Bài 8: Tính chất của dãy tỉ số bằng nhau

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

\(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+1+3y-2}{5+7}=\dfrac{2x+3y-1}{12}=\dfrac{2x+3y-1}{6x}\)

\(\Rightarrow6x=12\Rightarrow x=2\)

Thay vào biểu thức ta có:

\(2.2+\dfrac{1}{5}=\dfrac{3y-2}{7}\Rightarrow1=\dfrac{3y-2}{7}\Rightarrow3y-2=7\)

\(\Rightarrow3y=9\Rightarrow y=3\)

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

Chúc bạn học tốt!

Bổ sung bài làm bạn dưới thêm 1 trường hợp:

TH2: \(2x+3y-1=0\)

\(\Rightarrow\left\{{}\begin{matrix}2x+1=0\\3y-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-1}{2}\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy \(\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=\dfrac{-1}{2}\\x=\dfrac{2}{3}\end{matrix}\right.\)

Ta có:

\(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{a+b}{c+d}.\dfrac{a+b}{c+d}=\dfrac{a.b}{c.d}\Rightarrow\dfrac{a+b}{c+d}=\dfrac{a}{b}=\dfrac{c}{d}\)

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

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=\dfrac{a+b+c}{a+b+c}=1\)(do a+b+c\(\ne\)0)

\(\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=1\Rightarrow a=b=c\Rightarrow a=b=c=2012\)

Vậy b=c=2012

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

\(\dfrac{a}{c}=\dfrac{c}{b}=\dfrac{b}{d}=\dfrac{a^3}{c^3}=\dfrac{c^3}{b^3}=\dfrac{b^3}{d^3}=\dfrac{a^3+c^3-b^3}{c^3+b^3-d^3}\left(1\right)\)

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

Ta xét tích: \(\left(\dfrac{a}{c}\right)^3=\dfrac{a}{c}.\dfrac{a}{c}.\dfrac{a}{c}=\dfrac{a}{c}.\dfrac{c}{b}.\dfrac{b}{d}=\dfrac{a}{d}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\dfrac{a^3+c^3-b^3}{c^3+b^3-d^3}=\dfrac{a}{d}\left(dpcm\right)\)

a sai đề

b) Ta có:

\(\dfrac{3a-2b}{5}=\dfrac{2c-5a}{3}=\dfrac{5b-3c}{2}\Leftrightarrow\dfrac{5\left(3a-2b\right)}{25}=\dfrac{3\left(2c-5a\right)}{9}=\dfrac{2\left(5b-3c\right)}{4}\)Hay \(\dfrac{15a-10b}{25}=\dfrac{6c-15a}{9}=\dfrac{10b-6c}{4}\)

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

\(\dfrac{15a-10b}{25}=\dfrac{6c-15a}{9}=\dfrac{10b-6c}{4}=\dfrac{15a-10b+6c-15a+10b-6c}{25+9+4}=\dfrac{0}{25+9+4}=0\)

Nên

\(\left\{{}\begin{matrix}3a=2b\\2c=5a\\5b=3c\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{2}=\dfrac{b}{3}\\\dfrac{c}{5}=\dfrac{a}{2}\\\dfrac{b}{3}=\dfrac{c}{5}\end{matrix}\right.\Leftrightarrow\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{5}\)

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

\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{5}=\dfrac{a+b+c}{2+3+5}=\dfrac{-50}{10}=-5\)

\(\Rightarrow\left\{{}\begin{matrix}a=-5.2=-10\\b=-5.3=-15\\c=-5.5=-25\end{matrix}\right.\)

\(\dfrac{xy}{ay+bx}=\dfrac{yz}{bz+cy}=\dfrac{zx}{cx+az}=\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}\left(1\right)\)

Ta có: \(\dfrac{xy}{ay+bx}=\dfrac{yz}{bz+cy}=\dfrac{zx}{cx+az}\)

\(\Rightarrow\dfrac{xyz}{ayz+bxz}=\dfrac{xyz}{bxz+cxy}=\dfrac{xyz}{cxy+ayz}\)

\(\Rightarrow ayz+bxz=bxz+cxy=cxy+ayz\)

\(\Rightarrow\left\{{}\begin{matrix}ayz+bxz=bxz+cxy\\ayz+bxz=cxy+ayz\\bxz+cxy=cxy+ayz\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}ayz=cxy\\bxz=cxy\\bxz=ayz\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}az=cx\\bz=cy\\bx=ay\end{matrix}\right.\left(2\right)\)

Thay (2) vào (1) ta có :

\(\dfrac{xy}{2ay}=\dfrac{yz}{2bz}=\dfrac{xz}{2cx}=\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}\)

\(\Rightarrow\dfrac{x}{2a}=\dfrac{y}{2b}=\dfrac{z}{2c}=\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}\left(3\right)\)

\(\Rightarrow\dfrac{x^2}{4a^2}=\dfrac{y^2}{4b^2}=\dfrac{z^2}{4c^2}=\dfrac{\left(x^2+y^2+z^2\right)^2}{\left(a^2+b^2+c^2\right)^2}=\)\(\dfrac{x^2+y^2+z^2}{4a^2+4b^2+4c^2}\)

