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a) \(2x-3>5x+10\) \(\Leftrightarrow\) \(2x-5x>10 +3\Leftrightarrow-3x>13\Leftrightarrow x< \dfrac{13}{-3}\) vậy \(x< \dfrac{13}{-3}\)
b) \(2x^2-3x>x+7x\) \(\Leftrightarrow\) \(2x^2-3x-x-7x>0\)
\(\Leftrightarrow\) \(2x^2-11x>0\) \(\Leftrightarrow\) \(x\left(2x-11\right)>0\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x>0\\2x-11>0\end{matrix}\right.\)\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x>0\\x>\dfrac{11}{2}\end{matrix}\right.\)
\(\Rightarrow\) \(x>\dfrac{11}{2}\) vậy \(x>\dfrac{11}{2}\)
c) \(\left(x-1\right)\left(x+3\right)< 0\) \(\Leftrightarrow\) \(x^2+3x-x-3< 0\)
\(\Leftrightarrow\) \(x^2+2x-3>0\) \(\Leftrightarrow\) \(x^2-x+3x-3>0\)
\(\Leftrightarrow\) \(x\left(x-1\right)+3\left(x-1\right)\) \(\Leftrightarrow\) \(\left(x+3\right)\left(x-1\right)\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x-1>0\\x+3>0\end{matrix}\right.\)\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x>1\\x>-3\end{matrix}\right.\) \(\Rightarrow\) \(x>1\) vậy \(x>1\)
a: \(A\left(x\right)+B\left(x\right)\)
\(=-2x^3+11x^2-5x-\dfrac{1}{5}+2x^3-3x^2-7x+\dfrac{1}{5}\)
\(=8x^2-12x\)
b: C(x)=A(x)-B(x)
\(=-2x^3+11x^2-5x-\dfrac{1}{5}-2x^3+3x^2+7x-\dfrac{1}{5}\)
\(=-4x^3+14x^2+2x-\dfrac{2}{5}\)
/5x-4/=/x+2/
\(\orbr{\begin{cases}5x-4=x+2\\5x-4=-x+2\end{cases}}suyra\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{1}{2}\end{cases}}\)
vậy x=3/2 hoặc x=1/2
a: \(=2x^3:\dfrac{-3}{2}x+4x:\dfrac{3}{2}x-5:\dfrac{3}{2}\)
=-4/3x^2+8/3-10/3
=-4/3x^2-2/3
d: \(\dfrac{3x^3-5x+2}{x-3}=\dfrac{3x^3-9x^2+9x^2-27x+22x-66+68}{x-3}\)
\(=3x^2+9x+22+\dfrac{68}{x-3}\)
a) |x - 5| - 2x = 3
| x - 5| = 3 + 2x
=> x - 5 = 3 + 2x hoặc x - 5 = -3 - 2x
=> -5 - 3 = 2x - x -5 + 3 = -2x - x
=> x = -8 -2 = -3x
=> x = 2/3
b) |2x - 1| + 3x = 1
|2x - 1| = 1 - 3x
=> 2x - 1 = 1 - 3x hoặc 2x - 1 = -1 + 3x
=> -1 - 1 = -3x - 2x -1 + 1 = 3x - 2x
=> -2 = -5x 0 = x
=> x = 2/5
c) | x - 5| = 3x - 2
=> x - 5 = 3x - 2 hoặc x - 5 = -3x + 2
=> -5 + 2 = 3x - x -5 - 2 = -3x - x
=> -3 = 2x -7 = -4x
=> x = -3/2 x = 7/4
d) |9 - 7x| = 5x - 3
=> 9 - 7x = 5x - 3 hoặc 9 - 7x = -5x + 3
=> 9 + 3 = 5x + 7x 9 - 3 = -5x + 7x
=> 12 = 12x 6 = 2x
=> x = 1 x = 3
\(a.\)\(x^2+3x=0\)
\(\Leftrightarrow x\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-3\end{cases}}\)
\(b.\)\(5x^3-4x=0\)
\(\Leftrightarrow x\left(5x^2-4\right)=0\)
\(c.\)\(\left(x+2\right)\left(7-4x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\7-4x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{7}{4}\end{cases}}}\)
\(d.\)\(2x\left(x+1\right)-x-1=0\)
\(\Leftrightarrow2x\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{1}{2}\end{cases}}}\)
|5\(x\) - 4| = |\(x+2\)|
\(\left[{}\begin{matrix}5x-4=x+2\\5x-4=-x-2\end{matrix}\right.\)
\(\left[{}\begin{matrix}4x=6\\6x=2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
vậy \(x\in\) { \(\dfrac{1}{3};\dfrac{3}{2}\)}
|2\(x\) - 3| - |3\(x\) + 2| = 0
|2\(x\) - 3| = | 3\(x\) + 2|
\(\left[{}\begin{matrix}2x-3=3x+2\\2x-3=-3x-2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-5\\x=\dfrac{1}{5}\end{matrix}\right.\)
vậy \(x\in\){ -5; \(\dfrac{1}{5}\)}
a) |2x+3x|=|4x-3|
\(\Rightarrow\orbr{\begin{cases}2x+3x=4x-3\\2x+3x=-4x+3\end{cases}\Rightarrow\orbr{\begin{cases}5x-4x=-3\\5x+4x=3\end{cases}\Rightarrow}\orbr{\begin{cases}x=-3\\9x=3\end{cases}\Rightarrow}\orbr{\begin{cases}x=-3\\x=\frac{1}{3}\end{cases}}}\)
b) |7x|-|5x+6|=0
=>|7x|=|5x+6|
\(\Rightarrow\orbr{\begin{cases}7x=5x+6\\7x=-5x-6\end{cases}\Rightarrow\orbr{\begin{cases}7x-5x=6\\7x+5x=-6\end{cases}\Rightarrow}\orbr{\begin{cases}2x=6\\12x=-6\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\x=\frac{-1}{2}\end{cases}}}\)
c) |3/2+1/2|=|4x-1|
=>|4x-1|=2
\(\Rightarrow\orbr{\begin{cases}4x-1=2\\4x-1=-2\end{cases}\Rightarrow\orbr{\begin{cases}4x=3\\4x=-1\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{3}{4}\\x=\frac{-1}{4}\end{cases}}}\)
a: 2x-3>5x+10
=>-3x>13
hay x<-13/3
b: \(2x^2-3x>x+7x\)
\(\Leftrightarrow2x^2-10x>0\)
=>2x(x-5)>0
=>x>5 hoặc x<0
c: (x-1)(x+3)<0
=>x+3>0 và x-1<0
=>-3<x<1