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Làm hộ tui đi à,đây là Sol của thầy Sỹ,đọc là 1 chuyện nhưng hiểu mới là vấn đề.
BĐT đẹp vãi ra mà ối sồi ôi lời giải khủng VCL.Hóng cách nhẹ hơn...
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{3a-c}{3b-d}=\dfrac{3bk-dk}{3b-d}=k\)
\(\dfrac{2a+3c}{2b+3d}=\dfrac{2bk+3dk}{2b+3d}=k\)
Do đó: \(\dfrac{3a-c}{3b-d}=\dfrac{2a+3c}{2b+3d}\)
c: \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\dfrac{b^2}{d^2}\)
\(\dfrac{2ab+b^2}{2cd+d^2}=\dfrac{2\cdot bk\cdot b+b^2}{2\cdot dk\cdot d+d^2}=\dfrac{b^2}{d^2}\)
Do đó: \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{2ab+b^2}{2cd+d^2}\)
Bài 2:
Để \(x^4+ax^3+b\vdots x^2-1\) thì \(x^4+ax^3+b\) phải được viết dưới dạng :
\(x^4+ax^3+b=(x^2-1)Q(x)\) với $Q(x)$ là đa thức thương.
Thay $x=1$ và $x=-1$ lần lượt ta có:
\(\left\{\begin{matrix} 1+a+b=(1^2-1)Q(1)=0\\ 1-a+b=[(-1)^2-1]Q(-1)=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} a+b=-1\\ -a+b=-1\end{matrix}\right.\Rightarrow \left\{\begin{matrix} a=0\\ b=-1\end{matrix}\right.\)
PP 2 xin đợi bạn khác giải quyết :)
Bài 3:
Ta có: \(\frac{\sqrt{12}-\sqrt{27}-\sqrt{48}}{1-\sqrt{5}+9\sqrt{9-4\sqrt{5}}}=\frac{\sqrt{12}-\sqrt{27}-\sqrt{48}}{1-\sqrt{5}+9\sqrt{5+4-4\sqrt{5}}}\)
\(=\frac{\sqrt{12}-\sqrt{27}-\sqrt{48}}{1-\sqrt{5}+9\sqrt{(2-\sqrt{5})^2}}=\frac{\sqrt{12}-\sqrt{27}-\sqrt{48}}{1-\sqrt{5}+9(\sqrt{5}-2)}=\frac{\sqrt{3}(2-3-4)}{-17+8\sqrt{5}}=\frac{-5\sqrt{3}}{-17+8\sqrt{5}}\)
\(=\frac{5\sqrt{3}}{17-8\sqrt{5}}\)
a: \(=ab\cdot\dfrac{4}{3}a^2b^4\cdot7abc=\dfrac{28}{3}a^4b^6c\)
b: \(a^3b^3\cdot a^2b^2c=a^5b^5c\)
c: \(=\dfrac{2}{3}a^3b\cdot\dfrac{-1}{2}ab\cdot a^2b=\dfrac{-1}{3}a^6b^3\)
d: \(=-\dfrac{7}{3}a^3c^2\cdot\dfrac{1}{7}ac^2\cdot6abc=-2a^5bc^5\)
e: \(=\dfrac{-3}{2}\cdot\dfrac{1}{4}\cdot ab^2\cdot bca^2\cdot b=\dfrac{-3}{8}a^3b^4c\)
a) \(2\sqrt{a^2}=2\left|a\right|=2a\) (vì \(a\ge0\))
b) \(\sqrt{3a^2}=\left|a\right|\sqrt{3}=-a\sqrt{3}\) (vì \(a< 0\))
c) \(5\sqrt{a^4}=5\sqrt{\left(a^2\right)^2}=5\left|a^2\right|=5a^2\)
d) \(\dfrac{1}{3}\sqrt{c^6}=\dfrac{1}{3}\sqrt{\left(c^3\right)^2}=\dfrac{1}{3}\left|c^3\right|=\dfrac{1}{3}\left(-c^3\right)=-\dfrac{1}{3}c^3\) (vì \(c< 0\Rightarrow c^3< 0\))
\(a)2\sqrt{a^2}=2.\left|a\right|=2a\) ( vì \(a\ge0\) )
\(b)\sqrt{3a^2}=\left|a\right|\sqrt{3}=-a\sqrt{3}\) ( vì \(a< 0\) )
\(c)5\sqrt{a^4}=5\sqrt{\left(a^2\right)^2}=5\left|a^2\right|=5a^2\)
\(d)\dfrac{1}{3}\sqrt{c^6}=\dfrac{1}{3}\sqrt{\left(c^3\right)^2}=\dfrac{1}{3}\left|c^3\right|=\dfrac{1}{3}\left(-c^3\right)\) ( vì \(c< 0\Rightarrow c^3< 0\) )
Chúc bn học tốt!
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>\(a=bk;c=dk\)
1: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2\cdot bk+3\cdot dk}{2b+3d}=\dfrac{k\left(2b+3d\right)}{2b+3d}=k\)
\(\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k\left(2b-3d\right)}{2b-3d}=k\)
Do đó: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
2: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4\cdot bk-3b}{4\cdot dk-3d}=\dfrac{b\left(4k-3\right)}{d\left(4k-3\right)}=\dfrac{b}{d}\)
\(\dfrac{4a+3b}{4c+3d}=\dfrac{4bk+3b}{4dk+3d}=\dfrac{b\left(4k+3\right)}{d\left(4k+3\right)}=\dfrac{b}{d}\)
Do đó: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)
3: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3bk+5b}{3bk-5b}=\dfrac{b\left(3k+5\right)}{b\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)
\(\dfrac{3c+5d}{3c-5d}=\dfrac{3dk+5d}{3dk-5d}=\dfrac{d\left(3k+5\right)}{d\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)
Do đó: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
4: \(\dfrac{3a-7b}{b}=\dfrac{3bk-7b}{b}=\dfrac{b\left(3k-7\right)}{b}=3k-7\)
\(\dfrac{3c-7d}{d}=\dfrac{3dk-7d}{d}=\dfrac{d\left(3k-7\right)}{d}=3k-7\)
Do đó: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
\(3x^2+2xy+3y^2=\left(x+y\right)^2+2\left(x^2+y^2\right)\ge\left(x+y\right)^2+\left(x+y\right)^2=2\left(x+y\right)^2\)
\(\Rightarrow A\ge\sqrt{2}\left(a+b\right)+\sqrt{2}\left(b+c\right)+\sqrt{2}\left(c+a\right)\)
\(A\ge2\sqrt{2}\left(a+b+c\right)\ge\frac{2\sqrt{2}}{3}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=6\sqrt{2}\)
\(A_{min}=6\sqrt{2}\) khi \(a=b=c=1\)