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2 tháng 10 2016

a)\(\sqrt{\frac{2x-3}{x-1}}=2\RightarrowĐk:\frac{2x-3}{x-1}\ge0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x\ge\frac{3}{2}\\x< 1\end{array}\right.\)

\(\sqrt{\frac{2x-3}{x-1}}=2\Rightarrow\frac{2x-3}{x-1}=4\)

\(\Leftrightarrow2x-3=4\left(x-1\right)\Leftrightarrow2x-3=4x-4\)

\(\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)(nhận)

b)\(\frac{\sqrt{2x-3}}{\sqrt{x-1}}=2\RightarrowĐk:\begin{cases}2x-3\ge0\\x-1>0\end{cases}\)

\(\Leftrightarrow x\ge\frac{3}{2}\)

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

\(\Leftrightarrow2x-3=4x-4\)\(\Leftrightarrow x=\frac{1}{2}\)(loại)

c)\(\sqrt{\frac{4x+3}{x+1}}=3\RightarrowĐk:\frac{4x+3}{x+1}\ge0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x\ge\frac{-3}{4}\\x< -1\end{array}\right.\)

\(\sqrt{\frac{4x+3}{x+1}}=3\Rightarrow\frac{4x+3}{x+1}=9\)

\(\Leftrightarrow4x+3=9\left(x+1\right)\Leftrightarrow4x+3=9x+9\)

\(\Leftrightarrow5x=-6\Leftrightarrow x=\frac{-6}{5}\)(nhận)

c)\(\frac{\sqrt{4x+3}}{\sqrt{x+1}}=3\RightarrowĐk:\begin{cases}4x+3\ge0\\x+1>0\end{cases}\)

\(\Rightarrow x\ge\frac{-3}{4}\)

\(\frac{\sqrt{4x+3}}{\sqrt{x+1}}=3\Rightarrow\sqrt{4x+3}=3\sqrt{x+1}\)

\(\Leftrightarrow4x+3=9\left(x+1\right)\Leftrightarrow4x+3=9x+9\)

\(\Leftrightarrow x=\frac{-6}{5}\)(loại)

30 tháng 3 2020
https://i.imgur.com/iX7y3qX.jpg
30 tháng 3 2020
https://i.imgur.com/GMDpx0f.jpg
NV
6 tháng 3 2019

ĐKXĐ: \(x\ge\frac{3}{4}\)

\(\sqrt{4x-3}-\sqrt{2x-1}+\sqrt{\frac{x+3}{2x-1}}-\sqrt{\frac{x+3}{4x-3}}\ge0\)

\(\Rightarrow\sqrt{4x-3}-\sqrt{2x-1}+\sqrt{x+3}\left(\frac{\sqrt{4x-3}-\sqrt{2x-1}}{\sqrt{4x-3}.\sqrt{2x-1}}\right)\ge0\)

\(\Rightarrow\left(\sqrt{4x-3}-\sqrt{2x-1}\right)\left(1+\frac{\sqrt{x+3}}{\sqrt{4x-3}\sqrt{2x-1}}\right)\ge0\)

\(\Rightarrow\sqrt{4x-3}-\sqrt{2x-1}\ge0\) (do \(1+\frac{\sqrt{x+3}}{\sqrt{4x-3}\sqrt{2x-1}}>0\))

\(\Rightarrow\sqrt{4x-3}\ge\sqrt{2x-1}\)

\(\Rightarrow4x-3\ge2x-1\)

\(\Rightarrow x\ge1\)

a) \(\left\{{}\begin{matrix}x\ge0\\-\sqrt{x+7}< 0\\-5x-4\ne0\\-3x+2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x+7>0\\-5x\ne4\\-3x\ne-2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x>-7\\x\ne\frac{-4}{5}\\x\ne\frac{2}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne\frac{2}{3}\end{matrix}\right.\)

b) \(\left\{{}\begin{matrix}x\ge0\\x+4\ne0\\x-2\ge0\\-2x-10\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne-4\\x\ge2\\-2x\ne10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\ne-5\end{matrix}\right.\Leftrightarrow x\ge2\)

c) \(\left\{{}\begin{matrix}x\ge0\\-x-3\ne0\\2x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne-3\\x\ne-\frac{3}{2}\end{matrix}\right.\Leftrightarrow x\ge0\)

d) \(\left\{{}\begin{matrix}2x-7\ge0\\x\ge0\\3x-4\ne0\\x-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{7}{2}\\x\ge0\\x\ne\frac{4}{3}\\x\ne3\end{matrix}\right.\Leftrightarrow x\ge\frac{7}{2}\)

4 tháng 8 2020

em cảm ơn nhiều ạ

NV
27 tháng 10 2019

a/ ĐKXĐ: ...

