Điều kiện về x,y và yêu cầu đề bài là gì, bạn nên ghi rõ ra nhé.

\(-\dfrac{x^2}{2}+\dfrac{7}{2}x^2+x\\ =x^2\left(-\dfrac{1}{2}+\dfrac{7}{2}\right)+x\\ =3x^2+x\)

Thì bạn cứ làm thôi, đề bà đó !!!

A tăng 15% và B tăng 20% thì được tổng là 1160

=>1,15A+1,2B=1160

\(1,15A+1,15B=1,15\left(A+B\right)=1,15\cdot1000=1150\)

=>\(1,15A+1,2B-1,15A-1,15B=1160-1150\)

=>0,05B=10

=>B=10:0,05=200

=>A=1000-200=800

=>A=1000-200=800

\(\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)-28=0\)

=>\(\left(x^2+5x-6\right)\left(x^2+5x+6\right)-28=0\)

=>\(\left(x^2+5x\right)^2-36-28=0\)

=>\(\left(x^2+5x-8\right)\left(x^2+5x+8\right)=0\)

mà \(x^2+5x+8=x^2+5x+\dfrac{25}{4}+\dfrac{7}{4}=\left(x+\dfrac{5}{2}\right)^2+\dfrac{7}{4}>0\forall x\)

nên \(x^2+5x-8=0\)

\(\Delta=5^2-4\cdot1\cdot\left(-8\right)=25+32=57>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left[{}\begin{matrix}x=\dfrac{-5-\sqrt{57}}{2}\\x=\dfrac{-5+\sqrt{57}}{2}\end{matrix}\right.\)

\(\dfrac{3}{5}\cdot x^2y^5\cdot x^3y^2\cdot\dfrac{-2}{3}\)

\(=\dfrac{3}{5}\cdot\dfrac{-2}{3}\cdot x^2\cdot x^3\cdot y^5\cdot y^2\)

\(=-\dfrac{2}{5}x^5y^7\)

\(\dfrac{3}{5}x^2y^5x^3y^2\cdot\dfrac{-2}{3}\\ =\left(\dfrac{3}{5}\cdot\dfrac{-2}{3}\right)\cdot\left(x^2\cdot x^3\right)\cdot\left(y^5\cdot y^2\right)\\ =-\dfrac{2}{5}x^5y^7\)

\(A=\dfrac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^{10}\cdot6^{19}-7\cdot2^{29}\cdot27^6}\)

\(=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{5\cdot2^{29}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)

\(=\dfrac{2^{29}\cdot3^{18}\left(5\cdot2-3^2\right)}{2^{29}\cdot3^{18}\cdot\left(5\cdot3-7\right)}=\dfrac{10-9}{15-7}=\dfrac{1}{8}\)

\(B=\dfrac{1}{2}\cdot\left(1+\dfrac{1}{1\cdot3}\right)\cdot\left(1+\dfrac{1}{2\cdot4}\right)\cdot...\cdot\left(1+\dfrac{1}{2022\cdot2024}\right)\)

\(=\dfrac{1}{2}\cdot\left(1+\dfrac{1}{2^2-1}\right)\left(1+\dfrac{1}{3^2-1}\right)\cdot...\cdot\left(1+\dfrac{1}{2023^2-1}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2^2}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2}{\left(3-1\right)\left(3+1\right)}\cdot...\cdot\dfrac{2023^2}{\left(2023-1\right)\left(2023+1\right)}\)

\(=\dfrac{1}{2}\cdot\dfrac{2\cdot3\cdot4\cdot...\cdot2023}{1\cdot2\cdot3\cdot...\cdot2022}\cdot\dfrac{2\cdot3\cdot...\cdot2023}{3\cdot4\cdot...\cdot2024}\)

\(=\dfrac{1}{2}\cdot\dfrac{2023}{1}\cdot\dfrac{2}{2024}=\dfrac{1}{2024}\cdot\dfrac{2023}{1}=\dfrac{2023}{2024}\)

\(A=\dfrac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^{10}.6^{19}-7.2^{29}.27^6}\)

\(=\dfrac{5.\left(2^2\right)^{15}.\left(3^2\right)^9-2^2.3^{20}.\left(2^3\right)^9}{5.2^{10}.\left(2.3\right)^{19}-7.2^{29}.\left(3^3\right)^6}\)

\(=\dfrac{5.2^{30}.3^{18}-2^2.3^{20}.2^{27}}{5.2^{10}.2^{19}.3^{19}-7.2^{29}.3^{18}}\\ =\dfrac{5.2^{30}.3^{18}-2^{29}.3^{20}}{5.2^{29}.3^{19}-7.2^{29}.3^{18}}\\ =\dfrac{2^{29}.3^{18}\left(5.2-3^2\right)}{2^{29}.3^{18}.\left(5.3-7\right)}\\ =\dfrac{10-9}{15-7}=\dfrac{1}{8}\)

\(---\)

\(B=\dfrac{1}{2}.\left(1+\dfrac{1}{1.3}\right).\left(1+\dfrac{1}{2.4}\right).\left(1+\dfrac{1}{3.5}\right)......\left(1+\dfrac{1}{2022.2024}\right)\\ =\dfrac{1}{2}.\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}...\dfrac{2023^2}{2022.2024}\\ =\dfrac{1}{2}.\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}...\dfrac{2023^2}{2022.2024}\\ =\dfrac{1}{2}.\dfrac{\left(2.3.4...2023\right)^2}{\left(1.2.3...2022\right)\left(3.4.5...2024\right)}\\ =\dfrac{1}{2}.\dfrac{2023.2}{2024}=\dfrac{2023}{2024}\)

#$\mathtt{Toru}$

Ta có:

\(100=2^2.5^2\)

\(200=2^3.5^2\)

\(120=2^3.3.5\)

\(BCNN\left(100;200;120\right)=2^3.3.5^2=8.3.25=600\)

Ta có:

100 = 22.52

200 = 23 . 5

120 = 23.3.5

=> BCNN(100; 200; 120) = 2^2 . 2^3. 3 . 5^2 = 2400

Chu vi hình tròn là:

  23,6 x 2 x 3,14 = 148.208 (cm)

Diện tích hình tròn là:

  23,6 x 23,6  x 3.14 = 1748.8544(cm2)