Ta thấy \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}\ge\dfrac{x^2}{a^2+b^2+c^2}+\dfrac{y^2}{a^2+b^2+c^2}+\dfrac{z^2}{a^2+b^2+c^2}=\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}\).

Mà đẳng thức xảy ra nên ta phải có x = y = z = 0 (Do \(a^2,b^2,c^2>0\)).

Thay vào đẳng thức cần cm ta có đpcm.

Bạn muốn làm gì với đề bài này?

\(\frac{2021}{1\cdot2}+\frac{2021}{2\cdot3}+...+\frac{2021}{9\cdot10}=2021\cdot\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{9\cdot10}\right)\)

\(=2021\cdot\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)=2021\cdot\left(1-\frac{1}{10}\right)\)

\(=2021\cdot\frac{9}{10}=\frac{18189}{10}\)

Ta có : \(\frac{2021}{1.2}+\frac{2021}{2.3}+\frac{2021}{3.4}+...+\frac{2021}{9.10}=2021\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)

\(=2021\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=2021\left(1-\frac{1}{10}\right)=2021.\frac{9}{10}=1818,9\)

Ta có: \(b^2=ac\)

=>\(\frac{a}{b}=\frac{b}{c}\)

Đặt \(\frac{a}{b}=\frac{b}{c}=k\)

=>a=bk; b=ck

=>\(a=ck\cdot k=ck^2;b=ck\)

\(\frac{\left(a+b\right)^{2021}}{\left(b+c\right)^{2021}}=\frac{\left(ck^2+ck\right)^{2021}}{\left(ck+c\right)^{2021}}=\frac{\left\lbrack ck\left(k+1\right)\right\rbrack^{2021}}{\left\lbrack c\left(k+1\right)\right\rbrack^{2021}}=k^{2021}\)

\(\frac{a^{2021}+b^{2021}}{b^{2021}+c^{2021}}=\frac{\left(ck^2\right)^{2021}+\left(ck\right)^{2021}}{\left(ck\right)^{2021}+c^{2021}}\)

\(=\frac{c^{2021}\cdot k^{2021}\left(k^{2021}+1\right)}{c^{2021}\left(k^{2021}+1\right)}=k^{2021}\)

Do đó: \(\frac{\left(a+b\right)^{2021}}{\left(b+c\right)^{2021}}=\frac{a^{2021}+b^{2021}}{b^{2021}+c^{2021}}\)

Ta có: \(b^2=ac\)

=>\(\frac{a}{b}=\frac{b}{c}\)

Đặt \(\frac{a}{b}=\frac{b}{c}=k\)

=>a=bk; b=ck

=>\(a=ck\cdot k=ck^2;b=ck\)

\(\frac{\left(a+b\right)^{2021}}{\left(b+c\right)^{2021}}=\frac{\left(ck^2+ck\right)^{2021}}{\left(ck+c\right)^{2021}}=\frac{\left\lbrack ck\left(k+1\right)\right\rbrack^{2021}}{\left\lbrack c\left(k+1\right)\right\rbrack^{2021}}=k^{2021}\)

\(\frac{a^{2021}+b^{2021}}{b^{2021}+c^{2021}}=\frac{\left(ck^2\right)^{2021}+\left(ck\right)^{2021}}{\left(ck\right)^{2021}+c^{2021}}\)

\(=\frac{c^{2021}\cdot k^{2021}\left(k^{2021}+1\right)}{c^{2021}\left(k^{2021}+1\right)}=k^{2021}\)

Do đó: \(\frac{\left(a+b\right)^{2021}}{\left(b+c\right)^{2021}}=\frac{a^{2021}+b^{2021}}{b^{2021}+c^{2021}}\)

Sửa đề: \(2021^3+2021^4+2021^5+2021^6\)

Ta có: \(2021^3+2021^4+2021^5+2021^6\)

\(=2021^3\left(1+2021\right)+2021^5\left(1+2021\right)\)

\(=2022\left(2021^3+2021^5\right)\) ⋮2022

Các bạn đặt câu hỏi về đề Toán lớp 4 đi

Cậu trả lời đi, sáng mai tớ phải nộp rồi. Nhanh nhé, tớ tìm cho

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x= 2002/3000

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