a) \(\frac{2}{3a}-\frac{3}{a}=\frac{2}{3a}-\frac{9}{3a}=\frac{-7}{3a}=\frac{7}{15}\Leftrightarrow-3a=15\Leftrightarrow a=-5\)

b)\(2x^3-1=15\Leftrightarrow2x^3=16\Leftrightarrow x^3=8\Leftrightarrow x=2\)

\(\Rightarrow\frac{2+16}{9}=\frac{y-15}{16}=2\Leftrightarrow y-15=32\Leftrightarrow y=47\)

c) \(\left|x\right|=3\Rightarrow\orbr{\begin{cases}x=-3\\x=3\end{cases}}\) rồi xét 2 trường hợp để tính A nhé :)

Bài 1: ĐK của a: \(a\ne0\)

Quy đồng VT ta có: \(\frac{2a-9a}{3a^2}=\frac{7}{15}\)

                    \(\Leftrightarrow\frac{-7a}{3a^2}=\frac{7}{15}\)

                    \(\Leftrightarrow-7a.15=3a^2.7\)

                    \(\Leftrightarrow-105a=21a^2\)

                    \(\Leftrightarrow-105a-21a^2=0\)

                    \(\Leftrightarrow a\left(-105-21a\right)=0\)

                    \(\Leftrightarrow\hept{\begin{cases}a=0\left(l\right)\\-105-21a=0\end{cases}\Leftrightarrow a=-5\left(n\right)}\)

Vậy:..

Ta có \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{a+b+c}{b+c+d}\)

=> \(\left(\frac{a}{b}\right)^3=\left(\frac{b}{c}\right)^3=\left(\frac{c}{d}\right)^3=\left(\frac{a+b+c}{b+c+d}\right)^3\)

=> \(\left(\frac{a}{b}\right)^3=\left(\frac{a+b+c}{b+c+d}\right)^3\)

=> \(\frac{a}{b}.\frac{a}{b}.\frac{a}{b}=\left(\frac{a+b+c}{b+c+d}\right)^3\)

=> \(\frac{a}{b}.\frac{b}{c}.\frac{c}{d}=\left(\frac{a+b+c}{b+c+d}\right)^3\) (Vì \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\))

=> \(\frac{a}{d}=\left(\frac{a+b+c}{b+c+d}\right)^3\)(đpcm)

\(\frac{1}{4}+\frac{1}{3}:2x=-5\)

\(\frac{1}{3}:2x=-5-\frac{1}{4}\)

\(\frac{1}{3}:2x=\frac{-21}{4}\)

\(2x=\frac{1}{3}:\frac{-21}{4}\)

\(2x=\frac{-4}{63}\)

\(x=\frac{-4}{63}:2\)

\(x=\frac{-2}{63}\)

\(\)

\(\frac{1}{4}+\frac{1}{3}:2x=-5\)

\(\Rightarrow\frac{1}{3}:2x=-\frac{21}{4}\)

\(\Rightarrow2x=\frac{-4}{63}\)

\(\Rightarrow x=\frac{-2}{63}\)

\(\left(3x-\frac{1}{4}\right)\left(x+\frac{1}{2}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-\frac{1}{4}=0\\x+\frac{1}{2}=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{12}\\x=\frac{-1}{2}\end{cases}}}\)

\(\left(2x-5\right)\left(\frac{3}{2}x+9\right)\left(0,3x-12\right)=0\)

Th1 : \(2x-5=0\Rightarrow x=\frac{5}{2}\)

Th2 : \(\frac{3}{2}x+9=0\Rightarrow x=-6\)

Th3 : \(0,3x-12=0\Rightarrow x=\frac{12}{0,3}\)

\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\Rightarrow c=\frac{1}{\frac{1}{2a}+\frac{1}{2b}}=\frac{1}{\frac{2\left(a+b\right)}{4ab}}=\frac{4ab}{2\left(a+b\right)}=\frac{2ab}{a+b}\)

\(\frac{a-c}{c-b}=\frac{a-\frac{2ab}{a+b}}{\frac{2ab}{a+b}-b}=\frac{a\left(1-\frac{2b}{a+b}\right)}{b\left(\frac{2a}{a+b}-1\right)}=\frac{a\left(\frac{a-b}{a+b}\right)}{b\left(\frac{a-b}{a+b}\right)}=\frac{a}{b}\)

\(\RightarrowĐPCM\)

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

\(A=\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)

Vậy .......

Haiz, sao lại thiếu sự quan sát thế nhỉ?

TH1: \(a+b+c=0\)\(\Rightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)\(\Rightarrow A=\frac{a}{-a}=\frac{b}{-b}=\frac{c}{-c}=-1\)

TH2: \(a+b+c\ne0\)\(\Rightarrow A=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)

\(\frac{1}{3}y+\frac{2}{5}\left(y+1\right)=0\)

\(\frac{1}{3}y+\frac{2}{5}y+\frac{2}{5}=0\)

\(y\left(\frac{1}{3}+\frac{2}{5}\right)=-\frac{2}{5}\)

\(y\left(\frac{1.5+2.3}{15}\right)=\frac{-2}{5}\)

\(\frac{11}{15}y=\frac{-2}{5}\)

\(y=\frac{-2}{5}\div\frac{11}{15}\)

\(y=\frac{-2}{5}.\frac{15}{11}\)

\(y=\frac{-6}{11}\)

\(\frac{-15}{12}y+\frac{3}{7}=\frac{6}{5}y-\frac{1}{2}\)

\(\frac{6}{5}y-\frac{1}{2}=\frac{-15}{12}y+\frac{3}{7}\)

\(\frac{1}{2}=\frac{6}{5}y+\frac{15}{12}y+\frac{3}{7}\)

\(\frac{1}{2}-\frac{3}{7}=\frac{6}{5}y+\frac{15}{12}y\)

\(\frac{1}{14}=y\left(\frac{6}{5}+\frac{15}{12}\right)\)

\(\frac{1}{14}=\frac{49}{20}y\)

\(y=\frac{1}{14}\div\frac{49}{20}\)

\(y=\frac{10}{343}\)