Question

[Trần Anh Đại]

cho các số a.b.c;d nguyên dương đôi một khác nhau t/m:

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

cmr abcd là 1 số chính phương

Answer 1

[๖Fly༉Donutღღ]

Mình không chắc câu này lắm nhưng thôi giải dùm bạn vậy :((

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

\(\Leftrightarrow\)\(1+\frac{a}{a+b}+1+\frac{b}{b+c}+1+\frac{c}{c+d}+1+\frac{d}{d+a}=6\)

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

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

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

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

\(\Leftrightarrow\)\(b\left(c+d\right)\left(d+a\right)-d\left(a+b\right)\left(b+c\right)\)

\(\Leftrightarrow\)\(abc-acd+bd^2-b^2d=0\)

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

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

Vậy bạn tự kết luận nha

Answer 2

[zZz Cool Kid_new zZz]

\(\Leftrightarrow1+\frac{a}{a+b}+1+\frac{b}{b+c}+1+\frac{c}{c+d}+1+\frac{d}{d+a}=6\)

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

\(\Leftrightarrow1-\frac{a}{a+b}-\frac{b}{b+c}+1-\frac{c}{c+d}-\frac{d}{d+a}=0\)

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

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

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

\(\Leftrightarrow b\left(c-a\right)\left(c+d\right)\left(d+a\right)+d\left(a-c\right)\left(a+b\right)\left(b+c\right)=0\)

\(\Leftrightarrow b\left(c-a\right)\left(c+d\right)\left(d+a\right)-d\left(c-a\right)\left(a+b\right)\left(b+c\right)=0\)

\(\Leftrightarrow b\left(c+d\right)\left(d+a\right)-d\left(a+b\right)\left(b+c\right)=0\)

\(\Leftrightarrow\left(bc+bd\right)\left(d+a\right)-\left(da+db\right)\left(b+c\right)=0\)

\(\Leftrightarrow bcd+bca+bd^2+bda-abd-adc-db^2-dbc=0\)

\(\Leftrightarrow bca-acd+bd^2-b^2d=0\)

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

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

\(\Leftrightarrow ac-bd=0\)

\(\Leftrightarrow ac=bd\)

\(\Leftrightarrow\left(ac\right)^2=abcd\)\(\left(đpcm\right)\)

dành cho người không hiểu bài trên 

                                                                           \(#huybip#\)