bài này

nhìn

trông có vẻ hơi khó...

bài này hết sức đơn giản, hơn nữa nó cũng có trong sách những viên kim cương của trần phương

Trả lời:

\(\sqrt{2x+2\sqrt{2x-1}}+\sqrt{2x-2\sqrt{2x-1}}\)\(\left(ĐK:\frac{1}{2}\le x\le1\right)\)

\(=\sqrt{2x-1+2\sqrt{2x-1}+1}+\sqrt{2x-1-2\sqrt{2x-1}+1}\)

\(=\sqrt{\left(2x-1+1\right)^2}+\sqrt{\left(2x-1-1\right)^2}\)

\(=\sqrt{\left(2x\right)^2}+\sqrt{\left(2x-2\right)^2}\)

\(=\left|2x\right|+\left|2x-2\right|\)

\(=2x+2-2x\)

\(=2\)

Mk cx như bn

a) ĐK: \(x\ge\frac{1}{2}\).

\(\sqrt{2x-1}+\sqrt{x+4}=6\)

\(\Leftrightarrow\sqrt{2x-1}-3+\sqrt{x+4}-3=0\)

\(\Leftrightarrow\frac{2x-1-9}{\sqrt{2x-1}+3}+\frac{x+4-9}{\sqrt{x+4}+3}=0\)

\(\Leftrightarrow\left(x-5\right)\left(\frac{2}{\sqrt{2x-1}+3}+\frac{1}{\sqrt{x+4}+3}\right)=0\)

\(\Leftrightarrow x-5=0\)

\(\Leftrightarrow x=5\).

b) ĐK: \(x\ge\frac{1}{2}\).

 \(\sqrt{x+3}-\sqrt{2x-1}=1\)

\(\Leftrightarrow\sqrt{x+3}-2+1-\sqrt{2x-1}=0\)

\(\Leftrightarrow\frac{x+3-4}{\sqrt{x+3}+2}+\frac{1-\left(2x-1\right)}{1+\sqrt{2x-1}}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{\sqrt{x+3}+2}-\frac{2}{1+\sqrt{2x-1}}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\\frac{1}{\sqrt{x+3}+2}=\frac{2}{1+\sqrt{2x-1}}\left(1\right)\end{cases}}\)

\(\left(1\right)\Leftrightarrow2\sqrt{x+3}+4=1+\sqrt{2x-1}\)

Có \(4>1,2\sqrt{x+3}=\sqrt{4x+12}>\sqrt{2x-1}\)

do đó phương trình \(\left(1\right)\)vô nghiệm. 

a) ĐK : x >= 1/2

\(\Leftrightarrow\left(\sqrt{2x-1}-3\right)+\left(\sqrt{x+4}-3\right)=0\)

\(\Leftrightarrow\frac{2x-1-9}{\sqrt{2x-1}+3}+\frac{x+4-9}{\sqrt{x+4}+3}=0\)

\(\Leftrightarrow\left(x-5\right)\left(\frac{2}{\sqrt{2x-1}+3}+\frac{1}{\sqrt{x+4}+3}\right)=0\)(1)

Dễ thấy với x >= 1/2 thì \(\frac{2}{\sqrt{2x-1}+3}+\frac{1}{\sqrt{x+4}+3}>0\)

nên (1) <=> x - 5 = 0 <=> x = 5 (tm)

Vậy phương trình có nghiệm x = 5

\(VT=\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}}\)

\(\sqrt{2}VT=\sqrt{12+2\sqrt{11}}-\sqrt{12-2\sqrt{11}}\)

\(=\sqrt{\left(\sqrt{11}+1\right)^2}-\sqrt{\left(\sqrt{11}-1\right)^2}=\left|\sqrt{11}+1\right|-\left|\sqrt{11}-1\right|\)

\(=\sqrt{11}+1-\sqrt{11}+1=2\)

\(\Rightarrow VT=\frac{2}{\sqrt{2}}=\sqrt{2}=VP\)( đpcm ) 

a) A có nghĩa <=> \(\frac{3x-5}{x-1}\ge0\)

<=> \(\hept{\begin{cases}3x-5\ge0\\x-1>0\end{cases}}\) hoặc \(\hept{\begin{cases}3x-5\le0\\x-1< 0\end{cases}}\)

<=> \(\hept{\begin{cases}x\ge\frac{5}{3}\\x>1\end{cases}}\)hoặc \(\hept{\begin{cases}x\le\frac{5}{3}\\x< 1\end{cases}}\)

