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8.
ĐKXĐ: \(x\ge\frac{2}{3}\)
\(\Leftrightarrow\frac{9\left(x+3\right)}{\sqrt{4x+1}+\sqrt{3x-2}}=x+3\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\left(l\right)\\\frac{9}{\sqrt{4x+1}+\sqrt{3x-2}}=1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{4x+1}+\sqrt{3x-2}=9\)
\(\Leftrightarrow\sqrt{4x+1}-5+\sqrt{3x-2}-4=0\)
\(\Leftrightarrow\frac{4\left(x-6\right)}{\sqrt{4x+1}+5}+\frac{3\left(x-6\right)}{\sqrt{3x-2}+4}=0\)
\(\Leftrightarrow\left(x-6\right)\left(\frac{4}{\sqrt{4x+1}+5}+\frac{3}{\sqrt{3x-2}+4}\right)=0\)
\(\Leftrightarrow x=6\)
6.
ĐKXD: ...
\(\Leftrightarrow2\left(x^2-6x+9\right)+\left(x+5-4\sqrt{x+1}\right)=0\)
\(\Leftrightarrow2\left(x-3\right)^2+\frac{\left(x-3\right)^2}{x+5+4\sqrt{x+1}}=0\)
\(\Leftrightarrow\left(x-3\right)^2\left(2+\frac{1}{x+5+4\sqrt{x+1}}\right)=0\)
\(\Leftrightarrow x=3\)
7.
\(\sqrt{x-\frac{1}{x}}-\sqrt{2x-\frac{5}{x}}+\frac{4}{x}-x=0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x-\frac{1}{x}}=a\ge0\\\sqrt{2x-\frac{5}{x}}=b\ge0\end{matrix}\right.\) \(\Rightarrow a^2-b^2=\frac{4}{x}-x\)
\(\Rightarrow a-b+a^2-b^2=0\)
\(\Leftrightarrow\left(a-b\right)\left(a+b+1\right)=0\)
\(\Leftrightarrow a=b\Leftrightarrow x-\frac{1}{x}=2x-\frac{5}{x}\)
\(\Leftrightarrow x=\frac{4}{x}\Rightarrow x=\pm2\)
Thế nghiệm lại pt ban đầu để thử (hoặc là bạn tìm ĐKXĐ từ đầu)
1.
ĐK: \(-1\le x\le4\)
Đặt \(\sqrt{x+1}+\sqrt{4-x}=t\left(t\ge0\right)\)
\(\Leftrightarrow\sqrt{\left(x+1\right)\left(4-x\right)}=\frac{t^2-5}{2}\)
\(PT\Leftrightarrow t+\frac{t^2-5}{2}=5\Rightarrow t^2+2t-15=0\) \(\Rightarrow\left[{}\begin{matrix}t=3\\t=-5\left(l\right)\end{matrix}\right.\)
\(t=3\Rightarrow\sqrt{-x^2+3x+4}=2\) \(\Leftrightarrow-x^2+3x+4=4\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\) (tm)
2.
ĐK:\(x\ge4\)
Đặt \(\sqrt{x+4}+\sqrt{x-4}=t\left(t\ge0\right)\)
\(\Rightarrow2\sqrt{x^2-16}=t^2-2x\)
\(PT\Leftrightarrow t=2x-12+t^2-2x\)
\(\Leftrightarrow t^2-t-12=0\Rightarrow\left[{}\begin{matrix}t=4\\t=-3\left(l\right)\end{matrix}\right.\) Giải tiếp như trên.
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/ 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 \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow x+\frac{1}{4x}=a^2-1\)
\(\Rightarrow5a=2\left(a^2-1\right)+4\Leftrightarrow2a^2-5a+2=0\)
\(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+\frac{1}{2\sqrt{x}}=2\\\sqrt{x}+\frac{1}{2\sqrt{x}}=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-4\sqrt{x}+1=0\\2x-\sqrt{x}+1=0\left(vn\right)\end{matrix}\right.\)
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: ...
