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a) Ta có: \(2x+x^2=0\)
\(\Leftrightarrow x\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
b) Ta có: \(\left(2x+1\right)^2-25=0\)
\(\Leftrightarrow\left(2x-4\right)\left(2x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(5x+10y=5\left(x+2y\right)\)
b) \(3x^2y+9xy^2z=3xy\left(x+3yz\right)\)
g) \(x^2-x-6=\left(x-3\right)\left(x+2\right)\)
h) \(x^2+9x+8=\left(x+8\right)\left(x+1\right)\)
l) \(x^2-10x+9=\left(x-1\right)\left(x-9\right)\)
k) \(x^2+x-12=\left(x+4\right)\left(x-3\right)\)
l) \(3x^2+8x+4=\left(3x+2\right)\left(x+2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
Ta có số tiền tủ lạnh sau khi giảm lần 1 là: \(15000000-15000000.20\%=12000000đ\)
Số tiền tủ lạnh sau khi giảm lần 2 là: \(12000000-12000000.5\%=11400000đ\)
Số tiền cửa hàng thu được sau khi bán 5 cái tủ lạnh: \(11400000.5=57000000đ\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,=20x^2-15x+10x-20x^2=-5x\\ 2,=4x^2-20x+25-4x^2+18x-18=7-2x\\ 3,=\left(6x^3-4x^2-12x+8x+15x-10\right):\left(3x-2\right)\\ =\left(3x-2\right)\left(2x^2-4x+5\right):\left(3x-2\right)\\ =2x^2-4x+5\\ 4,=\dfrac{5x+25-2x+10+x^2+2x-35}{\left(x+5\right)\left(x-5\right)}=\dfrac{x\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}=\dfrac{x}{x-5}\\ 5,=\dfrac{3x-8-x-6}{x-7}+\dfrac{\left(x-2\right)\left(x+2\right)}{x-2}=\dfrac{2\left(x-7\right)}{x-7}+x+2=2+x+2=x+4\\ 6,=\dfrac{x^2+8x+16+2x-8-6x-8}{\left(x-4\right)\left(x+4\right)}=\dfrac{x\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{x}{x-4}\\ 7,=\dfrac{x\left(x-7\right)}{2x\left(x+3\right)}\cdot\dfrac{4\left(x-3\right)\left(x+3\right)}{\left(x-7\right)^2}=\dfrac{2\left(x-3\right)}{x-7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Xét tứ giác AHCE có
M là trung điểm của AC
M là trung điểm của HE
Do đó: AHCE là hình bình hành
mà \(\widehat{AHC}=90^0\)
nên AHCE là hình chữ nhật
![](https://rs.olm.vn/images/avt/0.png?1311)
a:\(x^4+x^3-6x^2+ax+b⋮x^2+x-1\)
\(\Leftrightarrow x^4+x^3-x^2-5x^2-5x+5+\left(a+5\right)x+\left(b-5\right)⋮x^2+x-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=-5\\b=5\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^4-8x=x\left(x^3-8\right)=x\left(x-2\right)\left(x^2+2x+4\right)\)
\(x^2-y^2-6x+9=\left(x^2-6x+9\right)-y^2=\left(x-3\right)^2-y^2=\left(x+y-3\right)\left(x-y-3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2
a) 3x(x - 1) - 3(x - 1) = 0
(x - 1)(3x - 3) = 0
3(x - 1)(x - 1) = 0
3(x - 1)² = 0
x - 1 = 0
x = 1
b) x² - x = 0
x(x - 1) = 0
x = 0 hoặc x - 1 = 0
*) x - 1 = 0
x = 1
Vậy x = 0; x = 1
c) 25x² - 100x = 0
25x(x - 4) = 0
25x = 0 hoặc x - 4 = 0
*) 25x = 0
x = 0
*) x - 4 = 0
x = 4
Vậy x = 0; x = 4
d) (2x - 1)² - 64 = 0
(2x - 1 - 8)(2x - 1 + 8) = 0
(2x - 9)(2x + 7) = 0
*) 2x - 9 = 0
2x = 9
x = 9/2
*) 2x + 7 = 0
2x = -7
x = -7/2
Vậy x = -7/2; x = 9/2
\(1.4.7+4.7.10+...+n\left(n+3\right)\left(n+6\right)\\ =\dfrac{n^2\left(n+1\right)^2}{4}+9\cdot\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}+18\cdot\dfrac{n\left(n+1\right)}{2}\)
\(=\dfrac{n\left(n+1\right)\left(n^2+13n+42\right)}{4}=\dfrac{n\left(n+1\right)\left(n+6\right)\left(n+7\right)}{4}\)
Áp dụng vào bài toán:
\(P=\dfrac{2021.2022.2027.2028}{4}=...\)
CM:
Với \(n=1\Leftrightarrow1.4.7=28\)
\(\dfrac{n\left(n+1\right)\left(n+6\right)\left(n+7\right)}{4}=\dfrac{2.7.8}{4}=28\)
Giả sử \(n=k\Leftrightarrow1.4.7+4.7.10+...+k\left(k+3\right)\left(k+6\right)=\dfrac{k\left(k+1\right)\left(k+6\right)\left(k+7\right)}{4}\)
Với \(n=k+1\), cần cm:
\(1.4.7+4.7.10+...+k\left(k+3\right)\left(k+6\right)+\left(k+1\right)\left(k+4\right)\left(k+7\right)=\dfrac{\left(k+1\right)\left(k+2\right)\left(k+7\right)\left(k+8\right)}{4}\)
Ta có \(VT=\dfrac{k\left(k+1\right)\left(k+6\right)\left(k+7\right)}{4}+\left(k+1\right)\left(k+4\right)\left(k+7\right)\)
\(=\left(k+1\right)\left(k+7\right)\left[\dfrac{k\left(k+6\right)}{4}+k+4\right]=\left(k+1\right)\left(k+7\right)\left(\dfrac{k^2+10k+16}{4}\right)\\ =\dfrac{\left(k+1\right)\left(k+7\right)\left(k+2\right)\left(k+8\right)}{4}=VP\)
Do đó theo pp quy nạp ta đc đpcm