A researcher has made a report that says “having a lot of money makes you a bad person”. Professor Paul Piff spent ten years looking at the personalities of rich people and found that their behaviour was very different to the behaviour of poor people. Dr Paul Piff said rich people are more likely to break rules, not follow the law, and not care about other people. He also said wealthy people cheat more at games and are less likely to help other people. Professor Paul Piff told the BBC that: "Wealth makes you more worried about your own interests, your own desires, your own welfare." He added that rich people think their own goals and needs are the most important things in their life.
Professor Paul Piff also found that poorer people are much more generous than rich people. The poor give a higher percentage of their money to help the others than the rich. He said poorer people give one – and – a – half times as much as richer people. He said: "The wealthier you are, the less generous you are. You give significantly smaller portions away to other people." However, he did say that rich people could change their behaviour and become nicer and more generous. He said rich people can help others more if you tell them that giving and helping is a good thing. He said: "Reminding people of the benefits of cooperation, or the advantages of community, causes wealthier individuals to be just as giving as poor people."
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`#3107.101107`
`x^3 - 3x^2 - 4x + 12 = 0`
`\Leftrightarrow (x^3 - 3x^2) - (4x - 12) = 0`
`\Leftrightarrow x^2(x - 3) - 4(x - 3) = 0`
`\Leftrightarrow (x^2 - 4)(x - 3) = 0`
`\Leftrightarrow (x - 2)(x + 2)(x - 3) = 0`
`\Leftrightarrow`\(\left[{}\begin{matrix}x-2=0\\x+2=0\\x-3=0\end{matrix}\right.\)
`\Leftrightarrow`\(\left[{}\begin{matrix}x=2\\x=-2\\x=3\end{matrix}\right.\)
Vậy, `x \in {-2; 2; 3}`
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Sử dụng HĐT: \(A^2-B^2=\left(A-B\right)\left(A+B\right).\)
`#3107.101107`
`(x - 1)(x + 2) - 2x - 4 = 0`
`\Leftrightarrow (x - 1)(x + 2) - (2x + 4) = 0`
`\Leftrightarrow (x - 1)(x + 2) - 2(x + 2) = 0`
`\Leftrightarrow (x + 2)(x - 1 - 2) = 0`
`\Leftrightarrow (x + 2)(x - 3) = 0`
`\Leftrightarrow`\(\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\)
`\Leftrightarrow`\(\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
Vậy, `x \in {-2; 3}.`
\(\left(x-1\right)\left(x+2\right)-2x-4=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)-\left(2x+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)-2\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-1-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0+3\\x=0-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy x=3 hoặc x=-2
Bài 1:
a. ĐKXĐ: $1-x\neq 0; 1+x\neq 0; 1-x^2\neq 0$
$\Leftrightarrow x\neq \pm 1$
b.
\(A=\frac{1+x}{(1-x)(1+x)}+\frac{2(1-x)}{(x+1)(1-x)}-\frac{5-x}{(1-x)(1+x)}\\ =\frac{1+x+(2-2x)-(5-x)}{(1-x)(1+x)}=\frac{-2}{(1-x)(1+x)}=\frac{-2}{1-x^2}\)
M H K P Q I O
a/ Xét tg MKP và tg MHQ có
\(MP=MH\left(gt\right);MK=MQ\left(gt\right)\) (1)
\(\widehat{KMP}=\widehat{HMP}+\widehat{HMK}=90^o+\widehat{HMK}\)
\(\widehat{HMQ}=\widehat{KMQ}+\widehat{HMK}=90^o+\widehat{HMK}\)
\(\Rightarrow\widehat{KMP}=\widehat{HMQ}\) (2)
Từ (1) và (2) => tg MKP = tg MHQ => PK=HQ
b/
Xét tg vuông HMP có
\(\widehat{MPH}+\widehat{MHP}=90^o\)
\(\Rightarrow\widehat{MPK}+\widehat{HPK}+\widehat{MHP}=90^o\)
Ta có
tg MKP = tg MHQ (cmt) \(\Rightarrow\widehat{MPK}=\widehat{MHQ}\)
\(\Rightarrow\widehat{MHQ}+\widehat{HPK}+\widehat{MHP}=90^o\)
\(\Rightarrow\left(\widehat{MHQ}+\widehat{MHP}\right)+\widehat{HPK}=90^o\)
\(\Rightarrow\widehat{PHQ}+\widehat{HPK}=90^o\)
Xét tg HPI có
\(\widehat{PHQ}+\widehat{HPK}=90^o\left(cmt\right)\Rightarrow\widehat{HIP}=90^o\Rightarrow PK\perp HQ\)
c/
Hội Gióng Phù Đổng chính thống được tổ chức vào hai ngày mùng 8 và 9 tháng 4 Âm lịch hàng năm tại xã Phù Đổng, nơi sinh ra Thánh Gióng với huyền thoại “Phù Đổng Thiên Vương“. Không đi Hội Gióng cũng hư mất người"… Lễ hội Gióng được bắt đầu tổ chức từ khoảng thế kỷ XI, vào đời Vua Lý Thái Tổ.
Lời giải:
a. Với $x,y$ nguyên thì $x-2, 2y+1$ nguyên.
Mà $(x-2)(2y+1)=8$ nên $2y+1$ là ước của $8$
$2y+1$ lẻ nên $2y+1=1$ hoặc $2y+1=-1$
Nếu $2y+1=1\Rightarrow x-2=8$
$\Rightarrow y=0; x=10$
Nếu $2y+1=-1\Rightarrow x-2=-8$
$\Rightarrow y=-1; x=-6$
b.
$8-x, 4y+1$ là số nguyên. Mà $(8-x)(4y+1)=20$ nên $4y+1$ là ước của $20$.
Mà $4y+1$ chia $4$ dư $1$ nên $4y+1\in \left\{1; 5\right\}$
Nếu $4y+1=1$ thì $8-x=20$
$\Rightarrow y=0; x=-12$
Nếu $4y+1=5$ thì $8-x=4$
$\Rightarrow y=1; x=4$
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