\(\Leftrightarrow x+1-\frac{15}{6}\sqrt{x}+\sqrt{x^2-4x+1}-\frac{1}{2}\sqrt{x}=0\)
\(\Leftrightarrow\frac{x^2-\frac{17}{4}x+1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{x^2-\frac{17}{4}x+1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}=0\)
\(\Leftrightarrow\left(x^2-\frac{17}{4}x+1\right)\left(\frac{1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}\right)=0\)
\(\Leftrightarrow x^2-\frac{17}{4}x+1=0\)
\(\Leftrightarrow4x^2-17x+4=0\)
a/ ĐKXĐ: ...
\(\Leftrightarrow2\sqrt{\frac{x}{x-1}}-\sqrt{\frac{x-1}{x}}=\frac{2\left(x-1\right)}{x}+3\)
Đặt \(\sqrt{\frac{x-1}{x}}=a>0\)
\(\frac{2}{a}-a=2a^2+3\Leftrightarrow2a^3+a^2+3a-2=0\)
\(\Leftrightarrow\left(2a-1\right)\left(a^2+a+2\right)=0\Leftrightarrow a=\frac{1}{2}\)
\(\Rightarrow\sqrt{\frac{x-1}{x}}=\frac{1}{2}\Leftrightarrow4\left(x-1\right)=x\)
b/ ĐKXĐ: ...
\(\Leftrightarrow3\sqrt{\frac{2x}{x-1}}+4\sqrt{\frac{x-1}{2x}}=\frac{3\left(x-1\right)}{2x}+10\)
Đặt \(\sqrt{\frac{x-1}{2x}}=a>0\)
\(\frac{3}{a}+4a=3a^2+10\Leftrightarrow3a^3-4a^2+10a-3=0\)
\(\Leftrightarrow\left(3a-1\right)\left(a^2-a+3\right)=0\Leftrightarrow a=\frac{1}{3}\)
\(\Leftrightarrow\sqrt{\frac{x-1}{2x}}=\frac{1}{3}\Leftrightarrow9\left(x-1\right)=2x\)
c/ ĐKXĐ: ...
\(\Leftrightarrow\sqrt{\frac{x}{3-2x}}+5\sqrt{\frac{3-2x}{x}}=\frac{4\left(3-2x\right)}{x}+5\)
Đặt \(\sqrt{\frac{3-2x}{x}}=a>0\)
\(\frac{1}{a}+5a=4a^2+5\Leftrightarrow4a^3-5a^2+5a-1=0\)
\(\Leftrightarrow\left(4a-1\right)\left(a^2-a+1\right)=0\Leftrightarrow a=\frac{1}{4}\)
\(\Leftrightarrow\sqrt{\frac{3-2x}{x}}=\frac{1}{4}\Leftrightarrow16\left(3-2x\right)=x\)
d/ ĐKXĐ: ...
Đặt \(\sqrt{\frac{x-1}{x}}=a>0\)
\(a^2-2a=3\Leftrightarrow a^2-2a-3=0\Rightarrow\left[{}\begin{matrix}a=-1\left(l\right)\\a=3\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{\frac{x-1}{x}}=3\Leftrightarrow x-1=9x\)
a, ĐK: \(6x^2-12x+7\ge0\) (*)
\(PT\Leftrightarrow\left\{{}\begin{matrix}x^2-2x\ge0\\6x^2-12x+7=x^4-4x^3+4x^2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x^2-2x\ge0\\x^4-4x^3-2x^2+12x-7=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-2x\ge0\\\left(x-1\right)^2\left(x^2-2x-7\right)=0\end{matrix}\right.\) \(\Rightarrow x=1\pm2\sqrt{2}\) (thỏa mãn ĐK)
Vậy...
