Đặt \(\left(\frac{a-b}{c},\frac{b-c}{a},\frac{c-a}{b}\right)\rightarrow\left(x,y,z\right)\)

Khi đó:\(\left(\frac{c}{a-b},\frac{a}{b-c},\frac{b}{c-a}\right)\rightarrow\left(\frac{1}{x},\frac{1}{y},\frac{1}{z}\right)\)

Ta có:

\(P\cdot Q=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3+\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\)

Mặt khác:\(\frac{y+z}{x}=\left(\frac{b-c}{a}+\frac{c-a}{b}\right)\cdot\frac{c}{a-b}=\frac{b^2-bc+ac-a^2}{ab}\cdot\frac{c}{a-b}\)

\(=\frac{c\left(a-b\right)\left(c-a-b\right)}{ab\left(a-b\right)}=\frac{c\left(c-a-b\right)}{ab}=\frac{2c^2}{ab}\left(1\right)\)

Tương tự:\(\frac{x+z}{y}=\frac{2a^2}{bc}\left(2\right)\)

\(=\frac{x+y}{z}=\frac{2b^2}{ac}\left(3\right)\)

Từ ( 1 );( 2 );( 3 ) ta có:
\(P\cdot Q=3+\frac{2c^2}{ab}+\frac{2a^2}{bc}+\frac{2b^2}{ac}=3+\frac{2}{abc}\left(a^3+b^3+c^3\right)\)

Ta có:\(a+b+c=0\)

\(\Rightarrow\left(a+b\right)^3=-c^3\)

\(\Rightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)

\(\Rightarrow a^3+b^3+c^3=3abc\)

Khi đó:\(P\cdot Q=3+\frac{2}{abc}\cdot3abc=9\)

Mách mk nốt 2 bài kia vs

chiju

Đặt\(\frac{a}{3}=\frac{b}{12}=\frac{c}{5}\)= k => a= 3k; b= 12k;c=5k

a.b.c = 22,5 => 3k.12k.5k = 22,5 = 180k3 = 22,5 => k3 = 0,125 => k = 0,5

Do đó:\(\frac{a}{3}=0,5=>a=1,5\)

          \(\frac{b}{12}=0,5=>b=6\)

          \(\frac{c}{5}=0,5=>c=2,5\)   

Vậy...

Đặt \(\frac{a}{3}=\frac{b}{12}=\frac{c}{5}\)= k => a = 3k ; b = 12k ; c = 5k 

a.b.c = 22,5 => 3k.12k.5k = 22,5 => 180k³ = 22,5 => k³ = 0,125 => k= 0,5 

Do đó : \(\frac{a}{3}=0,5\Rightarrow a=1,5\)

             \(\frac{b}{12}=0,5\Rightarrow b=6\)

             \(\frac{c}{5}=0,5\Rightarrow c=2,5\)

Vậy ...

Ta có: a/3=b/12=c/5. Đặt a=3k ; b=12k ; c=5k

=>3.12.5.k^3=22,5 => k^3=0,125=>k=0,5

=>a=3.0,5=1,5

    b=12.0,5=6

    c=5.0,5=2,5

Cho abc=0 thì không chứng minh được, a+b+c=0 là đủ rồi

Ta có: a+b+c=0 => a+b=-c

=>(a+b)²=(-c)²

=>a²+2ab+b²=c²

=>a²+b²-c²=-2ab

Tương tự ta có: b²+c²-a²=-2bc ; c²+a²-b²=-2ca

=>\(\frac{1}{b^2+c^2-a^2}+\frac{1}{c^2+a^2-b^2}+\frac{1}{a^2+b^2-c^2}=-\frac{1}{2bc}-\frac{1}{2ca}-\frac{1}{2ab}=\frac{a+b+c}{-2abc}=0\) (đpcm)

Cho \(abc=0\)thì không chứng minh được, \(a+b+c=0\)là đủ rồi.

