Câu 1: C

Câu 2: D

Câu 3: A

Câu 4: B

Thank Nguyễn Lê Phương Thịnh^^

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Sửa đề: \(D=\frac{1}{10\cdot11}+\frac{1}{11\cdot12}+\cdots+\frac{1}{99\cdot100}\)

\(=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\cdots+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{10}-\frac{1}{100}=\frac{10}{100}-\frac{1}{100}=\frac{9}{100}\)

=>D là số hữu tỉ

\(C=0,3\cdot12,8+0,3\cdot7,2\)

\(=0,3\cdot\left(12,8+7,2\right)\)

\(=0,3\cdot20=6\)

=>C là số hữu tỉ

\(E=\frac{\frac{4}{11}+\frac{4}{121}-\frac{4}{12321}}{\frac{9}{11}+\frac{9}{121}-\frac{9}{12321}}=\frac{4\left(\frac{1}{11}+\frac{1}{121}-\frac{1}{12321}\right)}{9\left(\frac{1}{11}+\frac{1}{121}-\frac{1}{12321}\right)}=\frac49\)

=>E là số hữu tỉ

Ta có: \(B=\frac{4}{15}\cdot\frac{-25}{24}=\frac{4\cdot\left(-25\right)}{15\cdot24}=\frac{-100}{360}=\frac{-5}{18}\)

=>B là số hữu tỉ

\(A=\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}\)NHÂN CẢ TỬ VÀ MẪU CỦA TỪNG P/S VỚI 2 TA ĐƯỢC:

\(A=\frac{2}{8}+\frac{2}{24}+\frac{2}{48}+\frac{2}{80}+\frac{2}{120}+\frac{2}{168}\)

\(A=\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{12.14}\)

\(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{12}-\frac{1}{14}\)

\(A=\frac{1}{2}-\frac{1}{14}\)

\(A=\frac{3}{7}\)

Giá trị của biểu thứu A là 3/7

\(\dfrac{12}{x}=\dfrac{-9}{15}\)
\(\dfrac{12}{x}=\dfrac{-3}{5}\)
\(x=-\dfrac{12\times5}{3}\)
\(x=-20\)

`12/x=-9/15`

`=>12*15=-9*x`

`=>-9*x=180`

`=>x=180:(-9)`

`=>x=-20`

`=>B`

Ta có: \(a^2\left(a+1\right)-b^2\left(b-1\right)-11ab+2024\) (1)

Lại có: \(a-b=\sqrt{29+12\sqrt{5}}-2\sqrt{5}\) 

\(=\sqrt{\left(2\sqrt{5}\right)^2+2\cdot2\sqrt{5}\cdot3+3^2}-2\sqrt{5}\)

\(=\sqrt{\left(2\sqrt{5}+3\right)^2}-2\sqrt{5}\)

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

\(=3\)

\(\Rightarrow a=b+3\)

Thay \(a=b+3\) vào (1), ta được:

\(\left(b+3\right)^2\left(b+3+1\right)-b^2\left(b-1\right)-11\left(b+3\right)b+2024\)

\(=\left(b^2+6b+9\right)\left(b+4\right)-b^3+b^2-11b^2-33b+2024\)

\(=b\left(b^2+6b+9\right)+4\left(b^2+6b+9\right)-b^3-10b^2-33b+2024\)

\(=b^3+6b^2+9b+4b^2+24b+36-b^3-10b^2-33b+2024\)

\(=\left(b^3-b^3\right)+\left(6b^2+4b^2-10b^2\right)+\left(9b+24b-33b\right)+\left(2024+36\right)\)

\(=2060\)

$\Rightarrow$ Chọn đáp án $C$.

Ta có : \(a-b=\sqrt{29+12\sqrt{5}}-2\sqrt{5}\)

\(\Rightarrow a-b=\sqrt{20+12\sqrt{5}+9}-2\sqrt{5}\)

\(\Rightarrow a-b=\sqrt{\left(2\sqrt{5}+3\right)^2}-2\sqrt{5}\)

\(\Rightarrow a-b=2\sqrt{5}+3-2\sqrt{5}\)

\(\Rightarrow a-b=3\)

Xét biểu thức : \(a^2\left(a+1\right)-b^2\left(b-1\right)-11ab+2024\)

\(=a^3+a^2-b^3+b^2-11ab+2024\)

\(=a^3-b^3+a^2+b^2-2ab-9ab+2024\)

\(=a^3-b^3-9ab+a^2-2ab+b^2+2024\)

\(=a^3-3ab\left(a-b\right)-b^3+\left(a-b\right)^2+2024\) vì \(a-b=3\)

\(=\left(a-b\right)^3+\left(a-b\right)^2+2024\)

\(=3^3+3^2+2024\)

\(=2060\)

\(\Rightarrow C\)

A = 1 + 2 -  3 - 4 + 5 + 6 - 7 - 8 + 9 + 10 - 11 - 12 + .........- 299 - 300 + 301 + 302

A = 1 + ( 2 - 3 - 4 + 5 ) + ( 6 - 7 -8 + 9 ) + ( 10 - 11 - 12 + 13 ) + ............- ( 298 - 299 - 300 + 301 ) + 302

A = 1 + 303 

A = 303

A = 1+ 2 - 3 - 4 - 5 + 6 - 7 -8 + 9 +......+ 298 - 299 - 300 + 301 + 302

A = 1 + ( 2 - 3 - 4 + 5 ) + ( 6 - 7 - 8 + 9 ) + ......+ ( 298 - 299 - 300 + 301 ) + 302

A = 1 + ( 2 + 5 - 3 - 4 ) + ( 6 + 9 - 7 - 8 ) + .......+ ( 298 + 301 - 299 - 300) + 302

A = 1 +          0            +        0            +  .......+               0                 +   302

A = 1 + 302

A =    303