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Từ \(b^2=ac\)\(\Rightarrow\frac{a}{b}=\frac{b}{c}\)
\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{2017b}{2017c}=\frac{a+2017b}{b+2017c}\)
\(\Rightarrow\left(\frac{a}{b}\right)^2=\left(\frac{a+2017b}{b+2017c}\right)^2=\frac{\left(a+2017b\right)^2}{\left(b+2017c\right)^2}\)(1)
Ta có: \(\left(\frac{a}{b}\right)^2=\frac{a}{b}.\frac{a}{b}=\frac{a}{b}.\frac{b}{c}=\frac{a}{c}\)(2)
Từ (1) và (2) \(\Rightarrow\frac{a}{c}=\frac{\left(a+2017b\right)^2}{\left(b+2017c\right)^2}=\left(\frac{a}{b}\right)^2\left(đpcm\right)\)
b^2 = a.c
=> a/b = b/c
Đặt a/b = b/c = k
=> a=bk ; b=ck
=> a = c.k.k = c.k^2 => a/c = k^2
Lại có : (a+2011b)^2/(b+2011c)^2
= (bk+2011b)^2/(ck+2011c)^2
= [b.(k+2011)]^2/[c.(k+2011)]^2
= b^2.(k+2011)^2/c^2.(k+2011)^2
= b^2/c^2
= (b/c)^2
= k^2
=> a/c = (a+2011)^2/(b+2011c)^2
Tk mk nha
\(1a,\) Ta có: \(\left(2x-6\right)^2\ge0\forall x\Rightarrow\left(2x-6\right)^2+36\ge36\forall x\)
\(\Rightarrow\frac{2016}{\left(2x-6\right)^2+63}\le\frac{2016}{63}=32\)
\(\Rightarrow\left|y+2015\right|+32\le32\)
\(\Rightarrow\left|y+2015\right|\le0\)
\(\Rightarrow\left|y+2015\right|=0\)
\(\Rightarrow y=-2015\)
\(\Rightarrow2x-6=0\Rightarrow x=3\)
Vậy \(x=3;y=-2015\)
b)
Ta có: \(b^2=ac.\)
\(\Rightarrow\frac{a}{b}=\frac{b}{c}.\)
\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{2017b}{2017c}.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{a}{b}=\frac{b}{c}=\frac{2017b}{2017c}=\frac{a+2017b}{b+2017c}.\)
\(\Rightarrow\frac{a}{b}=\frac{a+2017b}{b+2017c}\)
\(\Rightarrow\left(\frac{a}{b}\right)^2=\left(\frac{a+2017b}{b+2017c}\right)^2\)
\(\Rightarrow\left(\frac{a}{b}\right)^2=\frac{\left(a+2017b\right)^2}{\left(b+2017c\right)^2}.\)
\(\Rightarrow\frac{a}{b}.\frac{a}{b}=\frac{\left(a+2017b\right)^2}{\left(b+2017c\right)^2}\)
\(\Rightarrow\frac{a}{b}.\frac{b}{c}=\frac{\left(a+2017b\right)^2}{\left(b+2017c\right)^2}.\)
\(\Rightarrow\frac{a}{c}=\frac{\left(a+2017b\right)^2}{\left(b+2017c\right)^2}\left(đpcm\right).\)
Chúc bạn học tốt!
Bài 1:
Đặt \(\frac{a}{b}=\frac{c}{d}=t\Rightarrow a=bt; c=dt\). Khi đó:
a)
\(\frac{a^2}{a^2+b^2}=\frac{(bt)^2}{(bt)^2+b^2}=\frac{b^2t^2}{b^2(t^2+1)}=\frac{t^2}{t^2+1}(1)\)
\(\frac{c^2}{c^2+d^2}=\frac{(dt)^2}{(dt)^2+d^2}=\frac{d^2t^2}{d^2(t^2+1)}=\frac{t^2}{t^2+1}(2)\)
Từ $(1);(2)$ suy ra đpcm.
