Question

a) \(\dfrac{x-4}{2022}\)+\(\dfrac{x-3}{2021}\)+\(\dfrac{x-2}{2020}\)+\(\dfrac{x-1}{2019}\)

b) Cho C = 26a + b(-13) +72 . 87 ; Biết 2 . a - b = 72

c) 2\(^x\) + 2\(^{x+1}\) + ... + 2\(^{x+2018}\) = 2\(^{^{ }2021}\) - 4

d) 2 . 3\(^x\) + 5 . 3\(^{x+1}\) = 153

Answer 1

Lời giải:

a. Đề cần gì với biểu thức trên vậy bạn?

b.

$C=13(2a-b)+72.87=13.72+72.87=72(13+87)=72.100=7200$

c.

$2^{x}+2^{x+1}+....+2^{x+2018}=2^{2021}-4$
$2^x(1+2+2^2....+2^{2018})=2^{2021}-4$
Đặt $A=1+2+2^2....+2^{2018}$

$2A=2+2^2+2^3+...+2^{2019}$

$\Rightarrow 2A-A=2^{2019}-1$

$\Rightarrow A=2^{2019}-1$

Vậy: $2^x(2^{2019}-1)=2^{2021}-4$
$\Rightarrow 2^x(2^{2019}-1)=2^2(2^{2019}-1)$

$\Rightarrow 2^x=2^2$

$\Rightarrow x=2$

d.

$2.3^x+5.3^{x+1}=153$

$3^x(2+5.3)=153$

$3^x.17=153$

$3^x=153:17=9=3^2$

$\Rightarrow x=2$

Answer 2

a.

$\frac{x-4}{2022}+\frac{x-3}{2021}+\frac{x-2}{2020}+\frac{x-1}{2019}=-4$

$\Leftrightarrow \frac{x-4}{2022}+1+\frac{x-3}{2021}+1+\frac{x-2}{2020}+1+\frac{x-1}{2019}+1=0$

$\Leftrightarrow \frac{x+2018}{2022}+\frac{x+2018}{2021}+\frac{x+2018}{2020}+\frac{x+2018}{2019}=0$

$\Leftrightarrow (x+2018)(\frac{1}{2022}+\frac{1}{2021}+\frac{1}{2020}+\frac{1}{2019})=0$

$\Leftrightarrow x+2018=0$

$\Leftrightarrow x=-2018$ (tm)