\(x-\dfrac{x}{12}+\dfrac{x}{\dfrac{18}{7}}-\dfrac{7}{12}+\dfrac{7}{18}=-\dfrac{4}{7}\)

\(\Rightarrow x\left(1-\dfrac{1}{12}+\dfrac{1}{\dfrac{18}{7}}\right)=-\dfrac{4}{7}+\dfrac{7}{12}-\dfrac{7}{18}\)

\(\Rightarrow x\left(\dfrac{216-18+7}{12.18}\right)=\dfrac{-864+882-588}{7.12.18}\)

\(\Rightarrow x\left(\dfrac{205}{12.18}\right)=\dfrac{-570}{7.12.18}\)

\(\Rightarrow x=\dfrac{-570}{7.12.18}.\dfrac{12.18}{205}\)

\(\Rightarrow x=\dfrac{-114}{287}\)

Ta có :

\(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^o\)

\(\Rightarrow\widehat{A}+\widehat{B}=360^o-\left(\widehat{C}+\widehat{D}\right)\)

\(\Rightarrow\widehat{A}+\widehat{B}=360^o-\left(60+80\right)=220^o\)

mà \(\widehat{A}-\widehat{B}=10^o\)

\(\Rightarrow\widehat{A}=\left(220-10\right):2=105^o\)

\(\Rightarrow\widehat{B}=105-10=95^o\)

Vậy \(\left\{{}\begin{matrix}\widehat{A}=105^o\\\widehat{B}=95^o\end{matrix}\right.\)

\(\dfrac{x-2023}{6}+\dfrac{x-2023}{10}+\dfrac{x-2023}{15}+\dfrac{x-2023}{21}=\dfrac{8}{21}\)

\(\left(x-2023\right)\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}\right)=\dfrac{8}{21}\)

\(\left(x-2023\right).\dfrac{8}{21}=\dfrac{8}{21}\)

\(x-2023=1\)

\(x=2024\)

Vậy..............

\(...\Rightarrow\left(x-2023\right)\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}\right)=\dfrac{8}{21}\)

\(\Rightarrow\left(x-2023\right)\left(\dfrac{35+21+14+1}{210}\right)=\dfrac{8}{21}\)

\(\Rightarrow\left(x-2023\right).\dfrac{71}{210}=\dfrac{8}{21}\)

\(\Rightarrow\left(x-2023\right).\dfrac{71}{210}=\dfrac{8}{21}.\dfrac{210}{71}=\dfrac{80}{71}\)

\(\Rightarrow x-2023=\dfrac{80}{71}\Rightarrow x=\dfrac{80}{71}+2023=\dfrac{143713}{71}\)

`@` `\text {Ans}`

`\downarrow`

`c)`

\(2-3^{x-1}-7=11\)

`\Rightarrow`\(3^{x-1}-5=11\)

`\Rightarrow`\(3^{x-1}=11+5\)

`\Rightarrow`\(3^{x-1}=16\) 

Bạn xem lại đề

`d)`

\(\left(x-\dfrac{3}{5}\right)\div\dfrac{-1}{3}=-0,4\)

`\Rightarrow`\(x-\dfrac{3}{5}=-0,4\cdot\left(-\dfrac{1}{3}\right)\)

`\Rightarrow`\(x-\dfrac{3}{5}=\dfrac{2}{15}\)

`\Rightarrow`\(x=\dfrac{2}{15}+\dfrac{3}{5}\)

`\Rightarrow`\(x=\dfrac{11}{15}\)

Vậy, \(x=\dfrac{11}{15}\)

Giả sử a//BC. Theo đề ta có:

\(\widehat{A_1}=\widehat{C_1}\) (hai góc so le trong) (1)

\(\widehat{A_1}=\dfrac{1}{2}\widehat{ABC}+\widehat{BAC}\) (vì BD là tia phân giác của \(\widehat{ABC}\)) (2)

\(\widehat{C_1}=\widehat{ABC}+\widehat{BAC}\) (vì \(\widehat{C_1}\) là góc ngoài của \(\widehat{C}\) ) (3)

Từ (1); (2) và (3) suy ra \(\dfrac{1}{2}\widehat{ABC}=\widehat{ABC}\), hay \(\dfrac{1}{2}=1\) (vô lí)

Suy ra a không song song với BC, hay a cắt đường thẳng BC

\(100:\left(x-1\right)^2=4\)

\(\left(x-1\right)^2=100:4=25\)

\(\left(x-1\right)^2=5^2\)

\(\Rightarrow x-1=5\)

\(x=5+1=6\)

