giúp vs ạ

`@` `\text {Ans}`

`\downarrow`

Số `%` của Na phải là `56` chứ ạ?

Gọi ct chung: \(\text{Na}_{\text{x}}\text{S}_{\text{y}}\)

Ta có: \(\text{NTK - Na: 23 amu; S: 32 amu}\)

Lập tỉ lệ ta có:

\(\text{%Na = }59\div23\approx2,43\) làm tròn lên là `2`

Vậy, có `3` nguyên tử Na trong phân tử \(\text{Na}_{\text{x}}\text{S}_{\text{y}}\)

\(\%\text{S}=44\div32=1,375\) làm tròn lên là `1`

Vậy, có `1` nguyên tử S trong phân tử \(\text{Na}_{\text{x}}\text{S}_{\text{y}}\)

`=>`\(\text{CTHH: Na}_2\text{S}\)

Lời giải:

$S_{35}=1-2+3-4+5-....-34+35=(1-2)+(3-4)+(5-6)+....+(33-34)+35$

$=\underbrace{(-1)+(-1)+...+(-1)}_{17}+35$

$=(-1)\times 17+35=35-17=18$

$S_{60}=1-2+3-4+5-.....+59-60$

$=(1-2)+(3-4)+(5-6)+....+(59-60)=\underbrace{(-1)+(-1)+...+(-1)}_{30}=(-1).30=-30$

$S_{35}+S_{60}=18+(-30)=-12$

`2^x=32`

`2^x=2^5`

`x=5`

Ví dụ 9:

a.

$x-\sqrt{x}-\sqrt{xy}+\sqrt{y}=(x-\sqrt{xy})-(\sqrt{x}-\sqrt{y})$

$=\sqrt{x}(\sqrt{x}-\sqrt{y})-(\sqrt{x}-\sqrt{y})$

$=(\sqrt{x}-\sqrt{y})(\sqrt{x}-1)$

b.

$=\sqrt{(x-3)(x+3)}-2\sqrt{x-3}$

$=\sqrt{x-3}(\sqrt{x+3}-2)$
c.

$x-\sqrt{x}-6=(x+2\sqrt{x})-(3\sqrt{x}+6)$

$=\sqrt{x}(\sqrt{x}+2)-3(\sqrt{x}+2)=(\sqrt{x}+2)(\sqrt{x}-3)$

Bài 3: 

ĐKXĐ: $x>0; x\neq 1$

a. 

\(E=\frac{\sqrt{x}(\sqrt{x}+1)}{(\sqrt{x}-1)^2}:\left[\frac{(\sqrt{x}+1)(\sqrt{x}-1)}{\sqrt{x}(\sqrt{x}-1)}+\frac{\sqrt{x}}{\sqrt{x}(\sqrt{x}-1)}+\frac{2-x}{\sqrt{x}(\sqrt{x}-1)}\right]\)

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

\(=\frac{\sqrt{x}(\sqrt{x}+1)}{(\sqrt{x}-1)^2}:\frac{\sqrt{x}+1}{\sqrt{x}(\sqrt{x}-1)}=\frac{\sqrt{x}(\sqrt{x}+1)}{(\sqrt{x}-1)^2}.\frac{\sqrt{x}(\sqrt{x}-1)}{\sqrt{x}+1}=\frac{x}{\sqrt{x}-1}\)

b.

$E=\frac{x}{\sqrt{x}-1}>1$
$\Leftrightarrow \frac{x}{\sqrt{x}-1}-1>0$

$\Leftrightarrow \frac{x-\sqrt{x}+1}{\sqrt{x}-1}>0(*)$

Do $x-\sqrt{x}+1=(\sqrt{x}-\frac{1}{2})^2+\frac{3}{4}>0$ với mọi $x$ thuộc đkxđ nên $(*)\Leftrightarrow \sqrt{x}-1>0$

$\Leftrightarrow x>1$

Kết hợp đkxđ suy ra $x>1$

\(P=\sqrt{x}+1+\dfrac{1}{\sqrt{x}}>=2\cdot\sqrt{\sqrt{x}\cdot\dfrac{1}{\sqrt{x}}}+1=3\)

Dấu = xảy ra khi x=1

\(C=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\\ 3C=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}\)

Xét \(3C-C=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{9^{98}}-\dfrac{1}{3}-\dfrac{1}{3^2}-\dfrac{1}{3^3}-...-\dfrac{1}{3^{99}}\\ 2C=1-\dfrac{1}{3^{99}}\\ C=\dfrac{1-\dfrac{1}{3^{99}}}{2}=\dfrac{1}{2}-\dfrac{\dfrac{1}{3^{99}}}{2}< \dfrac{1}{2}\left(dpcm\right)\)

Lời giải:

ĐKXĐ: $x\geq 4$
PT $\Leftrightarrow x+4\sqrt{x-4}=4$

$\Leftrightarrow (x-4)+4\sqrt{x-4}=0$

$\Leftrightarrow \sqrt{x-4}(\sqrt{x-4}+4)=0$

Hiển nhiên $\sqrt{x-4}+4>0$ với mọi $x\geq 4$

$\Rightarrow \sqrt{x-4}=0$

$\Leftrightarrow x=4$ (tm)

\(\left(3x-1\dfrac{2}{3}\right)^2-\dfrac{5}{4}=\dfrac{1}{2}\\ \left(3x-\dfrac{5}{3}\right)^2=\dfrac{1}{2}+\dfrac{5}{4}\\ \left(3x-\dfrac{5}{3}\right)^2=\dfrac{7}{4}\\ \Rightarrow\left[{}\begin{matrix}3x-\dfrac{5}{3}=\dfrac{\sqrt{7}}{2}\\3x-\dfrac{5}{3}=\dfrac{-\sqrt{7}}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}3x=\dfrac{\sqrt{7}}{2}+\dfrac{5}{3}\\3x=\dfrac{-\sqrt{7}}{2}+\dfrac{5}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}3x=\dfrac{3\sqrt{7}+10}{6}\\3x=\dfrac{-3\sqrt{7}+10}{6}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3\sqrt{7}+10}{18}\\x=\dfrac{-3\sqrt{7}+10}{18}\end{matrix}\right.\)