\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

vãi cả loz sao lại sai ?

*\(\frac{a}{b}=\frac{c}{d}\)=> \(\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2-c^2}{b^2-d^2}=\frac{a}{b}.\left(\frac{a}{b}\right)=\frac{ac}{bd}\)(đpcm)

* \(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\)=> \(\frac{a^2}{b^2}=\frac{c^2}{d^2}=\left(\frac{a+c}{b+d}\right)^2\)(1)

Ta lại có \(\frac{a}{b}=\frac{c}{d}\)=>\(\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}\)(2)

Từ (1),(2) => đpcm

Mình lớp 5 nên mình không biết

đi mk làm cho

a: Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=>a=bk; c=dk

\(\frac{ab}{cd}=\frac{bk\cdot b}{dk\cdot d}=\frac{b^2}{d^2}\)

\(\frac{a^2-b^2}{c^2-d^2}=\frac{\left(bk\right)^2-b^2}{\left(dk\right)^2-d^2}=\frac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\frac{b^2}{d^2}\)

Do đó: \(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)

b: \(\frac{7a-4b}{3a+5b}=\frac{7\cdot bk-4b}{3\cdot bk+5b}=\frac{b\left(7k-4\right)}{b\left(3k+5\right)}=\frac{7k-4}{3k+5}\)

\(\frac{7c-4d}{3c+5d}=\frac{7\cdot dk-4d}{3\cdot dk+5d}=\frac{d\left(7k-4\right)}{d\left(3k+5\right)}=\frac{7k-4}{3k+5}\)

Do đó: \(\frac{7a-4b}{3a+5b}=\frac{7c-4d}{3c+5d}\)

c: \(\frac{ac}{bd}=\frac{bk\cdot dk}{bd}=k^2\)

\(\frac{a^2+c^2}{b^2+d^2}=\frac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\frac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\)

\(\frac{\left(c-a\right)^2}{\left(d-b\right)^2}=\frac{\left(dk-bk\right)^2}{\left(d-b\right)^2}=\frac{k^2\left(d-b\right)^2}{\left(d-b\right)^2}=k^2\)

Do đó; \(\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}=\frac{\left(c-a\right)^2}{\left(d-b\right)^2}\)

d: \(\frac{a^3+b^3}{c^3+d^3}=\frac{\left(bk\right)^3+b^3}{\left(dk\right)^3+d^3}=\frac{b^3\left(k^3+1\right)}{d^3\left(k^3+1\right)}=\frac{b^3}{d^3}\)

\(\frac{\left(a+b\right)^3}{\left(c+d\right)^3}=\frac{\left(bk+b\right)^3}{\left(dk+d\right)^3}=\frac{b^3\left(k+1\right)^3}{d^3\left(k+1\right)^3}=\frac{b^3}{d^3}\)

Do đó: \(\frac{a^3+b^3}{c^3+d^3}=\frac{\left(a+b\right)^3}{\left(c+d\right)^3}\)

Do đó:

b) ab(c^2+d^2)=ab.c^2+ab.d^2=(a.c)(b.c)+(a.d)(b.d)
cd(a^2+b^2)=cd.a^2+cd.b^2=(c.a)(d.a)+(c.b)(d.b)
(a.c)(b.c)+(a.d)(b.d)=(c.a)(d.a)+(c.b)(d.b) vì mỗi vế đều bằng nhau

a: Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=>a=bk; c=dk

\(\frac{a}{a-b}=\frac{bk}{bk-b}=\frac{bk}{b\left(k-1\right)}=\frac{k}{k-1}\)

\(\frac{c}{c-d}=\frac{dk}{dk-d}=\frac{dk}{d\left(k-1\right)}=\frac{k}{k-1}\)

Do đó: \(\frac{a}{a-b}=\frac{c}{c-d}\)

c: \(\frac{a}{3a+b}=\frac{bk}{3bk+b}=\frac{bk}{b\left(3k+1\right)}=\frac{k}{3k+1}\)

\(\frac{c}{3c+d}=\frac{dk}{3dk+d}=\frac{dk}{d\left(3k+1\right)}=\frac{k}{3k+1}\)

Do đó: \(\frac{a}{3a+b}=\frac{c}{3c+d}\)

e: \(\frac{a\cdot b}{c\cdot d}=\frac{bk\cdot b}{dk\cdot d}=\frac{b^2k}{d^2k}=\frac{b^2}{d^2}\)

\(\frac{a^2-b^2}{c^2-d^2}=\frac{\left(bk\right)^2-b^2}{\left(dk\right)^2-d^2}=\frac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\frac{b^2}{d^2}\)

Do đó: \(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)