Tìm x thuộc Q: x/3/4=—2/1/2
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a, 11/12 - ( 2/5 + x ) = 2/3
<=> \(\frac{2}{5}+x=\frac{11}{12}-\frac{2}{3}=\frac{1}{4}\)
=> x=\(\frac{1}{4}-\frac{11}{12}=-\frac{2}{3}\)
b, 2x . ( x - 1/7 ) = 0
<=>\(\left[\begin{array}{nghiempt}x=0\\x-\frac{1}{7}=0\end{array}\right.\)<=> \(\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{7}\end{array}\right.\)
vậy x={\(0;\frac{1}{7}\)}
c, 3/4 + 1/4 : x = 2/5
<=>\(\frac{1}{4}:x=\frac{2}{5}-\frac{3}{4}=-\frac{7}{20}\)
<=> \(x=\frac{1}{4}:\left(-\frac{7}{20}\right)=-\frac{5}{7}\)
vậy x=-5/7
a) \(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}\)
\(\Leftrightarrow\frac{11}{12}-\frac{2}{5}-x=\frac{2}{3}\)
\(\Leftrightarrow-x=\frac{2}{3}-\frac{11}{12}+\frac{2}{5}=\frac{3}{20}\)
\(\Leftrightarrow x=-\frac{3}{20}\)
b) \(2x\left(x-\frac{1}{7}\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-\frac{1}{7}=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{7}\end{array}\right.\)
c) \(\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\)
\(\Leftrightarrow\frac{1}{4x}=\frac{2}{5}-\frac{3}{4}=-\frac{7}{20}\)
\(\Leftrightarrow4x=\frac{-20}{7}\)
\(\Leftrightarrow x=-\frac{5}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2.\(P=\frac{x+1}{2x+5}+\frac{x+2}{2x+4}+\frac{x+3}{2x+3}\)
\(=\frac{x+1}{2x+5}+1+\frac{x+2}{2x+4}+1+\frac{x+3}{2x+3}+1-3\)
\(=\frac{3x+6}{2x+5}+\frac{3x+6}{2x+4}+\frac{3x+6}{2x+3}-3\)
\(=\left(3x+6\right)\left(\frac{1}{2x+5}+\frac{1}{2x+4}+\frac{1}{2x+3}\right)-3\)
Áp dụng BĐT Cô-si ta có:
\(a+b+c\ge3\sqrt[3]{abc}\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\)
Nhân vế với vế của 3 BĐT trên ta được:
\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge3\sqrt[3]{abc}.3\sqrt[3]{\frac{1}{abc}}=9\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\left(1\right)\)
Áp dụng BĐT \(\left(1\right)\)ta được:
\(\frac{1}{2x+5}+\frac{1}{2x+4}+\frac{1}{2x+3}\ge\frac{9}{6x+12}\)
\(\Leftrightarrow\left(3x+6\right)\left(\frac{1}{2x+5}+\frac{1}{2x+4}+\frac{1}{2x+3}\right)-3\ge3\left(x+2\right).\frac{9}{6\left(x+2\right)}-3\)
\(\Leftrightarrow P\ge\frac{3}{2}\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(x+1\right)\left(x-2\right)< 0\)
\(\Leftrightarrow\begin{cases}x+1< 0\\x-2>0\end{cases}\) hoặc \(\begin{cases}x+1>0\\x-2< 0\end{cases}\)
\(\Leftrightarrow\begin{cases}x< -1\\x>2\end{cases}\) (loại) hoặc \(\begin{cases}x>-1\\x< 2\end{cases}\)
\(\Leftrightarrow-1< x< 2\)
b)\(\left(x-2\right)\left(x+\frac{2}{3}\right)>0\)
\(\Leftrightarrow\begin{cases}x-2>0\\x+\frac{2}{3}>0\end{cases}\) hoặc \(\begin{cases}x-2< 0\\x+\frac{2}{3}< 0\end{cases}\)
\(\Leftrightarrow\begin{cases}x>2\\x>-\frac{2}{3}\end{cases}\) hoặc \(\begin{cases}x< 2\\x< -\frac{2}{3}\end{cases}\)
\(\Leftrightarrow x>2\) hoặc \(x< -\frac{2}{3}\)
a) \(\left(x+1\right)\left(x-2\right)< 0\)
\(\Rightarrow x+1\) và \(x-2\) trái dấu nhau.
Mà \(x-2< x+1\) với mọi x
\(\Rightarrow\begin{cases}x-2< 0\\x+1>0\end{cases}\Leftrightarrow\begin{cases}x< 2\\x>-1\end{cases}\Leftrightarrow-1< x< 2\)
\(\Rightarrow x\in\left\{0;1\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{4+x}{x+3}=\frac{x+3+1}{x+3}=1+\frac{1}{x+3}\)(x\(\ne\)-3)
de A thuoc Z ma x thuoc Z \(\Leftrightarrow x+3\in\)Ư(3)={1;-1;3;-3}
ta co bang
x+3 | 1 | -1 | 3 | -3 |
x | -2(tm) | -4(tm) | 0(tm) | -6(tm) |
vay de A thuoc Z khi x \(\in\){-2;-4;0;-6}
co \(|^{ }_{ }x+1|^{ }_{ }\ge0\)voi moi x
\(\Rightarrow|^{ }_{ }x+1|^{ }_{ }-2\ge-2\)hay B \(\ge\)-2
dau "=" xay ra khi x+1=0\(\Leftrightarrow\)x=-1
vay voi x=-1 thi B dat gia tri nho nhat la -2