b)
\(\left(\frac{a+c}{b+d}\right)^2=\left(\frac{bt+dt}{b+d}\right)^2=\left(\frac{t(b+d)}{b+d}\right)^2=t^2(3)\)
\(\frac{a^2+c^2}{b^2+d^2}=\frac{(bt)^2+(dt)^2}{b^2+d^2}=\frac{t^2(b^2+d^2)}{b^2+d^2}=t^2(4)\)
Từ $(3);(4)\Rightarrow \left(\frac{a+c}{b+d}\right)^2=\frac{a^2+c^2}{b^2+d^2}$ (đpcm)
Bài 2:
Từ $a^2=bc\Rightarrow \frac{a}{c}=\frac{b}{a}$
Đặt $\frac{a}{c}=\frac{b}{a}=t\Rightarrow a=ct; b=at$. Khi đó:
a)
$\frac{a^2+c^2}{b^2+a^2}=\frac{(ct)^2+c^2}{(at)^2+a^2}=\frac{c^2(t^2+1)}{a^2(t^2+1)}=\frac{c^2}{a^2}=(\frac{c}{a})^2=\frac{1}{t^2}(1)$
Và:
$\frac{c}{b}=\frac{a}{tb}=\frac{a}{t.at}=\frac{1}{t^2}(2)$
Từ $(1);(2)$ suy ra đpcm.
b)
$\left(\frac{c+2019a}{a+2019b}\right)^2=\left(\frac{c+2019a}{ct+2019at}\right)^2=\left(\frac{c+2019a}{t(c+2019a)}\right)^2=\frac{1}{t^2}(3)$
Từ $(2);(3)$ suy ra đpcm.
Ta co:\(b^2=ac\Leftrightarrow\frac{a}{b}=\frac{b}{c}\)
\(=\frac{2007b}{2007c}=\frac{a+2007b}{b+2007c}\)
\(\Rightarrow\left(\frac{a+2007b}{b+2007c}\right)^2=\left(\frac{a}{b}\right)^2=\frac{a}{b}\times\frac{b}{c}=\frac{a}{c}\)
Vậy \(\frac{a}{c}=\left(\frac{a+2007b}{b+2007c}\right)^2\left(đpcm\right)\)
b)Để N có giá trị nguyên thì căn x-5 EƯ(9)={1;-1;3;-3;9;-9}
=>căn x E{6;4;8;2;14;-4}
=>xE{36;24;64;4;196;16}
Vậy để N có giá trị nguyên thì x E{36;24;64;4;196;16}
Giải:
Ta có: \(b^2=ac\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\)
Đặt \(\dfrac{a}{b}=\dfrac{b}{c}=k\)
+) \(k^2=\dfrac{a}{b}.\dfrac{b}{c}=\dfrac{a}{c}\) (1)
+) \(k=\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{2011b}{2011c}=\dfrac{a+2011b}{b+2011c}\) ( t/c dãy tỉ số bằng nhau )
\(\Rightarrow k^2=\left(\dfrac{a+2011b}{b+2011c}\right)^2=\dfrac{\left(a+2011b\right)^2}{\left(b+2011c\right)^2}\) (2)
Từ (1), (2) \(\Rightarrow\dfrac{a}{c}=\dfrac{\left(a+2011b\right)^2}{\left(b+2011c\right)^2}\left(đpcm\right)\)
Giải:
Từ hằng đẳng thức: \(\left(a+b\right)^2=a^2+2ab+b\) ta có:
\(VP=\dfrac{\left(a+2011b\right)^2}{\left(b+2011c\right)^2}=\dfrac{a^2+2.2011ab+\left(2011b\right)^2}{b^2+2.2011bc+\left(2011c\right)^2}\)
\(=\dfrac{a^2+2.2011ab+2011^2ac}{ac+2.2011bc+2011^2c^2}\)
\(=\dfrac{a\left(a+2.2011b+2011^2c\right)}{c\left(a+2.2011b+2011^2c\right)}=\dfrac{a}{c}=VT\)
Vậy \(\dfrac{a}{c}=\dfrac{\left(a+2011b\right)^2}{\left(b+2011c\right)^2}\) (Đpcm)