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a) \(\left(x+1\right)\left(x-2\right)< 0\Rightarrow\hept{\begin{cases}x+1>0\\x-2< 0\end{cases}}\Rightarrow\hept{\begin{cases}x>-1\\x< 2\end{cases}}\Rightarrow x=\left\{1;0\right\}\)
b) Xét 2 trường hợp
+ TH1: \(\hept{\begin{cases}x-2< 0\\x+\frac{2}{3}< 0\end{cases}\Rightarrow\hept{\begin{cases}x< 2\\x< -\frac{2}{3}\end{cases}}}\)=> \(x< -\frac{2}{3}\)thỏa mãn đề bài
+ TH2: \(\hept{\begin{cases}x-2>0\\x+\frac{2}{3}>0\end{cases}\Rightarrow\hept{\begin{cases}x>2\\x>-\frac{2}{3}\end{cases}}}\)=> x > 2 thỏa mãn đề bài
Vậy \(\orbr{\begin{cases}x< -\frac{2}{3}\\x>2\end{cases}}\)thỏa mãn đề bài
\(\left|3x-4\right|-\left|y+3\right|=0\)
\(\Rightarrow\left|3x-4\right|+\left|3-y\right|=0\)
\(\Rightarrow\hept{\begin{cases}3x-4=0\\3-y=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{4}{3}\\y=3\end{cases}}}\)
\(3\sqrt{x}-2x=0\)
\(\Leftrightarrow3\sqrt{x}=2x\)
\(\Leftrightarrow\sqrt{x}=\frac{2x}{3}\)
\(\Leftrightarrow\left(\sqrt{x}\right)^2=\frac{4x^2}{9}\)
\(\Leftrightarrow x=\frac{4x^2}{9}\)
\(\Leftrightarrow\frac{4x^2}{x}=9\)
\(\Leftrightarrow4x=9\)
\(\Leftrightarrow x=\frac{9}{4}\)
\(3\sqrt{x}-2x=0\)
\(\Leftrightarrow9x-4x^2=0\)
\(\Leftrightarrow x\left(9-4x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\9-4x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{9}{4}\end{cases}}}\)
a) \(\left(x+5\right)\left(x-2\right)< 0\)
\(\Leftrightarrow\hept{\begin{cases}x+5>0\\x-2< 0\end{cases}}\) hoặc \(\hept{\begin{cases}x+5< 0\\x-2>0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x>-5\\x< 2\end{cases}}\) hoặc \(\hept{\begin{cases}x< -5\\x>2\end{cases}}\) (loại)
Vậy -5 < x < 2
b) \(\left(x+2\right)\left(x-\frac{3}{5}\right)>0\)
\(\Leftrightarrow\hept{\begin{cases}x+2>0\\x-\frac{3}{5}>0\end{cases}}\) hoặc \(\hept{\begin{cases}x+2< 0\\x-\frac{3}{5}< 0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x>-2\\x>\frac{3}{5}\end{cases}}\) hoặc \(\hept{\begin{cases}x< -2\\x< \frac{3}{5}\end{cases}}\)
Vậy x > 3/5 hoặc x < -2
a ) ( x + 5 )( x - 2 ) < 0
=> x + 5 duong va x - 2 am hoac x + 5 am va x - 2 duong
Neu x + 5 duong va x - 2 am thi
-5 < x < 2
=> x \(\in\left\{1;0;-1;-2;-3;-4\right\}\)
Neu x + 5 am va x - 2 duong thi :
x < -5 va x > 2
Vi 2 dieu kien tren mau thuan vs nhau nen x\(\varnothing\)trong truong hop nay
a.
\(-\frac{2}{3}-\frac{1}{3}\times\left(2x-5\right)=\frac{3}{2}\)
\(-\frac{2}{3}-\frac{2}{3}x+\frac{5}{3}=\frac{3}{2}\)
\(\left(-\frac{2}{3}+\frac{5}{3}\right)-\frac{2}{3}x=\frac{3}{2}\)
\(\frac{3}{3}-\frac{2}{3}x=\frac{3}{2}\)
\(1-\frac{2}{3}x=\frac{3}{2}\)
\(\frac{2}{3}x=1-\frac{3}{2}\)
\(\frac{2}{3}x=\frac{2}{2}-\frac{3}{2}\)
\(\frac{2}{3}x=-\frac{1}{2}\)
\(x=-\frac{1}{2}\div\frac{2}{3}\)
\(x=-\frac{1}{2}\times\frac{3}{2}\)
\(x=-\frac{3}{4}\)
b.
\(\frac{1}{3}x+\frac{2}{5}\times\left(x-1\right)=0\)
\(\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0\)
\(x\times\left(\frac{1}{3}+\frac{2}{5}\right)=\frac{2}{5}\)
\(x\times\left(\frac{5}{15}+\frac{6}{15}\right)=\frac{2}{5}\)
\(x\times\frac{11}{15}=\frac{2}{5}\)
\(x=\frac{2}{5}\div\frac{11}{15}\)
\(x=\frac{2}{5}\times\frac{15}{11}\)
\(x=\frac{6}{11}\)
Chúc bạn học tốt
a ) \(-\frac{2}{3}-\frac{1}{3}\left(2x-5\right)=\frac{3}{2}\)
\(\frac{1}{3}\left(2x-5\right)=-\frac{2}{3}-\frac{3}{2}\)
\(\frac{1}{3}\left(2x-5\right)=\frac{-13}{6}\)
\(\left(2x-5\right)=-\frac{13}{6}:\frac{1}{3}\)
\(\left(2x-5\right)=-\frac{13}{6}.\frac{3}{1}\)
\(\left(2x-5\right)=-\frac{13}{2}\)
\(2x=-\frac{13}{2}+5\)
\(2x=-\frac{3}{2}\)
\(\Rightarrow x=-\frac{3}{2}:2\)
\(\Rightarrow x=-\frac{3}{2}.\frac{1}{2}\)
\(\Rightarrow x=-\frac{3}{4}\)
| 2 - x | va | x - 3 | luon \(\ge\)0
ma | 2 - x | - | x - 3 | = 0
=> | 2 - x | = | x - 3 |
voi x\(\ge\)3
=> -2+x=x-3(vo ly)
voi x<3
=> 2-x=-x+3
=> 2x=-1
=> x=-1/2