Áp dụng t/c dãy tỉ số bằng nhau, ta có:
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{a+b+c+d}=1\)
\(\Rightarrow\hept{\begin{cases}a=b\\b=c\\c=d\end{cases}}\Rightarrow a=b=c=d\)
\(\Rightarrow M=\frac{\left(2a\right)^4}{a^4}=16\)
Cho \(a.b.c.d\ne0\) và \(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\). Tính \(P=\left(1+\frac{b}{a}\right)\left(1+\frac{c}{d}\right)\left(1+\frac{a}{c}\right)\)
cho a;b;c;d là các số thực khác 0 thảo mãn
\(\frac{a-b+c+d}{b}=\frac{a+b-c+d}{c}=\frac{a+b+c-d}{d}=\frac{b+c+d-a}{a}\)
Tính giá trị của biểu thức
\(M=\frac{\left(a+b+c\right)\left(a+b+d\right)\left(b+c+d\right)\left(c+d+a\right)}{abcd}\)
Cho dãy tỉ số bằng nhau \(\frac{3a+b+c+d}{a}=\frac{a+3b+c+d}{b}=\frac{a+b+3c+d}{c}=\frac{a+b+c+3d}{d}\)
Tính Q=\(\left(\frac{a+b}{c+d}\right)^2+\left(\frac{b+c}{a+d}\right)^2+\left(\frac{c+d}{a+b}\right)^2+\left(\frac{a+d}{b+c}\right)^2\)
Biet \(\frac{-a+b+c+d}{a}=\frac{a-b+c+d}{b}=\frac{a+b-c-d}{c}=\frac{a+b+c-d}{d}\)
Tinh gia tri bieu thuc \(\left(\frac{a}{b}+1\right).\left(\frac{b}{c}+1\right).\left(\frac{c}{d}+1\right).\left(1+\frac{d}{a}\right)\)
1/ cho \(\frac{a}{b}=\frac{c}{d}\)chứng minh rằng:
a) \(\frac{a.b}{c.d}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
b)\(\frac{a,d}{c.b}=\frac{\left(a+b\right).\left(a-b\right)}{\left(c+d\right).\left(c-d\right)}\)
2/ cho \(a.b=c^2\)chứng minh : \(\frac{a}{b}=\frac{\left(2a+3c\right)^2}{\left(2c+3b\right)^2}\)
Cho a,b,c,d thoả mãn:
\(\frac{a+b+c}{d}=\frac{b+c+d}{a}=\frac{a+c+d}{b}=\frac{d+a+b}{c}\)
Tìm: \(B=\left(1+\frac{a+b}{c+d}\right)\cdot\left(1+\frac{b+c}{d+d}\right)\cdot\left(1+\frac{c+d}{a+b}\right)\cdot\left(1+\frac{d+a}{b+c}\right)\)
cho a/b = c/d . tính \(\frac{a.b}{c.d}+\left[\left(\frac{a+b}{c+d}\right)^2:\left(\frac{a^2+b^2}{c^2+d^2}\right)\right]-\frac{a^2-b^2}{c^2-d^2}\)
Cho\(\frac{a}{b}\)=\(\frac{c}{d}\) chứng minh
1,\(\frac{a^2+c^2}{b^2+d^2}\)=\(\frac{a.c}{b.d}\)
2,\(\frac{a^2+c^2}{b^2+d^2}\)=\(\frac{a^2-c^2}{b^2-d^2}\)
\(3,\left(a+c\right).\left(b-d\right)=\left(a-c\right).\left(b+d\right)\)
\(4,\left(b+d\right).c=\left(c+c\right).d\)
\(5,\frac{4.a-12.b}{8.a+11.b}=\frac{4.c-12.d}{8.c+11.d}\)
\(6,\frac{\left(a+c\right)^2}{\left(b+d\right)^2}=\frac{\left(a+c\right)^2}{\left(b+d\right)^2}\)
\(7,\frac{a^{10}+b^{10}}{\left(a+b\right)^{10}}=\frac{c^{10}+d^{10}}{\left(c+d\right)^{10}}\)
Rút gọn biểu thức sau:
\(\frac{\left(a.b+b.c+c.d+d.a\right).a.b.c.d}{\left(c+d\right).\left(a+b\right)+\left(b-c\right).\left(a-d\right)}\)
