8 disappointing

12 preservation

13 seriously

Bài 13:

a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge1\\x\ne3\end{matrix}\right.\)

b: Ta có: \(P=\dfrac{x-3}{\sqrt{x-1}-\sqrt{2}}\)

\(=\dfrac{\left(\sqrt{x-1}-\sqrt{2}\right)\left(\sqrt{x-1}+\sqrt{2}\right)}{\sqrt{x-1}-\sqrt{2}}\)

\(=\sqrt{x-1}+\sqrt{2}\)

c: Thay \(x=12-6\sqrt{2}\) vào P, ta được:

\(P=\sqrt{11-6\sqrt{2}}+\sqrt{2}=3-\sqrt{2}+\sqrt{2}=3\)

d: Ta có: \(\sqrt{x-1}\ge0\forall x\) thỏa mãn ĐKXĐ

\(\Leftrightarrow\sqrt{x-1}+\sqrt{2}\ge\sqrt{2}\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi x=1

\(\dfrac{12}{16}=\dfrac{132}{176}\\ \dfrac{13}{16}=\dfrac{143}{176}\\ Ta.có:\dfrac{16}{22}< \dfrac{132}{176}< \dfrac{17}{22}< \dfrac{143}{176}< \dfrac{18}{22}\\ Vậy:Chọn.số.17\)

12 I consider that to manage to know what other people are thinking is quite impossible

12:

a: ĐKXĐ: -2x+3>=0

=>x<=3/2

b: ĐKXĐ: x^2<>0

=>x<>0

c: ĐKXĐ: x+3>0

=>x>-3

d:ĐKXĐ: -5/x^2+6>=0

=>x^2+6<0

=>x thuộc rỗng

4. 

a) Ta có: \(\sqrt{x}=3\)

\(\Leftrightarrow x=3^2=9\)

 Vậy \(x=9\)

b) Ta có: \(\sqrt{x}=\sqrt{5}\)

\(\Leftrightarrow x=\sqrt{5}^2=5\)

Vậy \(x=5\)

c) Ta có: \(\sqrt{x}=0\)

\(\Leftrightarrow x=0^2=0\)

Vậy \(x=0\)

d) Ta có: \(\sqrt{x}=-2\)

\(\Leftrightarrow x=-2^2=4\)

Vậy \(x=4\)

\(P=3^{10}+3^{11}+3^{12}\)

\(=3^{10}\cdot\left(1+3+3^2\right)\)

\(=3^{10}\cdot13\)

Vì \(13⋮13\) nên \(3^{10}\cdot13⋮13\)

hay \(P⋮13\)

Vậy ...

#\(Toru\)

P = 3¹⁰ + 3¹¹ + 3¹²

= 3¹⁰.(1 + 3 + 3²)

= 3¹⁰ . 13 ⋮ 13

Vậy P ⋮ 13