Các bạn oi giúp mình tí, cmr:
(A+b+c)^2= a2+b2+c2+2ab+2bc+2ac
(A+b+c)^2+a^2+b^2+c^2=(a+b)^2+(b+c)^2+(c+a)^2
Những câu hỏi liên quan
(1) (a+b+c)2a2+b2+c2+2ab+2bc+2ac(a+b+c)2a2+b2+c2+2ab+2bc+2ac(2) (a+b−c)2a2+b2+c2+2ab−2bc−2ac(a+b−c)2a2+b2+c2+2ab−2bc−2ac(3) (a−b−c)2a2+b2+c2−2ab−2ac+2bc(a−b−c)2a2+b2+c2−2ab−2ac+2bc(4) a3+b3(a+b)3−3ab(a+b)a3+b3(a+b)3−3ab(a+b)(5) a3−b3(a−b)3+3ab(a−b)a3−b3(a−b)3+3ab(a−b)(6) (a+b+c)3a3+b3+c3+3(a+b)(b+c)(c+a)(a+b+c)3a3+b3+c3+3(a+b)(b+c)(c+a)(7) a3+b3+c3−3abc(a+b+c)(a2+b2+c2−ab−bc−ac)a3+b3+c3−3abc(a+b+c)(a2+b2+c2−ab−bc−ac) (8) (a−b)3+(b−c)3+(c−a)33(a−b)(b−c)(c−a)(a−b)3+(b−c)3+(c−a)33(a−b)(b−c)(c−a)(9)...
Đọc tiếp
(1) (a+b+c)2=a2+b2+c2+2ab+2bc+2ac(a+b+c)2=a2+b2+c2+2ab+2bc+2ac
(2) (a+b−c)2=a2+b2+c2+2ab−2bc−2ac(a+b−c)2=a2+b2+c2+2ab−2bc−2ac
(3) (a−b−c)2=a2+b2+c2−2ab−2ac+2bc(a−b−c)2=a2+b2+c2−2ab−2ac+2bc
(4) a3+b3=(a+b)3−3ab(a+b)a3+b3=(a+b)3−3ab(a+b)
(5) a3−b3=(a−b)3+3ab(a−b)a3−b3=(a−b)3+3ab(a−b)
(6) (a+b+c)3=a3+b3+c3+3(a+b)(b+c)(c+a)(a+b+c)3=a3+b3+c3+3(a+b)(b+c)(c+a)
(7) a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ac)a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ac)
(8) (a−b)3+(b−c)3+(c−a)3=3(a−b)(b−c)(c−a)(a−b)3+(b−c)3+(c−a)3=3(a−b)(b−c)(c−a)
(9) (a+b)(b+c)(c+a)−8abc=a(b−c)2+b(c−a)2+c(a−b)2(a+b)(b+c)(c+a)−8abc=a(b−c)2+b(c−a)2+c(a−b)2
(10) (a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)−abc(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)−abc
(11) ab2+bc2+ca2−a2b−b2c−c2a=(a−b)3+(b−c)3+(c−a)33ab2+bc2+ca2−a2b−b2c−c2a=(a−b)3+(b−c)3+(c−a)33
(12)ab3+bc3+ca3−a3b−b3c−c3a=(a+b+c)[(a−b)3+(b−c)3+(c−a)3]3ab3+bc3+ca3−a3b−b3c−c3a=(a+b+c)[(a−b)3+(b−c)3+(c−a)3]3
Chứng minh giùm mik hằng đẳng thức kia vs
a, cho a=+b+c =1; a,b,c dương
tìm GTNN: A= a/b2+1 + b/c2+1 + c/a2+1
b, cho a,b,c dương có tổng =2
tìm GTNN; B= a/ab+2c + b/bc+2a + c/ca+2b
c, cho a,b,c dương và a+b+c<1
tìm GTNN: C= 1/a2+2bc + 1/ b2+2ac + 1/c2+2ab
Cho a, b, c đôi một khác nhau và khác 0 không thỏa mãn:
(a+b+c)2 = a2 + b2 + c2
Tính giá trị biểu thức: A = \(\dfrac{a^2}{a^2+2bc}\) + \(\dfrac{b^2}{b^2+2ca}\) + \(\dfrac{c^2}{c^2+2ab}\)
mk cần gấp mong mn giúp đỡ, cảm ơn mn rất nhiều.
