cho 1/a +1/b +1/c = 0. tính gia trị của biểu thức : (1/a2 +2bc) + (1/ b2+ 2ac ) + 1/c2+ 2ab
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6 tháng 11 2018
\(\text{Ta có: }\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0.\)
\(\Leftrightarrow bc+ac+ab=0\Rightarrow\hept{\begin{cases}bc=-ac-ab\\ac=-bc-ab\\ab=-bc-ac\end{cases}}\)
\(\Rightarrow BT\text{hức}=\frac{bc}{a^2+2bc}+\frac{ac}{b^2+2ac}+\frac{ab}{c^2+2ab}\)
\(=\frac{bc}{a^2-ac-ab+bc}+\frac{ac}{b^2-bc-ab+ac}+\frac{ab}{c^2-bc-ac+ab}\)
\(=\frac{bc}{a\left(a-b\right)-c\left(a-b\right)}+\frac{ac}{b\left(b-a\right)-c\left(b-a\right)}+\frac{ab}{c\left(c-a\right)-b\left(c-a\right)}\)
\(=\frac{bc}{\left(a-c\right)\left(a-b\right)}-\frac{ac}{\left(b-c\right)\left(a-b\right)}+\frac{ab}{\left(a-c\right)\left(b-c\right)}\)
\(=\frac{bc\left(b-c\right)-ac\left(a-c\right)+ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{b^2c-bc^2-a^2c+ac^2+ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\frac{c\left(b^2-a^2\right)-c^2\left(b-a\right)+ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{c^2\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+ab\left(a+b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\frac{\left(a-b\right)\left(c^2-ac-bc+ab\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{c\left(c-b\right)-a\left(c-b\right)}{\left(b-c\right)\left(a-c\right)}=\frac{\left(a-c\right)\left(b-c\right)}{....}=1\)
Lâu ko lm đổi dấu hơi thừa ra!! ko hiểu chỗ nào thì ib mk giải thích cho
9 tháng 12 2021
Vì a,b,c là 3 cạnh tam giác nên \(a+b>c\Leftrightarrow ac+bc>c^2\)
CMTT: \(ab+bc>b^2;ab+ac>a^2\)
Cộng vế theo vế \(\Leftrightarrow a^2+b^2+c^2< ab+bc+ca+ab+bc+ca\)
\(\Leftrightarrow a^2+b^2+c^2< 2ab+2bc+2ca\\ \Leftrightarrow a^2+b^2+c^2-2ab-2bc-2ca< 0\)
KN
16 tháng 9 2020
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(\Leftrightarrow\frac{bc+ca+ab}{abc}=0\)
\(\Leftrightarrow bc+ca+ab=0\)
\(\Leftrightarrow\hept{\begin{cases}bc=-ab-ca\\ca=-ab-bc\\ab=-ca-bc\end{cases}}\)
Ta có : \(A=\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ac}+\frac{c^2}{c^2+2ab}\)
\(\Leftrightarrow A=\frac{a^2}{a^2+bc-ab-ca}+\frac{b^2}{b^2+ac-ab-bc}+\frac{c^2}{c^2+ab-ca-bc}\)
\(\Leftrightarrow A=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-a\right)\left(b-c\right)}+\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)
\(\Leftrightarrow A=\frac{a^2}{\left(a-b\right)\left(a-c\right)}-\frac{b^2}{\left(b-c\right)\left(a-b\right)}+\frac{c^2}{\left(a-c\right)\left(b-c\right)}\)
\(\Leftrightarrow A=\frac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(\Leftrightarrow A=\frac{a^2\left(b-c\right)-b^2\left[\left(b-c\right)+\left(a-b\right)\right]+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(\Leftrightarrow A=\frac{a^2\left(b-c\right)-b^2\left(b-c\right)-b^2\left(a-b\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(\Leftrightarrow A=\frac{\left(a^2-b^2\right)\left(b-c\right)-\left(b^2-c^2\right)\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(\Leftrightarrow A=\frac{\left(a+b\right)\left(a-b\right)\left(b-c\right)-\left(b+c\right)\left(b-c\right)\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(\Leftrightarrow A=\frac{\left(a-b\right)\left(b-c\right)\left[\left(a+b\right)-\left(b+c\right)\right]}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(\Leftrightarrow A=\frac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=1\)
CM
11 tháng 3 2018
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
10 tháng 6 2021
mới ăn miếng cơm cà ngon nhức nách luôn ai thèm cơm cà không điểm danh nào
15 tháng 6 2023
1: (a-1)(a-3)(a-4)(a-6)+9
=(a^2-7a+6)(a^2-7a+12)+9
=(a^2-7a)^2+18(a^2-7a)+81
=(a^2-7a+9)^2>=0
b: \(A=\dfrac{a^4-4a^3+a^2+4a^3-16a+4+16a-3}{a^2}=\dfrac{16a-3}{a^2}\)
a^2-4a+1=0
=>a=2+căn 3 hoặc a=2-căn 3
=>A=11-4căn 3 hoặc a=11+4căn 3
