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Bài 2 )
\(a\left(y+z\right)=b\left(x+z\right)=c\left(x+y\right)\)
\(\Leftrightarrow\frac{a\left(y+z\right)}{abc}=\frac{b\left(x+z\right)}{abc}=\frac{c\left(x+y\right)}{abc}\)
\(\Leftrightarrow\frac{y+z}{bc}=\frac{x+z}{ac}=\frac{x+y}{ab}\)
\(\Leftrightarrow\frac{bc}{y+z}=\frac{ac}{x+z}=\frac{ab}{x+y}\)
Đặt \(\frac{bc}{y+z}=\frac{ac}{x+z}=\frac{ab}{x+y}=k\)
\(\Rightarrow\left\{\begin{matrix}bc=k\left(y+z\right)=ky+kz\\ac=k\left(x+z\right)=kx+kz\\ab=k\left(x+y\right)=kx+ky\end{matrix}\right.\) (1)
Gỉa sử điều cần chứng minh là đúng ta có
\(\frac{y-z}{a\left(b-c\right)}=\frac{z-x}{b\left(c-a\right)}=\frac{x-y}{c\left(a-b\right)}\)
\(\Leftrightarrow\frac{y-z}{ab-ac}=\frac{z-x}{bc-ab}=\frac{x-y}{ac-bc}\)
Thế (1) vào biểu thức
\(\frac{y-z}{kx+ky-\left(kx+kz\right)}=\frac{z-x}{ky+kz-\left(kx+ky\right)}=\frac{x-y}{kx+kz-\left(ky+kz\right)}\)
\(\Leftrightarrow\frac{y-z}{ky-kz}=\frac{z-x}{kz-kx}=\frac{x-y}{kx-ky}\)
\(\Leftrightarrow\frac{y-z}{k\left(y-z\right)}=\frac{z-x}{k\left(z-x\right)}=\frac{x-y}{k\left(x-y\right)}\)
\(\Leftrightarrow\frac{1}{k}=\frac{1}{k}=\frac{1}{k}\) ( điều này luôn luôn đúng )
\(\Rightarrow\) ĐPCM
\(x^2_2=x_1.x_3\Rightarrow\frac{x_2}{x_1}=\frac{x_3}{x_2},x^2_3=x_2.x_4\Rightarrow\frac{x_4}{x_3}=\frac{x_3}{x_2}\)\(\Rightarrow\frac{x_2}{x_1}=\frac{x_3}{x_2}=\frac{x_4}{x_3}\)
áp dụng t.c dãy tỉ số bằng nhau ta có:
\(\frac{x_2}{x_1}=\frac{x_3}{x_2}=\frac{x_4}{x_3}=\frac{x_2+x_3+x_4}{x_1+x_2+x_3}\Rightarrow\left(\frac{x_2}{x_1}\cdot\frac{x_3}{x_2}\cdot\frac{x_4}{x_3}\right)=\left(\frac{x_2+x_3+x_4}{x_1+x_2+x_3}\right)^3\Rightarrow\frac{x_4}{x_1}=\left(\frac{x_2+x_3+x_4}{x_1+x_2+x_3}\right)^3\)
\(\Rightarrow\frac{x_1}{x_4}=\left(\frac{x_1+x_2+x_3}{x_2+x_3+x_4}\right)^3\left(đpcm\right)\)
Từ \(X_2^2=X_1.X_3\)\(\Rightarrow\frac{X_1}{X_2}=\frac{X_2}{X_3}\)(1)
Từ \(X_3^2=X_2.X_4\)\(\Rightarrow\frac{X_2}{X_3}=\frac{X_3}{X_4}\)(2)
Từ (1) và (2) \(\Rightarrow\frac{X_1}{X_2}=\frac{X_2}{X_3}=\frac{X_3}{X_4}=\frac{X_1+X_2+X_3}{X_2+X_3+X_4}\)
\(\Rightarrow\left(\frac{X_1}{X_2}\right)^3=\left(\frac{X_1+X_2+X_3}{X_2+X_3+X_4}\right)^3\)(1)
Từ \(\left(\frac{X_1}{X_2}\right)^3=\frac{X_1}{X_2}.\frac{X_1}{X_2}.\frac{X_1}{X_2}=\frac{X_1}{X_2}.\frac{X_2}{X_3}.\frac{X_3}{X_4}=\frac{X_1}{X_4}\)(2)
Từ (1) và (2) \(\Rightarrowđpcm\)
Bài 1:
Ta có: \(\frac{\left(a+2012b\right)^2}{\left(b+2012c\right)^2}=\frac{a^2+2.2012.ab+2012^2.b^2}{b^2+2.2012.bc+2012^2.c^2}=\frac{a^2+2.2012.ab+2012^2.ac}{ac+2.2012.bc+2012^2.c^2}=\frac{a\left(a+2.2012.b+2012^2.c\right)}{c\left(a+2.2012.b+2012^2.c\right)}=\frac{a}{c}\)
Vậy...
