Bài 7: Tỉ lệ thức

a) \(x^2=\left(-15\right).\left(-60\right)=900=>x=\)\(\pm\)\(30\)

b) \(-x^2=\dfrac{-16}{25}=>x^2=\dfrac{16}{25}=>x=\)\(\pm\)\(\dfrac{4}{5}\)

*Ta có : 4.256=16.64 =1024

\(\Rightarrow\) Có thể lập được tất cả 4 tỉ lệ thức sau :

\(\dfrac{4}{16}=\dfrac{64}{256};\dfrac{4}{64}=\dfrac{16}{256};\dfrac{256}{16}=\dfrac{64}{4};\dfrac{256}{64}=\dfrac{16}{4}\)

*Ta có : 16.1024 = 64.256 = 16384

\(\Rightarrow\) Có thể lập được tất cả 4 tỉ lệ thức sau :

\(\dfrac{16}{64}=\dfrac{256}{1024};\dfrac{64}{16}=\dfrac{1024}{256};\dfrac{16}{256}=\dfrac{64}{1024}\); \(\dfrac{64}{1024}=\dfrac{16}{256}\)

*Ta có : 4.1024 = 16.256 = 4096

\(\Rightarrow\) Có thể lập được tất cả 4 tỉ lệ thức sau :

\(\dfrac{4}{16}=\dfrac{256}{1024};\dfrac{4}{256}=\dfrac{16}{1024};\dfrac{256}{4}=\dfrac{1024}{16}\); \(\dfrac{16}{4}=\dfrac{1024}{256}\)

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) thì \(a=b.k\) , \(c=d.k\)

Ta tính giá trị của các tỉ số \(\dfrac{a-b}{a};\dfrac{c-d}{c}\) theo \(k\)

\(\dfrac{a-b}{a}=\dfrac{b.k-b}{b.k}=\dfrac{b.\left(k-1\right)}{b.k}=\dfrac{k-1}{k}\left(1\right)\)

\(\dfrac{c-d}{c}=\dfrac{d.k-d}{d.k}=\dfrac{d\left(k-1\right)}{d.k}=\dfrac{k-1}{k}\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\) suy ra \(\dfrac{a-b}{a}=\dfrac{c-d}{c}\)

a)\(3,8:\left(2x\right)=\dfrac{1}{4}:2\dfrac{2}{3}\)

\(2x=3,8:\dfrac{3}{32}\)

\(x=\dfrac{608}{15}:2\)

\(x=\dfrac{304}{15}\)

b)\(\left(0,25x\right):3=\dfrac{5}{6}:0,125\)

\(0,25x=\dfrac{20}{3}\times3\)

\(x=20:0,25\)

\(x=80\)

c) \(0,01:2,5=\left(0,75\times x\right):0,75\)

\(0,004=0,75\times x:0,75\)

\(0,004=x\times0,75:0,75\)

\(0,004=x\times1\)

\(0,004=x\)

\(\Rightarrow x=0,004\)

d)\(1\dfrac{1}{3}:0,8=\dfrac{2}{3}:\left(0,1x\right)\)

\(\dfrac{5}{3}=\dfrac{2}{3}:0,1\times x\)

\(\dfrac{5}{3}=\dfrac{20}{3}\times x\)

\(\Rightarrow x=\dfrac{20}{3}:\dfrac{5}{3}\)

\(x=4\)

\(\dfrac{a}{b}=\dfrac{c}{d}=>ad=bc=>ab+ad=ab+bc\)

\(a\left(b+d\right)=b\left(a+c\right)\)

\(\dfrac{a}{b}=\dfrac{a+c}{b+d}\)

\(\dfrac{a}{b}=\dfrac{c}{d}=>\)\(ad=bc\)

\(\dfrac{a}{a-b}=\dfrac{ad}{d\left(a-b\right)}=\dfrac{bc}{ad-bd}=\dfrac{bc}{bc-bd}=\dfrac{bc}{b\left(c-d\right)}\dfrac{c}{c-d}\)

ta có:

\(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)

mà \(\dfrac{a}{b}=\dfrac{c}{d}\)\(\Rightarrow\)\(\dfrac{aa}{bb}=\dfrac{a^2+a^2}{b^2+b^2}\)

\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{a^2.2}{b^2.2}\)

\(\Rightarrow\)\(\dfrac{a^2}{b^2}=\dfrac{a^2}{b^2}\)

\(\Rightarrow\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}(ĐPCM)\)

B

Đặt k = \(\dfrac{x}{4}=\dfrac{y}{7}\Rightarrow x=4k,y=7k\)

Từ x.y = 112, ta có: 4k.7k = 112

\(\Rightarrow\) \(28k^2\) = 112

\(\Rightarrow k^2=4\)

\(\Rightarrow\left[{}\begin{matrix}k=-2\\k=2\end{matrix}\right.\)

Có 2 trường hợp xảy ra:

TH1: k = -2

\(\Rightarrow x=-8,y=-14\)

TH2: k = 2

\(\Rightarrow x=8,y=14\)

Vậy \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-8\\y=-14\end{matrix}\right.\\\left\{{}\begin{matrix}x=8\\y=14\end{matrix}\right.\end{matrix}\right.\)