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a: x<5 thì 5-x>0
A=5x+5-x+5=4x+10
b: Khi x>=0 thì \(B=5x+10+3x=8x+10\)
Khi x<0 thì B=5x+10-3x=2x+10
d: Khi x>=3 thì \(D=x-3-3x+15=-2x+12\)
Khi x<3 thì D=3-x-3x+15=-4x+18
\(x\left(3x-5\right)-9x+15=0\)
\(\Leftrightarrow x\left(3x-5\right)-3\left(3x-5\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(3x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\3x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{5}{3}\end{cases}}\)
\(3x\left(x-5\right)-2\left(5-x\right)=0\)
\(\Leftrightarrow3x\left(x-5\right)+2\left(x-5\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x+2=0\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-2}{3}\\x=5\end{cases}}\)
a) Ta có: \(3x\left(7x-2\right)-14x+4=0\)
\(\Leftrightarrow3x\left(7x-2\right)-2\left(7x-2\right)=0\)
\(\Leftrightarrow\left(7x-2\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}7x-2=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}7x=2\\3x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{7}\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{2}{7};\dfrac{2}{3}\right\}\)
b) ĐKXĐ: \(x\notin\left\{0;3\right\}\)
Ta có: \(\dfrac{2x+1}{x-3}+\dfrac{5-3x}{x}=\dfrac{2x^2-15}{x^2-3x}\)
\(\Leftrightarrow\dfrac{x\left(2x+1\right)}{x\left(x-3\right)}+\dfrac{\left(5-3x\right)\left(x-3\right)}{x\left(x-3\right)}=\dfrac{2x^2-15}{x\left(x-3\right)}\)
Suy ra: \(2x^2+x+5x-15-3x^2+9x-2x^2+15=0\)
\(\Leftrightarrow-3x^2+15x=0\)
\(\Leftrightarrow-3x\left(x-5\right)=0\)
mà -3<0
nên x(x-5)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=5\left(nhận\right)\end{matrix}\right.\)
Vậy: S={5}
a) (x + 2)(3x - 15) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\3x-15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)
b) |x - 5| = 3x + 1
\(\Leftrightarrow\left[{}\begin{matrix}x< 5\\x\ge5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\\text{ko có x thỏa mãn}\end{matrix}\right.\)
=> x = 1
1. 3y = 0
=> y = 0
2. 1+x = 0
<+ x = -1
3.
\(1-2t=0\)
\(\Leftrightarrow2t=1\)
\(\Leftrightarrow\dfrac{1}{2}\)
4. 2x +x + 3 =0
\(\Leftrightarrow3x+3=0\)
\(\Leftrightarrow x=-3\)
5.
\(25x-20=0\)
\(\Leftrightarrow25x=20\)
\(\Leftrightarrow x=\dfrac{4}{5}\)
7.
2x-3 = x+5
<=> 2x - x = 5+3
<=> x = 8
8.
x-8=2x+3
<=> x - 2x = 3+8
<=> -x = 11
<=> x = -11
9. 17-2x = 3x-5
<=> -2x-3x = -5-17
<=> -5x = -22
<=> x = \(\dfrac{22}{5}\)
10.
2x+x+22=0
<=> 3x+22=0
<=> 3x = -22
<=> x = \(\dfrac{-22}{3}\)
Mấy bài kia tự giải tương tự nhá!!!
\(3x\left(x+5\right)-3x-15=0\)
\(\Rightarrow3x\left(x+5\right)-3\left(x+5\right)=0\)
\(\Rightarrow3\left(x+5\right)\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)
\(3x\left(x+5\right)-3x-15=0\\ \Rightarrow3x\left(x+5\right)-3\left(x+5\right)=0\\ \Rightarrow\left(3x-3\right)\left(x+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x-3=0\\x+5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)