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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
Ta có:\(b^2=ac\Leftrightarrow\frac{a}{b}=\frac{b}{c}\Rightarrow\frac{a^2}{b^2}=\frac{b^2}{c^2}=\frac{a}{b}\cdot\frac{b}{c}=\frac{a}{c}\)
Mà\(\frac{a}{b}=\frac{b}{c}=\frac{2015b}{2015c}=\frac{a+2015b}{b+2015c}\)
Nên suy ra\(\frac{a}{c}=\frac{a^2}{b^2}=\left(\frac{a+2015b}{b+2015c}\right)^2=\frac{\left(a+2015b\right)^2}{\left(b+2015c\right)^2}\)
Vậy\(\frac{a}{c}=\frac{\left(a+2015b\right)^2}{\left(b+2015c\right)^2}\left(đpcm\right)\)
Ta có b^2=ac =>a/b=c/d. Đặt a/b=c/d=k(khác 0) =>a=bk;b=ck =>a/c=c.k^2/c=k^2 (1) (a+2015b)^2/(b+2015c)^2=(bk+2015b/ck+2015c)^2=(b(k+2015)/(c(k+2015))^2=(b/c)^2=(ck/c)^2=k^2 (2) Từ (1) và (2) => a/c=(a+2015b/b+2015c)^2 => (đpcm)
Ta có:
\(b^2=ac\rightarrow\frac{a}{b}=\frac{b}{c}\) ( \(b\ne0,c\ne0\)
\(c^2=bd\rightarrow\frac{b}{c}=\frac{c}{d}\) \(d\ne0\)
\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\rightarrow\frac{abc}{bcd}=\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}\) ( \(bcd\ne0\)vì \(b^3+c^3+d^3\ne0\))
áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{a+b+c}{b+c+d}\rightarrow\frac{abc}{bcd}=\left(\frac{a+b+c}{b+c+d}\right)^3\)
\(\frac{abc}{bcd}=\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\)
\(\Rightarrow\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\left(\frac{a+b+c}{b+c+d}\right)^3\left(đpcm\right)\)
Ta có: b2=ac\(\Rightarrow\frac{b}{c}=\frac{a}{b}\)
\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{2016.b}{2016.c}\)(1)
áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{a}{b}=\frac{2016.b}{2016.c}=\frac{a+2016.b}{b+2016.c}\)(2)
Từ (1) và (2) ta có: \(\frac{a}{b}=\frac{b}{c}=\frac{a+2016.b}{b+2016.c}\)
\(\Rightarrow\frac{\left(a+2016.b\right)^2}{\left(b+2016.c\right)^2}=\frac{a^2}{b^2}=\frac{a}{b}.\frac{a}{b}=\frac{a}{b}.\frac{b}{c}\)(vì \(\frac{a}{b}=\frac{b}{c}\))\(=\frac{a}{c}\)(điều phải chứng minh)
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