\(\hept{\begin{cases}ac=b^2\Rightarrow\frac{a}{b}=\frac{b}{c}\\ab=c^2\Rightarrow\frac{c}{a}=\frac{b}{c}\end{cases}\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{a}}\)

Theo t/c cuae dãy tỉ số bằng nhau ta có:

\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1\) (vì a+b+c khác 0)

=> a/b = 1 => a = b

b/c = 1 => b = c

=> a=b=c

=> \(\frac{b^{3333}}{a^{1111}.c^{2222}}=\frac{b^{3333}}{b^{1111}.b^{2222}}=1\)

cho ac=b2;ab=c2,a+b+ckhác 0 và a,b,clà các số khác 0.

tính;b3333a1111.c2222 

Toán lớp 7

{

⇒ab =bc =ca 

Theo t/c cuae dãy tỉ số bằng nhau ta có:

ab =bc =ca =a+b+cb+c+a =1 (vì a+b+c khác 0)

=> a/b = 1 => a = b

b/c = 1 => b = c

=> a=b=c

=> b3333a1111.c2222 =b3333b1111.b2222 =1

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

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

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

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

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

\(\Leftrightarrow a^2+ab+\frac{b^2}{3}=c^2+\frac{b^2}{3}+a^2+ac+c^2\)

\(\Leftrightarrow ab=2c^2+ca\Leftrightarrow ab+ac=2c^2+2ac\)

\(\Leftrightarrow a\left(b+c\right)=2c\left(a+c\right)\Rightarrow\frac{2c}{a}=\frac{b+c}{a+c}\rightarrowđpcm\)

\(\left\{{}\begin{matrix}a^2+ab+\dfrac{b^2}{3}=25\\c^2+\dfrac{b^2}{3}=9\\a^2+ac+c^2=16\end{matrix}\right.\)

\(\Rightarrow a^2+ab+\dfrac{b^2}{3}=c^2+\dfrac{b^2}{3}+a^2+ac+c^2\)

\(\Rightarrow ab=2c^2+ac\)

Biến đổi 1 chút là ra

\(\rightarrowđpcm\)

\(\hept{\begin{cases}a^2+ab+\frac{b^2}{3}=25\\c^2+\frac{b^2}{3}=9\end{cases}}\Rightarrow a^2+ac-c^2=16\)

\(\Rightarrow a^2+ab-c^2=a^2+ac+c^2\left(=16\right)\)

\(\Rightarrow ab-c^2=ac+c^2\)

\(\Rightarrow ab=ac+2c^2\)

\(\Rightarrow ab+ac=2ac+2c^2\)

\(\Leftrightarrow a\left(b+c\right)=2c\left(a+c\right)\)

\(\Leftrightarrow\frac{2c}{a}=\frac{b+c}{a+c}\left(đpcm\right)\)