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a, 7\(x\).(2\(x\) + 10) =0
\(\left[{}\begin{matrix}x=0\\2x+10=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\2x=-10\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Vậy \(x\in\) {-5; 0}
b, -9\(x\) : (2\(x\) - 10) = 0
9\(x\) = 0
\(x\) = 0
c, (4 - \(x\)).(\(x\) + 3) = 0
\(\left[{}\begin{matrix}4-x=0\\x+3=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
Vậy \(x\in\) {-3; 4}
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=>x^2-7x=0-10=10
=>x.(x-7)=10=1.10=(-1).(-10)=2.5=......(tự viết tiếp nha)
=> ta có bảng sau:
(cột 1:x ; x-7 ;
cột 2;3;4;....ghi các giá trị )
Kết luận
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\(x^2-7x+10=0\)
\(\Leftrightarrow x^2-2x-5x+10=0\)
\(\Leftrightarrow x\left(x-2\right)-5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\x-5=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=5\end{array}\right.\)
x2 -7x + 10 = 0
=> x2 - 2x - 5x + 10 = 0
=> x(x - 2) - 5(x - 2) = 0
=> (x - 5)(x - 2) = 0
=> x-5=0;x=5
x-2=0;x=2
Vậy x = 2 hoặc x = 5
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: x2 - 7x + 10 = 0
<=> x.(x-7) + 10 = 0
<=> x.(x - 7) = -10
=> x và x - 7 E Ư(-10)
=> Ư(-10) = { - 1;1;-2;2;-5;5;-10;10}
Ta có: nếu x = -1 thì x - 7 = 10
............................ Làm tương tự
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a) x - 96 = (433 - x) - 15
x - 96 = 433 - x - 15
2x = 433 - 15 + 96
2x = 514
x = 257
b) 7x (-x-10) = 0
\(\Leftrightarrow\orbr{\begin{cases}7x=0\\-x-10=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-10\end{cases}}\)
c) 17 (3x - 6) (2x - 8) = 0
(3x - 6) (2x - 8) = 0
\(\Leftrightarrow\orbr{\begin{cases}3x-6=0\\2x-8=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}3x=6\\2x=8\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=4\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, 7x . (x-10 ) = 0
=> \(\hept{\begin{cases}7x=0\\x-10=0\end{cases}}\) => \(\hept{\begin{cases}x=0\\x=10\end{cases}}\)
b,17 . (3x-6).(2x-8) = 0
(3x-6).(2x-8) = 0 : 17
(3x-6).(2x-8) = 0
=>\(\hept{\begin{cases}3x-6=0\\2x-8=0\end{cases}}\)=>\(\hept{\begin{cases}x=2\\x=4\end{cases}}\)
Nhớ nha
![](https://rs.olm.vn/images/avt/0.png?1311)
a, 7\(x\).(\(x\) - 10) = 0
\(\left[{}\begin{matrix}7x=0\\x-10=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\)
Vậy \(x\in\) {0; 10}
b, 17.(3\(x\) - 6).(2\(x\) - 18) = 0
\(\left[{}\begin{matrix}3x-6=0\\2x-18=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}3x=6\\2x-18=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=6:3\\x=18:2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2\\x=9\end{matrix}\right.\)
x2 -7x + 10 = 0
=> x2 - 2x - 5x + 10 = 0
=> x(x - 2) - 5(x - 2) = 0
=> (x - 5)(x - 2) = 0
=> \(\left[\begin{array}{nghiempt}x-5=0\\x-2=0\end{array}\right.\) => \(\left[\begin{array}{nghiempt}x=5\\x=2\end{array}\right.\)
Vậy x = 2 hoặc x = 5
x2 - 7x + 10 = 0
=> x2 - 2x - 5x + 10 = 0
=> ( x2 - 2x ) - ( 5x - 10 ) = 0
=> x( x - 2 ) - 5( x - 2 ) = 0
=> ( x - 2 ) . ( x - 5 ) = 0
Ta có: x - 2 = 0
=> x = 2
hoặc x - 5 = 0
=> x = 5
Vậy x = 2 hoặc x = 5