4x+2x=0 là 0
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a: =>(x-3)(x+1)=0
=>x=3 hoặc x=-1
b: =>x(x-3)=0
=>x=0 hoặc x=3
c: =>(x-5)(x+1)=0
=>x=5 hoặc x=-1
d: =>5x^2+7x-5x-7=0
=>(5x+7)(x-1)=0
=>x=1 hoặc x=-7/5
e: =>x^2-4=0
=>x=2 hoặc x=-4
h: =>x^2-4x+4-3=0
=>(x-2)^2=3
=>\(x=2\pm\sqrt{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x^2-4x=0\)
\(x\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}}\)
b) \(4x^2-9=0\)
\(\left(2x\right)^2-3^2=0\)
\(\left(2x+3\right)\left(2x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+3=0\\2x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-3}{2}\\x=\frac{3}{2}\end{cases}}}\)
c) \(2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\left(x-3\right)\left(2x+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\2x+5=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{-5}{2}\end{cases}}}\)
d) \(x\left(2x+9\right)-4x-18=0\)
\(x\left(2x+9\right)-2\left(2x+9\right)=0\)
\(\left(2x+9\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+9=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-9}{2}\\x=2\end{cases}}}\)
e) \(\left(2x-1\right)^2-\left(x+2\right)^2=0\)
\(\left(2x-1-x-2\right)\left(2x-1+x+2\right)=0\)
\(\left(x-3\right)\left(3x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\3x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{-1}{3}\end{cases}}}\)
\(x^2-4x=0\)
\(x.\left(x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-4=0\Leftrightarrow x=4\end{cases}}\)
\(4x^2-9=0\)
\(2^2x^2-9=0\)
\(\left(2x\right)^2-9=0\)
\(\left(2x\right)^2-3^2=0\)
\(\Rightarrow\orbr{\begin{cases}\left(2x\right)^2=\left(-3\right)^2\\\left(2x\right)^2=3^2\end{cases}\Rightarrow\orbr{\begin{cases}2x=-3\\2x=3\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-3}{2}\\x=\frac{3}{2}\end{cases}}}}\)
\(2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\left(x-3\right)\cdot\left(2x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-3\right)=0\\2x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0+3\\2x=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{-5}{2}\end{cases}}}\)
\(x\left(2x+9\right)-4x-18=0\)
\(x\left(2x+9\right)-\left(4x+18\right)=0\)
\(x\left(2x+9\right)-\left(2\cdot2x+2\cdot9\right)=0\)
\(x\left(2x+9\right)-2.\left(2x+9\right)=0\)
\(\left(2x+9\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}2x+9=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=-9\\x=0+2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-9}{2}\\x=2\end{cases}}}\)
\(\left(2x-1\right)^2-\left(x+2\right)^2=0\)
\(\Rightarrow\left(2x-1\right)^2=\left(x+2\right)^2\)
\(\Rightarrow\orbr{\begin{cases}2x-1=x+2\\2x-1=-x+2\end{cases}\Rightarrow\orbr{\begin{cases}2x=3+x\\2x=-x+3\end{cases}\Rightarrow\orbr{\begin{cases}2x-x=3\\2x+x=3\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}}}\)
\(\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Chọn C.
Đặt t = 2x > 0 phương trình đã cho thành : t2 + (x - 7) t - 4x + 10 = 0 (1)
Coi (1) là phương trình bậc hai ẩn t, ta có
∆ = (x - 7) 2 - 4( -4x + 12) = (x + 1) 2 ≥ 0
Do đó (1)
+ TH1. T = 4 thì 2x = 4 nên x = 2
+ TH2. T = 3 thì t = 3 - x và 2x = 3 - x, theo câu trên ta được x = 1
Tóm lại, phương trình đã cho có nghiệm là 1 và 2.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a.x\left(x^2-1\right)=0\\ \Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\\\Leftrightarrow \left[{}\begin{matrix}x=0\\x-1=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
\(b.\left(x-\frac{1}{2}\right)\left(2x+5\right)=0\\\Leftrightarrow \left[{}\begin{matrix}x-\frac{1}{2}=0\\2x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{5}{2}\end{matrix}\right. \)
Câu \(b\) thấy hơi kì nên chắc đề như này.
\(c.x-2\left(\frac{2}{3}x-6\right)=0\\\Leftrightarrow x-\frac{4}{3}x+12=0\\\Leftrightarrow -\frac{1}{3}x+12=0\\\Leftrightarrow -\frac{1}{3}x=-12\\\Leftrightarrow x=36\)
\(d.x^2-2x=0\\\Leftrightarrow x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
\(e.\left(x^2-2x+1\right)-4=0\\ \Leftrightarrow\left(x-1\right)^2-4=0\\\Leftrightarrow \left(x-1-2\right)\left(x-1+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+1\right)=0\\\Leftrightarrow \left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
\(f.x\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\2x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{1}{2}\end{matrix}\right.\)
\(g.4x^2+4x+1=0\\ \Leftrightarrow4\left(x^2+x+\frac{1}{4}\right)=0\\\Leftrightarrow x^2+x+\frac{1}{4}=0\\\Leftrightarrow \left(x+\frac{1}{2}\right)^2=0\\\Leftrightarrow x+\frac{1}{2}=0\\ \Leftrightarrow x=-\frac{1}{2}\)
\(h.x^2-5x+6=0\\ \Leftrightarrow x^2-2x-3x+6=0\\\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x-2\right)=0\\\Leftrightarrow \left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
\(i.2x^2+3x=0\\ \Leftrightarrow x\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\2x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-\frac{3}{2}\end{matrix}\right.\)
\(\begin{array}{l} a)x\left( {{x^2} - 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = 0\\ {x^2} - 1 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = 1\\ x = - 1 \end{array} \right.\\ b)\left( {x - \dfrac{1}{2}} \right)\left( {2x + 5} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x - \dfrac{1}{2} = 0\\ 2x + 5 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{1}{2}\\ x = - \dfrac{5}{2} \end{array} \right.\\ c)\left( {x - 2} \right)\left( {\dfrac{2}{3}x - 6} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x - 2 = 0\\ \dfrac{2}{3}x - 6 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 2\\ x = 9 \end{array} \right. \end{array}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. 3x(x-2)-x+2=0
3x(x-2)-(x-2)=0
(3x-1)(x-2)=0
=>\(\hept{\begin{cases}3x-1=0\\x-2=0\end{cases}}\)
=> \(\hept{\begin{cases}3x=1\\x=2\end{cases}}\)
=>\(\hept{\begin{cases}x=\frac{1}{3}\\x=2\end{cases}}\)
vậy x thuộc (1/3;2)
![](https://rs.olm.vn/images/avt/0.png?1311)
Suy ra (2x-4)-(3x-3×5)=1 Suy ra(2x-4)-3x+15=1 Suy ra 2x-4-3x+15=1 Suy ra (2x-3x)+(15-4)=1 -1x+11=1 1-11=-1x -1x=-10 X=10
4x + 2x =0
=> x = 0
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