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\(\sqrt{x+2}\) = \(\frac{5}{7}\)
\(\Rightarrow\) x + 2 = \(\left(\frac{5}{7}\right)^2\)
\(\Rightarrow\) x + 2 = \(\frac{25}{49}\)
\(\Rightarrow\) x = \(\frac{25}{49}\) - 2
\(\Rightarrow\) x = \(\frac{-73}{49}\)
Vậy x = \(\frac{-73}{49}\)
\(\sqrt{x+2}=\frac{5}{7}\)
\(\Rightarrow x+2=\left(\frac{5}{7}\right)^2\)
\(\Rightarrow x+2=\frac{25}{49}\)
\(x=\frac{25}{49}-2=-\frac{73}{49}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b, Đk : x >= -2
pt <=> x+2 = 25/49
<=> x = 25/49 - 2 = -73/49
Tk mk nha
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{x+2}=\frac{5}{7}\)
Đk:\(x\ge-2\)
Thấy 2 vế luôn dương, bình 2 vế ta có:
\(\sqrt{\left(x+2\right)^2}=\left(\frac{5}{7}\right)^2\)
\(\Leftrightarrow x+2=\frac{25}{49}\)
\(\Leftrightarrow x=-\frac{73}{49}\left(tm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{x+2}\)= \(\frac{5}{7}\) \(\Rightarrow\) x + 2 = \(\frac{5^2}{7^2}\)
x + 2 = \(\frac{25}{49}\)
x = \(\frac{25}{49}\) - 2 = \(\frac{-73}{49}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
4) mấy bài kia trình bày dài lắm!! (lười ý mà ahihi)
\(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+|x+y+z|=0.\)
\(\Leftrightarrow|x-\sqrt{2}|+|y+\sqrt{2}|+|x+y+z|=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-\sqrt{2}=0\\y+\sqrt{2}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{2}\\y=-\sqrt{2}\end{cases}}}\)
Tìm z thì dễ rồi
\(đk:x\ge-2\)
\(\sqrt{x+2}=\frac{5}{7}\Leftrightarrow x+2=\frac{25}{49}\Leftrightarrow x=\frac{25}{49}-2=\frac{-73}{49}\left(n\right)\)
vậy \(x=\frac{-73}{49}\)
\(\sqrt{x+2}=\frac{5}{7}\)
\(\Rightarrow x+2=\frac{25}{49}\)
\(\Rightarrow x=\frac{25}{49}-2\)
\(\Rightarrow x=\frac{-73}{49}\)
Vậy, \(x=\frac{-73}{49}\)