= 14,06 cm² nha bạn.

Gọi \(ƯCLN_{\left(12n+1\right)}\) và \(ƯCLN_{\left(30n+2\right)}\) là \(a\left(a\ne0\right)\)

\(\left(12n+1\right)⋮a\)

\(\Rightarrow5\left(12n+1\right)⋮a\)

\(\left(30n+2\right)⋮a\)

\(\Rightarrow2\left(30n+2\right)⋮a\)

\(\Rightarrow\left[5\left(12n+1\right)\right]-\left[2\left(30n+2\right)\right]⋮a\)

\(\Rightarrow\left(60n+5\right)-\left(60n+4\right)⋮a\)

\(\Rightarrow60n+5-60n-4⋮a\)

\(\Rightarrow\left(60n-60n\right)+\left(5-4\right)⋮a\)

\(\Rightarrow1⋮a\)

\(\Rightarrow a=1\)

Vậy \(\dfrac{12n+1}{30n+2}\) là phân số tối giản

\(3a\) Ta có

\(A_{\left(x\right)}=x^2+10x+36\)

\(A_{\left(x\right)}=x^2+5x+5x+25+11\)

\(A_{\left(x\right)}=\left(x^2+5x\right)+\left(5x+25\right)+11\)

\(A_{\left(x\right)}=\left(x+5\right)x+\left(x+5\right)5+11\)

\(A_{\left(x\right)}=\left(x+5\right)\left(x+5\right)+11\)

\(A_{\left(x\right)}=\left(x+5\right)^2+11\)

Mà \(\left(x+5\right)^2\ge0\)

\(\Rightarrow A_{\left(x\right)}=\left(x+5\right)^2+11\ge11>0\)

Vậy đa thức \(A_{\left(x\right)}\) không có nghiệm

\(2b\) Ta có:

\(\dfrac{x-y}{z}=\dfrac{3y}{x-z}=\dfrac{x}{y}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta được

\(\dfrac{x-y}{z}=\dfrac{3y}{x-z}=\dfrac{x}{y}=\dfrac{\left(x-y\right)+3y+x}{z+\left(x-z\right)+y}\)

\(=\dfrac{x-y+3y+x}{z+x-z+y}=\dfrac{\left(x+x\right)+\left(-y+3y\right)}{\left(z-z\right)+x+y}\)

\(=\dfrac{2x+2y}{x+y}=\dfrac{2\left(x+y\right)}{x+y}=2\)

\(\Rightarrow\dfrac{x}{y}=2\Rightarrow x=2y\)

\(\Rightarrow\dfrac{x-y}{z}=2\)

\(\Rightarrow x-y=2z\)

\(\Rightarrow2y-y=2z\)

Vậy \(x=2y;y=2z\)

1 Ta có

\(a\left(a+b+c\right)=-\dfrac{1}{24}\)

\(b\left(a+b+c\right)=\dfrac{1}{16}\)

\(c\left(a+b+c\right)=-\dfrac{1}{72}\)

\(\Rightarrow a\left(a+b+c\right)+b\left(a+b+c\right)+c\left(a+b+c\right)=-\dfrac{1}{24}+\dfrac{1}{16}+-\dfrac{1}{72}\)

\(\Rightarrow\left(a+b+c\right)\left(a+b+c\right)=\dfrac{1}{144}\)

\(\Rightarrow\left(a+b+c\right)^2=\dfrac{1}{144}\)

\(\Rightarrow a+b+c=\dfrac{1}{12}\)

\(\Rightarrow a\cdot\dfrac{1}{12}=-\dfrac{1}{24}\Rightarrow a=-\dfrac{1}{24}\div\dfrac{1}{12}=-\dfrac{1}{2}\)

\(b\cdot\dfrac{1}{12}=\dfrac{1}{16}\Rightarrow b=\dfrac{1}{16}\div\dfrac{1}{12}=\dfrac{3}{4}\)

Mình muốn đi

\(\dfrac{7}{4}-\left|\dfrac{3}{10}-\dfrac{7}{20}\right|-x=2-\left|\dfrac{1}{4}-\dfrac{9}{10}\right|\)

\(\Leftrightarrow\dfrac{7}{4}-\left|-\dfrac{1}{20}\right|-x=2-\left|-\dfrac{13}{20}\right|\)

\(\Leftrightarrow\dfrac{7}{4}-\dfrac{1}{20}-x=2-\dfrac{13}{20}\)

\(\Leftrightarrow\dfrac{17}{10}-x=\dfrac{27}{20}\)

\(\Leftrightarrow x=\dfrac{17}{10}-\dfrac{27}{20}\)

\(\Leftrightarrow x=\dfrac{7}{20}\)

Vậy \(x=\dfrac{7}{20}\)