a,ĐK: x≥-1

Đặt \(t=\sqrt{x^2+5x+4}\left(t\ge0\right)\)

  ⇒ \(t^2+t-6=0\)

  \(\Leftrightarrow\left(t+3\right)\left(t-2\right)=0\)

  \(\Leftrightarrow\left[{}\begin{matrix}t=-3\left(loại\right)\\t=2\end{matrix}\right.\)

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

  \(\Leftrightarrow x^2+5x+4=4\)

  \(\Leftrightarrow x\left(x+5\right)=0\)

  \(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=-5\left(loại\right)\end{matrix}\right.\)

b,ĐK: \(0\le x\le2\)

Ta có: \(\left(x+5\right)\left(2-x\right)=3\sqrt{x^2+3x}\)

    \(\Leftrightarrow-x^2-3x+10=3\sqrt{x^2+3x}\)     (1)

Đặt \(t=\sqrt{x^2+3x}\left(t\ge0\right)\)

  \(\Rightarrow\left(1\right)\Leftrightarrow-t^2+10-3t=0\)  

             \(\Leftrightarrow\left(t+5\right)\left(2-t\right)=0\)

             \(\Leftrightarrow\left[{}\begin{matrix}t=-5\left(loại\right)\\t=2\end{matrix}\right.\)

             \(\Leftrightarrow\sqrt{x^2+3x}=2\)

             \(\Leftrightarrow x^2+3x=4\)

             \(\Leftrightarrow\left(x+4\right)\left(x-1\right)=0\)

             \(\Leftrightarrow\left[{}\begin{matrix}x=-4\left(loại\right)\\x=1\left(tm\right)\end{matrix}\right.\)

ĐKXĐ: ...

\(\sqrt[3]{x^3+5x^2}-x-2+x+1-\sqrt{\dfrac{5x^2-2}{6}}=0\)

\(\Leftrightarrow\dfrac{x^3+5x^2-\left(x+2\right)^3}{\left(x+2\right)^2+\left(x+2\right)\sqrt[3]{x^3+5x^2}+\sqrt[3]{\left(x^3+5x^2\right)^2}}+\dfrac{\left(x+1\right)^2-\dfrac{5x^2-2}{6}}{x+1+\sqrt{\dfrac{5x^2-2}{6}}}=0\)

\(\Leftrightarrow\left(x^2+12x+8\right)\left(\dfrac{1}{6\left(x+1\right)+\sqrt{6\left(5x^2-2\right)}}-\dfrac{1}{\left(x+2\right)^2+\left(x+2\right)\sqrt[3]{x^3+5x^2}+\sqrt[3]{\left(x^3+5x^2\right)^2}}\right)=0\)

\(\Leftrightarrow x^2+12x+8=0\)

Đk: \(x\ge6\)

pt\(\Leftrightarrow\sqrt{5x^2+4x}=5\sqrt{x}+\sqrt{x^2-3x-18}\)

\(\Leftrightarrow5x^2+4x=25x+x^2-3x-18+10\sqrt{x\left(x^2-3x-18\right)}\)

\(\Leftrightarrow2x^2-9x+9=5\sqrt{x^3-3x^2-18x}\)

\(\Leftrightarrow4x^4+81x^2+81-36x^3-162x+36x^2=25\left(x^3-3x^2-18x\right)\)

\(\Leftrightarrow4x^4-61x^3+192x^2+288x+81=0\)

\(\Leftrightarrow\left(x-9\right)\left(4x+3\right)\left(x^2-7x-3\right)=0\)

\(\Leftrightarrow\left(4x+3\right)\left(x-9\right)\left(x-\dfrac{7+\sqrt{61}}{2}\right)\left(x-\dfrac{7-\sqrt{61}}{2}\right)=0\)

mà x \(\ge6\) \(\Rightarrow\left\{{}\begin{matrix}4x+3>0\\x-\dfrac{7-\sqrt{61}}{2}>0\end{matrix}\right.\)

\(\Rightarrow\left(x-9\right)\left(x-\dfrac{7+\sqrt{61}}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=\dfrac{7+\sqrt{61}}{2}\end{matrix}\right.\)

Vậy.....

Giải phương trình $x^2-4x+6=\sqrt{2x^2-5x+3}+\sqrt{-3x^2+9x-5}$ - Phương trình - hệ phương trình - bất phương trình - Diễn đàn Toán học

a:

\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=3\)

=>|x-3|=3

=>x-3=3 hoặc x-3=-3

=>x=0 hoặc x=6

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

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

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

=>\(\left[{}\begin{matrix}\sqrt{x-1}+1=2\\\sqrt{x-1}+1=-2\left(loại\right)\end{matrix}\right.\Leftrightarrow\sqrt{x-1}=1\)

=>x-1=1

=>x=2

c:

ĐKXĐ: x>4/5

PT \(\Leftrightarrow\sqrt{\dfrac{5x-4}{x+2}}=2\)

=>\(\dfrac{5x-4}{x+2}=4\)

=>5x-4=4x+8

=>x=12(nhận)

d: ĐKXĐ: x-4>=0 và x+1>=0

=>x>=4

PT =>\(\left(\sqrt{x-4}+\sqrt{x+1}\right)^2=5^2=25\)

=>\(x-4+x+1+2\sqrt{\left(x-4\right)\left(x+1\right)}=25\)

=>\(\sqrt{4\left(x^2-3x-4\right)}=25-2x+3=28-2x\)

=>\(\sqrt{x^2-3x-4}=14-x\)

=>x<=14 và x^2-3x-4=(14-x)^2=x^2-28x+196

=>x<=14 và -3x-4=-28x+196

=>x<=14 và 25x=200

=>x=8(nhận)

a) \(\sqrt{x^2-6x+9}=3\)

\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=3\)

\(\Leftrightarrow\left|x-3\right|=3 \)

TH1: \(\left|x-3\right|=x-3\) với \(x\ge3\)

Pt trở thành:

\(x-3=3\) (ĐK: \(x\ge3\))

\(\Leftrightarrow x=3+3\)

\(\Leftrightarrow x=6\left(tm\right)\)

TH2: \(\left|x-3\right|=-\left(x-3\right)\) với \(x< 3\)

Pt trở thành:

\(-\left(x-3\right)=3\) (ĐK: \(x< 3\))

\(\Leftrightarrow x-3=-3\)

\(\Leftrightarrow x=-3+3\)

\(\Leftrightarrow x=0\left(tm\right)\)

b) \(\sqrt{x+2\sqrt{x-1}}=2\) (ĐK: \(x\ge1\))

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

\(\Leftrightarrow2\sqrt{x-1}=4-x\)

\(\Leftrightarrow4\left(x-1\right)=16-8x+x^2\)

\(\Leftrightarrow4x-4=16-8x+x^2\)

\(\Leftrightarrow x^2-12x+20=0\)

\(\Leftrightarrow\left(x-10\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=10\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)

c) \(\dfrac{\sqrt{5x-4}}{\sqrt{x+2}}=2\) (ĐK: \(x\ge\dfrac{4}{5}\))

\(\Leftrightarrow\dfrac{5x-4}{x+2}=4\)

\(\Leftrightarrow5x-4=4x+8\)

\(\Leftrightarrow x=12\left(tm\right)\)