Bài 7: Tỉ lệ thức

ở câu hs tương tự có người trả lời giúp bn đấy

Ta có \(\dfrac{ab}{b}=\dfrac{bc}{c}\Rightarrow\dfrac{a}{b}\cdot\dfrac{b}{b}=\dfrac{b}{c}\cdot\dfrac{c}{c}\)

\(\Rightarrow\dfrac{a}{b}\cdot1=\dfrac{b}{c}\cdot1\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\)(đpcm)

Đề sai nhé!

Đức Mạnh Trần Tommy

Giải:

Có:

\(\dfrac{\overline{ab}}{b}=\dfrac{\overline{bc}}{c}\)

\(\Leftrightarrow abc=b^2c\)

Rút gọn c ở cả hai vế, ta được:

\(ab=b^2\)

\(\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{b}\) (đpcm)

Hình như là sai đó bạn Đức Mạnh Trần Tommy, thực sự mình không biết làm bài này.

Học tốt!

Cần điều kiên là a,b,c,d khác 0 và a+b+c+d khác 0;a-b-c+d khác 0;a-b+c-d khác 0 ; a+b-c-d khác 0 nha

Bài này sử dung tính chất dãy tỉ số bằng nhau và tính giao hoán của phép nhân để làm nhé

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

\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}=\dfrac{a+b-c+b+c-a+c+a-b}{c+a+b}=\dfrac{a+b+c}{a+b+c}=1\)\(\dfrac{a+b-c}{c}=1\Leftrightarrow\dfrac{a+b}{c}-\dfrac{c}{c}=1\Leftrightarrow\dfrac{a+b}{c}-1=1\Leftrightarrow\dfrac{a+b}{c}=2\)\(\dfrac{b+c-a}{a}=1\Leftrightarrow\dfrac{b+c}{a}-\dfrac{a}{a}=1\Leftrightarrow\dfrac{b+c}{a}-1=1\Leftrightarrow\dfrac{b+c}{a}=2\)\(\dfrac{c+a-b}{b}=1\Leftrightarrow\dfrac{c+a}{b}-\dfrac{b}{b}=1\Leftrightarrow\dfrac{c+a}{b}-1=1\Leftrightarrow\dfrac{c+a}{b}=2\)\(P=\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{c}{b}\right)\left(1+\dfrac{a}{c}\right)\\ =\dfrac{a+b}{a}\cdot\dfrac{b+c}{b}\cdot\dfrac{c+a}{c}\\ =\left(a+b\right)\cdot\dfrac{1}{a}\cdot\left(b+c\right)\cdot\dfrac{1}{b}\cdot\left(c+a\right)\cdot\dfrac{1}{c}\\ =\left(a+b\right)\cdot\dfrac{1}{c}\cdot\left(b+c\right)\cdot\dfrac{1}{a}\cdot\left(c+a\right)\cdot\dfrac{1}{b}\\ =\dfrac{a+b}{c}\cdot\dfrac{b+c}{a}\cdot\dfrac{c+a}{b}\\ =2\cdot2\cdot2\\ =8\)

Vậy \(P=8\)

Giải:
Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\)

Ta có: \(xy=6\)

\(\Rightarrow6k^2=6\)

\(\Rightarrow k^2=1\)

\(\Rightarrow k=\pm1\)

+) \(k=1\Rightarrow x=2,y=3\)

+) \(k=-1\Rightarrow x=-2,y=-3\)

Vậy...

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\)

Vì xy = 6 \(\Rightarrow\) 2k . 3k = 6

6k = 6

k = 1

Vậy \(\left\{{}\begin{matrix}x=2.1=2\\y=3.1=3\end{matrix}\right.\)

Xét: \(\dfrac{x}{2}=\dfrac{y}{3}=k\)

\(\Rightarrow x=2k;y=3k\)

\(\Rightarrow xy=2k.3k=6k^2\)

\(\Rightarrow6k^2=6\Rightarrow k^2=1\Rightarrow\left[{}\begin{matrix}k=-1\\k=1\end{matrix}\right.\)

TH1: \(x=-2;y=-3\) (loại)

TH2: \(x=2;y=3\) (chọn)

Vậy x=2; y=3

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

\(\dfrac{x-3}{7}=\dfrac{y-5}{5}=\dfrac{z-7}{3}=\dfrac{x-3+y-5+z-7}{7+5+3}\)

\(=\dfrac{45-15}{15}=\dfrac{30}{15}=2\) (do \(x+y+z=45\))

\(\Rightarrow\left\{{}\begin{matrix}x=2.7+3\\y=2.5+5\\z=2.3+7\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=17\\y=15\\z=13\end{matrix}\right.\)

Vậy.......

Chúc bạn học tốt!!!

