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a: \(=a^2+2a\left(b-c\right)+\left(b-c\right)^2+a^2-2a\left(b-c\right)+\left(b-c\right)^2-2\left(b-c\right)^2\)
\(=2a^2+2\left(b-c\right)^2-2\left(b-c\right)^2=2a^2\)
b: \(=a^2+2a\left(b+c\right)+\left(b+c\right)^2+a^2-2a\left(b+c\right)+\left(b+c\right)^2+\left(b-c-a\right)^2+\left(c-a-b\right)^2\)
\(=2a^2+2\left(b+c\right)^2+\left(a-b+c\right)^2+\left(a+b-c\right)^2\)
\(=2a^2+2\left(b+c\right)^2+a^2-2a\left(b-c\right)+\left(b-c\right)^2+a^2+2a\left(b-c\right)+\left(b-c\right)^2\)
\(=2a^2+2\left(b+c\right)^2+2a^2+2\left(b-c\right)^2\)
\(=4a^2+2\left(b^2+2bc+c^2+b^2-2bc+c^2\right)\)
\(=4a^2+4b^2+4c^2\)
a) \(4a^3b^3c^2x+12a^3b^4c^2-16a^4b^5cx\)
\(=4a^3b^3c\left(cx+3bc-4ab^2x\right)\)
b) \(\left(b-2c\right)\left(a-b\right)-\left(a+b\right)\left(2c-b\right)\)
\(=\left(b-2c\right)\left(a-b+a+b\right)=2a\left(b-2c\right)\)
c) \(3a\left(a+5\right)-2\left(5+a\right)=\left(a+5\right)\left(3a-2\right)\)
d) \(\left(x+1\right)^2-3\left(x+1\right)=\left(x+1\right)\left(x+1-3\right)\)
Ta có: \(a+b+c=0\)
\(\Rightarrow a+b=-c\)
\(\Rightarrow\left(a+b\right)^3=-c^3\)
\(\Rightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)
\(\Rightarrow a^3+b^3+c^3=-3ab\left(a+b\right)\)
\(\Rightarrow a^3+b^3+c^3=-3ab.\left(-c\right)=3abc\)
Mặt khác: \(a+b+c=0\Rightarrow a^2=\left(-b-c\right)^2=\left(b+c\right)^2\)
\(\Rightarrow a^2-b^2-c^2=\left(b+c\right)^2-b^2-c^2=2bc\)
Tương tự ta có: \(b^2-a^2-c^2=2ca\)
\(c^2-a^2-b^2=2ab\)
\(\Rightarrow B=\frac{a^2}{2ab}+\frac{b^2}{2ca}+\frac{c^2}{2ab}=\frac{a^3+b^3+c^3}{2abc}=\frac{3abc}{2abc}=\frac{3}{2}\)
a)\(ab\left(a+b\right)-bc\left(b+c\right)+ac\left(a-c\right)\)
\(=\left(a+b\right)\left(b+c\right)\left(a-c\right)\)
b)\((a+b)(a^2-b^2)+(b+c)(b^2-c^2)+(c+a)(c^2-a^2)\)
\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
c)\(a^2b^2(a-b)+b^2c^2(b-c)+c^2a^2(c-a)\)
\(=\left(a-b\right)\left(a-c\right)\left(b-c\right)\left(ab+bc+ca\right)\)
d)\(a^4(b-c)+b^4(c-a)+c^4(a-b)\)
\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a^2+b^2+c^2+ab+bc+ca\right)\)
\(A=\left[\left(a+b\right)+\left(c+d\right)\right]^2+\left[\left(a+b\right)-\left(c+d\right)\right]^2+\left[\left(a-b\right)+\left(c-d\right)\right]^2+\left[\left(a-b\right)-\left(c-d\right)\right]^2\)
Ta có
\(\left[\left(a+b\right)+\left(c+d\right)\right]^2=\left(a+b\right)^2+2\left(a+b\right)\left(c+d\right)+\left(c+d\right)^2\)
\(\left[\left(a+b\right)-\left(c+d\right)\right]^2=\left(a+b\right)^2-2\left(a+b\right)\left(c+d\right)+\left(c+d\right)^2\)
\(\left[\left(a-b\right)+\left(c-d\right)\right]^2=\left(a-b\right)^2+2\left(a-b\right)\left(c-d\right)+\left(c-d\right)^2\)
\(\left[\left(a-b\right)-\left(c-d\right)\right]^2=\left(a-b\right)^2-2\left(a-b\right)\left(c-d\right)+\left(c-d\right)^2\)
\(A=2\left(a+b\right)^2+2\left(a-b\right)^2+2\left(c+d\right)^2+2\left(c-d\right)^2\)
\(A=2\left(a^2+2ab+b^2+a^2-2ab+b^2+c^2+2cd+d^2+c^2-2cd+d^2\right)\)
\(A=4\left(a^2+b^2+c^2+d^2\right)\)
a)
\(A=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(A=100+99+98+97+...+2+1\)
\(A=\frac{100.101}{2}=5050\)
b)
\(B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)
\(B=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)
\(B=\left(2^8-1\right)...\left(2^{64}+1\right)+1\)
\(B=\left(2^{64}-1\right)\left(2^{64}+1\right)+1\)
\(B=2^{128}-1+1=2^{128}\)
c)
\(C=a^2+b^2+c^2+2\left(ab+bc+ca\right)+a^2+b^2+c^2+2ab-2ac-2bc-2a^2-4ab-2b^2\)
\(C=2c^2\)
Ta có \(\frac{a^2+b^2}{2}\ge ab\), \(\frac{b^2+c^2}{2}\ge bc\),\(\frac{a^2+d^2}{2}\ge ad\),\(\frac{c^2+d^2}{2}\ge cd\)
Cộng từng vế của bđt trên ta được
\(a^2+b^2+c^2+d^2\ge ab+bc+ad+cd\)
=>\(1\ge\left(a+c\right)\left(b+d\right)\)
Dấu "=" xảy ra khi \(a=b=c=d=\frac{1}{2}\)
d, = a2
e, = 4 (a2 + b2 + c2)
đề dài qua, phân tích xong nổ óc @@
Ahihi mình cũng ngồi làm rồi nát ốc luôn ^^