Monday , May 23,2016
Mathematics
####Day 1:
####Answer and Analysis
- 6.我们先把这个二项式给列出来(两个未知数,两个方程)
{2x−3y=0①2x+2y=5②\begin{cases}2x-3y=0& \text{①}\\2x+2y=5& \text{②} \end{cases}{2x−3y=02x+2y=5①②
由②-①得,
{2y−(−3y)=5−0③\left\{{2y-(-3y)=5-0}\right.\text{③}{2y−(−3y)=5−0③
5y=5,y=1④5y=5, y=1 ④5y=5,y=1④
将④带入②
2x+2=5,x=32 2x+2=5 , x ={{3}\over{2}} 2x+2=5,x=23
∴ A∩B={(32,1)}\therefore\ A\cap B=\{({{3}\over{2}},1)\}∴ A∩B={(23,1)}
-
7. 画数轴
A∩B={x∣−2x≤x≤4}A\cap B=\{x|-2x\leq x\leq 4\}A∩B={x∣−2x≤x≤4}
A∪B={x∣−3≤x≤5}A\cup B=\{x|-3\leq x\leq 5\}A∪B={x∣−3≤x≤5} -
**8. **
解:
A={2,4,8}A=\{2,4,8\}A={2,4,8}
U={1,2,3,4,5,6,7,8,9,10,11}U=\{1,2,3,4,5,6,7,8,9,10,11\}U={1,2,3,4,5,6,7,8,9,10,11}
∁UA={1,3,5,6,7,9,10,11}\complement_{U}A=\{1,3,5,6,7,9,10,11\}∁UA={1,3,5,6,7,9,10,11}
- **9. **
∁UA={x≤−1 or x≥4}\complement_{U}A=\{x\leq-1\ or\ x\geq 4\}∁UA={x≤−1 or x≥4}
PS:零不是整数
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Wednesday , May 25,2016
Mathematics
####Day 2:
- **1. **
x2−2x<3x2−2x−3<0{x^{2}-2x<3}\\ {x^{2}-2x-3<0}x2−2x<3x2−2x−3<0
方法① 求根公式
ax2+bx+c<0 (a>0)ax^{2}+bx+c<0\ (a>0)ax2+bx+c<0 (a>0)
得,有两相异的根
{x∣x1<x<x2}\{x|x_{1}<x<x_{2}\}{x∣x1<x<x2}
x=−b±△2a , △=b2−4ac ⇒x=−b±b2−4ac 2ax={{-b\pm\sqrt{\bigtriangleup}}\over{2a}}\ ,\ \ \bigtriangleup=b^{2}-4ac\ \Rightarrow x={{-b\pm\sqrt{b^{2}-4ac\ }}\over{2a}}x=2a−b±△ , △=b2−4ac ⇒x=2a−b±b2−4ac
a=1,b=−2,c=−3a=1, b=-2, c=-3 a=1,b=−2,c=−3
带入求根公式,
x=−(−2)±(−2)2−4×1−32×1=2±42x={{-(-2)\pm\sqrt{(-2)^{2}-4\times 1-3}}\over{2\times 1}}={{2\pm 4}\over{2}}x=2×1−(−2)±(−2)2−4×1−3=22±4
(−1,3)(-1,3)(−1,3)
方法②:交叉相乘
x−3x1\begin{array}{cc}{x}&{\mathrm{{-}}{3}}\\{x}&{1}\end{array}xx−31
得,有两相异的根 (−1,3)(-1,3)(−1,3)
- **2. **
需要蓉蓉背的表格( 大0或并),
x2+2x−3≥0ax2+bx+c≥0(a>0){x∣x<x1orx>x2} \begin{array}{l} {{x}^{2}\mathrm{{+}}{2}{x}\mathrm{{-}}{3}\mathrm{\geq}{0}}\\ {{ax}^{2}\mathrm{{+}}{bx}\mathrm{{+}}{c}\mathrm{\geq}{0}{\mathrm{(}}{a}{\mathrm{>}}{0}{\mathrm{)}}}\\ {{\mathrm{\{}}{x}{\mathrm{|}}{x}{\mathrm{<}}{x}_{1}\hspace{0.33em}{or}\hspace{0.33em}{x}{\mathrm{>}}{x}_{2}{\mathrm{\}}}} \end{array} x2+2x−3≥0ax2+bx+c≥0(a>0){x∣x<x1orx>x2}
x+3x−1\begin{array}{cc}{x}&{\mathrm{{+}}{3}}\\{x}&{-}{1}\end{array}xx+3−1
得,有两相异的根
$$
\begin{array}{l}
{{\mathrm{(}}{x}\mathrm{{+}}{3}{\mathrm{)(}}{x}\mathrm{{-}}{1}{\mathrm{)}}\mathrm{{=}}{0}}\
{{x}{1}\mathrm{{=}}\mathrm{{-}}{3}\hspace{0.33em}{\mathrm{,}}\hspace{0.33em}{x}{1}\mathrm{{=}}{1}}
\end{array}
(-\infty ,-3]\cup [1,+\infty )$$
- **3. **
这个解只有一个根,x=x1=x2x=x_{1}=x_{2}x=x1=x2
4x2−4x+1≤04x^{2}-4x+1\leq 04x2−4x+1≤0
2x−12x−1\begin{array}{cc}{2x}&{\mathrm{{-}}{1}}\\{2x}&{\mathrm{{-}}{1}}\end{array}2x2x−1−1
2x−1=02x-1=02x−1=0
2x=12x=12x=1
x=12x={{1}\over{2}}x=21
{12}\{{{1}\over{2}}\}{21}
- **4. 5. **
When Δ<0 , ax2+bx+c≥0 ⇒R When\ \Delta <0\ ,\ \ ax^{2}+bx+c\geq 0\ \Rightarrow R\ When Δ<0 , ax2+bx+c≥0 ⇒R - **6. **
0x⇒x=0{{0}\over{x}}\Rightarrow x=0x0⇒x=0
Additional,
For showing particular example of absolute value equation
∣x−1∣<1|x-1|<1∣x−1∣<1
1. x−1<11.\ x-1<11. x−1<1
2. x−1>−12.\ x-1>-12. x−1>−1
English
Reading comprehension page 135 paragraph 3
- among 在…之中(三者以上)
- between 在…之中(两者以上)
- a few 一小部分的… (可数的)
- found -> find 找,在 (左边是过去式)
- sent ->send 发送
- pass from… 从…传…
eg. 宾语从句(句子1+that+句子2修饰宾语)
rongrong love someone that just like Don.
someone/somebody/something 把that省略
rongrong love someone just like Don.
蓉蓉喜欢蛋蛋这样的人。
蓉蓉爱某人
I like eating foods that taste delicious.
like doing something. 喜欢做某事
I will threw a basketball.
will be (verb 动词原型) 将要…
I go home.
I love you.
behind/after 在…之后
before 在…之前
the day before yesterday 前天
the day after tomorrow 后天
表语:形容主语的
rongrong is beautiful .
Don is handsome .
the cup is adorable/cute/cutie .
is/are/was/were + adj.
spent on = spend on 花费…在…
commercial(商业广告) =advertisement=ad. 广告
subscribe to 订阅…
as … as 跟…一样
do
1.强调词
I do envy rongrong . 的确/确实/真的 很羡慕蓉蓉
2.疑问句
Do you love me ?
Do you have got money?
Does 第三人称
took = take 花费、拿、乘
travel from 从一个地方到另一个地方(旅行)
apart from 除…之外
of course 当然
sort of 一些
time 时间 times 次/次数