Ta có: \(\frac{a}{c}=\frac{a-b}{b-c}\)

=>a(b-c)=c(a-b)

=>ab-ac=ca-cb

=>ab+cb=2ac

\(\frac{1}{a}+\frac{1}{c}=\frac{a+c}{ac}=\frac{b\left(a+c\right)}{b\cdot a\cdot c}=\frac{2ac}{b\cdot ac}=\frac{2}{b}\)

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

\(=\frac{a-2b+c}{ab-ac-b^2+bc}=\frac{a-2b+c}{2ac-ac-b^2}=\frac{a-2b+c}{ac-b^2}\)

\(=\frac{b\left(a-2b+c\right)}{b\left(ac-b^2\right)}=\frac{b\left(a+c\right)-2b^2}{b\left(ac-b^2\right)}=\frac{2ac - 2b^2}{b(ac - b^2)}=\frac{2(ac - b^2)}{b(ac - b^2)}=\frac{2}{b}\)

Do đó: \(\frac{1}{a}+\frac{1}{c}=\frac{1}{b-c}-\frac{1}{a-b}\)

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

mk ko bt 123

\(\frac{a}{c}=\frac{a-b}{b-c}\Rightarrow a\left(b-c\right)=c\left(a-b\right)\)           (1)

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

\(\frac{1}{b-c}-\frac{1}{a}=\frac{a-b+c}{a\left(b-c\right)}\)                  (3)

\(Từ\left(1\right),\left(2\right),\left(3\right)\Rightarrow\)điều phải chứng minh

Lời giải:

Đặt $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=t$

$\Rightarrow x=at; y=bt; z=ct$. Ta có:

$(x+y+z)^2=(at+bt+ct)^2=t^2(a+b+c)^2=t^2(*)$

Mặt khác:

$x^2+y^2+z^2=(at)^2+(bt)^2+(ct)^2=t^2(a^2+b^2+c^2)=t^2(**)$

Từ $(*); (**)\Rightarrow (x+y+z)^2=x^2+y^2+z^2$ (đpcm)

Lời giải:

PT $\Leftrightarrow \frac{a+b-x}{c}+1+\frac{a+c-x}{b}+1+\frac{b+c-x}{a}+1+\frac{4x}{a+b+c}-4=0$

$\Leftrightarrow \frac{a+b+c-x}{c}+\frac{a+b+c-x}{b}+\frac{a+b+c-x}{a}-\frac{4(a+b+c-x)}{a+b+c}=0$

$\Leftrightarrow (a+b+c-x)(\frac{1}{c}+\frac{1}{b}+\frac{1}{a}-\frac{4}{a+b+c})=0$

$\Rightarrow a+b+c-x=0$ hoặc $\frac{1}{c}+\frac{1}{b}+\frac{1}{a}-\frac{4}{a+b+c}=0$
Nếu $\frac{1}{c}+\frac{1}{b}+\frac{1}{a}-\frac{4}{a+b+c}=0$, khi đó $x$ nhận mọi giá trị thực.

Nếu $\frac{1}{c}+\frac{1}{b}+\frac{1}{a}-\frac{4}{a+b+c}\neq 0$

$\Rightarrow a+b+c-x=0$
$\Rightarrow x=a+b+c$

Ta có \(\frac{a}{c}=\frac{b}{b}=\frac{a+b}{c+b}\) (tính chất tỉ lệ thức)

Vậy \(\frac{a}{c}=\frac{b}{b}=1\)

\(\Rightarrow a=c\)

Vậy \(\frac{c}{a}=1\)

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

a) \(ac=b^2\Rightarrow\frac{a}{b}=\frac{b}{c}\)

\(ab=c^2\Rightarrow\frac{a}{c}=\frac{c}{b}\)

Suy ra: \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{a+b+c}=1\)

\(P=\frac{b^{333}}{a^{111}.c^{222}}=\frac{b^{333}}{a^{111}.c^{111}.c^{111}}=\frac{b^{333}}{\left(ac\right)^{111}.c^{111}}=\frac{b^{333}}{\left(b^2\right)^{111}.c^{111}}=\frac{b^{333}}{b^{222}.c^{111}}=\frac{b^{111}}{c^{111}}=\left(\frac{b}{c}\right)^{111}\)

\(=1^{111}=1\)

Ta có a+b+c=0 => b+c=-a => a^2=b^2+2bc+c^2=> a^2-b^2-c^2=2bc

Tương tự ta có : b^2-c^2-a^2=2ca

c^2-a^2-b^2=2ab

=> a^2/2bc+b^2/2ca+c^2/2ab=(a^3+b^3+c^3)/2abc

=>Ta lại có a^3+b^3+c^3=(a+b+c)^3+

Ta có: \(\dfrac{a+3c}{b+3d}=\dfrac{a+c}{b+d}\left(b\ne-d;b\ne-3d;b\ne0;d\ne0\right)\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

+, \(\dfrac{a+3c}{b+3d}=\dfrac{a+c}{b+d}=\dfrac{a+3c-\left(a+c\right)}{b+3d-\left(b+d\right)}=\dfrac{a+3c-a-c}{b+3d-b-d}=\dfrac{2c}{2d}=\dfrac{c}{d}\)

Khi đó: \(\dfrac{a+c}{b+d}=\dfrac{c}{d}\)

+, \(\dfrac{a+c}{b+d}=\dfrac{c}{d}=\dfrac{a+c-c}{b+d-d}=\dfrac{a}{b}\) (đpcm)

Áp dụng t/c dãy tỉ số bằng nhau:

\(\dfrac{a+3c}{b+3d}=\dfrac{a+c}{b+d}=\dfrac{a+3c-\left(a+c\right)}{b+3d-\left(b+d\right)}=\dfrac{2c}{2d}=\dfrac{c}{d}\) (1)

\(\dfrac{a+3c}{b+3d}=\dfrac{a+c}{b+d}=\dfrac{3a+3c}{3b+3d}=\dfrac{a+3c-\left(3a+3c\right)}{b+3d-\left(3b+3d\right)}=\dfrac{-2a}{-2b}=\dfrac{a}{b}\) (2)

(1);(2) \(\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}\)