Question

a) (x + 1/5 )²  = 9/25

b) 3/7 . 5 - 9 = 3.x + 52 phan x + 16 1/7

Answer 1

a) (x + 1/5)² = 9/25

=> (x + 1/5)² = (3/5)²

=> \(\orbr{\begin{cases}x+\frac{1}{5}=\frac{3}{5}\\x+\frac{1}{5}=-\frac{3}{5}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{2}{5}\\x=-\frac{4}{5}\end{cases}}\)

Vậy ...

Answer 2

\(a,\text{ }\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)

\(\left(x+\frac{1}{5}\right)^2=\left(\pm\frac{3}{5}\right)^2\)

\(x+\frac{1}{5}=\pm\frac{3}{5}\)

\(\Rightarrow\orbr{\begin{cases}x+\frac{1}{5}=\frac{-3}{5}\\x+\frac{1}{5}=\frac{3}{5}\end{cases}}\)                               \(\Rightarrow\orbr{\begin{cases}x=\frac{-3}{5}-\frac{1}{5}\\x=\frac{3}{5}-\frac{1}{5}\end{cases}}\)                   \(\Rightarrow\orbr{\begin{cases}x=-\frac{4}{5}\\x=\frac{2}{5}\end{cases}}\)

                                \(\Rightarrow\text{ }x\in\left\{-\frac{4}{5}\text{ ; }\frac{2}{5}\right\}\)

Answer 3

\(b,\text{ }\frac{3}{7}\cdot5-9=3x+\frac{52}{x}+16\frac{1}{7}\)

\(\frac{15}{7}-\frac{63}{7}=\frac{3x\cdot x}{x}+\frac{52}{x}+\frac{113}{7}\)

\(\frac{-48}{7}=\frac{3x^2+52}{x}+\frac{113}{7}\)

\(-\frac{48}{7}-\frac{113}{7}=\frac{3x^2+52}{x}\)

\(-\frac{161}{7}=\frac{3x^2+52}{x}\)

\(\Rightarrow\text{ }-161x=7\left(3x^2+52\right)\)

Tự làm tiếp nhé .............

Answer 4

\(\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)

\(\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\)

\(\Rightarrow x+\frac{1}{5}=\frac{3}{5}\)

\(\Rightarrow x=\frac{3}{5}-\frac{1}{5}\)

\(\Rightarrow x=\frac{2}{5}\)

~ Hok tốt ~

Answer 5

Sr, mik làm thiếu nha !

\(\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)

\(\left(x+\frac{1}{5}\right)^2=\left(\frac{\pm3}{5}\right)^2\)

\(\Rightarrow\orbr{\begin{cases}x+\frac{1}{5}=\frac{3}{5}\\x+\frac{1}{5}=\frac{-3}{5}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{3}{5}+\frac{1}{5}\\x=\frac{-3}{5}+\frac{1}{5}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{4}{5}\\x=\frac{-2}{5}\end{cases}}}\)

Vậy ..........................

~ Hok tốt ~