Lời giải:

$2023xy+2024yz+4047xz=2023xy+2024y(-x-y)+4047x(-x-y)$

$=-2024y^2-4047x^2-4048xy$

$=-[4047x^2+2024y^2+4048xy]$

$=-[2024(x^2+y^2+2xy)+2023x^2]=-[2024(x+y)^2+2023x^2]$

Vì $2024(x+y)^2+2023x^2\geq 0$ với mọi $x,y$

$\Rightarrow -[2024(x+y)^2+2023x^2]\leq 0$ với mọi $x,y$

Do đó nó không thể nhận giá trị dương.

a. 

\(=(\frac{-5}{12}+\frac{17}{12})+(\frac{4}{37}+\frac{-41}{37})=\frac{12}{12}+\frac{-37}{37}=1+(-1)=0\)

b.

\(=\frac{-20}{24}+\frac{9}{24}-\frac{1}{10}=\frac{-11}{24}+\frac{1}{10}=\frac{-43}{120}\)

c.

\(=\frac{-25}{27}-\frac{31}{42}+\frac{7}{27}+\frac{3}{42}\\ =\frac{-25}{27}+\frac{7}{27}+(\frac{-31}{42}+\frac{3}{42})\\ =\frac{-2}{3}+\frac{-2}{3}=\frac{-4}{3}\)

d.

\(=\frac{5}{36}+\frac{-28}{36}-\frac{5}{4}=\frac{5}{36}+\frac{-28}{36}-\frac{45}{36}\\ =\frac{-68}{36}=\frac{-17}{9}\)

e. 

\(=\frac{3}{5}+\frac{2}{5}: 2=\frac{3}{5}+\frac{1}{5}=\frac{4}{5}\)

f.

\(=\frac{3}{10}(\frac{-23}{7}+\frac{13}{7})=\frac{3}{10}.\frac{-10}{7}=\frac{-3}{7}\)

g.

\(=\frac{3}{2}+\frac{1}{4}+(-1)=\frac{7}{4}-1=\frac{3}{4}\)

h.

\(=\frac{2^{2023}}{(2^2)^{1011}}=\frac{2^{2023}}{2^{2022}}=2^{2023-2022}=2\)

i.

\(=4-4+(-8).\frac{5}{4}=0+(-10)=-10\)

Không gửi linh tinh ạ.

Để A có giá trị là một số nguyên thì:

\(\left(\sqrt{x}+1\right)⋮\left(\sqrt{x}-3\right)\)

\(\Leftrightarrow\left(\sqrt{x}-3\right)+4⋮\left(\sqrt{x}-3\right)\)

\(\Leftrightarrow4⋮\left(\sqrt{x}-3\right)\)

Vì \(x\in Z\) nên \(\left(\sqrt{x}-3\right)\inƯ\left(4\right)=\left\{\pm1,\pm2,\pm4\right\}\)

Ta có bảng sau:

Vậy ....

Ta có: \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-3\right)+4}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3}{\sqrt{x}-3}=\dfrac{4}{\sqrt{x}-3}=1+\dfrac{4}{\sqrt{x}-3}\)

Để A có giá trị là một số nguyên khi:

\(4⋮\sqrt{x}-3\) hay \(\sqrt{x}-3\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

Do đó:

\(\sqrt{x}-3=-1\Rightarrow\sqrt{x}=-1+3=2\Rightarrow x=4\)

\(\sqrt{x}-3=1\Rightarrow\sqrt{x}=1+3=4\Rightarrow x=16\)

\(\sqrt{x}-3=-2\Rightarrow\sqrt{x}=-2+3=1\Rightarrow x=1\)

\(\sqrt{x}-3=2\Rightarrow\sqrt{x}=2+3=5\Rightarrow x=25\)

\(\sqrt{x}-3=-4\Rightarrow\sqrt{x}=-4+3=-1\)  ( loại )

\(\sqrt{x}-3=4\Rightarrow\sqrt{x}=4+3=7\Rightarrow x=49\)

Vậy để A là một số nguyên khi \(x\in\left\{4;16;1;25;49\right\}\)

Lời giải:
$\frac{55-x}{1963}+\frac{50-x}{1968}+\frac{45-x}{1973}+\frac{40-x}{1978}+4=0$

$\frac{55-x}{1963}+1+\frac{50-x}{1968}+1+\frac{45-x}{1973}+1+\frac{40-x}{1978}+1=0$

$\frac{2018-x}{1963}+\frac{2018-x}{1968}+\frac{2018-x}{1973}+\frac{2018-x}{1978}=0$

$(2018-x)(\frac{1}{1963}+\frac{1}{1968}+\frac{1}{1973}+\frac{1}{1978})=0$

$\Rightarrow 2018-x=0$

$\Rightarrow x=2018$.

\(\dfrac{1}{3}.\sqrt{\dfrac{9}{25}}\) - (\(\dfrac{1}{3}\) + \(\dfrac{1}{2}\))²

= \(\dfrac{1}{3}\).\(\dfrac{3}{5}\) - (\(\dfrac{5}{6}\))²

= \(\dfrac{1}{5}\) - \(\dfrac{25}{36}\)

= - \(\dfrac{89}{180}\)