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

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}

a) x² - 5x - 6 = 0

=> x² - 2x - 3x - 6 = 0

=> (x² - 2x) + (-3x - 6) = 0

=> x(x - 2) - 3 (x - 2) = 0

=> (x - 2) (x - 3) = 0

=> x - 2 = 0 => x = 2

     x - 3 = 0 => x = 3

còn lại tương tự nhé!! 46566578768698945635655675656788787868789789879789098089364556546

Bạn giải giúp mình mấy câu kia được k=)))

\(2x-12-x=0\)

\(\Leftrightarrow2x-x=12\Leftrightarrow x=12\)

\(\left(x-7\right)\left(2x-8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\2x-8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\2x=8\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=4\end{cases}}}\)

2x - 12 - x = 0

\(\Leftrightarrow\)2x - x = 12

\(\Leftrightarrow\)x=12

=))

a, 12 - (2\(x^2\) - 3) = 7

            2\(x^2\)  - 3  =  12  - 7

           2\(x^2\) - 3  = 5

           2\(x^2\)  = 8

             \(x^2\)   = 4

             \(\left[{}\begin{matrix}x=-2\\x=2\end{matrix}\right.\)

a) \(12-\left(2x^2-3\right)=7\\ 12-2x^2+3=7\\ 15-2x^2=7\\ 2x^2=15-7=8\\ x^2=8:2=4\\ x=\pm2\)

b) \(3x^2-12=2x^2+4\\ 3x^2-2x^2=12+4\\ x^2=16\\ x=\pm4\)

b, 3\(x^2\) - 12  = 2\(x^2\) + 4

    3\(x^2\) - 2\(x^2\) = 12 + 4

     \(x^2\) = 16

      \(\left[{}\begin{matrix}x=-4\\x=4\end{matrix}\right.\)

a. (3x - 1).(2x + 7) - (x + 1).(6x - 5) = 16
<=> 6x^2 + 19x - 7 - (6x^2 + x - 5) = 16
<=> 18x - 2 = 16
<=> 18x = 18
<=> x = 1
b. (10x + 9).x - (5x - 1).(2x + 3) = 8
<=> 10x^2 + 9x - (10x^2 + 13x - 3) = 8
<=> -4x + 3 = 8
<=> -4x = 5
<=> x = -5/4
c. (3x - 5).(7 - 5x) + (5x + 2).(3x - 2) - 2 = 0
<=> -15x^2 + 46x - 35 + 15x^2 - 4x - 4 - 2 = 0
<=> 42x - 41 = 0
<=> x = 41/42

Câu 1 :

\(\text{ a) }12-2x-x^2=0\\ \Leftrightarrow2\left(6-x-x^2\right)=0\\ \Leftrightarrow6-x-x^2=0\\ \Leftrightarrow6-3x+2x-x^2=0\\ \Leftrightarrow\left(6-3x\right)+\left(2x-x^2\right)=0\\ \Leftrightarrow3\left(2-x\right)+x\left(2-x\right)=0\\ \Leftrightarrow\left(3+x\right)\left(2-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3+x=0\\2-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

Vậy \(x=-3\) hoặc \(x=2\)

\(\text{b) }\left(x^2-\dfrac{1}{2}x\right):2x-\left(3x-1\right):\left(3x-1\right)=0\\ \Leftrightarrow\dfrac{1}{2}x-\dfrac{1}{4}-1=0\\ \Leftrightarrow\dfrac{1}{2}x-\dfrac{5}{4}=0\\ \Leftrightarrow\dfrac{1}{2}x=\dfrac{5}{4}\\ \Leftrightarrow x=\dfrac{5}{2}\)

Vậy \(x=\dfrac{5}{2}\)

Câu 2:

\(N=x^2+5y^2+2xy-2y+2005\\ N=x^2+4y^2+y^2+2xy-2y+1+2004\\ N=\left(x^2+2xy+y^2\right)+\left(4y^2-2y+1\right)+2004\\ N=\left(x+y\right)^2+\left(2y-1\right)^2+2004\\ \text{Do }\left(x+y\right)^2\ge0\forall x;y\\ \left(2y-1\right)^2\ge0\forall y\\ \Rightarrow\left(x+y\right)^2+\left(2y-1\right)^2\ge0\forall x;y\\ \Rightarrow N=\left(x+y\right)^2+\left(2y-1\right)^2+2004\ge0\forall x;y\\ \text{Dấu "=" xảy ra khi : }\left\{{}\begin{matrix}\left(x+y\right)^2=0\\\left(2y-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\2y-1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-y\\y=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{2}\end{matrix}\right.\)

Vậy \(N_{\left(Min\right)}=2004\) khi \(x=-\dfrac{1}{2};y=\dfrac{1}{2}\)