sao mà\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\) đó cách giải đó
sao mà\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\) đó cách giải đó
\(cho\frac{a}{b}=\frac{b}{c}=\frac{c}{d}.CM:\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a}{d}\)
\(Cho:\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)\(Cm:\frac{a}{d}=\left(\frac{a+b+c}{b+c+d}\right)^3\)
cho:\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}cm:\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a}{d}\)
1/ Biết \(\frac{a}{b}=\frac{c}{d}\), chứng minh
a) \(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
b) \(\left(\frac{a-d}{c-b}\right)^4=\frac{a^4+b^4}{c^4+d^4}\)
2/ Cho \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)
Chứng minh \(\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a}{b}\)
3/ Cho \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)
Chứng minh a=b=c
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 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=\(\left(\frac{a+b}{c+d}\right)^2+\left(\frac{b+c}{a+d}\right)^2+\left(\frac{c+d}{a+b}\right)^2+\left(\frac{a+d}{b+c}\right)^2\)
cho \(\frac{a}{b}=\frac{c}{d}\) .CM \(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{a^2-b^2}{c^2-d^2}\) (b,c,d khác 0,c+d khác 0, c-d khác 0)
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 \(\left(\frac{a}{b}+1\right).\left(\frac{b}{c}+1\right).\left(\frac{c}{d}+1\right).\left(1+\frac{d}{a}\right)\)
Cho hai số hữu tỉ \(\frac{a}{b},\frac{c}{d}\left(d,b>0\right)\)
CM\(\frac{a}{b}>\frac{c}{d}\Rightarrow\frac{a}{b}>\frac{a+c}{b+d}>\frac{c}{d}\)