Chung minh rang
:(a+b+c)^2+a^2+b^2+c^2=(a+b)^2+(b+c)^2+(C+a)^2
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Chung minh rang:
(a+b+c)^2+a^2+b^2+c^2=(a+b)^2+(b+c)^2+(c+a)^2
Ta có VT
\(\left(a+b+c\right)^2+a^2+b^2+c^2\)
\(=a^2+b^2+c^2+2ab+2bc+2ca+a^2+b^2+c^2\)
\(=\left(a^2+2ab+b^2\right)+\left(b^2+2bc+c^2\right)+\left(a^2+2ac+c^2\right)\)
\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2=VP\)\(\left(\text{đ}pcm\right)\)
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cho a/b=c/b chung minh rang a^2+c^2/b^2+c^2
chung minh rang: (a+b)2/(a-b)2 +(b+c)2/(b-c)2+(a+c)2/(c-a)2 >=2
cho a/c=c/b chung minh rang
a, a^2+c^2/b^2+c^2=a/b
b, b^2-a^2/a^2+c^2=b-a/a
a)
Ta có
\(\frac{a}{c}=\frac{c}{b}\Rightarrow\frac{a^2}{c^2}=\frac{c^2}{b^2}=\frac{a^2+c^2}{c^2+b^2}\)
\(\frac{a}{b}=\frac{a}{c}.\frac{c}{b}=\left(\frac{a}{c}\right)^2\)
Mà\(\frac{a^2+c^2}{b^2+c^2}=\left(\frac{a}{c}\right)^2=\frac{a}{b}\). Vậy \(\frac{a^2+c^2}{b^2+c^2}=\frac{a}{b}\)
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Xem thêm câu trả lời
cho a,b,c la 3 so khac 0 va a+b+c=0 chung minh rang 1/a^2+b^2-c^2+1/b^2+c^2-a^2+1/c^2+a^2-b^2=0
CHo ti le thuc a/b=c/d Chung minh rang (a+b/c+d)^2=a^2+b^2/c^2+d^2
cho a,b,c la cac so thuc duong. chung minh rang 2a/(b+c)+2b/(c+a)+2c/(a+b)>=((a-b)^2+(b-c)^2+(c-a)^2)/(a+b+c)^2
Cho a/(b+c)+b/(c+a)+c/(a+b)=1. chung minh rang
a^2/(b+c)+b^2/(c+a)+c^2/(a+b)=0
cho a,b,c la cac so nguyen. Chung minh rang: (a^2+b^2+c^2)*(a+b+c)^2+(ab+bc+ca)^2
Moi hoc lop 6 a!
Nen chang tra loi dc dau!
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