CT
0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
16 tháng 1 2021
a3 + b3 + c3 = 3abc
⇒ a3 + b3 + c3 - 3abc = 0
⇒ ( a3 + b3 ) + c3 - 3abc = 0
⇒ ( a + b )3 - 3ab( a + b ) + c3 - 3abc = 0
⇒ [ ( a + b )3 + c3 ] - [ 3ab( a + b ) + 3abc ] = 0
⇒ ( a + b + c )[ ( a + b )2 - ( a + b ).c + c2 ] - 3ab( a + b + c ) = 0
⇒ ( a + b + c )( a2 + b2 + c2 - ab - bc - ac ) = 0
Vì a + b + c ≠ 0
⇒ a2 + b2 + c2 - ab - bc - ac = 0
⇒ 2( a2 + b2 + c2 - ab - bc - ac ) = 0
⇒ 2a2 + 2b2 + 2c2 - 2ab - 2bc - 2ac = 0
⇒ ( a2 - 2ab + b2 ) + ( b2 - 2bc + c2 ) + ( a2 - 2ac + c2 ) = 0
⇒ ( a - b )2 + ( b - c )2 + ( a - c )2 = 0
Vì \(\hept{\begin{cases}\left(a-b\right)^2\\\left(b-c\right)^2\\\left(a-c\right)^2\end{cases}}\ge0\forall a,b,c\)⇒ ( a - b )2 + ( b - c )2 + ( a - c )2 ≥ 0 ∀ a,b,c
Dấu "=" xảy ra khi a = b = c
Khi đó \(N=\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{a^2+a^2+a^2}{\left(a+a+a\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)
NN
17 tháng 1 2021
Từ \(a^3+b^3+c^3=3abc\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)^3-3\left(a+b\right).c\left(a+b+c\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b+c\right)^2-3\left(a+b\right)c-3ab\right]=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2+2ab+2bc+2ca-3ab-3bc-3ca\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(2a^2+2b^2+2c^2-2ab-2bc-2ca\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)\right]=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
Vì \(a+b+c\ne0\)\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Vì \(\left(a-b\right)^2\ge0\), \(\left(b-c\right)^2\ge0\), \(\left(c-a\right)^2\ge0\)\(\forall a,b,c\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)\(\forall a,b,c\)
Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Leftrightarrow a=b=c\)
Thay \(a=b=c\)vào N ta có: \(N=\frac{3a^2}{\left(3a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)
Vậy \(N=\frac{1}{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 9 2020
\(\left(x+y\right)^3=x^3+3x^2y+3xy^2+y^3=\left(x^3-6x^2y+9xy^2\right)+\left(y^3-6xy^2+9x^2y\right)\)
\(=x\left(x^2-6xy+9y^2\right)+y\left(y^2-6xy+9x^2\right)=x\left(x-3y\right)^2+y\left(y-3x\right)^2\)
b/
\(\left(a+b\right)^3+\left(a-b\right)^3=a^3+3a^2b+3ab^2+b^3+a^3-3a^2b+3ab^2-b^3\)
\(=2a^3+6ab^2=2a\left(a^2+3b^2\right)\)
c/
\(\left(a+b\right)^3-\left(a-b\right)^3=a^3+3a^2b+3ab^2+b^3-\left(a^3-3a^2b+3ab^2-b^3\right)\)
\(=6a^2b+2b^3=2b\left(b^2+3a^2\right)\)
d/
\(a^3+b^3=a^3+3a^2b+3ab^2+b^3-\left(3a^2b+3ab^2\right)\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)\)
