\(x^2-2\left(m-2\right)x+m^2+2m-3=0\left(1\right)\)

Để phương trình có hai nghiệm phân biệt thì Δ' > 0

\(\Rightarrow\left(m-2\right)^2-m^2-2m+3>0\Leftrightarrow m^2-4m+4-m^2-2m+3>0\Leftrightarrow-6m+7>0\Leftrightarrow m< \dfrac{7}{6}\)\)

Theo viét : \(\(\left\{{}\begin{matrix}x_1+x_2=2\left(m-2\right)\\x_1x_2=m^2+2m-3\end{matrix}\right.\)\)

Lại có :\( \dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{x_1+x_2}{5}\)

\(\Leftrightarrow\dfrac{x_1+x_2}{x_1x_2}=\dfrac{x_1+x_2}{5}\)

\(\Rightarrow\left(x_1+x_2\right)\left(x_1x_2\right)=5\left(x_1+x_2\right)\)

\(\Leftrightarrow\left(2m-4\right)\left(m^2+2m-3\right)=5\left(2m-4\right)\)

\(\Leftrightarrow2m^3+4m^2-6m-4m^2-8m+12=10m-20\)

\(\Leftrightarrow2m^3-24m+32=0\) \(\Leftrightarrow\left[{}\begin{matrix}m=-4\left(n\right)\\m=2\left(l\right)\end{matrix}\right.\)

Vậy \(m=-4\) thì thỏa điều kiện

=>75:x=-15

hay x=-5

\(75:x=-15\\ x=\left(-75\right):15\\ x=-5\)

=> 75 : x = -15
=> x = 75 : (-15)
=> x = -5

Tham khảo:

I have just returned from a 1-week holiday with my friends in Da Nang and Quang Ngai. We stayed most of the time in Da Nang. Then we spent the last few nights in a very picturesque and historic place called Hoi An Old Town. We had such a relaxing time on our holiday. We enjoyed some wonderful local food. It’s really delicious. The activities we enjoyed on holiday were walking, swimming and go out at night. The water is so fresh and clean, and the temperature is just refreshing – not too hot and not too cold. We went hiking in the mountains nearly for some days, which was very good for our fitness. I really enjoyed this holiday. I loved the fresh air and the scenery, and of course the food. I will certainly be going back there another year, as I think it’s one of my favourite parts of the world.

Đề bài là: Tính giá trị biểu thức:

=(3.(2^2)^2.2^7)/(3^2.2^20)

=(3.2^11)/(3^2.2^20)

=1/3.2^9

=1/1536

Đây nè bạn

co 2n+1chia het cho n+1

suy ra 2 (n+1)-1 chia het cho n+1

suy ra 1 chia het cho n+1 (vi 2(n+1) chia het cho n+1)

suy ra n+1=1

suy ra n=0

\(a, 48.19+81.48\)

\(=49.(19+81)\)

\(=49.100=4900\)

\(b,23.134-34.23\)

\(=23.(134-34)\)

\(=23.100=2300\)

e: =49-18-60=31-60=-29

a, 48.19+81.48
= 48.(19+81)
= 48.100
= 4800
b, 23.134 - 34.23
= 23.(134 - 34)
= 23.100
= 2300 
c,27.63+27.37
= 27.(63+37)
= 27.100
= 2700
d,64+125.147 - 47.125
= 64+125.(147 - 47)
= 64+125.100
= 64+12500
= 12564
e,(7² - 2.3²) - 60
= (49 - 2.9) - 60
= (49 - 18) - 60
=31 - 60
= 31+(-60)
= - (60 - 34)
= -29
f, 10² - 60:(5⁶:5⁴ - 3.5)
= 100 - 60:(5² - 3.5)
= 100 - 60:(25 - 3.5)
= 100 - 60:(25 - 15)
= 100 - 60:10
= 100 - 6
= 94
g, 3⁴.5²+122 - 2⁴.7 - 5²
= 81.25+122 - 16.7 - 25
= 2025+122 - 112 - 25
= 2147 - 112 - 25
= 2035 - 25
= 2010
h, 7⁴:7²+[6² - (10² - 4.16)]
= 7²+[36 - (100 - 4.16])
= 49+[36 - (100 - 4.16)]
= 49+[36 - (100 - 64)]
= 49+[36 - 36]
= 49+0
= 49
i, - 2015+145+( - 145)+15
= - 2015+[145+( - 145)]+15
= - 2015+0+15
= - 2015+15
= - ( 2015-15)
= - 2000
j, 25 - 18+( - 43)
= 7+ ( - 43)
= - (43 - 7)
= - 36
k, ( - 297) + 630 + 297 + ( - 330)
= [( - 297)+297]+630+( - 330)
= 0+630+( - 330)
= 630 - 330
= 300
l, ( - 25)+34+( - 75)+66
= [( - 25)+( - 75)]+(34+66)
= [ - (25+75)]+100
= ( - 100)+100
= 0
m, 568 - {5.[143 - (4 - 1)² ]+10}:10
= 568 - {5.[143 - 3² ]+10}:10
= 568 - {5.[143 - 9]+10}:10
= 568 - {5.134+10}:10
= 568 - {670+10}:10
= 568 - 680:10
= 568 - 68
= 500
n, 107 - {38+[7.3² - 24:6+(9 - 7)³ ]}:15
= 107 - {38+[7.9 - 24:6+(9 - 7)³ ]}:15
= 107 - {38+[7.9 - 24:6+2³ ]}:15
= 107 - {38+[7.9 - 24:6+8]}:15
= 107 - {38+[63 - 4+8]}:15
= 107 - {38+[59+8]}:15
= 107 - {38+67}:15
= 107 - 105:15
= 107 - 7
= 100

\(\Rightarrow2A=8+2^3+...+2^{2022}\\ \Rightarrow2A-A=8+2^3+...+2^{2022}-4-2^2-...-2^{2021}\\ \Rightarrow A=8+2^{2022}-4-2^2=8-4-4+2^{2022}=2^{2022}\left(đpcm\right)\)

\(A=2^2+2^2+2^3+...+2^{2021}=2^3+2^4+...+2^{2021}=2^{2022}\left(đpcm\right)\)

A=2²+2²+2³+...+2²⁰²¹
A2=2(2²+2²+2³+...+2²⁰²¹)
A2=2³+2³+2⁴+...+2²⁰²²
A2-A= 2³+2³+2⁴+...+2²⁰²²-2²+2²+2³+...+2²⁰²¹
A=2³-2²+2²+2²⁰²²
A=8-8+2²⁰²²
A=2²⁰²²