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a/ ĐKXĐ: \(x^2+2x-6\ge0\)
\(\Leftrightarrow x^2+2x-6+\left(x-2\right)\sqrt{x^2+2x-6}=0\)
\(\Leftrightarrow\sqrt{x^2+2x-6}\left(\sqrt{x^2+2x-6}+x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+2x-6}=0\left(1\right)\\\sqrt{x^2+2x-6}=2-x\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x^2+2x-6=0\Rightarrow x=-1\pm\sqrt{7}\)
\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}2-x\ge0\\x^2+2x-6=\left(2-x\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le2\\6x=10\end{matrix}\right.\) \(\Rightarrow x=\frac{5}{3}\)
Câu b nhìn ko ra hướng, ko biết đề có nhầm đâu ko :(
c/ ĐKXĐ: \(\left[{}\begin{matrix}x\ge0\\x\le-1\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{\left(x^2+x\right)\left(x^2+x+2\right)}-\left(3-x\right)\sqrt{x^2+x}=0\)
\(\Leftrightarrow\sqrt{x^2+x}\left(\sqrt{x^2+x+2}-3+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x=0\left(1\right)\\\sqrt{x^2+x+2}=3-x\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}3-x\ge0\\x^2+x+2=\left(3-x\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le3\\7x=7\end{matrix}\right.\) \(\Rightarrow x=1\)
d/
Ta có \(\sqrt{x^2+3x+4}=\sqrt{\left(x+\frac{3}{4}\right)^2+\frac{7}{4}}>1\)
\(\Rightarrow\sqrt{x^2+3x+4}-1>0\)
Nhân 2 vế của pt với \(\sqrt{x^2+3x+4}-1\)
\(\left(\sqrt{x^2+3x+4}-1\right)\left(x^2+3x+3\right)=3x\left(x^2+3x+3\right)\)
\(\Leftrightarrow\left(x^2+3x+3\right)\left(\sqrt{x^2+3x+4}-1-3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+3x+3=0\left(vn\right)\\\sqrt{x^2+3x+4}=3x+1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Rightarrow\left\{{}\begin{matrix}x\ge-\frac{1}{3}\\x^2+3x+4=\left(3x+1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow8x^2+3x-3=0\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{-3+\sqrt{105}}{6}\\x=\frac{-3-\sqrt{105}}{6}\left(l\right)\end{matrix}\right.\)
a/ ĐKXĐ: \(x\ne\pm5\)
\(\Leftrightarrow\left(x+5\right)^2-\left(x-5\right)^2=20\)
\(\Leftrightarrow\left(x^2+10x+25\right)-\left(x^2-10x+25\right)=20\)
\(\Leftrightarrow20x=20\Rightarrow x=1\)
b/ ĐKXĐ: \(x\ne\pm4\)
\(\Leftrightarrow2x\left(x-4\right)-4x=0\)
\(\Leftrightarrow2x^2-12x=0\)
\(\Leftrightarrow2x\left(x-6\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
c/ ĐKXĐ: \(x\ne\pm\frac{2}{3}\)
\(\Leftrightarrow\left(3x+2\right)^2-6\left(3x-2\right)=9x^2\)
\(\Leftrightarrow9x^2+12x+4-18x+12=9x^2\)
\(\Leftrightarrow6x=16\Rightarrow x=\frac{8}{3}\)
a/
\(\Leftrightarrow4x^2-12x+9=\left(3x-2\right)^2\)
\(\Leftrightarrow5x^2-5=0\Rightarrow x=\pm1\)
b/
\(\Leftrightarrow25x^2-10x+1=\left(x+6\right)^2\)
\(\Leftrightarrow24x^2-22x-35=0\Rightarrow\left[{}\begin{matrix}x=\frac{7}{4}\\x=-\frac{5}{6}\end{matrix}\right.\)
c/
\(\Leftrightarrow16x^2-8x+1=\left(x-3\right)^2\)
\(\Leftrightarrow15x^2-2x-8=0\Rightarrow\left[{}\begin{matrix}x=\frac{4}{5}\\x=-\frac{2}{3}\end{matrix}\right.\)
d/ \(x\ge\frac{3}{2}\)
\(\Leftrightarrow\left(5x+1\right)^2=\left(2x-3\right)^2\)
\(\Leftrightarrow21x^2+22x-8=0\Rightarrow\left[{}\begin{matrix}x=\frac{2}{7}\\x=-\frac{4}{3}\end{matrix}\right.\)
e/
\(\Leftrightarrow\left[{}\begin{matrix}3x-4=x-2\\3x-4=2-x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=2\\4x=6\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\frac{3}{2}\end{matrix}\right.\)
f/
\(\Leftrightarrow\left[{}\begin{matrix}3x^2-2x=6-x^2\\3x^2-2x=x^2-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x^2-2x-6=0\\2x^2-2x+6=0\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=\frac{3}{2}\end{matrix}\right.\)
g/
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x=2x^2-x-2\\x^2-2x=-2x^2+x+2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-2=0\\3x^2-3x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=\frac{3\pm\sqrt{33}}{6}\\\end{matrix}\right.\)
28. \(x^2+\frac{9x^2}{\left(x-3\right)^2}=40\) DK: \(x\ne3\)
PT\(\Leftrightarrow\left(x+\frac{3x}{x-3}\right)^2-6\frac{x^2}{x-3}-40=0\)\(\Leftrightarrow\frac{x^4}{\left(x-3\right)^2}-6\frac{x^2}{x-3}-40=0\)
Dat \(\frac{x^2}{x-3}=a\). PTTT \(a^2-6a-40=0\)\(\Leftrightarrow\left(a-10\right)\left(a+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=10\\a=-4\end{matrix}\right.\)
giai tiep
14. \(\frac{1}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}=1\) DK: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
PT\(\Leftrightarrow\frac{\sqrt{x}-1+\sqrt{x}+1}{x-1}=1\Leftrightarrow2\sqrt{x}=x-1\)\(\Leftrightarrow x-2\sqrt{x}+1=2\Leftrightarrow\left(\sqrt{x}-1\right)^2=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3+2\sqrt{2}\\x=3-2\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left(x^2-3x+1\right)\left(x+1\right)\left(x+2\right)\left(x-4\right)\left(x-5\right)=-30\)
\(\Leftrightarrow\left(x^2-3x+1\right)\left(x^2-3x-4\right)\left(x^2-3x-5\right)=-30\)
Đặt x^2-3x=a
=>(a+1)(a-4)(a-5)=-30
=>\(\left(a^2-3a-4\right)\left(a-5\right)=-30\)
=>\(a^3-5a^2-3a^2+15a-4a+20+30=0\)
=>a^3-8a^2+11a+50=0
=>a=-1,77
=>x^2-3x=-1,77
=>x^2-3x+1,77=0
hay \(\left[{}\begin{matrix}x=\dfrac{15+4\sqrt{3}}{10}\\x=\dfrac{15-4\sqrt{3}}{10}\end{matrix}\right.\)