\(\Leftrightarrow-\dfrac{16}{279}< \dfrac{x}{9}< =\dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{x}{9}=0\)

hay x=0

Lời giải:
$\frac{1}{x}+\frac{1}{y}=\frac{1}{8}$

$\Rightarrow \frac{x+y}{xy}=\frac{1}{8}$

$\Rightarrow 8(x+y)=xy$
$\Rightarrow xy-8x-8y=0$

$\Rightarrow x(y-8)-8(y-8)=64$

$\Rightarrow (x-8)(y-8)=64$

Do $x,y$ tự nhiên nên $x-8,y-8\in\mathbb{Z}$

$\Rightarrow x-8$ là ước của $64$. Mà $x-8>-8$ với mọi $x\in\mathbb{N}^*$ nên:

$x-8\in\left\{1; 2; 4; 8; 16; 32; 64; -1; -2; -4\right\}$

Đến đây bạn chỉ cần chịu khó xét các TH là được.

a)ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

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

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

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

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

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

\(\Rightarrow A=\dfrac{\sqrt{x}}{\sqrt{x}+1}\)

b) \(x=9\Rightarrow A=\dfrac{3}{3+1}=\dfrac{3}{4}\)

\(x=7-4\sqrt{3}\Rightarrow A=\dfrac{\sqrt{7-4\sqrt{3}}}{\sqrt{7-4\sqrt{3}}+1}=\dfrac{\sqrt{7-2\sqrt{12}}}{\sqrt{7-2\sqrt{12}}+1}=\dfrac{\sqrt{4-2\sqrt{3}\sqrt{4}+3}}{\sqrt{4-2\sqrt{3}\sqrt{4}+3}+1}=\dfrac{2-\sqrt{3}}{2-\sqrt{3}+1}=\dfrac{2-\sqrt{3}}{3-\sqrt{3}}=\dfrac{\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)}=\dfrac{3-\sqrt{3}}{6}\)

\(\Leftrightarrow\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{\left(2x-2\right).2x}=\dfrac{11}{24}\)

\(\Leftrightarrow\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+...+\dfrac{2x-\left(2x-2\right)}{\left(2x-2\right).2x}=\dfrac{11}{24}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2x-2}-\dfrac{1}{2x}=\dfrac{11}{24}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2x}=\dfrac{11}{24}\)

\(\Leftrightarrow\dfrac{1}{2x}=\dfrac{1}{2}-\dfrac{11}{24}\)

\(\Leftrightarrow\dfrac{1}{2x}=\dfrac{1}{24}\)

\(\Rightarrow2x=24\)

\(\Rightarrow x=12\)

Bài 3 :

\(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}\)

\(\dfrac{1}{2!}=\dfrac{1}{2.1}=1-\dfrac{1}{2}< 1\)

\(\dfrac{1}{3!}=\dfrac{1}{3.2.1}=1-\dfrac{1}{2}-\dfrac{1}{3}< 1\)

\(\dfrac{1}{4!}=\dfrac{1}{4.3.2.1}< \dfrac{1}{3!}< \dfrac{1}{2!}< 1\)

.....

\(\)\(\dfrac{1}{2023!}=\dfrac{1}{2023.2022....2.1}< \dfrac{1}{2022!}< ...< \dfrac{1}{2!}< 1\)

\(\Rightarrow\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}< 1\)

Bạn xem lại đề 2, phần mẫu của N

@Nguyễn Đức Trí: Đề bài nó như vậy mà

\(\Leftrightarrow\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{x\left(x+3\right)}=\dfrac{375}{376}\)

\(\Leftrightarrow1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{375}{376}\)

\(\Leftrightarrow1-\dfrac{1}{x+3}=\dfrac{375}{376}\)

\(\Leftrightarrow\dfrac{1}{x+3}=1-\dfrac{375}{376}=\dfrac{1}{376}\)

\(\Rightarrow x+3=376\)

\(\Rightarrow x=373\)

Lời giải:

$\frac{5}{x}-\frac{y}{3}=\frac{1}{6}$

$\Rightarrow \frac{15-xy}{3x}=\frac{1}{6}$

$\Rightarrow \frac{2(15-xy)}{6x}=\frac{x}{6x}$

$\Rightarrow 2(15-xy)=x$

$\Rightarrow 30=2xy+x$

$\Rightarrow 30=x(2y+1)$

$\Rightarrow x=\frac{30}{2y+1}$

Vì $x$ nguyên nên $\frac{30}{2y+1}$ nguyên

$\Rightarrow 2y+1$ là ước của $30$

Vì $2y+1$ lẻ nên $2y+1\in\left\{\pm 1; \pm 3; \pm 5; \pm 15\right\}$

$\Rightarrow y\in\left\{-1; 0; -2; 1; -3; 2; -8; 7\right\}$

Tương ứng với các giá trị $y$ trên ta có: $x\in\left\{-30; 30; -10; 10; -6; 6; -2;2\right\}$

\(\dfrac{1}{2}+\dfrac{-1}{3}+\dfrac{-2}{3}\le x< \dfrac{-3}{5}+\dfrac{1}{6}+\dfrac{-2}{5}+\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{1}{2}+\left(\dfrac{-1}{3}+\dfrac{-2}{3}\right)\le x< \left(\dfrac{-3}{5}+\dfrac{-2}{5}\right)+\left(\dfrac{1}{6}+\dfrac{3}{2}\right)\)

\(\Leftrightarrow\dfrac{1}{2}+\left(-1\right)\le x< -1+\dfrac{5}{3}\)

\(\Leftrightarrow\dfrac{-1}{2}\le x< \dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{-3}{6}\le x< \dfrac{4}{6}\)

\(\Leftrightarrow x\in\left\{-3;-2;-1;0;1;2;3\right\}\)

⇔x∈{−3;−2;−1;0;1;2;3}

Câu 2

(xⁿ)^m=x^m.n

2:

(xⁿ)^m = x^{n . m}

Câu 1

Ta có: \(\dfrac{a}{3}=\dfrac{b}{2}=\dfrac{2c}{10}\) và a-b+2c=77

\(\dfrac{a-b+2c}{3-2+10}=\dfrac{77}{11}=7\)

\(\dfrac{a}{3}=7\) ⇒ a=21

\(\dfrac{b}{2}=7\) ⇒ b=14

\(\dfrac{c}{5}=7\) ⇒ c=35