4}-\frac{2}{5}\)
\(x=-\frac{3}{20}\)
Vậy \(x=-\frac{3}{20}\)
b)\(\frac{3}{4}+\frac{1}{4}:x=\frac{2}{3}\)
\(\frac{1}{4}:x=-\frac{1}{12}\)
\(x=\frac{1}{4}:\left(-\frac{1}{12}\right)\)
\(x=-3\)
Vậy \(x=-3\)
