3ˣ + 4² = 16

3ˣ + 16 = 16

3ˣ = 16 - 16

3ˣ = 0 (vô lý)

Vậy không tìm được x thỏa mabx yêu cầu

`3^x+4^2=16`

`=>3^x+16=16`

`=>3^x=16-16`

`=>3^x=0`

`=>x=1`

\(5x\left(x-3\right)=\left(x-2\right)\left(5x-1\right)-5\\ \Leftrightarrow5x^2-15x=5x^2-11x+2-5\\ \Leftrightarrow4x=3\\ \Leftrightarrow x=\dfrac{3}{4}\)

\(a)\) Ta có : \(M:N:P:Q=1:2:3:4\)

\(\Rightarrow\dfrac{M}{1}=\dfrac{N}{2}=\dfrac{P}{3}=\dfrac{Q}{4}\left(1\right)\) 

Áp dụng tính chất của dãy tỉ số bằng nhau ta có : 

\(\left(1\right)=\dfrac{M+N+P+Q}{1+2+3+4}=\dfrac{360}{10}=36\)

\(\Rightarrow\left\{{}\begin{matrix}M=36.1=36\\N=36.2=72\\P=36.3=108\\Q=36.4=144\end{matrix}\right.\)

\(b)\) Xét từ giác MNPQ có : \(gócM+gócQ=36+144=180độ\)

Mà : 2 góc ở vị trí trong cùng phía .

\(\Rightarrow MN//PQ\left(đpcm\right)\)

a) \(\dfrac{M}{1}=\dfrac{N}{2}=\dfrac{P}{3}=\dfrac{Q}{4}=\dfrac{M+N+P+Q}{1+2+3+4}=\dfrac{360}{10}=36\)

\(\Rightarrow\left\{{}\begin{matrix}M=36.1=36^o\\N=36.2=72^o\\P=36.3=108^o\\Q=36.4=144^o\end{matrix}\right.\)

  \(\left(\dfrac{-1}{3}\right)^{-1}=-3\)

Vì \(\left(\dfrac{-1}{3}\right)^{-1}=\left(\dfrac{1}{-3}\right)^{-1}=\left(-3^{-1}\right)^{-1}=-3^{-1\times\left(-1\right)}=-3^1=-3\) 

=> \(\left(\dfrac{-1}{3}\right)^{-1}=-3\)

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

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

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

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

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

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

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`\(3^{x-1}=16\) 

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`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, \(x=\dfrac{11}{15}\)