câu hỏi tương tự nhé 

\(\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}=>\frac{x}{\frac{3}{2}}=\frac{y}{\frac{4}{3}}=\frac{z}{\frac{5}{4}}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{x}{\frac{3}{2}}=\frac{y}{\frac{4}{3}}=\frac{z}{\frac{5}{4}}=\frac{x+y+z}{\frac{3}{2}+\frac{4}{3}+\frac{5}{4}}=\frac{-7}{\frac{49}{12}}=-\frac{12}{7}\)

=>\(x=-\frac{12}{7}.\frac{3}{2}=-\frac{18}{7}\)

\(y=-\frac{12}{7}.\frac{4}{3}=-\frac{16}{7}\)

\(z=-\frac{12}{7}.\frac{5}{4}=-\frac{15}{7}\)

\(\frac{2x+3y+4z}{3+4+5}=\frac{-7}{12}\)

=> 2x/3=-7=>x= -21

=>3y/4=-7=>y=-28

=>4z/5= -7=> z= -35

tick bạn

( 50² + 48² + 46² +...+ 4² + 2²) - (49² + 47² +...+ 3² + 1²)

=50²- 49² + 48² - 47² +...+ 4² - 3² + 2² - 1²

= ( 50 - 49 )( 50 + 49 ) + ( 48 - 47 )( 48 + 47 )+...+( 4 - 3 )( 4 + 3) + ( 2-1 ) (2+1)

=50 + 49 + 48 + 47+...+ 4 + 3 + 2 + 1

=( 50 + 1) . 50 : 2 =1275

2x¹⁰=x¹²

=>2x¹⁰-x¹²=0

=>x¹⁰.(2-x²)=0

=>x¹⁰=0=>x=0

hoặc 2-x²=0=>x²=2=>\(x=\sqrt{2},-\sqrt{2}\)

Vậy \(x=0,\sqrt{2},-\sqrt{2}\)

=>2x¹⁰-x¹²=0

=>x¹⁰.(2-x²)=0

=>Hoặc x¹⁰=0 =>x=0

Hoặc 2-x²=0 =>x²=2=>ko tồn tại x

Vậy x=0

5/25 = 25/125

5/125=25/625

5/625=25/3125

25/5=125/25

125/5=625/25

625/5=3125/35

125/25=3125/625

25/125=625/3125

25/625=125/3125

625/25=3125/125

còn nhiều lắm

M= \(\frac{100^{100}+1}{100^{99}+1}=\frac{100^{100}+100-99}{100^{99}+1}=\frac{100^{100}+100}{100^{99}+1}-\frac{99}{100^{99}+1}=\frac{100.\left(100^{99}+1\right)}{100^{99}+1}-\frac{99}{100^{99}+1}\)

\(=100-\frac{99}{100^{99}+1}\)

N= \(\frac{100^{101}+1}{100^{100}+1}=\frac{100^{101}+100-99}{100^{100}+1}=\frac{100^{101}+100}{100^{100}+1}-\frac{99}{100^{100}+1}\)

\(=\frac{100.\left(100^{100}+1\right)}{100^{100}+1}-\frac{99}{100^{100}+1}=100-\frac{99}{100^{100}+1}\)

Vi 100¹⁰⁰+1>100⁹⁹+1

=> \(\frac{99}{100^{99}+1}>\frac{99}{100^{100}+1}\)

=> \(100-\frac{99}{100^{99}+1}

uk ai cũng có lúc nhầm mà chẳng sao đâu bạn ak