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A) \(\left(x+1\right).\left(x-2\right)< 0\)
\(=x.\left(x-2\right)+1.\left(x-2\right)< 0\)
\(=x.\left(x-2\right)+\left(x-2\right)< 0\)
\(\Rightarrow x\in Z\)
Vậy \(x>2\)
B)\(\left(x-2\right).\left(x+\frac{2}{3}\right)>0\)
\(x.\left(x+\frac{2}{3}\right)-2\left(x\frac{2}{3}\right)\)
\(\Rightarrow x+\frac{2}{3}=sốnguyên\)
Nên \(x\)thuốc phân số.
Câu c) tự làm nha.
a: \(-\dfrac{3}{2}x+\dfrac{1}{4}=\dfrac{1}{2}\left(x+1\right)\)
=>\(-\dfrac{3}{2}x+\dfrac{1}{4}=\dfrac{1}{2}x+\dfrac{1}{2}\)
=>\(-\dfrac{3}{2}x-\dfrac{1}{2}x=\dfrac{1}{2}-\dfrac{1}{4}\)
=>\(-2x=\dfrac{1}{4}\)
=>\(2x=-\dfrac{1}{4}\)
=>\(x=-\dfrac{1}{4}:2=-\dfrac{1}{8}\)
b: ĐKXĐ: x>=0
\(\left(6-3\sqrt{x}\right)\left(\left|x\right|-7\right)=0\)
=>\(\left\{{}\begin{matrix}6-3\sqrt{x}=0\\\left|x\right|-7=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3\sqrt{x}=6\\\left|x\right|=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=2\\\left[{}\begin{matrix}x=7\left(nhận\right)\\x=-7\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=7\left(nhận\right)\\x=4\left(nhận\right)\end{matrix}\right.\)
a) \(x-\dfrac{2}{3}=\dfrac{3}{8}\Rightarrow x=\dfrac{3}{8}+\dfrac{2}{3}=\dfrac{25}{24}\)
b) \(x-\dfrac{3}{4}=\dfrac{13}{10}:\dfrac{26}{5}\Rightarrow x-\dfrac{3}{4}=\dfrac{1}{4}\Rightarrow x=\dfrac{1}{4}+\dfrac{3}{4}=1\)
c) \(\dfrac{3}{2}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\Rightarrow x+\dfrac{1}{2}=\dfrac{3}{2}-\dfrac{4}{5}=\dfrac{7}{10}\)
\(\Rightarrow x=\dfrac{7}{10}-\dfrac{1}{2}=\dfrac{1}{5}\)
d) \(\left|x-2\right|-1=0\Rightarrow\left|x-2\right|=1\)
\(\Rightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
a: Ta có: \(x-\dfrac{2}{3}=\dfrac{3}{8}\)
\(\Leftrightarrow x=\dfrac{3}{8}+\dfrac{2}{3}=\dfrac{9}{24}+\dfrac{16}{24}=\dfrac{25}{24}\)
b: Ta có: \(x-\dfrac{3}{4}=\dfrac{13}{10}:\dfrac{26}{5}\)
\(\Leftrightarrow x-\dfrac{3}{4}=\dfrac{13}{10}\cdot\dfrac{5}{26}=\dfrac{1}{4}\)
hay x=1
\(a,1-3\left|2x-3\right|=-\dfrac{1}{2}\\ 3\left|2x-3\right|=1+\dfrac{1}{2}\\ 3\left|2x-3\right|=\dfrac{3}{2}\\ \left|2x-3\right|=\dfrac{3}{2}:3\\ \left|2x-3\right|=\dfrac{9}{2}\\ \Rightarrow\left[{}\begin{matrix}2x-3=\dfrac{9}{2}\\2x-3=-\dfrac{9}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=\dfrac{15}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
Vậy `x in {15/4;-3/4}`
\(b,\left(\left|x\right|-0,2\right)\left(x^3-8\right)=0\\ \left(\left|x\right|-0,2\right)\left(x-2\right)\left(x^2+2x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}\left|x\right|-0,2=0\\x-2=0\\x^2+2x+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\left|x\right|=0,2\\x=2\\\left(x+1\right)^2+3=0\left(lọai\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0,2\\x=-0,2\\x=2\end{matrix}\right.\)
Vậy `x in {+-0,2;2}`
a) Có (x + 1) > (x - 2)
Để (x + 1)(x - 2) < 0
Thì 2 thừa số phải trái dấu
mà (x + 1) > (x - 2)
=> \(\hept{\begin{cases}x-2< 0\\x+1>0\end{cases}}\Rightarrow\hept{\begin{cases}x< 2\\x>-1\end{cases}}\Rightarrow-1< x< 2\)
a) \(\left(x-2\right)\left(x+\frac{2}{3}\right)>0\)
\(\Leftrightarrow\hept{\begin{cases}x-2>0\\x+\frac{2}{3}>0\end{cases}}\Rightarrow\hept{\begin{cases}x>2\\x>-\frac{2}{3}\end{cases}}\Rightarrow x>2\)
Hoặc \(\hept{\begin{cases}x-2< 0\\x+\frac{2}{3}< 0\end{cases}}\Rightarrow\hept{\begin{cases}x< 2\\x< -\frac{2}{3}\end{cases}}\Rightarrow x< \frac{-2}{3}\)
Vậy \(\orbr{\begin{cases}x>2\\x< -\frac{2}{3}\end{cases}}\)
a)\(\left(x+1\right)\left(x-2\right)< 0\)
\(\Rightarrow\) x + 1 và x - 2 khác dấu nhau
mà x + 1 > x - 2 với mọi x
\(\Rightarrow\begin{cases}x+1>0\\x-2< 0\end{cases}\)\(\Rightarrow\begin{cases}x=-1\\x=2\end{cases}\)\(\Rightarrow-1< x< 2\)
\(\Rightarrow x\in\left\{0;1\right\}\)
nhân vào được pt bật 2 rồi giải có gì đâu!!!!!
a) x=2;-1
b) a*b>0
thì xét 2 th a>và b> hặc a<0 và b<0
hết