\(\left(a+b+c\right)^2=a^2+b^2+c^2\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=a^2+b^2+c^2\)
\(\Leftrightarrow2\left(ab+bc+ac\right)=0\Leftrightarrow ab+bc+ac=0\Leftrightarrow bc=-ab-ac\)
\(\dfrac{a^2}{a^2+2bc}=\dfrac{a^2}{a^2+bc-ac-ab}=\dfrac{a^2}{\left(a-c\right)\left(a-b\right)}\)
CMTT: \(\left\{{}\begin{matrix}\dfrac{b^2}{b^2+2ca}=\dfrac{b^2}{\left(b-a\right)\left(b-c\right)}\\\dfrac{c^2}{c^2+2ab}=\dfrac{c^2}{\left(c-a\right)\left(c-b\right)}=\dfrac{c^2}{\left(a-c\right)\left(b-c\right)}\end{matrix}\right.\)
\(\Rightarrow A=\dfrac{a^2}{\left(a-c\right)\left(a-b\right)}+\dfrac{b^2}{\left(b-a\right)\left(b-c\right)}+\dfrac{c^2}{\left(a-c\right)\left(b-c\right)}=\dfrac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\dfrac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=1\)
Đúng 3
Bình luận (1)
(1) (a+b−c)2a2+b2+c2+2ab−2bc−2ac(a+b−c)2a2+b2+c2+2ab−2bc−2ac
(3) a3+b3(a+b)3−3ab(a+b)a3+b3(a+b)3−3ab(a+b)
(5) (a+b+c)3a3+b3+c3+3(a+b)(b+c)(c+a)(a+b+c)3a3+b3+c3+3(a+b)(b+c)(c+a)
(7) (a−b)3+(b−c)3+(c−a)33(a−b)(b−c)(c−a)(a−b)3+(b−c)3+(c−a)33(a−b)(b−c)(c−a)
(9) (a+b)(b+c)(c+a)(a+b+c)(ab+bc+ca)−abc(a+b)(b+c)(c+a)(a+b+c)(ab+bc+ca)−abc
(11) ab3+bc3+ca3−a3b−b3c−c3a(a+b+c)[(a−b)3+(b−c)3+(c−a)3]3ab3+bc3+ca3−a3b−b3c−c3a(a+b+c)[(a−b)3+(b−c)3+(c−a)3]3
Chứng minh giùm mik hằng đẳng thức...
Đọc tiếp
(1) (a+b−c)2=a2+b2+c2+2ab−2bc−2ac(a+b−c)2=a2+b2+c2+2ab−2bc−2ac
(3) a3+b3=(a+b)3−3ab(a+b)a3+b3=(a+b)3−3ab(a+b)
(5) (a+b+c)3=a3+b3+c3+3(a+b)(b+c)(c+a)(a+b+c)3=a3+b3+c3+3(a+b)(b+c)(c+a)
(7) (a−b)3+(b−c)3+(c−a)3=3(a−b)(b−c)(c−a)(a−b)3+(b−c)3+(c−a)3=3(a−b)(b−c)(c−a)
(9) (a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)−abc(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)−abc
(11) ab3+bc3+ca3−a3b−b3c−c3a=(a+b+c)[(a−b)3+(b−c)3+(c−a)3]3ab3+bc3+ca3−a3b−b3c−c3a=(a+b+c)[(a−b)3+(b−c)3+(c−a)3]3
(1) (a+b−c)2a2+b2+c2+2ab−2bc−2ac(a+b−c)2a2+b2+c2+2ab−2bc−2ac
(3) a3+b3(a+b)3−3ab(a+b)a3+b3(a+b)3−3ab(a+b)
(5) (a+b+c)3a3+b3+c3+3(a+b)(b+c)(c+a)(a+b+c)3a3+b3+c3+3(a+b)(b+c)(c+a)
(7) (a−b)3+(b−c)3+(c−a)33(a−b)(b−c)(c−a)(a−b)3+(b−c)3+(c−a)33(a−b)(b−c)(c−a)
(9) (a+b)(b+c)(c+a)(a+b+c)(ab+bc+ca)−abc(a+b)(b+c)(c+a)(a+b+c)(ab+bc+ca)−abc
(11) ab3+bc3+ca3−a3b−b3c−c3a(a+b+c)[(a−b)3+(b−c)3+(c−a)3]3ab3+bc3+ca3−a3b−b3c−c3a(a+b+c)[(a−b)3+(b−c)3+(c−a)3]3
Chứng minh giùm mik hằng đẳng thức...