\(\Leftrightarrow3\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)-7\)

Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow a^2=x+\frac{1}{4x}+1\)

\(\Rightarrow x+\frac{1}{4x}=a^2-1\)

Pt trở thành:

\(3a=2\left(a^2-1\right)-7\)

\(\Leftrightarrow2a^2-3a-9=9\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}=3\)

\(\Leftrightarrow2x-6\sqrt{x}+1=0\)

\(\Rightarrow\sqrt{x}=\frac{3+\sqrt{7}}{2}\Rightarrow x=\frac{8+3\sqrt{7}}{2}\)

b/ ĐKXĐ:

\(\Leftrightarrow5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)

Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow x+\frac{1}{4x}=a^2-1\)

\(\Rightarrow5a=2\left(a^2-1\right)+4\Leftrightarrow2a^2-5a+2=0\)

\(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+\frac{1}{2\sqrt{x}}=2\\\sqrt{x}+\frac{1}{2\sqrt{x}}=\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x-4\sqrt{x}+1=0\\2x-\sqrt{x}+1=0\left(vn\right)\end{matrix}\right.\)

NV
27 tháng 10 2019

c/ ĐKXĐ: ...

\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)

\(\Leftrightarrow\frac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)

\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\frac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)

\(\Leftrightarrow2x^2-8x+5=0\)

d/ ĐKXĐ: ...

\(\Leftrightarrow x+1-\frac{15}{6}\sqrt{x}+\sqrt{x^2-4x+1}-\frac{1}{2}\sqrt{x}=0\)

\(\Leftrightarrow\frac{x^2-\frac{17}{4}x+1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{x^2-\frac{17}{4}x+1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}=0\)

\(\Leftrightarrow\left(x^2-\frac{17}{4}x+1\right)\left(\frac{1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}\right)=0\)

\(\Leftrightarrow x^2-\frac{17}{4}x+1=0\)

\(\Leftrightarrow4x^2-17x+4=0\)

8 tháng 10 2016

Ta có:

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

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

  = \(\frac{1}{2}\)(\(\sqrt{2}\)-1)

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

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

=> 4x2= 2-2\(\sqrt{2}\)+1

=> 4x2= -2( \(\sqrt{2}\)-1)+1

=> 4x2= -4x +1 => 4x2+4x-1=0

Lại có:

A1= (\(4x^5\)+\(4x^4\)- \(x^3\)+1)19

   = [  x3( 4x2+4x-1) +1]19

   =1

    A2=( \(\sqrt{4x^5+4x^4-5x^3+5x+3}\))3

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

       = 23=8

  A3= \(\frac{1-\sqrt{2x}}{\sqrt{2x^2+2x}}\)

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

Cộng 3 số vào ta được A

6 tháng 10 2016

no biet

NV
26 tháng 11 2019

a/ ĐKXĐ: \(-\frac{3}{2}\le x\le4\)

\(\sqrt{2x+3}+\sqrt{4-x}=6x-3\left(x+7-2\sqrt{\left(2x+3\right)\left(4-x\right)}\right)-10\)

\(\Leftrightarrow\sqrt{2x+3}+\sqrt{4-x}=3\left(x+7+2\sqrt{\left(2x+3\right)\left(4-x\right)}\right)-52\)

Đặt \(\sqrt{2x+3}+\sqrt{4-x}=a>0\Rightarrow a^2=x+7+2\sqrt{\left(2x+3\right)\left(4-x\right)}\)

Phương trình trở thành:

\(a=3a^2-52\Leftrightarrow3a^2-a-52=0\Rightarrow\left[{}\begin{matrix}a=-4\left(l\right)\\a=\frac{13}{3}\end{matrix}\right.\)

\(\sqrt{2x+3}+\sqrt{4-x}=\frac{13}{3}\)

Phương trình này vô nghiệm nên ko muốn giải tiếp, bạn bình phương lên và chuyển vế thôi :(

b/ ĐKXĐ: \(-\frac{1}{4}\le x\le1\)

Đặt \(\sqrt{4x+1}+2\sqrt{1-x}=a>0\Rightarrow a^2=5+4\sqrt{-4x^2+3x+1}\)

\(\Rightarrow\sqrt{-4x^2+3x+1}=\frac{a^2-5}{4}\)

Pt trở thành:

\(a+10\left(\frac{a^2-5}{4}\right)=13\)

\(\Leftrightarrow5a^2+2a-51=0\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{17}{5}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{-4x^2+3x+1}=\frac{a^2-5}{4}=1\)

\(\Leftrightarrow-4x^2+3x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{3}{4}\end{matrix}\right.\)

NV
26 tháng 11 2019

c/ \(\Leftrightarrow x^2\left(x^2+2\right)=12-x\sqrt{2x^2+4}\)

\(\Leftrightarrow x^2\left(2x^2+4\right)=24-2x\sqrt{2x^2+4}\)

Đặt \(x\sqrt{2x^2+4}=a\) ta được:

\(a^2=24-2a\Leftrightarrow a^2+2a-24=0\Leftrightarrow\left[{}\begin{matrix}a=4\\a=-6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x\sqrt{2x^2+4}=4\left(x>0\right)\\x\sqrt{2x^2+4}=-6\left(x< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2\left(2x^2+4\right)=16\\x^2\left(2x^2+4\right)=36\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^4+2x^2-8=0\\x^4+2x^2-18=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2=2\\x^2=-4\left(l\right)\\x^2=\sqrt{19}-1\\x^2=-\sqrt{19}-1\left(l\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}< 0\left(l\right)\\x=-\sqrt{\sqrt{19}-1}\\x=\sqrt{\sqrt{19}-1}>0\left(l\right)\end{matrix}\right.\)