bn chuyển dọc đc ko

Đaay nhé

khó nhìn qué

\(\dfrac{BC}{x}=\dfrac{BC+6x}{BC}=>BC^2=BC.x+6x^2\)

\(=>6x^2+BC.x-BC^2=0\)

\(< =>6\left(x^2+\dfrac{1}{6}BCx-\dfrac{1}{6}BC^2\right)=0\)

\(=>x^2+\dfrac{1}{6}BCx-\dfrac{1}{6}BC^2=0\)

\(< =>x^2+2.\dfrac{1}{12}BC.x+\left(\dfrac{1}{12}BC^2\right)-\left(\dfrac{1}{12}BC\right)^2-\dfrac{1}{6}BC^2=0\)

\(< =>\left(x+\dfrac{1}{12}BC\right)^2-\left(\dfrac{5}{12}BC\right)^2=0\)

\(=>\left(x+\dfrac{1}{12}BC+\dfrac{5}{12}BC\right)\left(x+\dfrac{1}{12}BC-\dfrac{5}{12}BC\right)=0\)

\(< =>\left(x+\dfrac{1}{2}BC\right)\left(x-\dfrac{1}{3}BC\right)=0\)

\(=>\left[{}\begin{matrix}x+\dfrac{1}{2}BC=0\\x-\dfrac{1}{3}BC=0\end{matrix}\right.=>\left[{}\begin{matrix}BC=2x\\BC=3x\end{matrix}\right.\)

\(a,\dfrac{-3}{5}+\dfrac{7}{21}+\dfrac{-4}{5}+\dfrac{7}{5}\)

\(=\left(\dfrac{-3}{5}+\dfrac{7}{5}\right)+\dfrac{7}{21}+-\dfrac{4}{5}\)

\(=\dfrac{4}{5}+\dfrac{7}{21}+-\dfrac{4}{5}\)

\(=\left(\dfrac{4}{5}+\dfrac{-4}{5}\right)+\dfrac{7}{21}\)

\(=0+\dfrac{7}{21}=\dfrac{7}{21}\)

\(b,\dfrac{-3}{17}+\left(\dfrac{2}{3}+\dfrac{3}{17}\right)\)

\(=-\dfrac{3}{17}+\dfrac{2}{3}+\dfrac{3}{17}\)

\(=\left(\dfrac{-3}{17}+\dfrac{3}{17}\right)+\dfrac{2}{3}\)

\(=0+\dfrac{2}{3}=\dfrac{2}{3}\)

\(c,\dfrac{-5}{21}+\left(\dfrac{-16}{21}+1\right)\)

\(=\dfrac{-5}{21}+\dfrac{-16}{21}+1\)

\(=\dfrac{-21}{21}+1\)

\(=\left(-1\right)+1=0\)

Tick mình nha ^^

a.\(\dfrac{-3}{5}\)+\(\dfrac{7}{21}\)+\(\dfrac{-4}{5}\)+\(\dfrac{7}{5}\)= (\(\dfrac{-3}{5}\)+\(\dfrac{-4}{5}\)+\(\dfrac{7}{5}\)) + \(\dfrac{7}{21}\)=0+\(\dfrac{7}{21}\)=\(\dfrac{7}{21}\)

b.\(\dfrac{-3}{17}\)+(\(\dfrac{2}{3}\)+\(\dfrac{3}{17}\))= \(\dfrac{-3}{17}\)+\(\dfrac{3}{17}\)+\(\dfrac{2}{3}\)=0+\(\dfrac{2}{3}\)=\(\dfrac{2}{3}\)

c.\(\dfrac{-5}{21}\)+(\(\dfrac{-16}{21}\)+1)=\(\dfrac{-5}{21}\)+\(\dfrac{-16}{21}\)+1=(-1)+ 1 = 0

a: \(-\dfrac{3}{5}+\dfrac{7}{21}+\dfrac{-4}{5}+\dfrac{7}{5}=-1+\dfrac{1}{3}+\dfrac{7}{5}=\dfrac{11}{15}\)

b: \(\dfrac{-3}{17}+\left(\dfrac{2}{3}+\dfrac{3}{17}\right)=\dfrac{2}{3}\)

c: \(\dfrac{-5}{21}+\dfrac{-16}{21}+1=-1+1=0\)

\(\dfrac{7}{5}+\dfrac{4}{7}-\dfrac{9}{10}=\dfrac{49-20}{35}-\dfrac{9}{10}=\dfrac{19}{35}-\dfrac{9}{10}=\dfrac{190-315}{350}=\dfrac{-125}{350}\)

\(\dfrac{2}{1}+\dfrac{3}{4}\text{×}\dfrac{8}{5}=\dfrac{8+3}{4}\text{×}\dfrac{8}{5}=\dfrac{11\text{×}8}{4\text{×}5}=\dfrac{88}{20}\)

mấy câu kia áp dụng là dc!