Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Áp dụng bđt bunhiacopski cho 3 số ta có
\(\left(a\sqrt{1-b^2}+b\sqrt{1-c^2}+c\sqrt{1-a^2}\right)^2\le\left(a^2+b^2+c^2\right)\left(1-a^2+1-b^2+1-c^2\right)\Leftrightarrow\frac{9}{4}\le\left(a^2+b^2+c^2\right)\left[3-\left(a^2+b^2+c^2\right)\right]\)(1)
Đặt \(a^2+b^2+c^2=k\)
Vậy (1)\(\Leftrightarrow\frac{9}{4}\le k\left(3-k\right)\Leftrightarrow\frac{9}{4}\le3k-k^2\Leftrightarrow k^2-3k+\frac{9}{4}\le0\Leftrightarrow\left(k-\frac{3}{2}\right)^2\le0\)
Vì \(\left(k-\frac{3}{2}\right)^2\ge0\)
Suy ra \(\left(k-\frac{3}{2}\right)^2=0\Leftrightarrow k-\frac{3}{2}=0\Leftrightarrow k=\frac{3}{2}\)
Vậy \(a^2+b^2+c^2=\frac{3}{2}\)
\(A=\frac{\sqrt{2}-1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+...+\frac{\sqrt{100}-\sqrt{99}}{\left(\sqrt{100}-\sqrt{99}\right)\left(\sqrt{100}+\sqrt{99}\right)}\)
\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\)
\(=\sqrt{100}-1=9\)
\(B=\frac{2}{2}+\frac{2}{2\sqrt{2}}+\frac{2}{2\sqrt{3}}+...+\frac{2}{2\sqrt{35}}\)
\(B>\frac{2}{\sqrt{1}+\sqrt{2}}+\frac{2}{\sqrt{2}+\sqrt{3}}+...+\frac{2}{\sqrt{35}+\sqrt{36}}\)
\(B>2\left(\frac{\sqrt{2}-1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}+...+\frac{\sqrt{36}-\sqrt{35}}{\left(\sqrt{36}-\sqrt{35}\right)\left(\sqrt{36}+\sqrt{35}\right)}\right)\)
\(B>2\left(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{36}-\sqrt{35}\right)\)
\(B>2\left(\sqrt{36}-1\right)=10>9=A\)
\(\Rightarrow B>A\)
Để biểu thức B có nghĩa thì \(xy\ne0\)
Khi đó ta có:
\(x^3+y^3=2x^2y^2\)
\(\Leftrightarrow\left(x^3+y^3\right)^2=4x^4y^4\)
\(\Leftrightarrow x^6+y^6+2x^3y^3=4x^4y^4\)
\(\Leftrightarrow x^6+y^6-2x^3y^3=4x^4y^4-4x^3y^3\)
\(\Leftrightarrow\left(x^3-y^3\right)^2=4x^4y^4\left(1-\frac{1}{xy}\right)\)
\(\Leftrightarrow1-\frac{1}{xy}=\left(\frac{x^3-y^3}{2x^2y^2}\right)^2\)
\(\Rightarrow\sqrt{1-\frac{1}{xy}}=\left|\frac{x^3-y^3}{2x^2y^2}\right|\) là một số hữu tỉ
Ta có :(a+b-c)2 \(\ge\) 0
<=>a2+b2+c2 \(\ge\) 2(bc-ab+ac)
<=>\(\frac{5}{3}\ge\) 2(bc-ab+ac)
<=>bc+ac-ab \(\le\frac{5}{6}< 1\)
<=>\(\frac{bc+ac-ab}{abc}< \frac{1}{abc}\) (vì a,b,c>0 nên chia cả 2 vế cho abc)
<=>\(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}< 1\) (đpcm)
a. Ta có:\(\frac{x}{y}\sqrt{\frac{y^2}{x^4}=}\) \(\frac{x}{y}.\frac{\left|y\right|}{x^2}=\frac{x.y}{x^2y}\)\(=\frac{1}{x}\)(Vì \(x\ne0;y>0\))
b \(3x^2\sqrt{\frac{8}{x^2}}=3x^2\frac{2\sqrt{2}}{\left|x\right|}=\frac{6x^2\sqrt{2}}{-x}=-6x\sqrt{2}\)( Vì \(x< 0\))
1. ĐKXĐ: \(\left\{{}\begin{matrix}a;b\ge0\\a\ne9\end{matrix}\right.\)
\(A=\frac{2\sqrt{a}+3\sqrt{b}}{\sqrt{a}\left(\sqrt{b}+2\right)-3\left(\sqrt{b}+2\right)}-\frac{6-\sqrt{ab}}{\sqrt{a}\left(\sqrt{b}+2\right)+3\left(\sqrt{b}+2\right)}\)
\(=\frac{2\sqrt{a}+3\sqrt{b}}{\left(\sqrt{a}-3\right)\left(\sqrt{b}+2\right)}-\frac{6-\sqrt{ab}}{\left(\sqrt{a}+3\right)\left(\sqrt{b}+2\right)}=\frac{\left(\sqrt{a}+3\right)\left(2\sqrt{a}+3\sqrt{b}\right)+\left(\sqrt{ab}-6\right)\left(\sqrt{a}-3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)\left(\sqrt{b}+2\right)}\)
\(=\frac{2a+9\sqrt{b}+a\sqrt{b}+18}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)\left(\sqrt{b}+2\right)}=\frac{a\left(\sqrt{b}+2\right)+9\left(\sqrt{b}+2\right)}{\left(a-9\right)\left(\sqrt{b}+2\right)}\)
\(=\frac{\left(a+9\right)\left(\sqrt{b}+2\right)}{\left(a-9\right)\left(\sqrt{b}+2\right)}=\frac{a+9}{a-9}\)
b .
\(\frac{a+9}{a-9}=\frac{b+10}{b-10}\Leftrightarrow\frac{a-9+18}{a-9}=\frac{b-10+20}{b-10}\)
\(\Leftrightarrow1+\frac{18}{a-9}=1+\frac{20}{b-10}\Leftrightarrow\frac{18}{a-9}=\frac{20}{b-10}\)
\(\Leftrightarrow18\left(b-10\right)=20\left(a-9\right)\Leftrightarrow18b=20a\Leftrightarrow\frac{a}{b}=\frac{9}{10}\)
3.
\(x^2-4x+4-\left(x^2+6x+9\right)=2x-10\)
\(\Leftrightarrow-10x-5=2x-10\)
\(\Leftrightarrow12x=5\)
b. \(\Leftrightarrow\left\{{}\begin{matrix}17\left(x-y\right)+7\left(2x+y\right)=833\\19\left(4x+y\right)+5\left(y-7\right)=1425\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}31x-10y=833\\76x+24y=1460\end{matrix}\right.\)
Bấm máy