Qua C vẽ tia Cz // DE

⇒ ∠DCz = ∠CDE = 108⁰

⇒ ∠zCA = ∠DCz - ∠DCA

= 108⁰ - 48⁰

= 60⁰

⇒ ∠zCA = ∠CAB = 60⁰

Mà ∠zCA và ∠CAB là hai góc so le trong

⇒ Cz // AB

Lại có:

Cz // DE

⇒ AB // DE

\(\dfrac{-3}{4}< \dfrac{a}{12}< \dfrac{-5}{9}\)

\(\Rightarrow\dfrac{-27}{36}< \dfrac{3a}{36}< \dfrac{-20}{36}\)

\(\Rightarrow-27< 3a< -20\)

\(\Rightarrow a=\left\{-8;-7\right\}\)

\(\dfrac{1}{1-\dfrac{1}{1-2^{-1}}}+\dfrac{1}{1+\dfrac{1}{2^{-1}}}\)

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

\(=\dfrac{2}{\left(1-2\right)\left(1+2\right)}=\dfrac{2}{1-4}=-\dfrac{2}{3}\)

mũ 2

0,(6) mux bằng 0,(6)^x

\(6\left(x-\dfrac{1}{4}\right)=4\left(\dfrac{5}{6}+x\right)\)

\(6x-\dfrac{3}{2}=\dfrac{10}{3}+4x\)

\(2x=\dfrac{10}{3}+\dfrac{3}{2}=\dfrac{29}{6}\)

\(x=\dfrac{29}{12}\)

6.x-3/2=4.x+10/3

6.x-3/2+3/2=4.x+10/3+3/2

6.x=4.x+29/6

6.x-4.x=4.x+29/6-4.x

2.x=29/6

x=29/12

\(\left|x\right|+3=5\)

\(\Rightarrow\left|x\right|=5-3\)

\(\Rightarrow\left|x\right|=2\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

\(\left|x\right|+3=5\)

Vì \(5>3\) nên \(\left|x\right|>0\)

\(x+3=5\)

\(x=5-3\)

\(x=2\)

Ta có:

\(x^2+\dfrac{1}{x^2}=4\)\(\left(x\ne0\right)\)

\(\left(x^2+\dfrac{1}{x^2}\right)^2=16\)

\(x^4+\dfrac{2.x^2}{x^2}+\dfrac{1}{x^4}=16\)

\(x^4+\dfrac{1}{x^4}=16-2=14\)

x² + 1/x² = 4

⇒ (x² + 1/x²)² = 16

⇒ x⁴ + 2.x².1/x² + 1/x⁴ = 16

⇒ x⁴ + 1/x⁴ + 2 = 16

⇒ x⁴ + 1/x⁴ = 16 - 2

⇒ x⁴ + 1/x⁴ = 14

`@TH1: x < 1`

Ta có: `Bth=1-x+3-x+5-x+7-x=16-4x`

`@TH2: 1 <= x < 3`

Ta có: `Bth=x-1+3-x+5-x+7-x=14-2x`

`@TH3: 3 <= x <= 5`

Ta có: `Bth=x-1+x-3+5-x+7-x=8`

`@TH4: 5 < x <= 7`

Ta có: `Bth=x-1+x-3+x-5+7-x=2x-2`

`@TH5: x > 7`

Ta có: `Bth=x-1+x-3+x-5+x-7=4x-16`

mong các bạn giải giúp mình

Ta có:

\(\widehat{xOn}=180^o-\widehat{nOy}=180^o-30^o=150^o\)

\(\widehat{xOm}=\widehat{nOy}=30^o\)(đối đỉnh)

\(\widehat{mOy}=\widehat{xOn}=150^o\)(đối đỉnh)