Question

Ludwig van Beethoven was a German composer and pianist. He was born on December 17th 1770 in Bonn, but he (1).................................. to Vienna, Austria in 1792, and lived there until his death in 1827. Beethoven showed his (2).............................. talents at an early age and was taught by his father Johann van Beethoven. Beethoven is regarded to as (3)................................ of the "three Bs" (along with Bach and Brahms). His best-known (4).................................. include 9 symphonies, 5 concertos for piano, 32 piano sonatas, and 16 string quarters. He also (5)............................... chamber music, choral works (including the celebrated Missa solemis), and songs. He was also a key figure in the transition from the 18th century musicial classicism to 19th century romanticism.

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Answer 1

Ludwig van Beethoven was a German composer and pianist. He was born on December 17th 1770 in Bonn, but he (1)..................moved................ to Vienna, Austria in 1792, and lived there until his death in 1827. Beethoven showed his (2)................innate.............. talents at an early age and was taught by his father Johann van Beethoven. Beethoven is regarded to as (3)..............one.................. of the "three Bs" (along with Bach and Brahms). His best-known (4)...................products............... include 9 symphonies, 5 concertos for piano, 32 piano sonatas, and 16 string quarters. He also (5)...............has................ chamber music, choral works (including the celebrated Missa solemis), and songs. He was also a key figure in the transition from the 18th century musicial classicism to 19th century romanticism.

Answer 2

a) Phương trình đường tròn là: $(x+3)^2+(y-4)^2=81$
b) Bán kính đường tròn là: $R=I M=\sqrt{(4-5)^2+(-1+2)^2}=\sqrt{2}$
Phương trình đường tròn là: $(x-5)^2+(y+2)^2=2$
c) Bán kính đường tròn là: $R=\frac{|5.1-12 \cdot(-1)-1|}{\sqrt{5^2+(-12)^2}}=\frac{16}{13}$
Phương trình đường tròn là: $(x-1)^2+(y+1)^2=\left(\frac{16}{13}\right)^2$
d) Gọi $I(a ; b)$ là trung điểm AB. Vậy tọa độ điểm I là: $I(1 ; 1)$
Bán kính đường tròn là: $R=I A=\sqrt{(3-1)^2+(-4-1)^2}=\sqrt{29}$
Phương trình đường tròn là: $(x-1)^2+(y-1)^2=29$
e) Giả sử tâm đường tròn là điểm $I(a ; b)$. Ta có: $I A=I B=I C \Leftrightarrow I A^2=I B^2=I C^2$
Vì $I A^2=I B^2, I B^2=I C^2$ nên: $\left\{\begin{array}{l}(1-a)^2+(1-b)^2=(3-a)^2+(1-b)^2 \\ (3-a)^2+(1-b)^2=(0-a)^2+(4-b)^2\end{array} \Leftrightarrow\left\{\begin{array}{l}a=2 \\ b=3\end{array}\right.\right.$ b
Vậy $I(2 ; 3)$ và $R=I A=\sqrt{(-1)^2+(-2)^2}=\sqrt{5}$
Vậy phương trình đường tròn đi qua 3 điểm $\mathrm{A}, \mathrm{B}, \mathrm{C}$ là: $(x-2)^2+(y-3)^2=5$