Cho $$\dfrac{6}{1+\dfrac{4(x+3)}{7+x}}$$ Tính: `B=\frac{4A}{7+x}` 21/07/2021 Bởi Kylie Cho $$\dfrac{6}{1+\dfrac{4(x+3)}{7+x}}$$ Tính: `B=\frac{4A}{7+x}`
Đáp án+Giải thích các bước giải: `1+(4(x+3))/(7+x)` `=(7+x)/(7+x)+(4(x+3))/(7+x)` `=(7+x+4x+12)/(7+x)` `=(5x+19)/(7+x)` `\to 6:(5x+19)/(7+x)` `=6.(7+x)/(5x+19)` `\toA=6.(7+x)/(5x+19)` `\to 4.A=4.6.(7+x)/(5x+19)` `\to 4A=(24(7+x))/(5x+19)` `\to (4A)/(7+x)=(24(7+x))/(5x+19):(7+x)` `\to B=(24(7+x))/(5x+19).(1)/(7+x)` `\to B=24/(5x+19)` Vậy `B=24/(5x+19)` Bình luận
\(1+\dfrac{4(x+3)}{7+x}=\dfrac{7+x+4x+12}{7+x}=\dfrac{5x+19}{7+x}\\→A=\dfrac{6}{\dfrac{5x+19}{7+x}}=\dfrac{6(7+x)}{5x+19}\\B=\dfrac{4A}{7+x}\\↔B=\dfrac{24(7+x)}{5x+19}.\dfrac{1}{7+x}=\dfrac{24(7+x)}{(5x+19)(7+x)}\\=\dfrac{24}{5x+19}(x\ne -\dfrac{19}{5})\) Bình luận
Đáp án+Giải thích các bước giải:
`1+(4(x+3))/(7+x)`
`=(7+x)/(7+x)+(4(x+3))/(7+x)`
`=(7+x+4x+12)/(7+x)`
`=(5x+19)/(7+x)`
`\to 6:(5x+19)/(7+x)`
`=6.(7+x)/(5x+19)`
`\toA=6.(7+x)/(5x+19)`
`\to 4.A=4.6.(7+x)/(5x+19)`
`\to 4A=(24(7+x))/(5x+19)`
`\to (4A)/(7+x)=(24(7+x))/(5x+19):(7+x)`
`\to B=(24(7+x))/(5x+19).(1)/(7+x)`
`\to B=24/(5x+19)`
Vậy `B=24/(5x+19)`
\(1+\dfrac{4(x+3)}{7+x}=\dfrac{7+x+4x+12}{7+x}=\dfrac{5x+19}{7+x}\\→A=\dfrac{6}{\dfrac{5x+19}{7+x}}=\dfrac{6(7+x)}{5x+19}\\B=\dfrac{4A}{7+x}\\↔B=\dfrac{24(7+x)}{5x+19}.\dfrac{1}{7+x}=\dfrac{24(7+x)}{(5x+19)(7+x)}\\=\dfrac{24}{5x+19}(x\ne -\dfrac{19}{5})\)