Đặt \(\frac{a}{b}=\frac{c}{d}\)
=> a = bk ; c = dk
Ta có: \(\left(\frac{a+b}{c+d}\right)^2=\left(\frac{bk+b}{dk+d}\right)^2=9\left(\frac{b.\left(k+1\right)}{d.\left(k+1\right)}\right)=\left(\frac{b}{d}\right)^2=\frac{b^2}{d^2}\) ( 1 )
Lại có: \(\frac{a^2+b^2}{c^2+d^2}=\frac{bk^2+b^2}{dk^2+d^2}=\frac{b^2.\left(k^2+1\right)}{d^2.\left(k^2+1\right)}=\frac{b^2}{d^2}\) ( 2 )
Từ ( 1 ) và ( 2 ) => \(\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)
CMR nếu a/b=c/d thì\(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
cho \(\frac{a}{b}=\frac{c}{d}.chứngminhrằng:\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2+8d^2}\)
Cho \(\frac{a}{b}=\frac{c}{d}.CMR:\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
cho a/b= c/d thì
chứng minh\(\frac{7a^2-3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
\(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
cho \(\frac{a}{b}=\frac{c}{d}\)Chứng minh
cho \(\frac{a}{b}=\frac{c}{d}\)
CMR
\(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
C/m : nếu a/b = c/d thì
\(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
Cho tỉ số \(\frac{a}{b}=\frac{c}{d}\) .CM
\(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
Cho \(\frac{a}{b}=\frac{c}{d}\) . Chứng minh rằng:
\(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
