mình chịu

a) Ta có :  \(\left|2x-1\right|>\left|\dfrac{-3}{4}\right|\)

\(\Rightarrow\left|2x-1\right|>\dfrac{3}{4}\)\(\Rightarrow\left[{}\begin{matrix}2x-1>\dfrac{3}{4}\\2x-1>\dfrac{-3}{4}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}2x>\dfrac{7}{4}\\2x>\dfrac{1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>\dfrac{7}{8}\\x>\dfrac{1}{8}\end{matrix}\right.\Leftrightarrow x>\dfrac{7}{8}\)

b) Ta có : |5x-4|=|x+4|

\(\Rightarrow\left[{}\begin{matrix}5x-4=x+4\\5x-4=-x-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}5x=x+8\\5x=-x\left(\text{vô lý}\right)\end{matrix}\right.\Rightarrow4x=8\Rightarrow x=2\)

c) Ta có : \(\left|0,5x-2\right|-\left|x-\dfrac{2}{3}\right|=0\)

\(\Rightarrow\left|0,5x-2\right|=\left|x-\dfrac{2}{3}\right|\)

\(\Rightarrow\left[{}\begin{matrix}0,5x-2=x-\dfrac{2}{3}\\0,5x-2=\dfrac{2}{3}-x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\dfrac{-1}{2}x=\dfrac{4}{3}\\\dfrac{3}{2}x=\dfrac{8}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-8}{3}\\x=\dfrac{16}{9}\end{matrix}\right.\)

d) \(3x-\left|x+15\right|=\dfrac{5}{4}\)

\(\left|x+15\right|=3x-\dfrac{5}{4}\)

Vì \(\left|x+15\right|\) ≥ 0 ∀ x => \(3x-\dfrac{5}{4}\) ≥ 0 ∀ x

=> \(3x\ge\dfrac{5}{4}\) ∀ x => \(x\ge\dfrac{5}{12}\) ∀ x

=> x + 15 > 0

Với x \(\ge\dfrac{5}{12}\) ta được :

\(x+15=3x-\dfrac{5}{4}\Rightarrow x+15+\dfrac{5}{4}=3x\Rightarrow2x=\dfrac{65}{4}\Rightarrow x=\dfrac{65}{8}\)

Ta có :

( x¹⁰⁰ )²⁰ = x^100.20 = x²⁰⁰⁰

(x100)20=x100.20=x2000

Tính 2.B = 2 -2²+2³-2⁴+........+2²⁰²¹

Lấy 2B + B ta được 3B =  1 + 2²⁰²¹

Vậy B = ( 1+2²⁰²¹) : 3 

Ta có : B = 1 - 2 + 2² - 2³ + ... + 2²⁰²⁰

=> 2B = 2 - 2² + 2³ - 2⁴ + ... + 2²⁰²¹

=> 2B + B = ( 2 - 2² + 2³ - 2⁴ + ... + 2²⁰²¹ ) + ( 1 - 2 + 2² - 2³ + ... + 2²⁰²⁰ )

=> 3B  = 2²⁰²¹ + 1 => B= \(\dfrac{2^{2021}-1}{3}\)

`a)`Ta có: \(\left\{{}\begin{matrix}AB\perp AC\\HE\perp AC\end{matrix}\right.\) \(\Rightarrow\)`AB////HE`

`b)`Ta có: \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\)

\(\Rightarrow\widehat{C}=180^o-90^o-60^o=30^o\)

Xét tam giác AHC, có:

\(\widehat{HAC}=180^o-30^o-90^o=60^o\)

\(\widehat{A}=\widehat{BAH}+\widehat{HAC}\)

\(\Rightarrow\widehat{BAH}=90^o-60^o=30^o\)

Ta có: \(\widehat{BAH}=\widehat{AHE}=30^o\) ( so le trong )

Vì \(\widehat{xBA}=\widehat{BAD}\left(=50^o\right)\) mà \(\widehat{xBA}\text{ và }\widehat{BAD}\) là 2 góc so le trong

=> Bx//AD (1)

Vì \(\widehat{DAC}=\widehat{ACy}\left(=30^o\right)\) mà \(\widehat{DAC}\text{ và }\widehat{ACy}\) là 2 góc so le trong

=> AD // Cy (2)

Từ (1) và (2) => Bx // Cy

Ta có:

`@` \(\widehat{ABx}=\widehat{DAB}=50^o\)

`=>Bx////AD` ( 2 góc so le trong bằng nhau ) (1)

`@`\(\widehat{ACy}=\widehat{DAC}=30^o\)

`=>Cy////AD` ( 2 góc so le trong bằng nhau ) (2)

\(\left(1\right);\left(2\right)\Rightarrow\)`Bx////Cy`