Question

[Thư Đặng]

1. Tìm x, biết:

a) \(9^{x-1}=\frac{1}{9}\)

b) \(\frac{1}{3}:\sqrt{7-3x^2}=\frac{2}{15}\)

2. Tìm các số x,y,z thỏa mãn:

\(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+\left|x+y+z\right|=0\)

Answer 1

[Đức Hiếu]

Bài 1:

a, \(9^{x-1}=\dfrac{1}{9}\)

\(\Rightarrow9^{x-1}=9^{-1}\)

Vì \(9\ne-1;9\ne0;9\ne1\) nên

\(x-1=-1\Rightarrow x=0\)

Vậy \(x=0\)

b, \(\dfrac{1}{3}:\sqrt{7-3x^2}=\dfrac{2}{15}\)

\(\Rightarrow\sqrt{7-3x^2}=\dfrac{1}{3}:\dfrac{2}{15}\)

\(\Rightarrow\sqrt{7-3x^2}=\dfrac{5}{2}\)

\(\Rightarrow\left(\sqrt{7-3x^2}\right)^2=\left(\dfrac{5}{2}\right)^2\)

\(\Rightarrow7-3x^2=\dfrac{25}{4}\)

\(\Rightarrow3x^2=\dfrac{3}{4}\Rightarrow x^2=\dfrac{1}{4}\)

\(\Rightarrow x=\pm\dfrac{1}{2}\)

Vậy \(x=\pm\dfrac{1}{2}\)

Chúc bạn học tốt!!!

Answer 2

[Đức Hiếu]

Bài 2:

Với mọi giá trị của \(x;y;z\in R\) ta có:

\(\sqrt{\left(x-\sqrt{2}\right)^2}\ge0;\sqrt{\left(y+\sqrt{2}\right)^2\ge}0;\left|x+y+z\right|\ge0\)

\(\Rightarrow\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+\left|x+y+z\right|\ge0\) với mọi giá trị của \(x;y;z\in R\).

Để \(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+\left|x+y+z\right|=0\) thì

\(\left\{{}\begin{matrix}\sqrt{\left(x-\sqrt{2}\right)^2}=0\\\sqrt{\left(y+\sqrt{2}\right)^2}=0\\\left|x+y+z\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-\sqrt{2}=0\\y+\sqrt{2}=0\\x+y+z=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\sqrt{2}\\y=-\sqrt{2}\\\sqrt{2}-\sqrt{2}+z=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\sqrt{2}\\y=-\sqrt{2}\\z=0\end{matrix}\right.\)

Vậy \(x=\sqrt{2};y=-\sqrt{2};z=0\)

Chúc bạn học tốt!!!