Cho các số a, b, c, d khác 0 và a+b+c+d\(\ne\)0, 5c - d \(\ne\) 0, 2d - b \(\ne\) 0 thỏa mãn \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{a}\). Tính giá trị biểu thức M = \(\dfrac{3a+b}{5b-d}+\dfrac{2b+3c}{2d-b}\)
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d: Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có: \(\dfrac{3c^2+5a^2}{3d^2+5b^2}=\dfrac{3\cdot\left(dk\right)^2+5\cdot\left(bk\right)^2}{3d^2+5b^2}=k^2\)
\(\dfrac{c^2}{d^2}=\dfrac{\left(dk\right)^2}{d^2}=k^2\)
Do đó: \(\dfrac{3c^2+5a^2}{3d^2+5b^2}=\dfrac{c^2}{d^2}\)
a: \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Leftrightarrow\dfrac{a}{b}-1=\dfrac{c}{d}-1\)
hay \(\dfrac{a-b}{b}=\dfrac{c-d}{d}\)
\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{a+b+c}{2\left(a+b+c\right)}=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{a+b}{c}=2\)
\(\Rightarrow P=2+2+2=6\)
a) \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\dfrac{a+b}{b}=\dfrac{bk+b}{b}=\dfrac{b\left(k+1\right)}{b}=k+1\) và \(\dfrac{c+d}{d}=\dfrac{dk+d}{d}=\dfrac{d\left(k+1\right)}{d}=k+1\)
\(\Rightarrow\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
b) \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a-b}{b}=\dfrac{b\left(k-1\right)}{b}=k-1\\\dfrac{c-d}{d}=\dfrac{d\left(k-1\right)}{d}=k-1\end{matrix}\right.\)\(\Rightarrow\dfrac{a-b}{b}=\dfrac{c-d}{d}\)
c) \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\Rightarrow\dfrac{a}{c}=\dfrac{a+b}{c+d}\Rightarrow\dfrac{a+b}{a}=\dfrac{c+d}{c}\)
d) \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}\Rightarrow\dfrac{a}{c}=\dfrac{a-b}{c-d}\Rightarrow\dfrac{a-b}{a}=\dfrac{c-d}{c}\)
Bài 1: Ta có:
\(M=\frac{ad}{abcd+abd+ad+d}+\frac{bad}{bcd.ad+bc.ad+bad+ad}+\frac{c.abd}{cda.abd+cd.abd+cabd+abd}+\frac{d}{dab+da+d+1}\)
\(=\frac{ad}{1+abd+ad+d}+\frac{bad}{d+1+bad+ad}+\frac{1}{ad+d+1+abd}+\frac{d}{dab+da+d+1}\)
$=\frac{ad+abd+1+d}{ad+abd+1+d}=1$
Bài 2:
Vì $a,b,c,d\in [0;1]$ nên
\(N\leq \frac{a}{abcd+1}+\frac{b}{abcd+1}+\frac{c}{abcd+1}+\frac{d}{abcd+1}=\frac{a+b+c+d}{abcd+1}\)
Ta cũng có:
$(a-1)(b-1)\geq 0\Rightarrow a+b\leq ab+1$
Tương tự:
$c+d\leq cd+1$
$(ab-1)(cd-1)\geq 0\Rightarrow ab+cd\leq abcd+1$
Cộng 3 BĐT trên lại và thu gọn thì $a+b+c+d\leq abcd+3$
$\Rightarrow N\leq \frac{abcd+3}{abcd+1}=\frac{3(abcd+1)-2abcd}{abcd+1}$
$=3-\frac{2abcd}{abcd+1}\leq 3$
Vậy $N_{\max}=3$
\(\)Ta có: \(a+b+c=0 \Rightarrow b+c=-a \Rightarrow (b+c)^2=(-a)^2 \Leftrightarrow b^2+c^2+2bc=a^2 \Leftrightarrow a^2-b^2-c^2=2bc\)
Tương tự: \(b^2-c^2-a^2=2ca;c^2-a^2-b^2=2ab\)
\(P=...=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ca}+\dfrac{c^2}{2bc}=\dfrac{a^3+b^3+c^3}{2abc}=\dfrac{3abc}{2abc}=\dfrac{3}{2}\)
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Bổ đề \(a+b+c=0 \Leftrightarrow a^3+b^3+c^3\)
Ở đây ta c/m chiều thuận:
Với \(a+b+c=0 \Leftrightarrow a+b=-c \Rightarrow (a+b)^3=(-c)^3 \Leftrightarrow a^3+b^3+3ab(a+b)=-c^3 \Leftrightarrow a^3+b^3+c^3=3abc(QED)\)
Đặt \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{a}=k\)(k<>0)
=>\(\left\{{}\begin{matrix}d=a\cdot k\\c=d\cdot k=a\cdot k\cdot k=ak^2\\b=ck=ak^3\\a=bk=ak^4\end{matrix}\right.\)
\(a=ak^4\)
=>\(ak^4-a=0\)
=>\(a\left(k^4-1\right)=0\)
=>\(k^4-1=0\)
=>\(\left[{}\begin{matrix}k=1\\k=-1\end{matrix}\right.\)
\(M=\dfrac{3a+b}{5b-d}+\dfrac{2b+3c}{2d-b}\)
\(=\dfrac{3\cdot ak^4+ak^3}{5\cdot ak^3-ak}+\dfrac{2\cdot ak^3+3\cdot ak^2}{2\cdot ak-ak^3}\)
\(=\dfrac{ak^3\left(3k+1\right)}{ak\left(5k^2-1\right)}+\dfrac{ak^2\left(2k+3\right)}{ak\left(2-k^2\right)}\)
\(=\dfrac{k^2\left(3k+1\right)}{5k^2-1}+\dfrac{k\left(2k+3\right)}{2-k^2}\)
TH1: k=1
=>\(M=\dfrac{1^2\left(3\cdot1+1\right)}{5\cdot1^2-1}+\dfrac{1\left(2\cdot1+3\right)}{2-1^2}=\dfrac{4}{4}+\dfrac{5}{1}=6\)
TH2: k=-1
=>\(M=\dfrac{\left(-1\right)^2\cdot\left(-3+1\right)}{5\cdot\left(-1\right)^2-1}+\dfrac{\left(-1\right)\left(2\cdot\left(-1\right)+3\right)}{2-\left(-1\right)^2}\)
\(=\dfrac{-2}{4}+\dfrac{1}{1}=-\dfrac{1}{2}+1=\dfrac{1}{2}\)
cíu tuii
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