Chương I : Số hữu tỉ. Số thực - Hỏi đáp

Theo đề bài ta có:

\(\left\{{}\begin{matrix}\dfrac{A}{3}=\dfrac{B}{4}\\\dfrac{B}{5}=\dfrac{C}{6}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{A}{15}=\dfrac{B}{20}\\\dfrac{B}{20}=\dfrac{C}{24}\end{matrix}\right.\)

\(\Rightarrow\dfrac{A}{15}=\dfrac{B}{10}=\dfrac{C}{24}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{A}{15}=\dfrac{B}{10}=\dfrac{C}{24}=\dfrac{A-B}{15-10}=\dfrac{22}{5}\)

Mình nghĩ đề sai nhé

Theo đề bài:

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

\(\Rightarrow\left\{{}\begin{matrix}a=bh\\c=dh\end{matrix}\right.\)

Khi đó:

\(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bh+b}{dh+d}\right)^2=\left[\dfrac{b\left(h+1\right)}{d\left(h+1\right)}\right]^2=\dfrac{b^2}{d^2}=\dfrac{b}{d}\)

\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{bh^2+b^2}{dh^2+d^2}=\dfrac{b^2\left(h^2+1\right)}{d^2\left(h^2+1\right)}=\dfrac{b^2}{d^2}=\dfrac{b}{d}\)

Ta có điều phải chứng minh

Ta có \(P_1>0,P_2< 0,P_3=0\) (Vì có thừa số \(\dfrac{0}{11}=0\))

Do đó \(P_2< P_3< P_1\)

\(\dfrac{x-1}{101}-\dfrac{x-13}{103}=\dfrac{x-17}{107}-\dfrac{x-19}{109}\)

\(\Rightarrow\left(\dfrac{x-11}{101}+1\right)-\left(\dfrac{x-13}{103}+1\right)=\left(\dfrac{x-17}{107}+1\right)-\left(\dfrac{x-19}{109}+1\right)\)

\(\Rightarrow\dfrac{x+90}{101}-\dfrac{x+90}{103}=\dfrac{x+90}{107}-\dfrac{x+90}{109}\)

\(\Rightarrow\dfrac{x+90}{101}-\dfrac{x+90}{103}-\dfrac{x+90}{107}+\dfrac{x+90}{109}=0\)

\(\Rightarrow\left(x+90\right)\left(\dfrac{1}{101}-\dfrac{1}{103}-\dfrac{1}{107}+\dfrac{1}{109}\right)=0\)

Vì \(\dfrac{1}{101}-\dfrac{1}{103}-\dfrac{1}{107}+\dfrac{1}{109}\ne0\)

Nên \(x+90=0\Rightarrow x=-90\)

Theo đề ta có: \(7x=3y\) và \(x-y=16\)

\(\Rightarrow\dfrac{x}{3}=\dfrac{y}{7}\) và \(x-y=16\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{x-y}{3-7}\)\(=\dfrac{16}{-4}=-4\)

\(\Rightarrow\dfrac{x}{3}=-4=-4.3=-12\)\(\Rightarrow\dfrac{y}{7}=-4=-4.7=-28\)

Vậy \(x=-12;y=-28\)

7x = 3y

\(\Rightarrow\) x/3 = y/7 và x - y = 16

Theo t/c của dãy tỉ số bằng nhau ta có

x/3 = y/7 = x - y/3-7 = 16/-4 = -4

Suy ra :

x = (-4).3 = -12

y = (-4).7 = -28

Bài 1:

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

a, Ta có: \(\dfrac{a+c}{c}=\dfrac{bk+dk}{dk}=\dfrac{\left(b+d\right)k}{dk}=\dfrac{b+d}{d}\)

\(\Rightarrowđpcm\)

b, Ta có: \(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=\dfrac{k\left(b+d\right)}{b+d}=k\) (1)

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

Từ (1), (2) \(\Rightarrowđpcm\)

c, Ta có: \(\dfrac{a-c}{a}=\dfrac{bk-dk}{bk}=\dfrac{k\left(b-d\right)}{bk}=\dfrac{b-d}{b}\)

\(\Rightarrowđpcm\)

d, Ta có: \(\dfrac{3a+5b}{2a-7b}=\dfrac{3bk+5b}{2bk-7b}=\dfrac{b\left(3k+5\right)}{b\left(2k-7\right)}=\dfrac{3k+5}{2k-7}\)(1)

\(\dfrac{3c+5d}{2c-7d}=\dfrac{3dk+5d}{2dk-7d}=\dfrac{d\left(3k+5\right)}{d\left(2k-7\right)}=\dfrac{3k+5}{2k-7}\) (2)

Từ (1), (2) \(\Rightarrowđpcm\)

e, Sai đề

f, \(\left(\dfrac{a-b}{c-d}\right)^{2012}=\left(\dfrac{bk-b}{dk-d}\right)^{2012}=\left[\dfrac{b\left(k-1\right)}{d\left(k-1\right)}\right]^{2012}=\dfrac{b^{2012}}{d^{2012}}\)(1)

\(\dfrac{a^{2012}+b^{2012}}{c^{2012}+d^{2012}}=\dfrac{b^{2012}k^{2012}+b^{2012}}{d^{2012}k^{2012}+d^{2012}}=\dfrac{b^{2012}\left(k^{2012}+1\right)}{d^{2012}\left(k^{2012}+1\right)}=\dfrac{b^{2012}}{d^{2012}}\) (2)

Từ (1), (2) \(\Rightarrowđpcm\)

2. a) 3/x-4=x+4/3

<=> 9=x^2-16

<=> x^2=25

<=> x=5 hoặc x=-5

b)

1)

a) \(1\dfrac{5}{6}=\dfrac{-x}{5}\)

\(\Rightarrow\dfrac{11}{6}=\dfrac{-x}{5}\)

\(\Rightarrow-x=\dfrac{5.11}{6}=\dfrac{55}{6}\)

\(\Rightarrow x=-\dfrac{55}{6}\)

b) 4,25 : 8 = -3,5 : x

\(\dfrac{4,25}{8}=\dfrac{-3,5}{x}\)

\(x=\dfrac{-3,5.8}{4,25}\)

\(x=\dfrac{-28}{4,25}\)

2.

\(-\dfrac{12}{1,6}=\dfrac{55}{-7\dfrac{1}{3}}\)

\(\Rightarrow-\dfrac{12}{1,6}=\dfrac{55}{-\dfrac{22}{3}}\)

Ta có thể lặp đc các tỉ lệ thức sau:

\(-\dfrac{12}{1,6}=\dfrac{55}{-\dfrac{22}{3}}\)

\(\dfrac{-\dfrac{22}{3}}{1,6}=\dfrac{55}{-12}\)

\(-\dfrac{12}{55}=\dfrac{1,6}{-\dfrac{22}{3}}\)

\(\dfrac{1,6}{-12}=\dfrac{-\dfrac{22}{3}}{55}\)

https://hoc24.vn/hoi-dap/ti-le-thuc.4497/

\(Từ\) \(\dfrac{-1,2}{1,6}=\dfrac{55}{-7\dfrac{1}{3}}\Rightarrow\) \(\dfrac{1,6}{-1,2}=\dfrac{-7\dfrac{1}{3}}{55}\)

\(\Rightarrow\dfrac{-7\dfrac{1}{3}}{55}=\dfrac{1,6}{-1,2}\)

\(\Rightarrow\dfrac{55}{-7\dfrac{1}{3}}=\dfrac{-1,2}{1,6}\)