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\(2x.\left(x-\frac{1}{7}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x=0\\x-\frac{1}{7}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{7}\end{cases}}}\)
Vậy x = 0 hoặc x = 1/7
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\(\left(x-2\right).\left(y-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\y-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\y=1\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=2\\y=1\end{cases}}\)
Ủng hộ nha Nguyen Phuong Thao
1)(x-2)(y-1)=0
=> \(\left[{}\begin{matrix}x-2=0\\y-1=0\end{matrix}\right. \)=>\(\left[{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
Vậy x,y\(\in\){2;1}
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\(2x-12-x=0\)
\(\Leftrightarrow2x-x=12\Leftrightarrow x=12\)
\(\left(x-7\right)\left(2x-8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\2x-8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\2x=8\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=4\end{cases}}}\)
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a)4x+4-3x+1=14
x+5=14
x=11
b)trường hợp 1 x2-9=0
x2=9
->x=3;-3
-trường hợp 2: x+2=0
x=-2
c)-th1:x2+9=0
x2=-9
->x rỗng
d)xy+2x-y-2=0
(xy-y)+(2x-2)=0
y(x-1)+2(x-1)=0
(y+2)(x-1)=0
th1: y+2=0
y=-2
th2:x-1=0
x=1
(th1: trường hợp 1)
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(2x + 1) + (2x + 2) + ... + (2x + 2015) = 0
=> 2015.2x + (1 + 2 + 3 + ... + 2015) = 0
=> 4030x + (2015 + 1).2015 : 2 = 0
=> 4030x = -2031120
=> x = -504
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\(\text{2^5.3 - 3.(x+1)=42}\)
\(3\left(x+1\right)=54\)
\(x+1=18\)
\(x=17\)
\(\text{(4x + 5) : 3 - 11^2:11=2^2}\)
\((4x+5):3-11=4\)
\((4x+5):3=15\)
\((4x+5)=5\)
\(4x=0\)
\(x=0\)
\(\text{3x - 15 . 4 = 6.(19 - x)}\)
\(\text{3x - 60 = 6.(19 - x)}\)
\(\text{3x - 60 :(19-x)= 6}\)
\(\Rightarrow x=6\)
\(\text{(x - 7).(2x - 16)=0}\)
\(\Rightarrow\hept{\begin{cases}x-7=0\\2x-16=0\end{cases}\Rightarrow\hept{\begin{cases}x=7\\8\end{cases}}}\)
7.(x - 1) + 2x.(x - 1) = 0
(x - 1).(7 + 2x) = 0
=> x - 1 = 0 hoặc 7 + 2x = 0
=> x = 1 hoặc 2x = -7
=> x = 1 hoặc x = -7/2
Vậy x thuộc {1 ; -7/2}
7( x - 1 ) + 2( x - 1 ) = 0
7x - 7 + 2x - 2 = 0
=> x = 1