Tìm x
c/ | 2x - 10 | + 2x -10 = 0
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3x(x - 10) = x - 10
(x - 10)(3x - 1) = 0
Th1:
x - 10 = 0
x = 10
TH2:
3x - 1 = 0
3x = 1
x = 1/3
Vậy x = 10 hoặc x = 1/3
x(x + 7) - (4x + 28) = 0
x(x + 7) - 4(x + 7) = 0
(x + 7)(x - 4) = 0
Th1:
x + 7 = 0
x = - 7
Th2:
x - 4 = 0
x = 4
Vậy x = - 7 hoặc x = 4
x(x - 4) = 2x - 8
x(x - 4) - 2(x - 4) = 0
(x - 2)(x - 4) = 0
Th1:
x - 2 = 0
x = 2
Th2:
x - 4 = 0
x = 4
Vậy x = 2 hoặc x = 4
(2x + 3)(x - 1) + (2x - 3)(x - 1) = 0
(x - 1)(2x + 3 + 2x - 3) = 0
4x(x - 1) = 0
Th1:
x = 0
Th2:
x - 1 = 0
x = 1
Vậy x = 0 hoặc x = 1
a)
\(3x\left(x-10\right)=x-10\)
\(\Rightarrow3x\left(x-10\right)-\left(x-10\right)=0\)
\(\Rightarrow\left(3x-1\right)\left(x-10\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}3x-1=0\\x-10=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{1}{3}\\x=10\end{array}\right.\)
b)
\(x\left(x+7\right)-\left(4x+28\right)=0\)
\(\Rightarrow x\left(x+7\right)-4\left(x+7\right)=0\)
\(\Rightarrow\left(x-4\right)\left(x+7\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=4\\x=-7\end{array}\right.\)
c)
\(x\left(x-4\right)=2x-8\)
\(\Rightarrow x\left(x-4\right)-2\left(x-4\right)=0\)
\(\Rightarrow\left(x-4\right)\left(x-2\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=4\\x=2\end{array}\right.\)
d)
\(\left(2x+3\right)\left(x-1\right)+\left(2x+3\right)\left(x-1\right)=0\)
\(\Rightarrow2\left(2x+3\right)\left(x-1\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x+3=0\\x-1=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{3}{2}\\x=1\end{array}\right.\)
a: Ta có: \(x\left(2x-3\right)-\left(2x-1\right)\left(x+5\right)=17\)
\(\Leftrightarrow2x^2-3x-2x^2-10x+x+5=17\)
\(\Leftrightarrow-12x=12\)
hay x=-1
|2x - 10| + 10 - 2x= 0
<=> | 2x - 10 | = 2x - 10
<=> 2x -10 ≥ 0
<=> 2x ≥ 10
<=> x ≥ 5
mà x thuộc Z
=> x thuộc Ơ 5;6;7;8;9;...Ư
Tìm x biết :a) ( 2x - 3 ).( x +1 ) > 0b) ( x + 5 ).(x-7) < 0c) | 2x - 3 | + 8 = 10d) ( 2x + 5 ) . | x -8 | . ( x2 + 1 ) = 0
=>|2x-10|=-2x+10
=>2x-10<=0
=>x<=5