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1.
a) 13\(\frac{1}{3}\) : 1\(\frac{1}{3}\) = 26 : (2x - 1)
<=> \(\frac{40}{3}:\frac{4}{3}\) = 13x - 26
<=> 10 + 26 = 13x
<=> 13x = 36
<=> x = \(\frac{36}{13}\)
b) 0,2 : 1\(\frac{1}{5}\) = \(\frac{2}{3}\) : (6x + 7)
<=> \(\frac{1}{5}:\frac{6}{5}\) = \(\frac{1}{9}x\) : \(\frac{2}{21}\)
<=> \(\frac{1}{6}\) = \(\frac{1}{9}x\) : \(\frac{2}{21}\)
<=> \(\frac{1}{9}x\) = \(\frac{2}{21}.\frac{1}{6}\) = \(\frac{1}{63}\)
<=> x = \(\frac{1}{7}\)
c) \(\frac{37-x}{x+13}\) = \(\frac{3}{7}\)
<=> (37 - x) . 7 = 3.(x + 13)
<=> 119 - 7x = 3x + 39
<=> -7x - 3x = 39 - 119
<=> -10x = -80
<=> x = 8
d) \(\frac{x-1}{x+5}=\frac{6}{7}\)
<=> 7(x - 1) = 6(x + 5)
<=> 7x - 7 = 6x + 30
<=> 7x - 6x = 30 + 7
<=> x = 37
e)
2\(\frac{2}{\frac{3}{0,002}}\) = \(\frac{1\frac{1}{9}}{x}\)
<=> \(\frac{1501}{750}\) = \(\frac{10}{9}:x\)
<=> x = \(\frac{10}{9}:\frac{1501}{750}\) = \(\frac{2500}{4503}\)
Bài 2. đề sai
Bài 3.
a) 6,88 : x = \(\frac{12}{27}\)
<=> x = 6,88 : \(\frac{12}{27}\)
<=> x = 15,48
b) 8\(\frac{1}{3}\) : \(11\frac{2}{3}\) = 13 : 2x
<=> \(\frac{25}{3}:\frac{35}{3}\) = 13 : 2x
<=> \(\frac{5}{7}=13:2x\)
<=> 2x = \(13:\frac{5}{7}\) = \(\frac{91}{5}\)
<=> x = 9,1
a) i)\(\frac{7\cdot25-7\cdot7}{7\cdot24+7\cdot3}=\frac{7\left(25-7\right)}{7\left(24+3\right)}=\frac{18}{27}=\frac{2}{3}\) ii)\(\frac{2\cdot\left(-1\right)\cdot13\cdot\left(-3\right)^2\cdot\left(-2\right)\cdot\left(-5\right)}{\left(-3\right)\cdot2\cdot2\cdot\left(-5\right)\cdot13\cdot2}=\frac{-3}{2}\)
b) i)\(\frac{3}{-4}< 0;\frac{-1}{-4}>0=>\frac{3}{-4}< \frac{-1}{-4}\)
ii) ta có \(\frac{15}{17}+\frac{2}{17}=1;\frac{25}{27}+\frac{2}{27}=1\)
mà \(\frac{2}{17}>\frac{2}{27}\) =>\(\frac{15}{17}< \frac{25}{27}\)
1.
\(f\left(x\right)=\frac{x-7}{\left(x-4\right)\left(4x-3\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định tại \(x=\left\{\frac{3}{4};4\right\}\)
\(f\left(x\right)=0\Rightarrow x=7\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}\frac{3}{4}< x< 4\\x>7\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< \frac{3}{4}\\4< x< 7\end{matrix}\right.\)
2.
\(f\left(x\right)=\frac{11x+3}{-\left(x-\frac{5}{2}\right)^2-\frac{3}{4}}\)
Vậy:
\(f\left(x\right)=0\Rightarrow x=-\frac{3}{11}\)
\(f\left(x\right)>0\Rightarrow x< -\frac{3}{11}\)
\(f\left(x\right)< 0\Rightarrow x>-\frac{3}{11}\)
3.
\(f\left(x\right)=\frac{3x-2}{\left(x-1\right)\left(x^2-2x-2\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định khi \(x=\left\{1;1\pm\sqrt{3}\right\}\)
\(f\left(x\right)=0\Rightarrow x=\frac{2}{3}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< 1-\sqrt{3}\\\frac{2}{3}< x< 1\\x>1+\sqrt{3}\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}1-\sqrt{3}< x< \frac{2}{3}\\1< x< 1+\sqrt{3}\end{matrix}\right.\)
4.
\(f\left(x\right)=\frac{\left(x-2\right)\left(x+6\right)}{\sqrt{6}\left(x+\frac{\sqrt{6}}{4}\right)^2+\frac{8\sqrt{2}-3\sqrt{6}}{8}}\)
Vậy:
\(f\left(x\right)=0\Rightarrow x=\left\{-6;2\right\}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< -6\\x>2\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow-6< x< 2\)