\(\sqrt{12-\frac{12}{x^2} }+\sqrt{x^2+\frac{12}{x} }=x^2+\frac{25}{2} \)
\(\frac{\sqrt[3]{10-x}+\sqrt[3]{8-x}}{\sqrt[3]{10-x}-\sqrt[3]{8-x}}=9-x \)
\(4x^2+\sqrt{2x+1}+5=12x\)
Những câu hỏi liên quan
bài 1) rút gọn
1) 5√frac{1}{5} 2)frac{12}{5}√frac{5}{4} 3)frac{30}{5sqrt{6}} 4) frac{20}{2sqrt{5}} 5)frac{2-sqrt{2}}{sqrt{2}} 6) frac{11+sqrt{11}}{1+sqrt{ }11} 7) frac{sqrt{21-sqrt{7}}}{1-sqrt{3}} 8)frac{sqrt{2+sqrt{3}}}{2+sqrt{6}} 9)frac{sqrt{10-sqrt{2}}}{sqrt{5-}1} 10)frac{2sqrt{3}-3sqrt{2}}{sqrt{3}-sqrt[]{2}}
bài 2) với các biểu thức đã cho là có nghĩa và rút gọn
1)frac{x-sqrt{x}}{sqrt{x}-1} 2)frac{xsqrt{x}-2x}{2-sqrt{x}} 3) frac{xsqrt{y}-...
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bài 1) rút gọn
1) 5√\(\frac{1}{5}\) 2)\(\frac{12}{5}\)√\(\frac{5}{4}\) 3)\(\frac{30}{5\sqrt{6}}\) 4) \(\frac{20}{2\sqrt{5}}\) 5)\(\frac{2-\sqrt{2}}{\sqrt{2}}\) 6) \(\frac{11+\sqrt{11}}{1+\sqrt{ }11}\) 7) \(\frac{\sqrt{21-\sqrt{7}}}{1-\sqrt{3}}\) 8)\(\frac{\sqrt{2+\sqrt{3}}}{2+\sqrt{6}}\) 9)\(\frac{\sqrt{10-\sqrt{2}}}{\sqrt{5-}1}\) 10)\(\frac{2\sqrt{3}-3\sqrt{2}}{\sqrt{3}-\sqrt[]{2}}\)
bài 2) với các biểu thức đã cho là có nghĩa và rút gọn
1)\(\frac{x-\sqrt{x}}{\sqrt{x}-1}\) 2)\(\frac{x\sqrt{x}-2x}{2-\sqrt{x}}\) 3) \(\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\) 4) \(\frac{a\sqrt{b}-\sqrt{a}}{\sqrt{b}-b\sqrt{a}}\) 5) \(\frac{a-1}{\sqrt{a}+1}\) 6) \(\frac{4-x}{2\sqrt{x}-x}\) 7)\(\frac{a+1+2\sqrt{a}}{1+\sqrt{a}}\) 8)\(\frac{3\sqrt{x}-x}{3+2\sqrt{3x}-x}\) 9)\(\frac{y+12-4\sqrt{3y}}{y-12}\) 10)\(\frac{4\sqrt{x}-x-4}{x-4}\) 11)\(\frac{x+y-2\sqrt{xy}}{x\sqrt{y}-y\sqrt{x}}\)
1 tháng 10 2019 lúc 22:41
giải pt
a) \(x\sqrt{x^2-4x+3}=x^2+x\)
b) \(x^2+x-12-\left(x-3\right)\sqrt{10-x^2}=0\)
c) \(\sqrt{6+x-x^2}=\frac{\left(2x+5\right)\sqrt{6+x-x^2}}{x+4}\)
d) \(\sqrt{\frac{12+x-x^2}{2x+9}}-\frac{\sqrt{12+x-x^2}}{x+3}=0\)
e) \(\sqrt{x^3}+\sqrt{x^3+x^2+2x}=3\sqrt{x}\)
a, ĐK:\(x^2-4x+3\ge0\Rightarrow\left[{}\begin{matrix}x\le1\\3\le x\end{matrix}\right.\)
\(PT\Leftrightarrow x\sqrt{x^2-4x+3}=x\left(x+1\right)\)
Với x = 0 \(\Rightarrow pttm\)
Với \(x\ne0\) \(\Rightarrow\sqrt{x^2-4x+3}=x+1\)
\(\Rightarrow\left\{{}\begin{matrix}x\ge-1\\x^2-4x+3=x^2+2x+1\end{matrix}\right.\)\(\Rightarrow x=\frac{1}{3}\left(tm\right)\)
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b,ĐK: \(-\sqrt{10}\le x\le\sqrt{10}\)
