123654321

123654321

cach giai

goi a la so can tim ta co

a chia het cho 2,3,4,5,6 suy ra cstc la 0

tiep theo a chia het cho 7,8,9 suy ra 

mk hết bít xin lỗi bn

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)    =>a=bk và c=dk

\(\Rightarrow\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2k}{d^2k}=\frac{b^2}{d^2}\)(1)

Mặt khác: \(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\frac{b^2\left(k+1\right)^2}{d^2\left(k+1\right)^2}=\frac{b^2}{d^2}\)(2)

Từ (1) và (2) =>\(\frac{ab}{cd}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)(đpcm)

a)Ta thấy:\(\left|x+3,5\right|\ge0\)

\(\Rightarrow\left|x+3,5\right|+1\frac{1}{4}\ge0+1\frac{1}{4}=\frac{5}{4}\)

\(\Rightarrow A\ge\frac{5}{4}\)

Dấu "=" xảy ra <=>x=-3,5

Vậy....

b)Đề sai 

c)Áp dụng BĐT |a|+|b|>=|a+b|,ta có:

\(\left|x-\frac{1}{2}\right|+\left|x-\frac{3}{4}\right|\ge\left|x-\frac{1}{2}+\frac{3}{4}-x\right|=\frac{1}{4}\)

\(\Rightarrow C\ge\frac{1}{4}\)

Dấu "=" xảy ra <=>x=1/2 hoặc 3/4

Vậy...

d)Ta thấy:\(\left|x+\frac{1}{3}\right|\ge0\)

\(\Rightarrow\left|x+\frac{1}{3}\right|+\frac{1}{7}\ge0+\frac{1}{7}=\frac{1}{7}\)

\(\Rightarrow D\ge\frac{1}{7}\)

Dấu "=" xảy ra <=>x=-1/3

Vậy...

2x² - 5x + 2 = 0

=> 2x² - 4x - x + 2 = 0

=> 2x(x - 2) - (x - 2) = 0

=> (2x - 1)(x - 2) = 0 

=> 2x - 1 = 0 => x = 1/2

hoặc x - 2 = 0 => x = 2

    Vậy x = 1/2 , x = 2

2x²-5x+2=0

=>2x²-4x-x+2=0

=>2x(x-2)-(x-2)=0

=>(2x-1)(x-2)=0

=>2x-1=0 hoặc x-2=0

+)Nếu 2x-1=0

=>2x=0+1=1

=>x=1:2=\(\frac{1}{2}\)

+)Nếu x-2=0

=>x=0+2=2

Vậy x=\(\frac{1}{2}\) hoặc x=2

Vì \(\frac{1}{3^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{2002^2}< \frac{1}{2001.2002}\)

\(\Rightarrow A=\frac{1}{3^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2002^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2001.2002}\)

mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2001.2002}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2001}-\frac{1}{2002}\)\(=1-\frac{1}{2002}< 1\)

\(\Rightarrow A=\frac{1}{3^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2002^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2001.2002}< 1\)

\(\Rightarrow A=\frac{1}{3^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2002^2}< 1\)(đpcm)

Nếu mà chỗ 3² ở phân số đầu tiên sửa thành 2² thì trông sẽ đẹp hơn nhé