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Ta có: \(a+\dfrac{1}{b}=-4\)
\(\Rightarrow\left(a+\dfrac{1}{b}\right)^3=\left(-4\right)^3\)
\(\Rightarrow a^3+3.a^2.\dfrac{1}{b}+3.a.\dfrac{1}{b^2}+\dfrac{1}{b^3}=-64\)
\(\Rightarrow a^3+\dfrac{3a^2}{b}+\dfrac{3a}{b^2}+\dfrac{1}{b^3}=-64\)
\(\Rightarrow a^3+\dfrac{1}{b^3}=-64-\dfrac{3a^2}{b}-\dfrac{3a}{b^2}\)
\(\Rightarrow a^3+\dfrac{1}{b^3}=-64-\dfrac{3a}{b}\left(a+\dfrac{1}{b}\right)\)
\(\Rightarrow a^3+\dfrac{1}{b^3}=-64-3.\left(-4\right).\left(-4\right)\)
\(\Rightarrow a^3+\dfrac{1}{b^3}=-112\)
\(P=a^3+\dfrac{1}{b^3}\\ =\left(a+\dfrac{1}{b}\right)\left(a^2+\dfrac{a}{b}+\dfrac{1}{b^2}\right)\\ =-4\left(a^2+\dfrac{2a}{b}+\dfrac{1}{b^2}-\dfrac{a}{b}\right)\\ =-4\left[\left(a+\dfrac{1}{b}\right)^2-\dfrac{a}{b}\right]\\ =-4\left[\left(-4\right)^2-\left(-4\right)\right]\\ =-80\)
(a+1/b)2=16 <=> a2+2a/b+1/b2=16 <=> a2+1/b2=24 (1)
Từ giả thiết và (1) suy ra: (a+1/b)(a2+1/b2)= -96 rồi tính đc cái cần tính
1/(a+b) + 1/(b+c) + 1/(c+a) = 4/(a+b+c)
=> [1/(a+b) + 1/(b+c) + 1/(c+a)](a+b+c) = 4
=> 3 + c/(a+b) +a/(b+c) + b/(c+a) = 4
=> [3 + c/(a+b) + a/(b+c) + b/(c+a)](a+b+c) = 4(a+b+c)
=> 3(a+b+c) + c + c2(a+b) + a + a2(b+c) + b + b2(c+a) = 4(a+b+c)
=> a2(b+c) + b2(c+a) + c2(a+b) = 0
Ko cần cảm ơn, mik giúp bạn chỉ vì mik đang sắp rơi vào danh sách học sinh dốt của hoc24h ^^
Ta có : a/b + b/c = 1 <=> (ac+b2)/(bc) (1)
c/a=-1 <=> c= -a => -3abc = +3c2b2 = 3(bc)2(2)
Ta có :
M = [(ac)3+(b2)3]/(bc) 3
<=> [(ac+b2)((ac)2-acb2+(b2)2]/(bc)3
<=> [( ac+b2)((ac) 2+2acb2+(b2)2 -3acb2]/(bc)3
<=> [(ac+b2)*((ac+b2)-3acb2)]/(bc)3
<=> [(ac+b2)/bc)] *[ (ac+b2)-3acb2)]/(bc)2
Từ( 1),(2) thay vào bt trên ta có
<=>1*[ (ac+b2)+3(cb)2]/(bc)2]
<=> 3+ [(ac+b) 2/(bc) 2]
<=> 3+[(ac+b )/(bc )] 2
<=> 3+12=4
Vậy M =4
Câu trả lời sai là:
(C) Giá trị của Q tại \(x=3\) là \(\dfrac{3-3}{3+3}=0\)
Do ĐKXĐ của phương trình
\(Q=\dfrac{x^2-6x+9}{x^2-9}\) là \(x\ne\pm3\)
á mk xl nhá mk ko đọc kĩ đề mk làm nhầm rùi bài mk làm là tìm GTNN nhá bạn ( mất công quá
)
ta có A= a+b+c+\(\dfrac{3}{a}+\dfrac{9}{2b}+\dfrac{4}{c}\)
= \(\dfrac{3a}{4}+\dfrac{a}{4}+\dfrac{b}{2}+\dfrac{b}{2}+\dfrac{c}{4}+\dfrac{3c}{4}+\dfrac{3}{a}+\dfrac{9}{2b}+\dfrac{4}{c}\)
=\(\left(\dfrac{3a}{4}+\dfrac{3}{a}\right)+\left(\dfrac{b}{2}+\dfrac{9}{2b}\right)+\left(\dfrac{c}{4}+\dfrac{4}{c}\right)+\dfrac{a}{4}+\dfrac{b}{2}+\dfrac{3c}{4}\)
vì a,b,c >0 ===> \(\dfrac{3a}{4}>0,\dfrac{3}{a}>0,\dfrac{b}{2}>0,\dfrac{9}{2b}>0,\dfrac{c}{4}>0,\dfrac{4}{c}>0\)
áp dụng BĐT côsi cho các cặp số dương ta đc:
\(\dfrac{3a}{4}+\dfrac{3}{a}>=2.\sqrt{\dfrac{3a}{4}.\dfrac{3}{a}}=3\)
\(\dfrac{b}{2}+\dfrac{9}{2b}>=3\)(làm như trên nhá)
\(\dfrac{c}{4}+\dfrac{4}{c}>=2\)
===> \(\dfrac{3a}{4}+\dfrac{3}{a}+\dfrac{b}{2}+\dfrac{9}{2b}+\dfrac{c}{4}+\dfrac{4}{c}>=8\left(1\right)\)
