cho a, b ,c ,d thoa man \(^{b^2=ac}\)\(^{c^2=bd}\)chung minh \(\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\dfrac{a}{d}\)
cho a, b, c, d khac 0 va thoa man
ac=b^2; bd=c^2
chung minh \(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{d}\)
Cho b^2 = ac ; c^2 = bd với b, c, d ≠ 0; b+c ≠ 0; b^3+c^3≠ d^3 3. Chứng minh rằng:
a) \(\dfrac{a^3+b^3-c^3}{b^3+c^3-d^3}=\left(\dfrac{a+b-c}{b+c-d}\right)^3\)
b) \(\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\dfrac{a}{d}\)
Cho a, b, c, d là 4 số khác 0 thỏa mãn \(b^2\) = ac; \(c^2\) = bd và \(b^3+c^3+d^3\ne0\)
Chứng minh rằng: \(\dfrac{a}{d}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\)
Cho a, b, c, d là 4 số khác 0 thỏa mãn: \(b^2=ac;c^2=bd\) và \(b^3+c^3+d^3\ne0\)
Chứng minh rằng: \(\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\) = \(\left(\dfrac{a+b+c}{b+c+d}\right)^3\)
\(\left\{{}\begin{matrix}b^2=ac\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\\c^2=bd\Rightarrow\dfrac{b}{c}=\dfrac{c}{d}\end{matrix}\right.\)\(\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
Áp dụng t/c dtsbn:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\Rightarrow\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a^3}{b^3}\left(1\right)\)
Và \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\Rightarrow\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\left(đpcm\right)\)
Cho a, b, c, d \(\ne0\) và \(b^2=ac,c^2=bd,b^3+c^3+d^3\ne0\)
Chứng minh: \(\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\dfrac{a}{d}\)
cho b2 =ac,c2=bd(b,c,d khác 0,b+c khác d, b3+c3 khác d
chứng minh rằng: \(\dfrac{a^3+b^3-c^3}{b^3+c^3+d^3}=\left(\dfrac{a+b-c}{b+c-d}\right)^3\)
Ta có: \(b^2=ac;c^2=bd\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{b}{c}\\\dfrac{b}{c}=\dfrac{c}{d}\end{matrix}\right.\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b-c}{b+c-d}\)
Đặt: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b-c}{b+c-d}=l\) ta có:
\(\left\{{}\begin{matrix}\left(\dfrac{a+b-c}{b+c-d}\right)^3=l^3\\\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3-c^3}{b^3+c^3-d^3}=l^3\end{matrix}\right.\Rightarrowđpcm\)
cho a , b, c la cac so thuc duong thoa man he thuc a+b+c=6abc
Chung minh rang \(\dfrac{bc}{a^3\left(c+2b\right)}+\dfrac{ac}{b^3\left(a+2c\right)}+\dfrac{ab}{c^3\left(b+2a\right)}\ge2\)
Cho 4 so a,b,c,d khac 0 thoa man;b^2=ac,c^2=bd,b^3+c^3+d^3 khac 0
CMR;a^3+b^3+c^3/b^3+c^3+d^3=a/d
Cho b2 = ac; c2 = bd. Với b,c,d khác 0; b+c khác d; b3 + c3 khác d3. Chứng minh rằng \(\dfrac{a^3+b^3-c^3}{b^3+c^3-d^3}=\left(\dfrac{a+b-c}{b+c-d}\right)^3\)
\(\left\{{}\begin{matrix}b^2=ac\\c^2=bd\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{b}{c}\\\dfrac{b}{c}=\dfrac{c}{d}\end{matrix}\right.\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
Đặt:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=k\) \(\Rightarrow\left\{{}\begin{matrix}a=bk\\b=ck\\c=dk\end{matrix}\right.\)
Thay vào r tính