Ta có :

A = 2²⁰²²-2²⁰²¹-2²⁰²⁰-...-2-1

=> A = 2²⁰²²-(2²⁰²¹+2²⁰²⁰+...+2+1)

=> A = 2²⁰²²-B ( với B=2²⁰²¹+2²⁰²⁰+...+2+1)

Lại có :

B = 2²⁰²¹+2²⁰²⁰+...+2+1

=> 2B=2²⁰²²+2²⁰²¹+...+2²+2

=> 2B-B=(2²⁰²²+2²⁰²¹+...+2²+2)-(2²⁰²¹+2²⁰²⁰+...+2+1)

=> B = 2²⁰²²-1 , thay vào A

=> A = 2²⁰²²-(2²⁰²²-1)

=> A = 1

=> 2023A = 2023

Vì \((2x+\dfrac{1}{3})^2 \ge 0 \forall x\)

\(<=>(2x+\dfrac{1}{3})^2-1 \ge -1 \forall x\)

 Hay \(A \ge -1 \forall x\)

Dấu "`=`" xảy ra \(<=>(2x+\dfrac{1}{3})^2=0<=>x=\dfrac{-1}{6}\)

\(A=\left(2x+\dfrac{1}{3}\right)^2-1\)

Vì : \(\left(2x+\dfrac{1}{3}\right)^2\ge0\forall x\\ =>A=\left(2x+\dfrac{1}{3}\right)^2-1\ge-1\)

Dấu ''='' xảy ra khi : \(\left(2x+\dfrac{1}{3}\right)^2=0\\ =>2x+\dfrac{1}{3}=0\\ =>x=-\dfrac{1}{6}\)

Vậy GTNN của `A` là : `-1` khi \(x=-\dfrac{1}{6}\)

(7-23-14)-(43-104)-(54-13)

=7-23-13-43+104-54+13

=-9

`(-2/5)^{x}-1=-21/25

`(-2/5)^{x}=-21/25+1

`(-2/5)^{x}=4/25`

`(-2/5)^{x}=(-2/5)^2`

`x=2`

\(\widehat{AOM}+\widehat{MOC}=\widehat{BON}+\widehat{NOC}\\ \Rightarrow\widehat{AOC}=\widehat{BOC}\\ \)

Vì \(\widehat{AOC}+\widehat{BOC}=180^0\Rightarrow\widehat{AOC}=\widehat{BOC}=\dfrac{180^0}{2}=90^0\\ \Rightarrow OC\perp AB\)

\(\sqrt{11^2-72}-1\dfrac{1}{2}:\sqrt{\dfrac{25}{4}}-\left(\dfrac{2021}{2022}\right)^0\)

\(=\sqrt{121-72}-\dfrac{3}{2}:\dfrac{5}{2}-1\)

\(=\sqrt{49}-\dfrac{3}{5}-1\)

\(=7-\dfrac{3}{5}-1\)

\(=\dfrac{32}{5}-1\\ \)

\(=\dfrac{27}{5}\)

\(5-3x+0,25=\dfrac{3}{4}-1,25\\ \Leftrightarrow5,25-3x=0,75-1,25\\ \Leftrightarrow3x=5,25-0,75+1,25\\ \Leftrightarrow3x=5,75\\ \Rightarrow x=5,75:3=\dfrac{575}{100}:3=\dfrac{23}{4}:3=\dfrac{23}{12}\)

\(\dfrac{1}{2}x+\dfrac{3}{2}x+\dfrac{7}{3}=-\dfrac{4}{3}\\ =>\left(\dfrac{1}{2}+\dfrac{3}{2}\right)x=-\dfrac{4}{3}-\dfrac{7}{3}\\ =>\dfrac{4}{2}x=-\dfrac{11}{3}\\ =>x=\left(-\dfrac{11}{3}\right):2\\ =>x=-\dfrac{11}{6}\)

244.244=59536

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