1. (x+1).(x-2)<0
2. (x-2).(x+\(\frac{2}{3}\))>0
3. 2.x (x-\(\frac{1}{7}\))=0
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1) \(x^2+\frac{8}{9}=\frac{41}{36}\)\(\Leftrightarrow x^2=\frac{1}{4}\Leftrightarrow x=\pm\frac{1}{2}\)
2) \(\left(x-3\right)^2+-\frac{9}{25}=\frac{2}{5}.\frac{8}{5}\)
\(\Leftrightarrow\left(x-3\right)^2=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=1\\x-3=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=2\end{cases}}\)
3) \(\frac{3}{11}.\frac{22}{6}-\left(x-1\right)^2=\frac{7}{16}\)
\(\Leftrightarrow\left(x-1\right)^2=\frac{9}{16}\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=\frac{3}{4}\\x-1=-\frac{3}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{4}\\x=\frac{1}{4}\end{cases}}\)
4) \(1+\left(x+1\right)^3=\frac{37}{64}\)
\(\Leftrightarrow\left(x+1\right)^3=-\frac{27}{64}\)
\(\Rightarrow x+1=-\frac{3}{4}\)
\(\Leftrightarrow x=-\frac{7}{4}\)
5) \(\left(x-\frac{1}{2}\right)^2-\frac{9}{16}=1\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=\frac{25}{16}\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{2}=\frac{5}{4}\\x-\frac{1}{2}=-\frac{5}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{4}\\x=-\frac{3}{4}\end{cases}}\)
6) Bn ghi rõ đề nha mk ko hiểu
6 ) \(\frac{3-x}{5-x}=\left(\frac{-3}{5}\right)^2\)
\(\frac{3-x}{5-x}=\frac{9}{25}\Leftrightarrow\frac{3-x}{5-x}-\frac{9}{25}\Leftrightarrow\frac{75-25x}{125-25x}-\frac{45-9x}{125-5x}=0\)
\(\Rightarrow\frac{75-25-45+9x}{125-25x}=0\Leftrightarrow5+9x=0\Leftrightarrow x=\frac{-5}{9}\)
\(\left(\dfrac{1}{2}+1\right)\times\left(\dfrac{1}{3}+1\right)\times\left(\dfrac{1}{4}+1\right)\times\left(\dfrac{1}{5}+1\right)\times\left(\dfrac{1}{6}+1\right)\)
\(=\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\times\dfrac{6}{5}\times\dfrac{7}{6}\)
\(=\dfrac{3\times4\times5\times6\times7}{2\times3\times4\times5\times6}\)
\(=\dfrac{7}{2}\)
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a, Vì 3 chia hết cho x-1 => x-1 thuộc Ư(-3)=1,3,-1,-3
Ta có bảng
x-1 | 1 | 3 | -1 | -3 |
x | 2 | 4 | 0 | -2 |
Vậy x thuộc 2 ; 4;0;-2
b, Vì -4 chia hết cho 2x - 1 nên 2x-1 ϵ Ư(-4) = 1;2;4;-1;-2;-4
Ta có bảng :
2x-1 | 1 | 2 | 4 | -1 | -2 | -4 |
x | 1 | vô lý | vô lý | 0 | vô lý | vô lỹ |
Vây x= 1 và 0
Ta có:
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\frac{2x-2}{4}=\frac{3y-6}{9}\)
Áp dụng tính chất của dãy tỉ số = nhau ta có:
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{\left(2x-2\right)+\left(3y-6\right)-\left(z-3\right)}{4+9-4}\)
\(=\frac{\left(2x+3y-z\right)-5}{9}=\frac{50-5}{9}=\frac{45}{9}=5\)
\(\Rightarrow\begin{cases}x-1=2.5=10\\y-2=3.5=15\\z-3=4.5=20\end{cases}\)\(\Rightarrow\begin{cases}x=11\\y=17\\z=23\end{cases}\)
Vậy x = 11; y = 17; z = 23
\(1.\\ A=\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\\ =\left|2+\sqrt{3}\right|+\left|2-\sqrt{3}\right|\\ =2+\sqrt{3}+2-\sqrt{3}=4\)
\(2.\\a.\\ P=3x-\sqrt{\left(x-5\right)^2}=3x-\left|x-5\right|\\ b.\\ x=2\Rightarrow P=3\)
\(3.\\ M=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{\left|x-1\right|}{x-1}\)
\(\cdot x>1\Rightarrow M=1\\ \cdot x=1\Rightarrow M=0\\\cdot x< 1\Rightarrow M=-1\)
B1.
Ta có:A\(=\sqrt{3+4\sqrt{3}+4}+\sqrt{3-4\sqrt{3}+4}\)
\(=\sqrt{\left(\sqrt{3}+2\right)^2}+\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\sqrt{3}+2+\sqrt{3}-2=2\sqrt{3}\)
1.x<-1 hoặc x=1
2.x>2
3.x=0 hoặc x=1/7