ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne9\end{matrix}\right.\)

Để A là số nguyên thì \(\sqrt{x}+2⋮\sqrt{x}-3\)

=>\(\sqrt{x}-3+5⋮\sqrt{x}-3\)

=>\(5⋮\sqrt{x}-3\)

=>\(\sqrt{x}-3\in\left\{1;-1;5;-5\right\}\)

=>\(\sqrt{x}\in\left\{4;2;8\right\}\)

=>\(x\in\left\{16;4;64\right\}\)

Thể tích phần bể chứa nước ban đầu là:

\(80\cdot50\cdot35=140000\left(cm^3\right)\)

Thể tích phần bể chứa nước lúc này sau khi thêm hòn đá là:

\(140000+20000=160000\left(cm^3\right)\)

Mực nước trong bể lúc này cao là:

\(160000:80:50=40\left(cm\right)\)

Thể tích ban đầu: 80 x 50 x 35 = 140.000 cm3

Sau khi thêm hòn đá: 140.000+20.000 =  160.000 cm3

=> Chiều cao mực nước = 160.000 / (80x50) = 40 cm

Hình đâu em?

Độ dài cạnh huyền là:

\(\sqrt{3^2+7^2}=\sqrt{9+49}=\sqrt{58}\left(cm\right)\)

Bình phương cạnh huyền là:

 3² + 7² =  58(cm²)

Cạnh huyền là:  \(\sqrt{58}\) m

1: \(\left(x^2+2xy-3\right)\left(-xy^2\right)\)

\(=-xy^2\cdot x^2-xy^2\cdot2xy+3\cdot xy^2\)

\(=-x^3y^2-2x^2y^3+3xy^2\)

2: \(3x\left(x+2\right)-3x^2-12=0\)

=>\(3x^2+6x-3x^2-12=0\)

=>6x-12=0

=>6x=12

=>x=2

3: \(\left(2x^3-\dfrac{9}{2}x^2+\dfrac{1}{xy}\right)\cdot x^2y^3\)

\(=2x^3\cdot x^2y^3-\dfrac{9}{2}x^2\cdot x^2y^3+\dfrac{x^2y^3}{xy}\)

\(=2x^5y^3-\dfrac{9}{2}x^4y^3+xy^2\)

2; 3\(x\)(\(x+2\)) - 3\(x^2\) - 12 = 0

    3\(x^2\) + 6\(x\) - 3\(x^2\) - 12 = 0

  (3\(x^2\) - 3\(x^2\)) + 6\(x\) - 12 = 0

   0 + 6\(x\) - 12 = 0

          6\(x\)  = 12

           \(x\) = 12 : 6

           \(x=2\) 

Vậy \(x=2\)

Lấy mẫu số, chứ không phải lấy tử số em nhé!

tất cả các số âm hay dương gì đều lấy mẫu số ạ?

a: \(-\dfrac{15}{19}=-1+\dfrac{4}{19}\)

\(-\dfrac{37}{41}=-1+\dfrac{4}{41}\)

\(-\dfrac{5}{9}=-1+\dfrac{4}{9}\)

\(\dfrac{23}{-27}=-\dfrac{23}{27}=-1+\dfrac{4}{27}\)

\(-\dfrac{7}{11}=-1+\dfrac{4}{11}\)

mà \(\dfrac{4}{41}< \dfrac{4}{27}< \dfrac{4}{19}< \dfrac{4}{11}< \dfrac{4}{9}\)

nên \(-\dfrac{37}{41}< -\dfrac{23}{27}< -\dfrac{15}{19}< -\dfrac{7}{11}< -\dfrac{5}{9}\)

mà \(-\dfrac{37}{41}< -\dfrac{76}{89}< -\dfrac{23}{27}\)

nên  \(-\dfrac{37}{41}< -\dfrac{76}{89}< -\dfrac{23}{27}< -\dfrac{15}{19}< -\dfrac{7}{11}< -\dfrac{5}{9}\)

Sửa đề: \(\dfrac{1}{5}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}\)

Đặt \(A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)

\(\dfrac{1}{5}-\dfrac{1}{6}< \dfrac{1}{5\cdot6}< \dfrac{1}{5^2}< \dfrac{1}{4\cdot5}=\dfrac{1}{4}-\dfrac{1}{5}\)

\(\dfrac{1}{6}-\dfrac{1}{7}< \dfrac{1}{6\cdot7}< \dfrac{1}{6^2}< \dfrac{1}{5\cdot6}=\dfrac{1}{5}-\dfrac{1}{6}\)

...