e/
\(a^3-b^3=a^3-3a^2b+3ab^2-b^3+3a^2b-3ab^2\)
\(=\left(a-b\right)^3+3ab\left(a-b\right)\)
MT
3 tháng 10 2015
a3-3ab2=2 và b3-3a2b=-11
=>(a3-3ab2)2=4 và (b3-3a2b)=121
=>a6-6a4b2+9a2b4=4 và b6-6a2b4+9a4b2=121
=>a6-6a4b2+9a2b4+b6-6a2b4+9a4b2=4+121
=>a6+3a4b2+3a2b4+b6=125
=>(a2+b2)3=125
=>a2+b2=5
24 tháng 10 2019
Sửa đề: Cho \(a^2+b^2+c^2=m\)
Tính: \(A=\left(2a+2b-c\right)^2+\left(2b+2c-a\right)^2+\left(2c+2a-b\right)^2\)
Giải:
Ta có: \(\left(x+y-z\right)^2=\left(x+y\right)^2-2\left(x+y\right).z+z^2=x^2+y^2+z^2+2xy-2xz-2yz\)
Ứng dụng vào bài trên:
\(A=\left[\left(2a\right)^2+\left(2b\right)^2+c^2+2\left(2a\right)\left(2b\right)-2\left(2a\right)c-2\left(2b\right)c\right]\)
\(+\left[\left(2b\right)^2+\left(2c\right)^2+a^2+2\left(2b\right)\left(2c\right)-2\left(2b\right)a-2\left(2c\right)a\right]\)
\(+\left[\left(2c\right)^2+\left(2a\right)^2+b^2+2\left(2c\right)\left(2a\right)-2\left(2c\right)b-2\left(2a\right)b\right]\)
\(=4a^2+4b^2+c^2+8ab-4ac-4bc\)
\(+4b^2+4c^2+a^2+8bc-4ba-4ca\)
\(+4c^2+4a^2+b^2+8ca-4cb-4ab\)
\(=9a^2+9b^2+9c^2=9\left(a^2+b^2+c^2\right)\)
\(=9m\).
26 tháng 5 2022
a: Đặt \(a^2+b^2=x\)
Ta có: \(M=\left(a^2+b^2+2\right)^3-\left(a^2+b^2-2\right)^3-12\left(a^2+b^2\right)^2\)
\(=\left(x+2\right)^3-\left(x-2\right)^3-12x^2\)
\(=x^3+6x^2+12x+8-\left(x^3-6x^2+12x-8\right)-12x^2\)
\(=x^3-6x^2+12x+8-x^3+6x^2-12x+8\)
\(=8\)
b: \(N=a^3+b^3+3ab\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\)
\(=1^3-3ab+3ab=1\)
OP
16 tháng 8 2016
\(M=a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)\)
\(=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left(a^2+2ab+b^2-2ab\right)+6a^2b^2\)
\(=\left(a^2+2ab+b^2-3ab\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2\)
\(=\left(a+b\right)^2-3ab+3ab\times\left(-2ab\right)+6a^2b^2\)
\(=-3ab-6a^2b^2+6a^2b^2\)
= - 3ab
6 tháng 12 2017
Câu 1:
Theo bài ra ta có:
\(a^{12}+b^{12}=a^{12}+a^{11}b-a^{11}b-ab^{11}+ab^{11}+b^{12}\)
\(=a^{11}\left(a+b\right)-ab\left(a^{10}+b^{10}\right)+b^{11}\left(a+b\right)\)
\(=\left(a+b\right)\left(a^{11}+b^{11}\right)-ab\left(a^{10}+b^{10}\right)\)
\(=\left(a+b\right)\left(a^{12}+b^{12}\right)-ab\left(a^{12}+b^{12}\right)\)(gt cho rồi nhé)
\(=\left(a^{12}+b^{12}\right)\left(a+b-ab\right)\)
\(\Rightarrow a+b-ab=1\)
\(\Leftrightarrow a+b-ab-1=0\)
\(\Leftrightarrow a\left(1-b\right)-\left(1-b\right)=0\)
\(\Leftrightarrow\left(1-b\right)\left(a-1\right)=0\)
\(\)\(\Leftrightarrow\left[{}\begin{matrix}b=1\\a=1\end{matrix}\right.\)
=> a^20 + b^20 = 2
:)) đừng ném đá nhá