\(\left(85\dfrac{7}{30}-83\dfrac{5}{18}\right):2\dfrac{2}{3}=0,01x:4\)

\(\Leftrightarrow[\left(85-83\right)+\left(\dfrac{7}{30}-\dfrac{5}{18}\right)]:2\dfrac{2}{3}=0,01x:4\)

\(\Leftrightarrow1\dfrac{43}{45}:2\dfrac{2}{3}=0,01x:4\)

\(\Leftrightarrow\dfrac{88}{45}:\dfrac{8}{3}=0,01x:4\)

\(\Leftrightarrow\dfrac{88}{45}.\dfrac{3}{8}=0,01x:4\)

\(\Leftrightarrow\dfrac{11}{15}=0,01x:4\)

\(\Leftrightarrow0,01x:4=\dfrac{11}{15}\)

\(\Leftrightarrow x=\dfrac{11}{15}.4:0,01\)

\(\Leftrightarrow x=\dfrac{880}{3}\)

Vậy x = \(\dfrac{880}{3}\)

Ta có:

\(\left(85\dfrac{7}{30}-83\dfrac{5}{18}\right):2\dfrac{2}{3}=0.01x:4\\ \left(\left(85-83\right)+\left(\dfrac{7}{30}-\dfrac{5}{18}\right)\right):\dfrac{8}{3}=0.01x:4\\ \dfrac{88}{45}:\dfrac{8}{3}=0.01x:4\\ \dfrac{11}{15}=0.01x:4\\ \dfrac{44}{15}=0.01x\\ x=\dfrac{44}{15}:0.01\\ x=\dfrac{880}{3}\)

Vậy \(x=\dfrac{880}{3}\)

a) \(\left\{{}\begin{matrix}3x=4y\\x+y=-56\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}3x-4y=0\\x+y=-56\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}3x-4y=0\\4x+4y=-224\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}7x=-224\\x+y=-56\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{-224}{7}=-32\\x+y=-56\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-32\\-32+y=-56\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-32\\y=-24\end{matrix}\right.\) vậy \(x=-32;y=-24\)

b/ Theo đề ta có: \(\dfrac{x}{4}=\dfrac{x}{5}\Rightarrow\dfrac{x}{12}=\dfrac{x}{15}\)

\(\dfrac{x}{3}=\dfrac{y}{2}\Rightarrow\dfrac{x}{12}=\dfrac{y}{8}\)

\(\Rightarrow\dfrac{x}{12}=\dfrac{y}{8}=\dfrac{z}{15}\) và \(x+y-z=10\)

Áp dụng t/c của dãy tỉ số = nhau ta có:

\(\dfrac{x}{12}=\dfrac{y}{8}=\dfrac{z}{15}=\dfrac{x+y-z}{12+8-15}=\dfrac{10}{5}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=2\cdot12=24\\y=2\cdot8=16\\z=2\cdot15=30\end{matrix}\right.\)

Vậy.............

a) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}3x=4y\\x+y=-56\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}3x-4y=0\\x+y=-56\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}3x-4y=0\\4x+4y=-224\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}7x=-224\\x+y=-56\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=\dfrac{-224}{7}=-32\\x+y=-56\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=-32\\-32+y=-56\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=-32\\y=-24\end{matrix}\right.\) vậy x= -32; y= -24

\(\dfrac{x-1}{x+2}=\dfrac{x-2}{x+3}\)

\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=\left(x+2\right)\left(x-2\right)\)

\(\Leftrightarrow x^2+2x-3=x^2-4\)

\(\Leftrightarrow2x=-1\)

\(\Leftrightarrow x=-0,5\)

Vậy...

\(\dfrac{x-1}{x+2}=\dfrac{x-2}{x+3}\)

\(\Rightarrow\left(x-1\right)\left(x+3\right)=\left(x+2\right)\left(x-2\right)\)

\(x^2+3x-x-3=x^2-2x+2x-4\)

\(x^2-2x-3=x^2-4\)

\(2x-3=4\)

\(2x=7\)

\(x=3,5\)

Ta có:

\(\dfrac{a}{b}=\dfrac{9}{4}\Rightarrow\dfrac{a}{9}=\dfrac{b}{4}\Rightarrow\dfrac{a}{27}=\dfrac{b}{12}\)(1)

\(\dfrac{b}{c}=\dfrac{3}{5}\Rightarrow\dfrac{b}{3}=\dfrac{c}{5}\Rightarrow\dfrac{b}{12}=\dfrac{c}{20}\)(2)

Từ (1) và (2) suy ra:

\(\dfrac{a}{27}=\dfrac{b}{12}=\dfrac{c}{20}\)

Đặt \(\dfrac{a}{27}=\dfrac{b}{12}=\dfrac{c}{20}=k\)

\(\Rightarrow\left\{{}\begin{matrix}a=27k\\b=12k\\c=20k\end{matrix}\right.\)(*)

Thay (*) vào biểu thức \(\dfrac{a-b}{b-c}\) ta có:

\(\dfrac{27k-12k}{12k-20k}=\dfrac{15k}{-8k}=\dfrac{-15}{8}\)

Vậy.............

Chúc bạn học tốt!!!