Bài 2:
\(\frac{x}{a+2b+c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}\Rightarrow\frac{a+2b+c}{x}=\frac{2a+b-c}{y}=\frac{4a-4b+c}{z}\)
\(\Rightarrow\frac{a+2b+c}{x}=\frac{2\left(2a+b-c\right)}{2y}=\frac{4a-4b+c}{z}=\frac{a+2b+c+4a+2b-2c+4a-4b+c}{x+2y+z}=\frac{a}{x+2y+z}\)(1)
\(\frac{2\left(a+2b+c\right)}{2x}=\frac{2a+b-c}{y}=\frac{4a-4b+c}{z}=\frac{2a+4b+2c+2a+b-c-4a+4b-c}{2x+y-z}=\frac{b}{2x+y-z}\) (2)
\(\frac{4\left(a+2b+c\right)}{4x}=\frac{4\left(2a+b-c\right)}{4y}=\frac{4a-4b+c}{z}=\frac{4a+8b+c-8a-4b+c+4a-4b+c}{4x-4y+z}=\frac{c}{4x-4y+z}\) (3)
Từ (1),(2),(3) suy ra \(\frac{a}{x+2y+z}=\frac{b}{2x+y-z}=\frac{c}{4x-4y+z}\)
bạn trên nhầm -4b thành +4b ở bài 2 ở phần (1) nha bạn, nhưng mình cũng cảm ơn
Sửa lại đề \(CM\)\(\frac{a}{c}=\frac{\left(a+20112b\right)^2}{\left(b+2012c\right)^2}\)
Có \(a,b,c\in R;a,b,c\ne0\)và \(b^2=ac\)
Ta có \(b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c}\)
Lại có \(\frac{a}{b}=\frac{b}{c}=\frac{2012b}{2012c}\Rightarrow\frac{a}{b}=\frac{a+2012b}{b+2012c}\)
\(\Rightarrow\frac{a^2}{b^2}=\frac{\left(a+2012b\right)^2}{\left(b+2012c\right)^2}\Rightarrow\frac{a^2}{ac}=\frac{\left(a+2012b\right)^2}{\left(b+2012c\right)^2}\)
Hay \(\frac{a}{c}=\frac{\left(a+2012b\right)^2}{\left(b+2012c\right)^2}\)
\(\frac{\left(a+2012.b\right)^2}{\left(b+2012.c\right)^2}=\frac{a^2+2.2012.a.b+2012^2.b^2}{b^2+2.2012.b.c+2012^2.c^2}=\frac{a^2+2.2012.a.b+2012^2.a.c}{a.c+2.2012.b.c+2012^2.c^2}=\)
\(=\frac{a\left(a+2.2012.b+2012^2.c\right)}{c\left(a+2.2012.b+2012^2.c\right)}=\frac{a}{c}\)
Xem lại đề bài
\(b^2=ac\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}\)
Đặt: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{2018b}{2018c}=t\)
tính chất dãy tỉ số bằng nhau: \(\dfrac{a}{b}=\dfrac{2018b}{2018c}=\dfrac{a+2018b}{b+2018c}\)
Ta có: \(\left\{{}\begin{matrix}\dfrac{a}{b}.\dfrac{b}{c}=\dfrac{a}{c}=t^2\\\left(\dfrac{a+2018b}{b+2018c}\right)^2=t^2\end{matrix}\right.\Leftrightarrowđpcm\)
Ta có
\(\hept{\begin{cases}a_2^2=a_1.a_3\\a_3^2=a_2.a_4\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{a_1}{a_2}=\frac{a_2}{a_3}\\\frac{a_2}{a_3}=\frac{a_3}{a_4}\end{cases}}}\)
\(\Rightarrow\frac{a_1}{a_2}=\frac{a_2}{a_3}=\frac{a_3}{a_4}\)
\(\Rightarrow\frac{a_1^3}{a_2^3}=\frac{a_2^3}{a_3^3}=\frac{a_3^3}{a_4^3}=\frac{a_1^3+a_2^3+a_3^3}{a_2^3+a_3^3+a_4^3}\left(1\right)\)
Ta lại có
\(\frac{a_2^2}{a_3^2}=\frac{a_1.a_3}{a_2.a_4}\)
\(\frac{a_2^3}{a_3^3}=\frac{a_1}{a_4}\left(2\right)\)
Từ (1) và (2)
\(\frac{a_1^3+a_2^3+a_3^3}{a_2^3+a_3^3+a_4^3}=\frac{a_1}{a_4}\)
\(b^2=ac\)
\(\Rightarrow\frac{a}{b}=\frac{b}{c}\)
Đặt \(\frac{a}{b}=\frac{b}{c}=k\), ta có: \(a=bk;b=ck\)
\(\frac{a}{c}=\frac{bk}{c}=\frac{ck\times k}{c}=k^2\) (1)
\(\left(\frac{a+2012b}{b+2012c}\right)^2=\left(\frac{bk+2012b}{ck+2012}\right)^2=\left(\frac{b\left(k+2012\right)}{c\left(k+2012\right)}\right)^2=\left(\frac{b}{c}\right)^2=k^2\) (2)
Từ (1) và (2)
=> \(\frac{a}{c}=\left(\frac{a+2012b}{b+2012c}\right)^2\left(\text{đ}pcm\right)\)
mk cũng đang cần. thanks