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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 $N_{\max}=3$
https://hoc24.vn/cau-hoi/cho-abc-0-thoa-man-abbcca3-tim-gia-tri-nho-nhat-cua-pdfrac13a1b2dfrac13b1c2dfrac13c1a2.6181078378966
\(P=\dfrac{1}{bc\left(b+c\right)+2023}+\dfrac{1}{ca\left(c+a\right)+2023}+\dfrac{1}{ab\left(a+b\right)+2023}\left(abc=2023\right)\)
\(\Leftrightarrow P=\dfrac{1}{bc\left(b+c\right)+abc}+\dfrac{1}{ca\left(c+a\right)+abc}+\dfrac{1}{ab\left(a+b\right)+abc}\)
\(\Leftrightarrow P=\dfrac{1}{bc\left(a+b+c\right)}+\dfrac{1}{ca\left(a+b+c\right)}+\dfrac{1}{ab\left(a+b+c\right)}\)
\(\Leftrightarrow P=\dfrac{1}{\left(a+b+c\right)}\left(\dfrac{1}{bc}+\dfrac{1}{ca}+\dfrac{1}{ab}\right)\)
\(\Leftrightarrow P=\dfrac{1}{\left(a+b+c\right)}\left[\dfrac{a^2bc+b^2ca+c^2ab}{\left(abc\right)^2}\right]\)
\(\Leftrightarrow P=\dfrac{1}{\left(a+b+c\right)}\left[\dfrac{abc\left(a+b+c\right)}{\left(abc\right)^2}\right]\)
\(\Leftrightarrow P=\dfrac{1}{abc}=\dfrac{1}{2023}\)
\(A=\dfrac{a}{abc+ab+a+1}+\dfrac{ba}{abcd+abc+ab+a}+\dfrac{\dfrac{c}{cd}}{\dfrac{acd}{cd}+\dfrac{cd}{cd}+\dfrac{c}{cd}+\dfrac{1}{cd}}+\dfrac{\dfrac{d}{d}}{\dfrac{dab}{d}+\dfrac{ad}{d}+\dfrac{d}{d}+\dfrac{1}{d}}\)
\(A=\dfrac{a}{abc+ab+a+1}+\dfrac{ab}{1+abc+ab+a}+\dfrac{\dfrac{1}{d}}{a+1+\dfrac{1}{d}+\dfrac{1}{cd}}+\dfrac{1}{ab+a+1+\dfrac{1}{d}}\)
Mà \(abcd=1\Rightarrow\dfrac{1}{d}=abc;\dfrac{1}{cd}=ab\)
\(\Rightarrow A=\dfrac{a}{abc+ab+a+a}+\dfrac{ab}{abc+ab+a+1}+\dfrac{abc}{a+1+abc+ab}+\dfrac{1}{ab+a+1+abc}\)
\(\Rightarrow A=\dfrac{a+ab+abc+1}{abc+ab+a+1}=1\)
\(A=\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}\)
\(A=\dfrac{a^2bc}{ab+a^2bc+abc}+\dfrac{b}{bc+b+abc}+\dfrac{c}{ac+c+1}\)
\(A=\dfrac{a^2bc}{ab\left(1+ac+c\right)}+\dfrac{b}{b\left(c+1+ac\right)}+\dfrac{c}{ac+c+1}\)
\(A=\dfrac{ac+1+c}{ac+c+1}\)
\(A=1\)
\(A=\dfrac{ab}{ab+a+1}+\dfrac{bc}{bc+b+1}+\dfrac{ca}{ca+c+1}\)
\(A=\dfrac{abc}{abc+ac+c}+\dfrac{bc}{bc+b+abc}+\dfrac{ca}{ca+c+1}\)
\(A=\dfrac{1}{1+ac+c}+\dfrac{c}{c+1+ac}+\dfrac{ca}{ca+c+1}\)
\(A=1\)
Lười đánh máy thật sự, buốt tay lắm:((
Ta có: \(Q=\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}\)
\(Q=\dfrac{ac}{c\left(ab+a+1\right)}+\dfrac{abc}{ac\left(bc+b+1\right)}+\dfrac{c}{ac+c+1}\)
\(Q=\dfrac{ac}{abc+ac+c}+\dfrac{abc}{abc^2+abc+ac}+\dfrac{c}{ac+c+1}\)
\(Q=\dfrac{ac}{1+ac+c}+\dfrac{1}{c+a+ac}+\dfrac{c}{ac+c+1}\)
\(Q=\dfrac{ac+1+c}{1+ac+c}=1\)
Vậy Q=1
Q=ab+a+1a+bc+b+1b+ac+c+1c
Q=\dfrac{ac}{c\left(ab+a+1\right)}+\dfrac{abc}{ac\left(bc+b+1\right)}+\dfrac{c}{ac+c+1}Q=c(ab+a+1)ac+ac(bc+b+1)abc+ac+c+1c
Q=\dfrac{ac}{abc+ac+c}+\dfrac{abc}{abc^2+abc+ac}+\dfrac{c}{ac+c+1}Q=abc+ac+cac+abc2+abc+acabc+ac+c+1c
Q=\dfrac{ac}{1+ac+c}+\dfrac{1}{c+a+ac}+\dfrac{c}{ac+c+1}Q=1+ac+cac+c+a+ac1+ac+c+1c
Q=\dfrac{ac+1+c}{1+ac+c}=1Q=1+ac+cac+1+c=1
chúc bạn thi tốt
ta có : \(M=\dfrac{1}{abc+ab+a+1}+\dfrac{1}{bcd+bc+b+1}+\dfrac{1}{acb+cd+c+1}+\dfrac{1}{abd+ad+d+1}\)
\(\Leftrightarrow M=\dfrac{abcd}{abcd+abc+ab+a}+\dfrac{1}{bcd+bc+b+1}+\dfrac{1}{acb+cd+c+1}+\dfrac{1}{abd+ad+d+1}\) \(\Leftrightarrow M=\dfrac{bcd}{bcd+bc+b+1}+\dfrac{1}{bcd+bc+b+1}+\dfrac{1}{acb+cd+c+1}+\dfrac{1}{abd+ad+d+1}\) \(\Leftrightarrow M=\dfrac{bcd+1}{bcd+bc+b+1}+\dfrac{1}{acb+cd+c+1}+\dfrac{1}{abd+ad+d+1}\) \(\Leftrightarrow M=\dfrac{abcd+bcd}{abcd+bcd+bc+b}+\dfrac{1}{acb+cd+c+1}+\dfrac{1}{abd+ad+d+1}\) \(\Leftrightarrow M=\dfrac{acd+cd}{acd+cd+c+1}+\dfrac{1}{acb+cd+c+1}+\dfrac{1}{abd+ad+d+1}\) \(\Leftrightarrow M=\dfrac{acd+cd+1}{acd+cd+c+1}+\dfrac{1}{abd+ad+d+1}\) \(\Leftrightarrow M=\dfrac{abcd+acd+cd}{abcd+acd+cd+c}+\dfrac{1}{abd+ad+d+1}\) \(\Leftrightarrow M=\dfrac{abd+ad+d}{abd+ad+d+1}+\dfrac{1}{abd+ad+d+1}\)