\(PT\Leftrightarrow\left(x-3\right)\left(x+4\right)-\left(x-3\right)\sqrt{10-x^2}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x+4-\sqrt{10-x^2}=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=3\\x+4=\sqrt{10-x^2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x^2+8x+16=10-x^2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x^2+4x+3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\end{matrix}\right.\)(tm)
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c/ ĐKXĐ: \(-2\le x\le3\)
\(\Leftrightarrow\left(x+4\right)\sqrt{6+x-x^2}-\left(2x+5\right)\sqrt{6+x-x^2}=0\)
\(\Leftrightarrow\sqrt{6+x-x^2}\left(x+4-2x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x^2+x+6=0\\-x-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-2\\x=-1\\x=3\end{matrix}\right.\)
d/ ĐKXĐ: \(3< x\le4\)
\(\Leftrightarrow\sqrt{-x^2+x+12}\left(\frac{1}{\sqrt{2x+9}}-\frac{1}{x+3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x^2+x+12=0\\\sqrt{2x+9}=x+3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-x^2+x+12=0\\2x+9=x^2+6x+9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-x^2+x+12=0\\x^2+4x=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-3\left(l\right)\\x=4\\x=0\\x=-4\left(l\right)\end{matrix}\right.\)
e/ ĐKXĐ: \(x\ge0\)
\(\Leftrightarrow\sqrt{x}\left(x+\sqrt{x^2+x+2}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\Rightarrow x=0\\\sqrt{x^2+x+2}=3-x\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}3-x\ge0\\x^2+x+2=\left(3-x\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le3\\7x=7\end{matrix}\right.\) \(\Rightarrow x=1\)
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Tìm điều kiện có nghĩa:1) sqrt{2x^2}2) sqrt{-x}3) sqrt{-x^2-3}4) sqrt{x^2+2x+3}5) sqrt{-a^2+8a-16}6) sqrt[]{16x^2-25}7) sqrt{4x^2-49}8) sqrt{8-x^2}9) sqrt{x^2-12}10) sqrt{x^2+2x-3}11) sqrt{2x^2+5x+3}12) sqrt{dfrac{4}{x-1}}13) sqrt{dfrac{-1}{x-3}}14) sqrt{dfrac{-3}{x+2}}15) sqrt{dfrac{1}{2a-1}}16) sqrt{dfrac{2}{3-2a}}17) sqrt{dfrac{-1}{2a-5}}18) sqrt{dfrac{-2}{3-5a}}19) sqrt{dfrac{-a}{5}}20) dfrac{1}{sqrt{-3a}}
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Tìm điều kiện có nghĩa:
1) \(\sqrt{2x^2}\)
2) \(\sqrt{-x}\)
3) \(\sqrt{-x^2-3}\)
4) \(\sqrt{x^2+2x+3}\)
5) \(\sqrt{-a^2+8a-16}\)
6) \(\sqrt[]{16x^2-25}\)
7) \(\sqrt{4x^2-49}\)
8) \(\sqrt{8-x^2}\)
9) \(\sqrt{x^2-12}\)
10) \(\sqrt{x^2+2x-3}\)
11) \(\sqrt{2x^2+5x+3}\)
12) \(\sqrt{\dfrac{4}{x-1}}\)
13) \(\sqrt{\dfrac{-1}{x-3}}\)
14) \(\sqrt{\dfrac{-3}{x+2}}\)
15) \(\sqrt{\dfrac{1}{2a-1}}\)
16) \(\sqrt{\dfrac{2}{3-2a}}\)
17) \(\sqrt{\dfrac{-1}{2a-5}}\)
18) \(\sqrt{\dfrac{-2}{3-5a}}\)