100 : (x - 1)² = 4

(x - 1)² = 100 : 4

(x - 1)² = 25

x - 1 = 5 hoặc x - 1 = -5

*) x - 1 = 5

x = 5 + 1

x = 6

*) x - 1 = -5

x = -5 + 1

x = -4

Vậy x = -4; x = 6

a) BA=BC(gt)

⇒B thuộc đường trung trực AC

DA=DC(gt)

⇒D thuộc đường trung trực AC

B và D là đường phân biệt cùng thuộc 1 đường trung trực AC nên đường thẳng BD là đường trung trực của AC

b) Xét △BAD và △BCD,có:

BA=BC

DA=DC

BC chung

⇒△BAD=△BCD(ccc)⇒góc BAD= góc BCD

Ta có BAD+BCD+ABC+ADC=360

          2BAD=360-ABC-ADC

          2BAD=360-100-80

          2BAD=180

⇒BAD=BCD=180/2=80

Bạn xem lại đề

Ta có:

∠M + ∠N + ∠P + ∠Q = 360⁰ (tổng các góc trong tứ giác MNPQ)

⇒ ∠M + ∠N + ∠P + (∠P + 10⁰) = 360⁰

⇒ ∠M + ∠N + (∠N + 10⁰) + (∠N + 10⁰ + 10⁰) = 360⁰

⇒ ∠M + (∠M + 10⁰) + (∠M + 10⁰ + 10⁰) + (∠M + 10⁰ + 10⁰ + 10⁰)

⇒ ∠M + ∠M + 10⁰ + ∠M + 20⁰ + ∠M + 30⁰ = 360⁰

⇒ 4∠M + 60⁰ = 360⁰

⇒ 4∠M = 360⁰ - 60⁰

⇒ 4∠M = 300⁰

⇒ ∠M = 300⁰ : 4

⇒ ∠M = 75⁰

⇒ ∠N = 75⁰ + 10⁰ = 85⁰

⇒ ∠P = 85⁰ + 10⁰ = 95⁰

⇒ ∠Q = 95⁰ + 10⁰ = 105⁰

\(\widehat{M}+\widehat{N}+\widehat{P}+\widehat{Q}=360^o\)

\(\widehat{M}+\widehat{M}+10+\widehat{M}+20+\widehat{M}+30=360\)

\(4\widehat{M}=360-60=300\Rightarrow M=75^o\)

D = \(\dfrac{1}{1\times1981}\) + \(\dfrac{1}{2\times1982}\)+...+ \(\dfrac{1}{25\times2005}\)

D =\(\dfrac{1}{1980}\times\)( \(\dfrac{1980}{1\times1981}\)+ \(\dfrac{1980}{2\times1982}\)+....+ \(\dfrac{1980}{25\times2005}\))

D = \(\dfrac{1}{1980}\) \(\times\)(\(\dfrac{1}{1}\) - \(\dfrac{1}{1981}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{1982}\)+....+ \(\dfrac{1}{25}\) \(\times\) \(\dfrac{1}{2005}\))

D= \(\dfrac{1}{1980}\)[( \(\dfrac{1}{1}\) + \(\dfrac{1}{2}\) +....+ \(\dfrac{1}{25}\)) - ( \(\dfrac{1}{1981}\)+ \(\dfrac{1}{1982}\)+...+ \(\dfrac{1}{2005}\))]

E =\(\dfrac{1}{25}\times\)( \(\dfrac{1}{1\times26}\)+ \(\dfrac{1}{2\times27}\)+...+ \(\dfrac{1}{1980\times2005}\))

E =  \(\dfrac{1}{25}\). (\(\dfrac{25}{1\times26}\) + \(\dfrac{25}{2\times27}\)+....+ \(\dfrac{25}{1980\times2005}\))

E = \(\dfrac{1}{25}\).(\(\dfrac{1}{1}\)-\(\dfrac{1}{26}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{27}\)+...+\(\dfrac{1}{1980}\)-\(\dfrac{1}{2005}\))

E=\(\dfrac{1}{25}\)[\(\dfrac{1}{1}\)+...+ \(\dfrac{1}{25}\)+ (\(\dfrac{1}{26}\)+...+\(\dfrac{1}{1980}\)) - (\(\dfrac{1}{26}\)+...+\(\dfrac{1}{1980}\)) - (\(\dfrac{1}{1981}\)+..\(\dfrac{1}{2005}\))]

E = \(\dfrac{1}{25}\) .[\(\dfrac{1}{1}\)+\(\dfrac{1}{2}\)+...+\(\dfrac{1}{25}\) - (\(\dfrac{1}{1981}\)+\(\dfrac{1}{1982}\)+...+ \(\dfrac{1}{2005}\))]

\(\dfrac{D}{E}\) = \(\dfrac{\dfrac{1}{1980}}{\dfrac{1}{25}}\) = \(\dfrac{5}{396}\)