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\(\left(1+\sqrt{1993}\right).\sqrt{1994-2\sqrt{1993}}\)
\(=\left(1+\sqrt{1993}\right).\sqrt{\left(\sqrt{1993}\right)^2-2.\sqrt{1993}+1}\)
\(=\left(1+\sqrt{1993}\right).\sqrt{\left(\sqrt{1993}-1\right)^2}\)
\(=\left(1+\sqrt{1993}\right).\left(\sqrt{1993}-1\right)\)
\(=1992\)
Bài 1:
\(P=x\sqrt{3-x^2}=\sqrt{x^2}\cdot\sqrt{3-x^2}\)
\(=\sqrt{x^2\left(3-x^2\right)}\)\(\le\frac{x^2+3-x^2}{2}=\frac{3}{2}\)
Dấu = khi \(x=\sqrt{\frac{3}{2}}\)
Vậy MaxP=\(\frac{3}{2}\Leftrightarrow x=\sqrt{\frac{3}{2}}\)
P/s : sửa đề
ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)
a) \(P=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(P=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(P=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(P=\frac{-3\sqrt{x}-3x}{x-9}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(P=\frac{-3\sqrt{x}\left(1+\sqrt{x}\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)
\(P=\frac{-3\sqrt{x}}{\sqrt{x}+3}\)
b) \(P< -\frac{1}{2}\)
\(\Leftrightarrow\frac{-3\sqrt{x}}{\sqrt{x}+3}+\frac{1}{2}< 0\)
\(\Leftrightarrow\frac{-6\sqrt{x}+\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}< 0\)
\(\Leftrightarrow\frac{-5\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}< 0\)
Mà \(2\left(\sqrt{x}+3\right)>0\)
\(\Rightarrow-5\sqrt{x}+3< 0\)
\(\Leftrightarrow-5\sqrt{x}< -3\)
\(\Leftrightarrow\sqrt{x}>\frac{3}{5}\)
\(\Leftrightarrow x>\frac{9}{25}\)
Vấy .................
c) \(P.\left(\sqrt{x}+3\right)+2\sqrt{x}-2+x=2\)
\(\Leftrightarrow\frac{-3\sqrt{x}}{\sqrt{x}+3}\left(\sqrt{x}+3\right)+2\sqrt{x}-2+x=2\)
\(\Leftrightarrow-3\sqrt{x}+2\sqrt{x}-2-2+x=0\)
\(\Leftrightarrow-\sqrt{x}-4+x=0\)
\(\Leftrightarrow-\sqrt{x}\left(1-\sqrt{x}\right)=4\)
Còn lại lập bảng tự tìm giá trị của x là ra .( Chú ý : đối chiếu ĐKXĐ )
d)
\(P.\left(\sqrt{x}+3\right)+x\left(\sqrt{x}-m\right)=x-\sqrt{x}\left(3+m\right)\)
\(\Leftrightarrow\frac{-3\sqrt{x}}{\sqrt{x}+3}\left(\sqrt{x}+3\right)+x\sqrt{x}-xm=x-3\sqrt{x}-m\sqrt{x}\)
\(\Leftrightarrow-3\sqrt{x}+x\sqrt{x}-xm-x+3\sqrt{x}+m\sqrt{x}=0\)
\(\Leftrightarrow\sqrt{x}\left(x+m\right)-x\left(m+1\right)=0\)
\(\Leftrightarrow\sqrt{x}\left[x+m-m\sqrt{x}-\sqrt{x}\right]=0\)
\(\Leftrightarrow\sqrt{x}\left[m\left(1-\sqrt{x}\right)-\sqrt{x}\left(1-\sqrt{x}\right)\right]=0\)
\(\Leftrightarrow\sqrt{x}=0;m-\sqrt{x}=0;1-\sqrt{x}=0\)
+) \(\sqrt{x}=0\Leftrightarrow x=0\left(TM\right)\)
+) \(1-\sqrt{x}=0\)
\(\Leftrightarrow x=1\left(TM\right)\)
+) \(m-\sqrt{x}=0\)
\(\Leftrightarrow\orbr{\begin{cases}m-\sqrt{0}=0\\m-\sqrt{1}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}m=0\\m=1\end{cases}}}\)
Vậy ..................
