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a) ĐKXĐ: x\(\ge\)-3
PT\(\Leftrightarrow\sqrt{\left(x+7\right)\left(x+3\right)}=3\sqrt{x+3}+2\sqrt{x+7}-6\)
Đặt \(\left(\sqrt{x+3},\sqrt{x+7}\right)=\left(a,b\right)\) \(\left(a,b\ge0\right)\)
PT\(\Leftrightarrow ab=3a+2b-6\Leftrightarrow a\left(b-3\right)-2\left(b-3\right)=0\)
\(\Leftrightarrow\left(a-2\right)\left(b-3\right)=0\Leftrightarrow\orbr{\begin{cases}a=2\\b=3\end{cases}}\)(TM ĐK)
TH 1: a=2\(\Leftrightarrow\sqrt{x+3}=2\Leftrightarrow x+3=4\Leftrightarrow x=1\)(tm)
TH 2: b=3\(\Leftrightarrow\sqrt{x+7}=3\Leftrightarrow x+7=9\Leftrightarrow x=2\)(tm)
Vậy tập nghiệm phương trình S={1; 2}
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:
\(a+10\left(\frac{a^2-5}{4}\right)=13\)
\(\Leftrightarrow5a^2+2a-51=0\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{17}{5}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{-4x^2+3x+1}=\frac{a^2-5}{4}=1\)
\(\Leftrightarrow-4x^2+3x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{3}{4}\end{matrix}\right.\)
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:
\(a^2=24-2a\Leftrightarrow a^2+2a-24=0\Leftrightarrow\left[{}\begin{matrix}a=4\\a=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\sqrt{2x^2+4}=4\left(x>0\right)\\x\sqrt{2x^2+4}=-6\left(x< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2\left(2x^2+4\right)=16\\x^2\left(2x^2+4\right)=36\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^4+2x^2-8=0\\x^4+2x^2-18=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2=2\\x^2=-4\left(l\right)\\x^2=\sqrt{19}-1\\x^2=-\sqrt{19}-1\left(l\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}< 0\left(l\right)\\x=-\sqrt{\sqrt{19}-1}\\x=\sqrt{\sqrt{19}-1}>0\left(l\right)\end{matrix}\right.\)
a/ ĐKXĐ: \(x>3\)
\(\Leftrightarrow\sqrt{2\left(x^2-16\right)}+x-3=7-x\)
\(\Leftrightarrow\sqrt{2\left(x^2-16\right)}=10-2x\) (\(x\le5\))
\(\Leftrightarrow2\left(x^2-16\right)=\left(10-2x\right)^2\)
\(\Leftrightarrow x^2-20x+66=0\)
b/ ĐKXĐ: \(x>0\)
\(\Leftrightarrow\sqrt{\frac{\left(x+1\right)\left(x^2-x+1\right)}{x}}-\sqrt{x+1}-\left(\sqrt{x^2-x+1}-\sqrt{x}\right)=0\)
\(\Leftrightarrow\sqrt{\frac{x+1}{x}}\left(\sqrt{x^2-x+1}-\sqrt{x}\right)-\left(\sqrt{x^2-x+1}-\sqrt{x}\right)=0\)
\(\Leftrightarrow\left(\sqrt{\frac{x+1}{x}}-1\right)\left(\sqrt{x^2-x+1}-\sqrt{x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{\frac{x+1}{x}}=1\\\sqrt{x^2-x+1}=\sqrt{x}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\frac{x+1}{x}=1\\x^2-x+1=x\end{matrix}\right.\)
c/ĐKXĐ: \(x\ge-1\)
\(\Leftrightarrow\sqrt{\frac{\left(x+1\right)\left(x^2-x+1\right)}{\sqrt{x+3}}}+\sqrt{x+1}-\left(\sqrt{x^2+x+1}+\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\sqrt{\frac{x+1}{x+3}}\left(\sqrt{x^2-x+1}+\sqrt{x+3}\right)-\left(\sqrt{x^2-x+1}+\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\left(\sqrt{\frac{x+1}{x+3}}-1\right)\left(\sqrt{x^2-x+1}+\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\sqrt{\frac{x+1}{x+3}}=1\Leftrightarrow x+1=x+3\)