Đọc tiếp
(1) (a+b−c)2=a2+b2+c2+2ab−2bc−2ac(a+b−c)2=a2+b2+c2+2ab−2bc−2ac
(3) a3+b3=(a+b)3−3ab(a+b)a3+b3=(a+b)3−3ab(a+b)
(5) (a+b+c)3=a3+b3+c3+3(a+b)(b+c)(c+a)(a+b+c)3=a3+b3+c3+3(a+b)(b+c)(c+a)
(7) (a−b)3+(b−c)3+(c−a)3=3(a−b)(b−c)(c−a)(a−b)3+(b−c)3+(c−a)3=3(a−b)(b−c)(c−a)
(9) (a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)−abc(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)−abc
(11) ab3+bc3+ca3−a3b−b3c−c3a=(a+b+c)[(a−b)3+(b−c)3+(c−a)3]3ab3+bc3+ca3−a3b−b3c−c3a=(a+b+c)[(a−b)3+(b−c)3+(c−a)3]3
16 tháng 8 2021 lúc 21:33
Rút gọn:
\(\dfrac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}\)
với: c2+2ab-2ac-2bc=0; b\(\ne\)c; a+b\(\ne\)c
Tham khảo:
Cho a≠b≠c, a+b≠c và c2+2ab-2ac-2bc=0 Hãy rút gọn \(B=\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}\) - Hoc24
Đúng 2
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\(\dfrac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}\)
\(=\dfrac{a^2+a^2-2ac+c^2}{b^2+b^2-2bc+c^2}\)
\(=\dfrac{2a^2-2ac+c^2}{2b^2-2bc+c^2}\)
Đúng 2
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cmr (a+b+c)^2 = a^2+b^2+c^2+2ab+2ac+2bc
cmr ( a+b+c)^2= a^2+b^2+c^2+2ab+2ac+2bc
Biến đổi vế trái ta có
(a+b+c)^2 = (a+b + c)( a+b+c) = a(a+b + c) + b(a+b+c ) + c (a+b+c )
= a^2 + ab +ac + ab + b^2 + bc + ac + bc + c^2
= a^2 + b^2 + c^2 + 2ab + 2bc + 2ac => ĐPCM
Đúng 0
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cho a^2+b^2+c^2=(a+b+c)^2. CMR: a^2/(a^2+2bc)+b^2/(b^2+2ac)+c^2/(c^2+2ab)=1
Ta có a2 + b2 + c2 = (a + b + c)2
<=> ab + bc + ca = 0
<=> \(\hept{\begin{cases}ab=-bc-ca\\bc=-ac-ab\\ca=-ab-bc\end{cases}}\)
Khi đó a2 + 2bc = a2 + bc + bc = a2 + bc - ac - ab = (a - b)(a - c)
Tương tư b2 + 2ac = (b - a)(b - c)
c2 + ab = (c - a)(c - b)
Khi đó \(\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ac}+\frac{c^2}{c^2+2ab}\)
\(=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-a\right)\left(b-c\right)}+\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)
\(=\frac{-a^2\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{-b^2\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{-c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{-a^2b+a^2c-b^2c+b^2a-c^2a+c^2b}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=1\)(đpcm)