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+ \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)
\(\Rightarrow\frac{a}{a-b}=\frac{c}{c-d}\)
+ \(\frac{a}{c}=\frac{3a}{3c}=\frac{b}{d}=\frac{3a+b}{3c+d}\) \(\Rightarrow\frac{a}{3a+b}=\frac{c}{3c+d}\)
+ \(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{a^2}{c^2}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
\(\Rightarrow\frac{a\cdot b}{c\cdot d}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}\)
\(\Rightarrow\frac{a}{b}\cdot\frac{a}{b}=\frac{a^2+c^2}{b^2+d^2}\Rightarrow\frac{a\cdot c}{b\cdot d}=\frac{a^2+c^2}{b^2+d^2}\)
câu cuối lm tương tự
Từ \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{c}\cdot\frac{a}{c}=\frac{a}{c}\cdot\frac{b}{d}=\frac{b}{d}\cdot\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{ab}{cd}\)
a)Áp dụng t/c của dãy tỉ số bằng nhau ta có:
\(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)
b)\(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{ab}{cd}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{ab}{cd}=\frac{2ab}{2cd}\)
Áp dụng t/c của dãy tỉ số bằng nhau ta có:
\(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{ab}{cd}=\frac{2ab}{2cd}=\frac{a^2+b^2-2ab}{c^2+d^2-2cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
a) Ta có:\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
=> \(\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2\) => \(\frac{a^2}{c^2}=\frac{b^2}{d^2}\)
Áp dụng t/c dãy tỉ số = nhau được: \(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}\)
Mặt khác, \(\frac{a^2}{c^2}=\frac{a}{c}.\frac{a}{c}=\frac{a}{c}.\frac{b}{d}=\frac{ab}{cd}\)
Vậy \(\frac{a^2-b^2}{c^2-d^2}=\frac{ab}{cd}\left(=\left(\frac{a}{c}\right)^2\right)\)
b) \(\frac{a}{c}=\frac{b}{d}\)(câu a) => \(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\) (t/c dãy tỉ số = nhau)
=> \(\left(\frac{a}{c}\right)^2=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
Mặt khác, \(\left(\frac{a}{c}\right)^2=\frac{ab}{cd}\)(câu a) nên \(\frac{ab}{cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\left(\frac{a}{c}\right)^2\)
\(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\)
\(\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}=\frac{a^2-b^2}{c^2-d^2}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\left(\frac{a+b}{c+d}\right)^2=\frac{ab}{cd}\)
Vậy \(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)và \(\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
=>a=b.k
c=d.k
ta có Vế Phải : \(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(b.k\right)^2+b^2}{\left(d.k\right)^2+d^2}=\frac{b^2.k^2+b^2}{d^2.k^2+d^2}=\frac{b^2.\left(k^2+1\right)}{d^2.\left(k^2+1\right)}=\frac{b^2}{d^2}\)
Vế Trái :\(\frac{ab}{cd}=\frac{b.k.b}{d.k.d}=\frac{b^2.k}{d^2.k}=\frac{b^2}{d^2}\)
vì \(\frac{b^2}{d^2}=\frac{b^2}{d^2}\)
=>VP=VT
=>\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
=> a=b.k; c=d.k
Suy ra:
\(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(b.k\right)^2+b^2}{\left(d.k\right)^2+d^2}=\frac{b^2}{d^2}\) (1)
\(\frac{ab}{cd}=\frac{b.k.b}{d.k.d}=\frac{b^2}{d^2}\) (2)
Từ (1) và (2) => \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
Chúc bạn học tốt!
Áp dụng tính chất của dãy tỉ số bằng nhau
a^2+b^2/c^2+d^2 = a^2/c^2 = b^2 / d^2
=>a/c = b/d
=>a/b = c/d
Chúc bạn học tốt nha
1, \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{3a}{3c}=\frac{b}{d}=\frac{3a+b}{3c+d}\Rightarrow\frac{a}{c}=\frac{3a+b}{3c+d}\Rightarrow\frac{a}{3a+b}=\frac{c}{3c+d}\)
2, a, Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{c}\cdot\frac{a}{c}=\frac{a}{c}\cdot\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{ab}{cd}\)
\(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{b}{d}\cdot\frac{b}{d}\Rightarrow\frac{ab}{cd}=\frac{b^2}{d^2}\)
\(\Rightarrow\frac{ab}{cd}=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}\)
b, Ta có: \(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{a-b}{c-d}\cdot\frac{a-b}{c-d}\Rightarrow\frac{ab}{cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
Ta có\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\) (1)
Ta lại có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\frac{a}{c}.\frac{b}{d}=\frac{ab}{cd}\)(2)
Từ 1 và 2 \(\Rightarrow\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
Ta có: \(\frac{a}{b}=\frac{c}{d}=>\frac{a}{c}=\frac{b}{d}\)
=>\(\frac{a}{c}.\frac{b}{d}=\frac{a}{c}.\frac{a}{c}=>\frac{ab}{cd}=\frac{a^2}{c^2}\)
\(\frac{a}{c}.\frac{b}{d}=\frac{b}{d}.\frac{b}{d}=>\frac{ab}{cd}=\frac{b^2}{d^2}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{ab}{cd}=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\)
=>\(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}\)