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cái này đưa sang vật lí sẽ hợp lí hơn

a , \(\dfrac{x+5}{x}=\dfrac{4}{3}\Leftrightarrow3\left(x+5\right)=4x\)

<=> 3x+15=4x

<=> x= 15

b , \(\dfrac{x-20}{x-10}=\dfrac{x+40}{x+70}\)

<=> \(\dfrac{x-10}{x-10}-\dfrac{10}{x-10}=\dfrac{x+70}{x+70}-\dfrac{30}{x+70}\)

<=> \(1-\dfrac{10}{x-10}=1-\dfrac{30}{x+70}\)

<=> \(\dfrac{10}{x-10}=\dfrac{30}{x+70}\Leftrightarrow\dfrac{1}{x-10}=\dfrac{3}{x+70}\)

<=> (x+70)=3(x-10)

<=> x+70 = 3x-30

<=> 100=2x

<=> x= 50

Cách nhanh nhất :

Theo bài ra :

2a=3b=>2a-3b=0 (1)

5b=6c=> 5b-6c=0 (2)

a+3b-2c=-5 (3)

Từ (1) , (2) , (3) ta có hệ :

Ý tính tổng c1 mik gọi la câu a , câu b sẽ là ý còn lại nha

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Câu 1 :

a , Ta có : \(\left(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{7}}\right)+\dfrac{0,6-\dfrac{3}{25}-\dfrac{3}{125}-\dfrac{3}{625}}{\dfrac{4}{5}-0,16-\dfrac{4}{125}-\dfrac{4}{625}}\)

= \(\left(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}\right)+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)

= \(\dfrac{1}{4}+\dfrac{3}{4}=\dfrac{4}{4}=1\)

b , Theo bài ra ta có : \(\dfrac{a_1-1}{9}=\dfrac{a_2-2}{8}=...=\dfrac{a_9-9}{1}\)= \(\dfrac{\left(a_1+a_2+...+a_9\right)-\left(1+2+..+9\right)}{\left(1+2+...+9\right)}=\dfrac{90-\dfrac{9\left(9+1\right)}{2}}{\dfrac{9\left(9+1\right)}{2}}=\dfrac{90-45}{45}=\dfrac{45}{45}=1\)

=> \(\dfrac{a_1-1}{9}=1\) => a₁ = 10

tương tự , ta suy ra : a₂ =a₃=a₄ =a₅ = a₆=a₇= a₈=a₉= 10

Vậy a₁=a₂ =a₃=a₄ =a₅ = a₆=a₇= a₈=a₉= 10

Cách làm :

Ta có : A =

<=> A = ( biến đổi)

Vậy A =

Tổ Quốc Việt Nam hay Trung Quốc Việt Nam

Ta có :

* 100A = 10²A = \(10^2\dfrac{\left(10^{49}+1\right)}{10^{51}+1}\) = \(\dfrac{10^{51}+100}{10^{51}+1}=\dfrac{10^{51}+1}{10^{51}+1}+\dfrac{99}{10^{51}+1}\) = \(1+\dfrac{99}{10^{51}+1}\)

* 100B = 10²B = \(10^2\dfrac{\left(10^{48}+1\right)}{10^{50}+1}=\dfrac{10^{50}+100}{10^{50}+1}=\dfrac{10^{50}+1}{10^{50}+1}+\dfrac{99}{10^{50}+1}\)

= \(1+\dfrac{99}{10^{50}+1}\)

Vì \(\dfrac{99}{10^{50}+1}>\dfrac{99}{10^{51}+1}\Rightarrow1+\dfrac{99}{10^{50}+1}>1+\dfrac{99}{10^{51}+1}\)

Hay 100B > 100A => B>A