Pt vô nghiệm
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\(\Leftrightarrow\frac{\sqrt{x}-1+\sqrt{x}+1}{x-1}=1\Leftrightarrow2\sqrt{x}=x-1\)\(\Leftrightarrow x-2\sqrt{x}+1=2\Leftrightarrow\left(\sqrt{x}-1\right)^2=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3+2\sqrt{2}\\x=3-2\sqrt{2}\end{matrix}\right.\)
\(A=\left(\frac{2}{\sqrt{x}-2}+\frac{3}{2\sqrt{x}+1}-\frac{5\sqrt{x}-7}{2x-3\sqrt{x}-2}\right):\)\(\frac{2\sqrt{x}+3}{5x-10\sqrt{x}}\)
\(=\left(\frac{2}{\sqrt{x}-2}+\frac{3}{2\sqrt{x}+1}-\frac{5\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\right)\)\(:\frac{2\sqrt{x}+3}{5x-10\sqrt{x}}\)
\(=\frac{2\left(2\sqrt{x}+1\right)+3\left(\sqrt{x}-2\right)-5\sqrt{x}+7}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\)\(:\frac{2\sqrt{x}+3}{5\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\)\(.\frac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\frac{2\sqrt{x}+3}{2\sqrt{x}+1}.\frac{5\sqrt{x}}{2\sqrt{x}+3}=\frac{5\sqrt{x}}{2\sqrt{x}+1}\)
\(A\in Z\Leftrightarrow\frac{5\sqrt{x}}{2\sqrt{x}+1}\in Z\Leftrightarrow\frac{10\sqrt{x}}{2\sqrt{x}+1}\in Z\)
\(\Rightarrow\frac{10\sqrt{x}+5-5}{2\sqrt{x}+1}\in Z\Leftrightarrow5-\frac{5}{2\sqrt{x}+1}\in Z\)
\(\Rightarrow\frac{5}{2\sqrt{x}+1}\in Z\Rightarrow2\sqrt{x}+1\inƯ_5\)
Mà \(Ư_5=\left\{\pm1;\pm5\right\}\)
Nhưng \(2\sqrt{x}+1\ge1\)
\(\Rightarrow\orbr{\begin{cases}2\sqrt{x}+1=1\\2\sqrt{x}+1=5\end{cases}\Rightarrow\orbr{\begin{cases}2\sqrt{x}=0\\2\sqrt{x}=4\end{cases}}}\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}=2\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}}\)
Vậy \(x\in\left\{0;4\right\}\)
P=\(\left(\frac{3\sqrt{x}\left(\sqrt{x}-2\right)+\sqrt{x}\left(\sqrt{x}+2\right)-(x-\sqrt{x})}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right):\left(\frac{3\sqrt{x}}{\sqrt{x}+2}\right)=\left(\frac{3x-6\sqrt{x}+x+2\sqrt{x}-x+\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right):\left(\frac{3\sqrt{x}}{\sqrt{x}+2}\right)=\left(\frac{3x-3\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right).\frac{\sqrt{x}+2}{3\sqrt{x}}=\frac{3\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}+2}{3\sqrt{x}}=\frac{\sqrt{x}-1}{\sqrt{x}-2}\)
Lời giải:
ĐK: $x\geq 0; x\neq 1$
Ta có:
$P(x)=\frac{15\sqrt{x}-11}{(\sqrt{x}-1)(\sqrt{x}+3)}-\frac{(3\sqrt{x}-2)(\sqrt{x}+3)}{(\sqrt{x}-1)(\sqrt{x}+3)}-\frac{(2\sqrt{x}-3)(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+3)}$
$=\frac{15\sqrt{x}-11-(3\sqrt{x}-2)(\sqrt{x}+3)-(2\sqrt{x}-3)(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+3)}$
$=\frac{-5x+13\sqrt{x}-8}{(\sqrt{x}-1)(\sqrt{x}+3)}=\frac{(8-5\sqrt{x})(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+3)}$
$=\frac{8-5\sqrt{x}}{\sqrt{x}+3}$
Với $P=\frac{8-5\sqrt{x}}{\sqrt{x}+3}$ thì chưa đủ cơ sở để khẳng định $P(x)\leq \frac{2}{3}$