\(\sqrt{\left(2011-x\right)^2+\left(x-1\right)^2}\ge\sqrt{\frac{1}{2}\left(2011-x+x-1\right)^2}=1005\sqrt{2}\)
a. ĐKXĐ \(x\ge0\)và \(x\ne9\)
Ta có \(K=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{3x-6\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(x-2\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}\)
b. Để \(K< -1\Rightarrow\frac{3\sqrt{x}-9+\sqrt{x}+3}{\sqrt{x}+3}< 0\Rightarrow\frac{4\sqrt{x}-6}{\sqrt{x}+3}< 0\Rightarrow4\sqrt{x}-6< 0\)vì \(\sqrt{x}+3\ge3\)
\(\Rightarrow0\le x< \frac{9}{4}\left(tm\right)\)
Vậy với \(0\le x< \frac{9}{4}\)thì K<-1
c. \(K=\frac{3\sqrt{x}-9}{\sqrt{x}+3}=3+\frac{-18}{\sqrt{x}+3}\)
Ta có \(\sqrt{x}+3\ge3\Rightarrow\frac{1}{\sqrt{x}+3}\le\frac{1}{3}\Rightarrow-\frac{18}{\sqrt{x}+3}\ge-6\Rightarrow3+\frac{-18}{\sqrt{x}+3}\ge-3\)
\(\Rightarrow K\ge-3\)
Vậy \(MinK=-3\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\)
Đặt \(x+\sqrt{1+x^2}=a\Rightarrow a-x=\sqrt{1+x^2}\Rightarrow a^2-2ax+x^2=1+x^2\)
=> \(a^2-1=2ax\Rightarrow x=\frac{1}{2}\left(a-\frac{1}{a}\right)\)
Tương tự, đặt \(y+\sqrt{1+y^2}=b\Rightarrow y=\frac{1}{2}\left(b-\frac{1}{b}\right)\)
=> x+y=\(\frac{1}{2}\left(a+b-\frac{1}{a}-\frac{1}{b}\right)=\frac{1}{2}\left(a+b-\frac{3}{3a}+\frac{3}{3b}\right)=\frac{1}{2}\left(a+b-\frac{1}{3}a-\frac{1}{3}b\right)\)(vì ab=3)
=\(\frac{1}{2}.\frac{2}{3}\left(a+b\right)=\frac{1}{3}\left(a+b\right)\)
Mà \(\left(a+b\right)^2\ge2ab=6\Rightarrow a+b\ge\sqrt{6}\Rightarrow\frac{1}{3}\left(a+b\right)\ge\frac{\sqrt{6}}{3}\)
dấu = xảy ra <=> a=b<=> x=y bạn tự thay vào và tự tìm nhá
^_^
Ta có: \(M=\sqrt{\left(1993-x\right)^2}+\sqrt{\left(1994-x\right)^2}>0\)
ĐKXĐ: \(\sqrt{\left(1993-x\right)^2}\ge0,\sqrt{\left(1994-x\right)^2}\ge0\forall x\inℝ\)
\(M=|1993-x|+|1994-x|\)
Ta có: GTNN của \(\sqrt{\left(1993-x\right)^2}=0\left(\sqrt{\left(1993-x\right)^2}\ge0\right)\)
GTNN của \(\sqrt{\left(1994-x\right)^2}=0\left(\sqrt{\left(1994-x\right)^2}\ge0\right)\)
=> GTNN của \(M=|1993-1994|hay|1994-1993|=1\)
Ta có: M = \(\sqrt{\left(1993-x\right)^2}+\sqrt{\left(1994-x\right)^2}\)
\(\Leftrightarrow\)M = \(\left|1993-x\right|+\left|1994-x\right|\)
= \(\left|x-1993\right|+\left|1994-x\right|\)
\(\ge\left|x-1993+1994-x\right|\)\(=\left|1\right|\)= 1
\(\Rightarrow M\ge1\)
Dấu "=" xảy ra khi: \(\left(x-1993\right)\left(1994-x\right)\ge0\)
\(\Leftrightarrow\hept{\begin{cases}x-1993\ge0\\1994-x\ge0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ge1993\\x\le1994\end{cases}}\)
\(\Leftrightarrow1993\le x\le1994\)
Vậy: min M = 1 \(\Leftrightarrow1993\le x\le1994\)