\(\Rightarrow\dfrac{x^2+y^2+y^2}{a^2+b^2+c^2}=\dfrac{1}{4}\left(4\right).\)Thay (3) vào (2) ta có :

\(\dfrac{x}{2a}=\dfrac{y}{2b}=\dfrac{z}{2c}=\dfrac{1}{4}\Rightarrow\left\{{}\begin{matrix}x=\dfrac{a}{2}\\y=\dfrac{b}{2}\\z=\dfrac{c}{2}\end{matrix}\right.\)

CD+CH+CA=3C

Đâu lak Đạt hả Đẹp Trai Không Bao Giờ Sai???

ta có : \(\dfrac{x}{3}=\dfrac{y}{7}\Leftrightarrow7x=3y\Leftrightarrow7x-3y=0\) (1)

ta có : \(\dfrac{y}{2}=\dfrac{z}{5}\Leftrightarrow5y=2z\Leftrightarrow5y-2z=0\) (2)

từ : \(x+y+z=-110\) và (1) ; (2)

ta có hệ phương trình : \(\left\{{}\begin{matrix}x+y+z=-110\\7x-3y=0\\5y-2z=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-12\\y=-28\\z=-70\end{matrix}\right.\)

vậy \(x=-12;y=-28;z=-70\)

Ta có:\(\dfrac{b+c-3a}{a}=\dfrac{a+c-3b}{b}=\dfrac{a+c-3c}{c}\)

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

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

\(\dfrac{b+c}{a}=\dfrac{a+c}{b}=\dfrac{a+b}{c}=\dfrac{2\left(a+b+c\right)}{a+b+c}\)

Nếu a+b+c=0

\(\Rightarrow\)\(\left\{{}\begin{matrix}b+c=a\\a+c=b\\a+b=c\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=-\left(b+c\right)\\b=-\left(a+c\right)\\c=-\left(a+b\right)\end{matrix}\right.\)

\(\Rightarrow M=-1\)

Nếu a+b+c\(\ne\)0

\(\Rightarrow\dfrac{b+c}{a}=\dfrac{a+c}{b}=\dfrac{a+b}{c}=2\)

\(\Rightarrow\left\{{}\begin{matrix}b+c=2a\\a+c=2b\\a+b=2c\end{matrix}\right.\)

\(\Rightarrow M=8\)

Tính \(M=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\left(1\right)\)

Ta có :

\(\dfrac{b+c-3a}{a}=\dfrac{a+c-3b}{b}=\dfrac{a+b-3c}{c}\)

\(\Rightarrow\dfrac{b+c-3a}{a}+3=\dfrac{a+c-3b}{b}+3=\dfrac{a+b-3c}{c}+3\)

\(\Rightarrow\dfrac{b+c-3a}{a}+\dfrac{3a}{a}=\dfrac{a+c-3b}{b}+\dfrac{3b}{b}=\dfrac{a+b-3c}{c}+\dfrac{3c}{c}\)

\(\Rightarrow\dfrac{b+c-3a+3a}{a}=\dfrac{a+c-3b+3b}{b}=\dfrac{a+b-3c+3c}{c}\)

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

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

\(\Rightarrow\dfrac{b+c}{a}=\dfrac{a+c}{b}=\dfrac{a+b}{c}=\dfrac{2a+2b+2c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)Suy ra :

\(\left\{{}\begin{matrix}\dfrac{b+c}{a}=2\\\dfrac{a+c}{b}=2\\\dfrac{a+b}{c}=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}b+c=2a\\a+c=2b\\a+b=2c\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{b+c}{2}=a\left(2\right)\\\dfrac{a+c}{2}=b\left(3\right)\\\dfrac{a+b}{2}=c\left(4\right)\end{matrix}\right.\)

THAY (2),(3),(4)vào(1) ta được :

M=\(\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\dfrac{b+c}{2}.\dfrac{a+c}{2}.\dfrac{a+b}{2}}\)

\(=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\dfrac{\left(b+c\right)\left(a+c\right)\left(a+b\right)}{2.2.2}}=2.2.2=2^3=8\)

Chúc bạn thành công trên con đường học tập

Mình bổ sung

Mặt khác nếu a+b+c=0 thì

\(\dfrac{b+c}{a}=\dfrac{a+c}{b}=\dfrac{a+b}{c}=\dfrac{b+c+a+c+a+b}{a+b+c}=0\)

ta chia ra 3 trường hợp sau

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{b+c}{a}=0\\\dfrac{a+c}{b}=0\\\dfrac{a+b}{c}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}b+c=a\\a+c=b\\a+b=c\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}c=a-b\\a=b-c\\b=c-a\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}c=-\left(a+b\right)\\a=-\left(b+c\right)\\b=-\left(c+a\right)\end{matrix}\right.\)

\(\)Ta thay thế vào M được

\(\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{a+b+c}=\)\(\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{-\left(b+c\right).-\left(c+a\right).-\left(a+b\right)}=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{-[\left(b+c\right)\left(c+a\right)\left(a+b\right)]}=-1\)

VẬY M có hai giá trị là 8 và -1

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

Ta có : \(\left\{{}\begin{matrix}\left(\dfrac{a+b+c}{b+c+d}\right)^3=t^3\\\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}=t^3\end{matrix}\right.\)

Ta có đpcm