có: \(\dfrac{a}{4}+\dfrac{b}{2}+\dfrac{3c}{4}=\dfrac{a+2b+3c}{4}\)
mà a+2b+3c >= 20
===> \(\dfrac{a+2b+3c}{4}>=\dfrac{20}{4}=5\)
===> \(\dfrac{a}{4}+\dfrac{b}{2}+\dfrac{3c}{4}>=5\left(2\right)\)
từ (1) và(2)===> a+b+c+\(\dfrac{3}{a}+\dfrac{9}{2b}+\dfrac{4}{c}>=13\)
===> A >= 13
Dấu ''='' xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{3a}{4}=\dfrac{3}{a}\\\dfrac{b}{2}=\dfrac{9}{2b}\\\dfrac{c}{4}=\dfrac{4}{c}\\a+2b+3c=20\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}a=2\\b=3\\c=4\end{matrix}\right.\)
Vậy Min A=13 <=>\(\left\{{}\begin{matrix}a=2\\b=3\\c=4\end{matrix}\right.\)
a ) \(a+b+c=0\)
\(\Leftrightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2+2.0=0\)
\(\Leftrightarrow a^2+b^2+c^2=0\)
Do \(a^2\ge0;b^2\ge0;c^2\ge0\)
\(\Rightarrow a^2+b^2+c^2\ge0\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c=0\) ( * )
Thay * vào biểu thức M , ta được :
\(M=\left(0-1\right)^{1999}+0^{2000}+\left(0+1\right)^{2001}\)
\(=-1^{1999}+0+1^{2001}\)
\(=-1+0+1\)
\(=0\)
Vậy \(M=0\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{abc}\)
\(\Leftrightarrow\dfrac{bc}{abc}+\dfrac{ac}{abc}+\dfrac{ab}{abc}=\dfrac{1}{abc}\)
\(\Leftrightarrow\dfrac{bc+ac+ab-1}{abc}=0\)
\(\Leftrightarrow bc+ac+ab-1=0\)
\(\Leftrightarrow bc+ac+ab=1\)
Mà \(a^2+b^2+c^2=1\)
\(\Rightarrow bc+ac+ab=a^2+b^2+c^2\)
\(\Rightarrow2bc+2ac+2ab=2a^2+2b^2+2c^2\)
\(\Rightarrow2a^2+2b^2+2c^2-2bc-2ac-2ab=0\)
\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
Do \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(a-c\right)^2\ge0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c\)
Mà \(P=\dfrac{a+b}{b+c}+\dfrac{b+c}{c+a}+\dfrac{c+a}{a+b}\)
\(\Rightarrow P=\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{a+c}{a+c}\)
\(\Rightarrow P=1+1+1=3\)
Vậy \(P=3\)
\(a+\dfrac{1}{b}=\dfrac{a}{b}\Leftrightarrow\dfrac{ab+1}{b}=\dfrac{a}{b}\Leftrightarrow ab+1=a\left(1\right)\)
\(\dfrac{a}{b}=-4\Leftrightarrow a=-4b\left(2\right)\)
Thay (2) vào (1), ta được:
\(-4b^2+1=-4b\)
\(\Rightarrow-4b^2+4b+1=0\)
\(\Rightarrow-4\left(b^2+b-\dfrac{1}{4}\right)=0\)
\(\Rightarrow-4\left(b^2+2\cdot b\cdot\dfrac{1}{2}+\dfrac{1}{4}\right)+2=0\)
\(\Rightarrow-4\left(b+\dfrac{1}{2}\right)^2=-2\)
\(\Rightarrow\left(b+\dfrac{1}{2}\right)^2=\dfrac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}b+\dfrac{1}{2}=\sqrt{\dfrac{1}{2}}\\b+\dfrac{1}{2}=-\sqrt{\dfrac{1}{2}}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}b=\dfrac{-1+\sqrt{2}}{2}\\b=\dfrac{-1-\sqrt{2}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}b=\dfrac{-1+\sqrt{2}}{2}\\a=2-2\sqrt{2}\end{matrix}\right.\\\left\{{}\begin{matrix}b=\dfrac{-1-\sqrt{2}}{2}\\a=2+2\sqrt{2}\end{matrix}\right.\end{matrix}\right.\)
Vậy ..................................