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Ta có : 3x(2x - 7) - (6x + 1)(x - 15) - 2010 = 0
=> 6x2 - 21x - (6x2 + x - 90x - 15) - 2010 = 0
=> 6x2 - 21x - 6x2 + 89x + 15 - 2010 = 0
=> 68x - 1995 = 0
?
b) 2x(x - 2012) - x + 2012 = 0
=> 2x(x - 2012) - (x - 2012) = 0
=> (x - 2012) (2x - 1) = 0
⇔[
| x−2012=0 |
| 2x−1=0 |
⇔[
| x=2012 |
| 2x=1 |
⇔[
| x=2012 |
| x=12 |
Vậy x = {2012;12 }
Ta có : 3x(2x - 7) - (6x + 1)(x - 15) - 2010 = 0
=> 6x2 - 21x - (6x2 + x - 90x - 15) - 2010 = 0
=> 6x2 - 21x - 6x2 + 89x + 15 - 2010 = 0
=> 68x - 1995 = 0
?
b) 2x(x - 2012) - x + 2012 = 0
=> 2x(x - 2012) - (x - 2012) = 0
=> (x - 2012) (2x - 1) = 0
\(\Leftrightarrow\orbr{\begin{cases}x-2012=0\\2x-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2012\\2x=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2012\\x=\frac{1}{2}\end{cases}}\)
Vậy x = \(\left\{2012;\frac{1}{2}\right\}\)
b) 2x(x - 2012) - x + 2012 = 0
<=> 2x(x - 2012) - (x - 2012)= 0
<=> (x - 2012)(2x - 1) = 0
<=> \(\begin{bmatrix} x - 2012= 0 & & \\ 2x - 1 = 0 & & \end{bmatrix} \)
<=> \(\begin{bmatrix} x = 2012 & & \\ x = \frac{1}{2} & & \end{bmatrix} \)
Vậy x = 2012 và x= \(\frac{1}{2}\)
pn bỏ dấu ngoặc vuông bên phải nhé
a)
PT <=> \(\left(\frac{x-1}{2012}-1\right)+\left(\frac{x-2}{2011}-1\right)+...+\left(\frac{x-2012}{1}-1\right)=0\)
<=> \(\frac{x-2013}{2012}+\frac{x-2013}{2011}+...+\frac{x-2013}{1}=0\)
<=> \(\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+...+\frac{1}{1}\right)=0\)
Mà \(\frac{1}{2012}+\frac{1}{2011}+...+\frac{1}{1}\ne0\)
<=> x - 2013 = 0
<=> x = 2013
KL: ...
b) PT <=> \(\left(x^4-5x^3\right)+\left(5x^3-25x^2\right)-\left(5x^2-25x\right)+\left(6x-30\right)=0\)
<=> \(x^3\left(x-5\right)+5x^2\left(x-5\right)-5x\left(x-5\right)+6\left(x-5\right)=0\)
<=> \(\left(x-5\right)\left(x^3+5x^2-5x+6\right)=0\)
<=> \(\left(x-5\right)\left[\left(x^3+6x^2\right)-\left(x^2+6x\right)+\left(x+6\right)\right]=0\)
<=> \(\left(x-5\right)\left[x^2\left(x+6\right)-x\left(x+6\right)+\left(x+6\right)\right]=0\)
<=> \(\left(x-5\right)\left(x+6\right)\left(x^2-x+1\right)=0\)
<=> \(\left[{}\begin{matrix}x=5\\x=-6\\x=\varnothing\end{matrix}\right.\)
KL: ...
a) Ta có: \(\frac{x-1}{2012}+\frac{x-2}{2011}+\frac{x-3}{2010}+...+\frac{x-2012}{1}=2012\)
\(\Leftrightarrow\frac{x-1}{2012}+\frac{x-2}{2011}+\frac{x-3}{2010}+...+\frac{x-2012}{1}-2012=0\)
\(\Leftrightarrow\frac{x-1}{2012}-1+\frac{x-2}{2011}-1+\frac{x-3}{2010}-1+...+\frac{x-2012}{1}-1=0\)
\(\Leftrightarrow\frac{x-2013}{2012}+\frac{x-2013}{2011}+\frac{x-2013}{2010}+...+\frac{x-2013}{1}=0\)
\(\Leftrightarrow\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+1\right)=0\)
mà \(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+1>0\)
nên x-2013=0
hay x=2013
Vậy: Tập nghiệm S={2013}
b) Ta có: \(x^4-30x^2+31x-30=0\)
\(\Leftrightarrow x^4+x-30x^2+30x-30=0\)
\(\Leftrightarrow\left(x^4+x\right)-\left(30x^2-30x+30\right)=0\)
\(\Leftrightarrow x\left(x^3+1\right)-30\left(x^2-x+1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left[x\left(x+1\right)-30\right]=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(x^2+x-30\right)=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(x^2+6x-5x-30\right)=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left[x\left(x+6\right)-5\left(x+6\right)\right]=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(x+6\right)\left(x-5\right)=0\)(1)
Ta có: \(x^2-x+1\)
\(=x^2-2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)
\(=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\)
Ta có: \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)
hay \(x^2-x+1>0\forall x\)(2)
Từ (1) và (2) suy ra (x+6)(x-5)=0
\(\Leftrightarrow\left[{}\begin{matrix}x+6=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=5\end{matrix}\right.\)
Vậy: Tập nghiệm S={-6;5}
Chọn A
*Ta có:
* Xét hàm số
ta có 
.
Theo định nghĩa đạo hàm ta có: