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?

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

yynbjkgyg

x= 2002/3000

ko bt đúng ko mong bn nhắc nhở

Xét khai triển:

\(2^{2021}=\left(1+1\right)^{2021}=C_{2021}^0+C_{2021}^1+...+C_{2021}^{2020}+C_{2021}^{2021}\) (1)

\(0=\left(1-1\right)^{2021}=C_{2021}^0-C_{2021}^1+C_{2021}^2+...+C_{2021}^{2020}-C_{2021}^{2021}\) (2)

Trừ vế cho vế (1) và (2):

\(2^{2021}=2.C_{2021}^1+2.C_{2021}^3+...+2C_{2021}^{2021}\)

\(\Rightarrow C_{2021}^1+C_{2021}^3+...+C_{2021}^{2019}+C_{2021}^{2021}=\dfrac{2^{2021}}{2}=2^{2020}\)

\(\Rightarrow C_{2021}^1+C_{2021}^3+...+C_{2021}^{2019}+1=2^{2020}\)

\(\Rightarrow C_{2021}^1+C_{2021}^3+...+C_{2021}^{2019}=2^{2020}-1\)