<=> \(\orbr{\begin{cases}x\ge\frac{5}{3}\\x< 1\end{cases}}\)

b) Với \(\orbr{\begin{cases}x\ge\frac{5}{3}\\x< 1\end{cases}}\)ta có:

A = 2 <=> \(\sqrt{\frac{3x-5}{x-1}}=3\)

<=> \(\frac{3x-5}{x-1}=9\)

=> \(3x-5=9\left(x-1\right)\)

<=> \(3x-5=9x-9\)

<=> \(6x=4\)

<=> \(x=\frac{2}{3}\)(tm)

\(a,\frac{3x-5}{x-1}\ge0;x-1\ne0\)

lập TH ra đc :

\(TH1:x\ge\frac{5}{3}\)

\(TH2:x\le1;x\ne1< =>x< 1\)

vậy với \(\orbr{\begin{cases}x\ge5\\x< 1\end{cases}}\)thì A có nghĩa

\(b,A=\sqrt{\frac{3x-5}{x-1}}=3\)

\(\frac{3x-5}{x-1}=9\)

\(3x-5=9x-9\)

\(x=\frac{2}{3}\left(TM\right)\)

\(\)

\(B=\frac{\sqrt{4+2\sqrt{3}}}{2}-\frac{\sqrt{4-2\sqrt{3}}}{2}\)

\(B=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}}{2}\)

\(B=\frac{\left|\sqrt{3}+1\right|-\left|\sqrt{3}-1\right|}{2}\)

\(B=\frac{\sqrt{3}+1-\sqrt{3}+1}{2}\)

\(B=\frac{2}{2}=1\)

\(\sqrt{11-2\sqrt{30}}:\left(1-\frac{\sqrt{5}}{\sqrt{6}}\right)\)

\(\sqrt{11-2\sqrt{3}\sqrt{5}\sqrt{2}}:\frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}}\)

\(\sqrt{\left(\sqrt{3}\sqrt{2}\right)^2-2\sqrt{3}\sqrt{5}\sqrt{2}+\sqrt{5}^2}.\frac{\sqrt{6}}{\sqrt{6}-\sqrt{5}}\)

\(\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}.\frac{\sqrt{6}}{\sqrt{6}-\sqrt{5}}\)

\(\left(\sqrt{6}-\sqrt{5}\right).\frac{\sqrt{6}}{\sqrt{6}-\sqrt{5}}\)

\(=\sqrt{6}\)

\(\sqrt{11-2\sqrt{30}}:\left(1-\frac{\sqrt{5}}{\sqrt{6}}\right)\)

\(=\sqrt{6-2\sqrt{30}+5}:\left(1-\frac{\sqrt{5}}{\sqrt{6}}\right)\)

\(=\sqrt{\left(\sqrt{6}\right)^2-2\sqrt{30}+\left(\sqrt{5}\right)^2}:\left(1-\frac{\sqrt{5}}{\sqrt{6}}\right)\)

\(=\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}:\left(1-\frac{\sqrt{5}}{\sqrt{6}}\right)\)

\(=\left|\sqrt{6}-\sqrt{5}\right|:\left(\frac{\sqrt{6}}{\sqrt{6}}-\frac{\sqrt{5}}{\sqrt{6}}\right)\)

\(=\left(\sqrt{6}-\sqrt{5}\right):\left(\frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}}\right)\)

\(=\left(\sqrt{6}-\sqrt{5}\right).\frac{\sqrt{6}}{\sqrt{6}-\sqrt{5}}\)

\(=\sqrt{6}\)

\(x^2-2\left(m+2\right)x-4m-12=0\)

\(\Delta'=\left(m+2\right)^2+4m+12=m^2+8m+16=\left(m+4\right)^2\)

Để phương trình có hai nghiệm phân biệt thì \(\Delta'>0\Leftrightarrow\left(m+4\right)^2>0\Leftrightarrow m\ne-4\).

Với \(m\ne-4\)phương trình có hai nghiệm phân biệt \(x_1,x_2\).

Theo định lí Viete ta có: 

\(\hept{\begin{cases}x_1+x_2=2\left(m+2\right)\\x_1x_2=-4m-12\end{cases}}\)

\(\left|x_1-x_2\right|=\sqrt{\left(x_1-x_2\right)^2}=\sqrt{\left(x_1+x_2\right)-4x_1x_2}\)

\(=\sqrt{4\left(m+2\right)^2+4\left(4m+12\right)}=2\sqrt{\left(m+4\right)^2}=2\left(m+4\right)=3\)

\(\Leftrightarrow m=-\frac{5}{2}\left(tm\right)\).