Bạn chưa đăng nhập. Vui lòng đăng nhập để hỏi bài

Những câu hỏi liên quan
nguyen hoang long
Xem chi tiết
Nhi Đào Quỳnh
Xem chi tiết
Pham Van Hung
17 tháng 11 2018 lúc 17:59

\(a^3+b^3=c\left(3ab-c^2\right)\Rightarrow a^3+b^3+c^3-3abc=0\)

\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

\(\Rightarrow\left(a+b+c\right)\left[2a^2+2b^2+2c^2-2ab-2bc-2ca\right]=0\)

\(\Rightarrow\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)

\(\Rightarrow\orbr{\begin{cases}a+b+c=0\left(loai\right)\\a=b=c\end{cases}}\)

Mà a + b + c = 3 nên a = b = c = 1

Khi đó \(A=672.\left(1+1+1\right)+2=672.3+2=2018\)

Trương Quân Bảo
Xem chi tiết
mã thị hằng
Xem chi tiết
Nguyễn Duyên
Xem chi tiết
Ngô Phúc Dương
Xem chi tiết
Luân Đào
4 tháng 1 2019 lúc 11:45

\(P=\left(b^2c+abc\right)\left(a^2b+abc\right)\left(c^2a+abc\right)\)

\(=bc\left(a+b\right)\cdot ab\left(c+a\right)\cdot ca\left(b+c\right)\)

\(=\left(abc\right)^2\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

Lại có:

\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc=0\)

\(\Leftrightarrow\left(a^2b+abc+a^2c\right)+\left(ab^2+b^2c+abc\right)+\left(bc^2+c^2a+abc\right)-abc=0\)

\(\Leftrightarrow a^2b+ca^2+ab^2+2abc+ac^2+b^2c+bc^2=0\)

\(\Leftrightarrow a^2\left(b+c\right)+a\left(b^2+2bc+c^2\right)+bc\left(b+c\right)=0\)

\(\Leftrightarrow a^2\left(b+c\right)+a\left(b+c\right)^2+bc\left(b+c\right)=0\)

\(\Leftrightarrow\left(b+c\right)\left(a^2+ab+ca+bc\right)=0\)

\(\Leftrightarrow\left(b+c\right)\left[a\left(a+b\right)+c\left(a+b\right)\right]=0\)

\(\Leftrightarrow\left(b+c\right)\left(a+b\right)\left(c+a\right)=0\)

\(\Rightarrow P=0\)

mã hằng
Xem chi tiết
Phi Hoàng
Xem chi tiết
You Are Mine
Xem chi tiết
Nhan Mạc Oa
5 tháng 11 2018 lúc 21:23

a) \(\dfrac{2a+3c}{2b+3d}\) = \(\dfrac{2a-3c}{2b-3d}\)

Từ \(\dfrac{a}{b}\) = \(\dfrac{c}{d}\) = k ( k \(\in\) Q, k \(\ne\) 0 )

=> \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

VP = \(\dfrac{2a+3c}{2b+3d}\) = \(\dfrac{2.b.k+3.d.k}{2b+3d}\) = \(\dfrac{k.\left(2b+3d\right)}{2b+3d}\) = k (1)

VT = \(\dfrac{2a-3c}{2b-3d}\) = \(\dfrac{2.b.k-3.d.k}{2b-3d}\) = \(\dfrac{k.\left(2b-3d\right)}{2b-3d}\) = k (2)

Từ (1) và (2) ta có: \(\dfrac{2a+3c}{2b+3d}\) = \(\dfrac{2a-3c}{2b-3d}\)

hay: (2a+3c).(3b-3d) = (2a-3c).(2b+3d)

Nguyễn Lê Phước Thịnh
17 tháng 11 2022 lúc 20:15

b: Đặt a/b=c/d=k

=>a=bk; c=dk

\(\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}=\dfrac{\left(bk+dk\right)^2}{\left(b+d\right)^2}=k^2\)

\(\dfrac{\left(a-c\right)^2}{\left(b-d\right)^2}=\dfrac{\left(bk-dk\right)^2}{\left(b-d\right)^2}=k^2\)

Do đó: \(\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}=\dfrac{\left(a-c\right)^2}{\left(b-d\right)^2}\)

c: \(\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{b^3k^3+b^3}{d^3k^3+d^3}=\dfrac{b^3}{d^3}\)

\(\dfrac{a^3-b^3}{c^3-d^3}=\dfrac{b^3k^3-b^3}{d^3k^3-d^3}=\dfrac{b^3}{d^3}\)

Do đó: \(\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{a^3-b^3}{c^3-d^3}\)

d: \(\dfrac{a^{2018}-b^{2018}}{a^{2018}+b^{2018}}=\dfrac{b^{2018}k^{2018}-b^{2018}}{b^{2018}k^{2018}+b^{2018}}=\dfrac{k^{2018}-1}{k^{2018}+1}\)

\(\dfrac{c^{2018}-d^{2018}}{c^{2018}+d^{2018}}=\dfrac{k^{2018}-1}{k^{2018}+1}\)

Do đó: \(\dfrac{a^{2018}-b^{2018}}{a^{2018}+b^{2018}}=\dfrac{c^{2018}-d^{2018}}{c^{2018}+d^{2018}}\)