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NT
4
17 tháng 8 2020
bn ơi cho mk hỏi bn lm tiếng anh hay toán mà chủ đề là tiếng anh mà bài lại là toán vậy alo????
17 tháng 8 2020
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OY
0
9 tháng 12 2017
a. \(\dfrac{\left(x^2+2x\right)}{\left(x+2\right)^2}=\dfrac{x\left(x+2\right)}{\left(x+2\right)\left(x+2\right)}=\dfrac{x}{x+2}\)
b. \(\dfrac{x^2-7x+12}{x^2-6x+9}=\dfrac{x^2-3x-4x+12}{\left(x-3\right)^2}\)
\(=\)\(\dfrac{x\left(x-3\right)-4\left(x-3\right)}{\left(x-3\right)\left(x-3\right)}=\dfrac{\left(x-4\right)\left(x-3\right)}{\left(x-3\right)\left(x-3\right)}=\dfrac{x-4}{x-3}\)
c. \(\dfrac{x^2-5x+6}{x^2-x-2}=\dfrac{x^2-2x-3x+6}{x^2-2x+x-2}\)
\(=\dfrac{x\left(x-2\right)-3\left(x-2\right)}{x\left(x-2\right)+\left(x-2\right)}=\dfrac{\left(x-3\right)\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{x-3}{x+1}\)
d. \(\dfrac{\left(x+y\right)^2-z^2}{2\left(x+y+z\right)}=\dfrac{\left(x+y-z\right)\left(x+y+z\right)}{2\left(x+y+z\right)}=\dfrac{x+y-z}{2}\)
24 tháng 7 2019
(x - 1)2 - (x + 1)2 = 5 - 6x
<=> x2 - 2x + 1 - x2 - 2x - 1 = 5 - 6x
<=> -4x = 5 - 6x
<=> -4x + 6x = 5
<=> 2x = 5
<=> x = \(\frac{5}{2}\)
27 tháng 7 2018
h) Ta có: \(\left\{{}\begin{matrix}\left|x-7\right|=\left|7-x\right|\ge7-x\\\left|x+5\right|\ge x+5\end{matrix}\right.\)
\(\Rightarrow\left|7-x\right|+\left|x+5\right|\ge\left(7-x\right)+\left(x+5\right)\)
\(\Rightarrow\left|x-7\right|+\left|x+5\right|\ge12\)
\(\Rightarrow H\ge12\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}7-x\ge0\\x+5\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le7\\x\ge-5\end{matrix}\right.\)
\(\Leftrightarrow-5\le x\le7\)
Vậy, MinH = 12 \(\Leftrightarrow-5\le x\le7\)
27 tháng 7 2018
a) Ta có: \(A=2x^2-8x+10\)
\(=2\left(x^2-4x+5\right)\)
\(=2\left(x^2-4x+2^2+1\right)\)
\(2\left[\left(x-2\right)^2+1\right]\)
Ta lại có: \(\left(x-2\right)^2\ge0\)
\(\Rightarrow2\left[\left(x-2\right)^2+1\right]\ge2\)
\(\Rightarrow A\ge2\)
Dấu bằng xảy ra \(\Leftrightarrow\left(x-2\right)^2=0\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
Vậy MinA = 2 \(\Leftrightarrow x=2\)
16 tháng 7 2018
\(a,\)Mình làm theo kiểu lược đồ
Nhẩm nghiệm của đa thức trên ta đc : 2
Có lược đồ sau :(dòng trên ghi các hệ số)
| 1 | -2 | -6 | 12 | |
| 2 | 1 | 0 | -6 | 0 |
Ta phân tích đc thành :\(\left(x-2\right)\left(x^2-6\right)\)
\(c,x^2-5x+4\)
\(=x^2-4x-x+4\)
\(=x\left(x-4\right)-\left(x-4\right)\)
\(=\left(x-1\right)\left(x-4\right)\)
\(d,3x^2+5x+2\)
\(=3x^2+3x+2x+2\)
\(=3x\left(x+1\right)+2\left(x+1\right)\)
\(=\left(x+1\right)\left(3x+2\right)\)
\(e,x^3-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x^3+y^3\right)+3xy\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)+3xy\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left[\left(x^2-xy+y^2\right)+3xy-1\right]\)
16 tháng 7 2018
\(x^3-2x^2-6x+12\)
\(=x^2.\left(x-2\right)-6\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2-6\right)\)
\(x^4-7x^2+12\)
\(=\left[\left(x^2\right)^2-2.3,5x+3,5^2\right]-0,25\)
\(=\left(x^2-3,5\right)^2-0,5^2\)
\(=\left(x^2-3,5-0,5\right)\left(x^2-3,5+0,5\right)\)
\(=\left(x^2-4\right)\left(x^2-3\right)\)
Câu c tương tự câu b
\(4x^2-6x-16⋮x-3\)
\(\Leftrightarrow4x^2-12x+6x-18+2⋮x-3\)
\(\Leftrightarrow4x\left(x-3\right)+6\left(x-3\right)+2⋮x-3\)
\(\Leftrightarrow\left(x-3\right)\left(4x+6\right)+2⋮x-3\)
Mà \(\left(x-3\right)\left(4x+6\right)⋮x-3\)
\(\Rightarrow2⋮x-3\)
\(\Rightarrow x-3\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
làm nốt
cách 2:
Để \(4x^2-6x-16\)chia hết cho x-3
\(\Leftrightarrow2⋮x-3\)
\(\Leftrightarrow x-3\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
Làm nốt