a/ ĐKXĐ: \(\left\{{}\begin{matrix}x\ge2\\x\ne\left\{3;11\right\}\end{matrix}\right.\)
Đặt \(\sqrt{x-2}=t\ge0\)
\(\Rightarrow\frac{3}{t-1}\ge\frac{5}{t-3}\)
\(\Leftrightarrow\frac{3}{t-1}-\frac{5}{t-3}\ge0\)
\(\Leftrightarrow\frac{3t-9-5t+5}{\left(t-1\right)\left(t-3\right)}\ge0\)
\(\Leftrightarrow\frac{-2t-4}{\left(t-1\right)\left(t-3\right)}\ge0\)
\(\Leftrightarrow\frac{t+2}{\left(t-1\right)\left(t-3\right)}\le0\)
\(\Leftrightarrow1< t< 3\)
\(\Rightarrow1< \sqrt{x-2}< 3\)
\(\Leftrightarrow1< x-2< 9\Rightarrow3< x< 11\)
b/
ĐKXĐ: \(x\ge3\)
- Với \(x=3\) BPT thỏa mãn
- Với \(x>3\Rightarrow\sqrt{x-3}>0\) BPT tương đương
\(x-\frac{1}{2-x}\le0\Leftrightarrow x+\frac{1}{x-2}\le0\)
\(\Leftrightarrow\frac{x^2-2x+1}{x-2}\le0\)
\(\Leftrightarrow\frac{\left(x-1\right)^2}{x-2}\le0\Rightarrow\) không tồn tại x thỏa mãn
Vậy BPT có nghiệm duy nhất \(x=3\)
\( 1)\sqrt[3]{{12 - x}} + \sqrt[3]{{14 + x}} = 2\\ \Leftrightarrow 12 - x + 3\sqrt[3]{{{{\left( {12 - x} \right)}^2}.\left( {14 + x} \right)}} + 3\sqrt[3]{{\left( {12 - x} \right){{\left( {14 + x} \right)}^2}}} + 14 + x = 8\\ \Leftrightarrow 3\sqrt[3]{{\left( {12 - x} \right)\left( {14 + x} \right)}}\left( {\sqrt[3]{{12 - x}} + \sqrt[3]{{14 + x}}} \right) = - 18\\ \Leftrightarrow 3\sqrt[3]{{\left( {12 - x} \right)\left( {14 + x} \right)}}.2 = - 18\\ \Leftrightarrow \sqrt[3]{{\left( {12 - x} \right)\left( {14 + x} \right)}} = - 3\\ \Leftrightarrow \left( {12 - x} \right)\left( {14 + x} \right) = {\left( { - 3} \right)^3}\\ \Leftrightarrow 168 - 2x - {x^2} = - 27\\ \Leftrightarrow {x^2} + 2x - 195 = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = - 15\\ x = 13 \end{array} \right. \)
Vậy...
1.
Đặt\(\left\{{}\begin{matrix}u=\sqrt[3]{12-x}\\v=\sqrt[3]{14+x}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u^3=12-x\\v^3=14+x\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}u^3+v^3=26\\u+v=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left(u+v\right)\left(u^2-uv+v^2\right)=26\\u+v=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u^2-uv+v^2=13\\v=2-u\end{matrix}\right.\)
\(\Rightarrow u^2-u\left(2-u\right)+\left(2-u\right)^2=13\) \(\Leftrightarrow3u^2-6u-9=0\) \(\Rightarrow\left[{}\begin{matrix}u=3\Rightarrow v=-1\\u=-1\Rightarrow v=3\end{matrix}\right.\) Tìm x.
2.ĐK: \(-40\le x\le57\)
Đặt \(\left\{{}\begin{matrix}\sqrt[4]{57-x}=u\\\sqrt[4]{x+40}=v\end{matrix}\right.\) \(\left(u,v\ge0\right)\) \(\Rightarrow\left\{{}\begin{matrix}u^4=57-x\\v^4=x+40\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u+v=5\\u^4+v^4=97\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}u^2+v^2=25-2uv\\\left(u^2+v^2\right)^2-2u^2v^2=97\end{matrix}\right.\) \(\Rightarrow\left(25-2uv\right)^2-2u^2v^2=97\)
\(\Leftrightarrow2u^2v^2-100uv+528=0\) \(\Rightarrow\left[{}\begin{matrix}uv=44\\uv=6\end{matrix}\right.\) Kết hợp \(u+v=5\) giải 2 trường hợp.
3.
ĐK: \(-\sqrt{17}\le x\le\sqrt{17}\)
Đặt \(x+\sqrt{17-x^2}=t\) \(\Rightarrow\frac{t^2-17}{2}=x\sqrt{17-x^2}\)
\(PT\Leftrightarrow t+\frac{t^2-17}{2}=9\) \(\Leftrightarrow t^2+2t-35=0\Rightarrow\left[{}\begin{matrix}t=5\\t=-7\end{matrix}\right.\) Giải tiếp.