Ta có: \(a+b+c=0\Rightarrow a+b=-c\)

\(\Rightarrow\left(a+b\right)^2=\left(-c\right)^2\)

\(\Rightarrow a^2+2ab+b^2=c^2\)

\(\Rightarrow a^2+b^2-c^2=-2ab\)

Tương tự ta có: \(b^2+c^2-a^2=-2ab;c^2+a^2-b^2=-2ca\)

\(\Rightarrow\frac{1}{b^2+c^2-a^2}+\frac{1}{c^2+a^2-b^2}+\frac{1}{a^2+b^2-c^2}=-\frac{1}{2bc}-\frac{1}{2ca}-\frac{1}{2ab}=\frac{a+b+c}{-2abc}=0\)

Đặt: \(\frac{a}{3}=\frac{b}{4}=\frac{c}{5}=k\)

Suy ra: a=3k, b=4k, c=5k

a.b.c=480 suy ra 3k.4k.5k=480

suy ra: 60.k^3=480

            k^3=480:60=8

Vậy k=2

Thay vào ta có:a=6, b=8,c=10

Đặt: \(\frac{a}{3}=\frac{b}{4}=\frac{c}{5}=k\)

\(\Rightarrow\)  a = 3k, b = 4k, c = 5k

a.b.c=480 \(\Rightarrow\) 3k.4k.5k=480

\(\Rightarrow\) 60.k^3=480

            k^3 = 480:60 = 8

Vậy  k= 2

Thay vào ta có:a = 6 , b = 8, c = 10

1, \(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}=\dfrac{a+b+c+d}{3\left(a+b+c+d\right)}=\dfrac{1}{3}\)

Do đó \(\left\{{}\begin{matrix}3a=b+c+d\left(1\right)\\3b=a+c+d\left(2\right)\\3c=a+b+d\left(3\right)\\3d=a+b+c\left(4\right)\end{matrix}\right.\)

Từ (1) và (2) \(\Rightarrow3\left(a+b\right)=a+b+2c+2d\Leftrightarrow2\left(a+b\right)=2\left(c+d\right)\Leftrightarrow a+b=c+d\Leftrightarrow\dfrac{a+b}{c+d}=1\)

Tương tự cũng có: \(\dfrac{b+c}{a+d}=1;\dfrac{c+d}{a+b}=1;\dfrac{d+a}{b+c}=1\)

\(\Rightarrow A=4\)

2, Có \(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}\)\(\Leftrightarrow\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{36}=\dfrac{x^2+y^2+z^2}{4+16+36}=\dfrac{14}{56}=\dfrac{1}{4}\)

Do đó \(\dfrac{x^2}{4}=\dfrac{1}{4};\dfrac{y^2}{16}=\dfrac{1}{4};\dfrac{z^2}{36}=\dfrac{1}{4}\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=1\\y^2=4\\z^2=9\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\pm1\\y=\pm2\\z=\pm3\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)=\left(1;2;3\right),\left(-1;-2;-3\right)\)

Bài 2 :

a, Ta có : \(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\)

\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}\)

\(\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{36}=\dfrac{x^2+y^2+z^2}{4+16+36}=\dfrac{1}{4}\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=1\\y^2=4\\z^2=9\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\pm1\\y=\pm2\\z=\pm3\end{matrix}\right.\)

Vậy ...

b, Ta có : \(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+3y-1}{5+7}=\dfrac{2x+3y-1}{6x}\)

\(\Rightarrow6x=12\)

\(\Rightarrow x=2\)

\(\Rightarrow y=3\)

Vậy ...

Cho a,b,c thuộc R và a.b.c=1.chứng minh  \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}=1\)

Giải:Ta có:\(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\)

\(=\frac{a.c}{abc+ac+c}+\frac{b}{bc+b+abc}+\frac{c}{ca+c+1}\)

\(=\frac{ac}{ac+c+1}+\frac{1}{c+1+ac}+\frac{c}{ca+c+1}\)

\(=\frac{ac+1+c}{ac+c+1}=1\)

Suy ra điều phải chứng minh

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(...=\frac{a+b-c+b+c-a+c+a-b}{c+a+b}=\frac{a+b+c}{a+b+c}=1\)( a, b, c khác 0 )

=> \(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}=1\)

=> \(\hept{\begin{cases}a+b-c=c\\b+c-a=a\\c+a-b=b\end{cases}}\Leftrightarrow\hept{\begin{cases}a+b=2c\\b+c=2a\\c+a=2b\end{cases}}\)

Thế vào P ta được :

\(P=\frac{2c}{a}\cdot\frac{2a}{b}\cdot\frac{2b}{c}=\frac{8abc}{abc}=8\)