\(\dfrac{1}{100}-\dfrac{1}{101}=\dfrac{1}{100\cdot101}< \dfrac{1}{100^2}< \dfrac{1}{100\cdot99}=\dfrac{1}{99}-\dfrac{1}{100}\)

Do đó: \(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{101}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

=>\(\dfrac{1}{5}-\dfrac{1}{101}< A< \dfrac{1}{4}-\dfrac{1}{100}\)

=>\(\dfrac{1}{5}< A< \dfrac{1}{4}\)

A = \(\dfrac{1}{5^2}\) + \(\dfrac{1}{6^2}\) + \(\dfrac{1}{7^2}\) + ... + \(\dfrac{1}{100^2}\)

\(\dfrac{1}{5.6}\) < \(\dfrac{1}{5^2}\) < \(\dfrac{1}{4.5}\)

\(\dfrac{1}{6.7}\) < \(\dfrac{1}{6^2}\) < \(\dfrac{1}{5.6}\)

\(\dfrac{1}{7.8}\) < \(\dfrac{1}{7^2}\) < \(\dfrac{1}{6.7}\)

......................

\(\dfrac{1}{100.101}\) < \(\dfrac{1}{100^2}\) < \(\dfrac{1}{99.100}\)

Cộng vế với vế ta có:

\(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\) + ... + \(\dfrac{1}{100.101}\)< \(\dfrac{1}{5^2}\)+\(\dfrac{1}{6^2}\)+...+\(\dfrac{1}{100^2}\)<\(\dfrac{1}{4.5}\)+\(\dfrac{1}{5.6}\)+...+\(\dfrac{1}{99.100}\)

\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{100}\)-\(\dfrac{1}{101}\) < \(\dfrac{1}{5^2}\)+\(\dfrac{1}{6^2}\)+...+\(\dfrac{1}{100^2}\)< \(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+...+\(\dfrac{1}{99}\)-\(\dfrac{1}{100}\)

\(\dfrac{1}{5}\) - \(\dfrac{1}{101}\) < \(\dfrac{1}{5^2}\) + \(\dfrac{1}{6^2}\)+...+\(\dfrac{1}{100^2}\) < \(\dfrac{1}{4}\) - \(\dfrac{1}{100}\)

\(\dfrac{6}{30}\) - \(\dfrac{1}{101}\) < \(\dfrac{1}{5^2}\) + \(\dfrac{1}{6^2}\)+ .... + \(\dfrac{1}{100^2}\) < \(\dfrac{1}{4}\) - \(\dfrac{1}{100}\) < \(\dfrac{1}{4}\)

\(\dfrac{5}{30}\) +( \(\dfrac{1}{30}\) - \(\dfrac{1}{101}\)) < \(\dfrac{1}{5^2}\) + \(\dfrac{1}{6^2}\) + ... + \(\dfrac{1}{100^2}\) < \(\dfrac{1}{4}\)

\(\dfrac{1}{6}\) + (\(\dfrac{1}{30}\) - \(\dfrac{1}{101}\)) < \(\dfrac{1}{5^2}\) + \(\dfrac{1}{6^2}\) + ... + \(\dfrac{1}{100^2}\) < \(\dfrac{1}{4}\)

Vì \(\dfrac{1}{30}\) > \(\dfrac{1}{101}\) ⇒  \(\dfrac{1}{30}\) - \(\dfrac{1}{101}\) > 0 ⇒ \(\dfrac{1}{6}\) + (\(\dfrac{1}{30}\) - \(\dfrac{1}{101}\)) > \(\dfrac{1}{6}\)

Vậy  \(\dfrac{1}{6}\) < \(\dfrac{1}{5^2}\) + \(\dfrac{1}{6^2}\) + ... + \(\dfrac{1}{100^2}\) < \(\dfrac{1}{4}\) (đpcm)

\(\left|\dfrac{4}{3}x-\dfrac{1}{4}\right|>=0\forall x\)

=>\(\left|\dfrac{4}{3}x-\dfrac{1}{4}\right|-\dfrac{2}{11}>=-\dfrac{2}{11}\forall x\)

Dấu '=' xảy ra khi \(\dfrac{4}{3}x-\dfrac{1}{4}=0\)

=>\(\dfrac{4}{3}x=\dfrac{1}{4}\)

=>\(x=\dfrac{1}{4}:\dfrac{4}{3}=\dfrac{3}{16}\)

  A =  |\(\dfrac{4}{3}\)\(x\) - \(\dfrac{1}{4}\)| - \(\dfrac{2}{11}\)

Vì |\(\dfrac{4}{3}\)\(x\) - \(\dfrac{1}{4}\)| ≥ 0 ∀ \(x\)

   |\(\dfrac{4}{3}x\) - \(\dfrac{1}{4}\)| - \(\dfrac{2}{11}\) ≥ - \(\dfrac{2}{11}\) dấu bằng xảy ra khi : \(\dfrac{4}{3}x\) - \(\dfrac{1}{4}\) = 0 

⇒ \(\dfrac{4}{3}\)\(x\) = \(\dfrac{1}{4}\) ⇒ \(x\) = \(\dfrac{1}{4}\) : \(\dfrac{4}{3}\) ⇒ \(x\) = \(\dfrac{3}{16}\)

Vậy giá  trị nhỏ nhất của biểu thức là - \(\dfrac{2}{11}\) khi \(x=\dfrac{3}{16}\)