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b) Ta có: \(a\left(b^2-c^2\right)+b\left(c^2-a^2\right)+c\left(a^2-b^2\right)\)
\(=ab^2-ac^2+bc^2-ba^2+ca^2-cb^2\)
\(=\left(ab^2-cb^2\right)+\left(ca^2-c^2a\right)+\left(bc^2-ba^2\right)\)
\(=b^2\left(a-c\right)+ca\left(a-c\right)+b\left(c^2-a^2\right)\)
\(=\left(a-c\right)\left(b^2+ca\right)-b\left(a-c\right)\left(a+c\right)\)
\(=\left(a-c\right)\left(b^2+ca-ba-bc\right)\)
\(=\left(a-c\right)\left[b\left(b-a\right)+c\left(a-b\right)\right]\)
\(=\left(a-c\right)\left[b\left(b-a\right)-c\left(b-a\right)\right]\)
\(=\left(a-c\right)\left(b-a\right)\left(b-c\right)\)
Ta có: a+b+c=0
nên a+b=-c
Ta có: \(a^2-b^2-c^2\)
\(=a^2-\left(b^2+c^2\right)\)
\(=a^2-\left[\left(b+c\right)^2-2bc\right]\)
\(=a^2-\left(b+c\right)^2+2bc\)
\(=\left(a-b-c\right)\left(a+b+c\right)+2bc\)
\(=2bc\)
Ta có: \(b^2-c^2-a^2\)
\(=b^2-\left(c^2+a^2\right)\)
\(=b^2-\left[\left(c+a\right)^2-2ca\right]\)
\(=b^2-\left(c+a\right)^2+2ca\)
\(=\left(b-c-a\right)\left(b+c+a\right)+2ca\)
\(=2ac\)
Ta có: \(c^2-a^2-b^2\)
\(=c^2-\left(a^2+b^2\right)\)
\(=c^2-\left[\left(a+b\right)^2-2ab\right]\)
\(=c^2-\left(a+b\right)^2+2ab\)
\(=\left(c-a-b\right)\left(c+a+b\right)+2ab\)
\(=2ab\)
Ta có: \(M=\dfrac{a^2}{a^2-b^2-c^2}+\dfrac{b^2}{b^2-c^2-a^2}+\dfrac{c^2}{c^2-a^2-b^2}\)
\(=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2ab}\)
\(=\dfrac{a^3+b^3+c^3}{2abc}\)
Ta có: \(a^3+b^3+c^3\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ca-cb+c^2\right)-3ab\left(a+b\right)\)
\(=-3ab\left(a+b\right)\)
Thay \(a^3+b^3+c^3=-3ab\left(a+b\right)\) vào biểu thức \(=\dfrac{a^3+b^3+c^3}{2abc}\), ta được:
\(M=\dfrac{-3ab\left(a+b\right)}{2abc}=\dfrac{-3\left(a+b\right)}{2c}\)
\(=\dfrac{-3\cdot\left(-c\right)}{2c}=\dfrac{3c}{2c}=\dfrac{3}{2}\)
Vậy: \(M=\dfrac{3}{2}\)
Do a+b+c= 0
<=> a+b= -c
=> (a+b)2= c2
Tương tự: (c+a)2= b2, (c+b)2= a2
Ta có: \(A=\frac{1}{b^2+c^2-a^2}+\frac{1}{c^2+a^2-b^2}+\frac{1}{a^2+b^2-c^2}\)
\(=\frac{1}{b^2+c^2-\left(b+c\right)^2}+\frac{1}{c^2+a^2-\left(c+a\right)^2}+\frac{1}{a^2+b^2-\left(a+b\right)^2}\)
\(=\frac{1}{-2bc}+\frac{1}{-2ca}+\frac{1}{-2ab}\)
\(=\frac{a+b+c}{-2abc}=0\)
Ta có
D = a ( b 2 + c 2 ) – b ( c 2 + a 2 ) + c ( a 2 + b 2 ) – 2 a b c = a b 2 + a c 2 – b c 2 – b a 2 + c a 2 + c b 2 – 2 a b c = ( a b 2 – a 2 b ) + ( a c 2 – b c 2 ) + ( a 2 c – 2 a b c + b 2 c ) = a b ( b – a ) + c 2 ( a – b ) + c ( a 2 – 2 a b + b 2 ) = - a b ( a – b ) + c 2 ( a – b ) + c ( a – b ) 2 = ( a – b ) ( - a b + c 2 + c ( a – b ) ) = ( a – b ) ( - a b + c 2 + a c – b c ) = ( a – b ) [ ( - a b + a c ) + ( c 2 – b c ) ]
= (a – b)[a(c – b) + c(c – b)]
= (a – b)(a + c)(c – b)
Với a = 99; b = -9; c = 1, ta có
D = (99 - (-9))(99 + 1) (1 - (-9)) = 108.100.10 = 108000
Đáp án cần chọn là: B
mới ăn miếng cơm cà ngon nhức nách luôn ai thèm cơm cà không điểm danh nào
