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\({A^2} - {B^2} = \left( {A - B} \right)\left( {A + B} \right)\)
Chọn D.
a) \(a^2+b^2=\left(a+b\right)^2-2ab\)
\(VP=\left(a+b\right)^2-2ab=a^2+2ab+b^2-2ab\)\(=a^2+b^2=VT\)
\(\Rightarrowđpcm\)
b)\(a^4+b^4=\left(a^2+b^2\right)^2-2a^2b^2\)
\(VP=a^4+b^4+2a^2b^2-2a^2b^2=a^4+b^4=VT\)\(\Rightarrowđpcm\)
c) \(a^6+b^6=\left(a^2+b^2\right)\left[\left(a^2+b^2\right)^2-3a^2b^2\right]\)
\(VP=\left(a^2+b^2\right)\left(a^4-a^2b^2+b^4\right)=a^6+b^6\)
\(VP=VT\Rightarrowđpcm\)
d)\(a^6-b^6=\left(a^2-b^2\right)[\left(a^2+b^2\right)^2-a^2b^2]\)
\(VP=\left(a^2-b^2\right)\left(a^4+a^2b^2+b^4\right)=a^6-b^6=VT\)
\(VP=VT\Rightarrowđpcm\)
BĐT cần CM tương đương:
\(3-VT\ge1\)
\(\Leftrightarrow\frac{a^2+2bc-a\left(b+c\right)}{a^2+2bc}+...\ge1\) (1)
\(VT\left(1\right)=\frac{\left[a^2+2bc-a\left(b+c\right)\right]^2}{\left(a^2+2bc\right)\left[a^2+2bc-a\left(b+c\right)\right]}+...\)
\(\ge\frac{\left[a^2+2bc-a\left(b+c\right)+b^2+2ca-b\left(c+a\right)+c^2+2ab-c\left(a+b\right)\right]^2}{\left(a^2+2bc\right)\left[a^2+2bc-a\left(b+c\right)\right]+...}\)
\(=\frac{\left(a^2+b^2+c^2\right)^2}{\left(a^2+2bc\right)\left[a^2+2bc-a\left(b+c\right)\right]+...}\) (2)
Ta cần chứng minh mẫu của (2) \(\le\left(a^2+b^2+c^2\right)^2\)
... Tự biến đổi ra thôi thi ta được 1 biểu thức không âm luôn đúng
=> BĐT trên đúng
=> đpcm
Dấu "=" xảy ra khi: a = b = c
a: \(\left(a^2-b^2\right)^2+\left(2ab\right)^2\)
\(=a^4-2a^2b^2+b^4+4a^2b^2\)
\(=a^4+2a^2b^2+b^4=\left(a^2+b^2\right)^2\)
b: \(\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
\(=a^2c^2+b^2d^2+a^2d^2+b^2c^2\)
\(=c^2\left(a^2+b^2\right)+d^2\left(a^2+b^2\right)\)
\(=\left(a^2+b^2\right)\left(c^2+d^2\right)\)
c: \(\left(ax+b\right)^2+\left(a-bx\right)^2+c^2x^2\)
\(=a^2x^2+b^2+a^2+b^2x^2+c^2x^2\)
\(=a^2\left(x^2+1\right)+b^2\left(x^2+1\right)+c^2x^2\)
\(=\left(x^2+1\right)\left(a^2+b^2\right)+c^2x^2\)
a) Ta có : \(a^2+1=a^2+ab+bc+ac=a\left(a+b\right)+c\left(a+b\right)=\left(a+b\right)\left(a+c\right)\)
Tương tự : \(b^2+1=\left(b+a\right)\left(b+c\right)\) ; \(c^2+1=\left(c+a\right)\left(c+b\right)\)
Suy ra \(\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)=\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\)
Vậy \(A=\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}=1\)
b) Ta có ; \(a^2+2bc-1=a^2+2bc-\left(ab+bc+ac\right)=a^2-ab+bc-ac=a\left(a-b\right)-c\left(a-b\right)\)
\(=\left(a-b\right)\left(a-c\right)\)
Tương tự : \(b^2+2ac-1=\left(a-b\right)\left(c-b\right)\) ; \(c^2+2ab-1=\left(a-c\right)\left(b-c\right)\)
Suy ra \(\left(a^2+2bc-1\right)\left(b^2+2ac-1\right)\left(c^2+2ab-1\right)=\left(a-b\right)^2.\left(c-a\right)^2.\left[-\left(b-c\right)^2\right]\)
Vậy : \(B=\frac{-\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)}=-1\)
ta có : a+b+c=0=>a+b=-c ; b+c=-a ; a+c=-b
ta có: M= \(\frac{2ab}{a^2+\left(b+c\right)\left(b-c\right)}+\frac{2bc}{b^2+\left(c+a\right)\left(c-a\right)}+\frac{2ca}{c^2+\left(a+b\right)\left(a-b\right)}\)
M=\(\frac{2ab}{a^2-a\left(b-c\right)}+\frac{2bc}{b^2-b\left(c-a\right)}+\frac{2ca}{c^2-c\left(a-b\right)}\)
M=\(\frac{2ab}{a\left(a-b+c\right)}+\frac{2bc}{b\left(b-c+a\right)}+\frac{2ca}{c\left(c-a+b\right)}\)
M=\(\frac{2ab}{-ab+\left(a+c\right)}+\frac{2bc}{-bc+\left(a+b\right)}+\frac{2ac}{-ac+\left(b+c\right)}\)
M=\(\frac{2ab}{-2ab}+\frac{2bc}{-2bc}+\frac{2ca}{-2ca}\)
M=-1-1-1=-3
Vậy với a+b+c=0 thì M=-3
\(a,b)\)Ta có: \(\left(a\pm b\right)^2\)
\(=\left(a\pm b\right)\left(a\pm b\right)\)
\(=a^2\pm ab\pm ab+b^2\)
\(=a^2\pm ab+b^2\)
\(c)\)\(\left(a+b\right)\left(a-b\right)=a^2-ab+ab-b^2=a^2-b^2\)
2ab*