Giải giúp mình với

Đặt: (a;b;c;d)→(2016;x;y;2015)(a;b;c;d)→(2016;x;y;2015)

Phương trình trở thành:

∑ab+c=2∑ab+c=2

Đây chính là bất đẳng thức NesbitNesbit 4 biến.

Suy ra x=2015;y=2016x=2015;y=2016.

Đặt: (a; b; c; d) --> (2016; x; y; 2015)

Phương trình trở thành: \(\text{∑}\frac{a}{b+c}=2\)

=> x = 2015; y = 2016

ai giúp với đáp án là x=2015;y=2016 cách giải làm sao

đặt 2016=a;x=b;y=c;2015=d

pt trở thành:

\(\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}=2\)

đến đấy là bđt nesbit 4 số,dễ rồi

Cậu xem đề có còn gì nữa không ạ ? 

a) Ta có:

\(\frac{2a+b}{a+b}+\frac{2b+c}{b+c}+\frac{2c+d}{c+d}+\frac{2d+a}{d+a}=6\)

\(\Leftrightarrow\left[\left(\frac{2a+b}{a+b}-1\right)+\left(\frac{2b+c}{b+c}-1\right)-1\right]+\left[\left(\frac{2c+d}{c+d}-1\right)+\left(\frac{2d+a}{d+a}-1\right)-1\right]=0\)

\(\Leftrightarrow\left(\frac{a}{a+b}+\frac{b}{b+c}-1\right)+\left(\frac{c}{c+d}+\frac{d}{d+a}-1\right)=0\)

\(\Leftrightarrow\left(\frac{a.\left(b+c\right)}{\left(a+b\right).\left(b+c\right)}+\frac{b.\left(a+b\right)}{\left(a+b\right).\left(b+c\right)}-\frac{\left(a+b\right).\left(b+c\right)}{\left(a+b\right).\left(b+c\right)}\right)+\left(\frac{c.\left(d+a\right)}{\left(c+d\right).\left(d+a\right)}+\frac{d.\left(c+d\right)}{\left(c+d\right).\left(d+a\right)}-\frac{\left(c+d\right).\left(d+a\right)}{\left(c+d\right).\left(d+a\right)}\right)=0\)

\(\Leftrightarrow\left(\frac{ab+ac}{\left(a+b\right).\left(b+c\right)}+\frac{ab+b^2}{\left(a+b\right).\left(b+c\right)}-\frac{ab+ac+b^2+bc}{\left(a+b\right).\left(b+c\right)}\right)+\left(\frac{cd+ac}{\left(c+d\right).\left(d+a\right)}+\frac{cd+d^2}{\left(c+d\right).\left(d+a\right)}-\frac{cd+ac+d^2+ad}{\left(c+d\right).\left(d+a\right)}\right)=0\)

\(\Leftrightarrow\left(\frac{ab+ac+ab+b^2-ab-ac-b^2-bc}{\left(a+b\right).\left(b+c\right)}\right)+\left(\frac{cd+ac+cd+d^2-cd-ac-d^2-ad}{\left(c+d\right).\left(d+a\right)}\right)=0\)

\(\Leftrightarrow\frac{ab-bc}{\left(a+b\right).\left(b+c\right)}+\frac{cd-ad}{\left(c+d\right).\left(d+a\right)}=0\)

\(\Leftrightarrow\frac{ab-bc}{\left(a+b\right).\left(b+c\right)}=-\frac{cd-ad}{\left(c+d\right).\left(d+a\right)}\)

\(\Leftrightarrow\frac{ab-bc}{\left(a+b\right).\left(b+c\right)}=\frac{ad-cd}{\left(c+d\right).\left(d+a\right)}\)

\(\Leftrightarrow\frac{b.\left(a-c\right)}{\left(a+b\right).\left(b+c\right)}=\frac{d.\left(a-c\right)}{\left(c+d\right).\left(d+a\right)}\)

\(\Leftrightarrow\frac{b}{\left(a+b\right).\left(b+c\right)}=\frac{d}{\left(c+d\right).\left(d+a\right)}\) (vì \(a;b;c;d\) là số nguyên dương).

\(\Leftrightarrow b\left(c+d\right).\left(d+a\right)=d\left(a+b\right).\left(b+c\right)\)

\(\Leftrightarrow\left(bc+bd\right).\left(d+a\right)=\left(ad+bd\right).\left(b+c\right)\)

\(\Leftrightarrow bcd+abc+bd^2+abd=abd+acd+b^2d+bcd\)

\(\Leftrightarrow bd^2+abc=b^2d+acd\)

\(\Leftrightarrow bd^2-b^2d=acd-abc\)

\(\Leftrightarrow bd.\left(d-b\right)=ac.\left(d-b\right)\)

\(\Leftrightarrow bd.\left(d-b\right)-ac.\left(d-b\right)=0\)

\(\Leftrightarrow\left(d-b\right).\left(bd-ac\right)=0\)

Vì \(a;b;c;d\) là số nguyên dương.

\(\Rightarrow d-b>0\)

\(\Rightarrow d-b\ne0.\)

\(\Leftrightarrow bd-ac=0\)

\(\Leftrightarrow bd=ac.\)

Lại có:

\(A=abcd\)

\(\Rightarrow A=ac.bd\)

\(\Rightarrow A=ac.ac\)

\(\Rightarrow A=\left(ac\right)^2.\)

\(\Rightarrow A=abcd\) là số chính phương (đpcm).

Chúc bạn học tốt!