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a) (x-2)*(-5-x^2)>0
\(\Rightarrow\orbr{\begin{cases}x-2>0\\-5-x^2>0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x^2=-5\end{cases}}\)
=>x=2 (vì x2\(\ge0\))
Vậy....
a) \(\left(x-5\right).x=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-5=0\\x=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=5\\x=0\end{cases}}\)
Vậy....
b) \(\left(x-2\right)\left(1-x\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-2=0\\1-x=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2\\x=1\end{cases}}\)
Vậy...
c) \(\left(x+1\right)\left(x^2+4\right)=0\)
Ta thấy \(x^2\ge0\) \(\forall x\)
nên \(x^2+4>0\)
\(\Rightarrow\)\(x+1=0\)
\(\Leftrightarrow\)\(x=-1\)
Vậy...
d) \(\left(2x-4\right)\left(9-3x\right)=0\)
\(\Leftrightarrow\)\(6\left(x-2\right)\left(3-x\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-2=0\\3-x=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
Vậy....
a)\(\orbr{\begin{cases}x-5=0\\x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}}\)
b)\(\orbr{\begin{cases}x-2=0\\1-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=1\end{cases}}}\)
c)\(\orbr{\begin{cases}x+1=0\\x^2+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x^2=-4\end{cases}}}\Leftrightarrow x=-1\)( DO \(x^2\ge0\)mà\(4\le0\))
d)\(\orbr{\begin{cases}2x-4=0\\9-3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}}\)
a: \(\Leftrightarrow x\cdot\dfrac{1}{2}=\dfrac{1}{4}\cdot\dfrac{-7}{2}+\dfrac{5}{3}=\dfrac{19}{24}\)
hay x=19/12
b: \(\Leftrightarrow\left(\dfrac{45}{11}-3x\right)\cdot\dfrac{11}{5}=\dfrac{21}{5}\)
\(\Leftrightarrow-3x+\dfrac{45}{11}=\dfrac{21}{11}\)
=>-3x=-24/11
=>x=8/11
c: \(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\5-\dfrac{1}{2}x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=10\end{matrix}\right.\)
d: \(\Leftrightarrow\left(2x+\dfrac{3}{5}\right)^2=\dfrac{16}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{3}{5}=\dfrac{4}{5}\\2x+\dfrac{3}{5}=-\dfrac{4}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{10}\\x=-\dfrac{7}{10}\end{matrix}\right.\)
\(-\frac{25}{35}\cdot\frac{5}{6}< x< (-2)^2\)
\(\Rightarrow-\frac{25}{7}\cdot\frac{1}{6}< x< 4\)
\(\Rightarrow-\frac{25}{42}< x< 4\)
Tự tìm x nhé :v nếu x không có thì ghi : x không thỏa mãn điều kiện
a) \(\left(x^2+1\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x^2-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=-1\\x^2=4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\varnothing\\x=\pm2\end{cases}}}\)
Vậy x=\(\pm2\)
b) \(\left(x^3-27\right)\left(x^3+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^3-27=0\\x^3+8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^3=27\\x^3=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3\\x=-2\end{cases}}}\)
Vậy x=3; x=-2
d) \(|3x+8|-|x-4|=0\)
\(\Leftrightarrow|3x+8|=|x-4|\)
\(\Leftrightarrow\orbr{\begin{cases}3x+8=x-4\\-3x-8=x-4\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=-12\\-4x=4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-6\\x=-1\end{cases}}}\)
Vậy x=-6; x=-1
ta có: (x - 7)(2x - 8) = 0
=> \(\hept{\begin{cases}x-7=0\\2x-8=0\end{cases}}\)
=> \(\hept{\begin{cases}x=7\\x=4\end{cases}}\)
chúc bạn học giỏi!! ^^
ok mk nhé!!! 64467657678563463465757575686876679646346454564556757567568568973453434
\(\left(x+1\right)^2+\left(x^2+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)^2=0\\\left(x^2+1\right)=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x+1=0\\x^2+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-1\\x^2=-1\end{cases}\Leftrightarrow}x=-1}\)
Vậy x=-1
\(\Rightarrow\left[{}\begin{matrix}x+2=0\\x-4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=4\end{matrix}\right.\)