19) \(\sqrt{\dfrac{-a}{5}}\)
20) \(\dfrac{1}{\sqrt{-3a}}\)
1) \(ĐK:x\in R\)
2) \(ĐK:x< 0\)
3) \(ĐK:x\in\varnothing\)
4) \(=\sqrt{\left(x+1\right)^2+2}\)
\(ĐK:x\in R\)
5) \(=\sqrt{-\left(a-4\right)^2}\)
\(ĐK:x\in\varnothing\)
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7 tháng 11 2019 lúc 15:14
giải pt
a) \(x+\sqrt{x+8}\left(1-\sqrt{x+8}\right)=\sqrt{x}+\sqrt{x+3}-8\)
b) \(2\left(2-x\right)=\sqrt{2x-4}\left(\sqrt{5-x}-\sqrt{3x-3}\right)\)
c) \(\sqrt[3]{24+x}.\sqrt{12-x}-6\sqrt{12-x}=x-12\)
d) \(\frac{x-1}{2\sqrt{3-2x}-3}=\frac{x-1}{3-2\sqrt[3]{5+3x}}\)
a/ ĐKXĐ: ...
\(\Leftrightarrow x+8+\sqrt{x+8}-\left(x+8\right)=\sqrt{x}+\sqrt{x+3}\)
\(\Leftrightarrow\sqrt{x+8}=\sqrt{x}+\sqrt{x+3}\)
\(\Leftrightarrow x+8=2x+3+2\sqrt{x^2+3x}\)
\(\Leftrightarrow5-x=2\sqrt{x^2+3x}\) (\(x\le5\))
\(\Leftrightarrow x^2-10x+25=4\left(x^2+3x\right)\)
\(\Leftrightarrow...\)
b/ ĐKXĐ: \(2\le x\le5\)
\(\Leftrightarrow2\left(x-2\right)+\sqrt{2\left(x-2\right)}\left(\sqrt{5-x}-\sqrt{3x-3}\right)=0\)
\(\Leftrightarrow\sqrt{2\left(x-2\right)}\left(\sqrt{2x-4}+\sqrt{5-x}-\sqrt{3x-3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\\sqrt{2x-4}+\sqrt{5-x}=\sqrt{3x-3}\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x+1+2\sqrt{\left(2x-4\right)\left(5-x\right)}=3x-3\)
\(\Leftrightarrow\sqrt{\left(2x-4\right)\left(5-x\right)}=x-2\)
\(\Leftrightarrow\left(2x-4\right)\left(5-x\right)=\left(x-2\right)^2\)
\(\Leftrightarrow...\)
c/ ĐKXĐ: \(x\le12\)
\(\Leftrightarrow\sqrt[3]{24+x}\sqrt{12-x}-6\sqrt{12-x}+12-x=0\)
\(\Leftrightarrow\sqrt{12-x}\left(\sqrt[3]{24+x}-6+\sqrt{12-x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=12\\\sqrt[3]{24+x}+\sqrt{12-x}=6\left(1\right)\end{matrix}\right.\)
Xét (1):
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{24+x}=a\\\sqrt{12-x}=b\ge0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=6\\a^3+b^2=36\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=6-a\\a^3+b^2=36\end{matrix}\right.\)
\(\Leftrightarrow a^3+\left(6-a\right)^2=36\)
\(\Leftrightarrow a^3+a^2-12a=0\)
\(\Leftrightarrow a\left(a^2+a-12\right)=0\Rightarrow\left[{}\begin{matrix}a=0\\a=3\\a=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt[3]{24+x}=0\\\sqrt[3]{24+x}=3\\\sqrt[3]{24+x}=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}24+x=0\\24+x=27\\24+x=-64\end{matrix}\right.\)
d/ ĐKXĐ: \(x\le\frac{3}{2}\) ; \(x\ne\frac{3}{8};x\ne-\frac{13}{24}\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{2\sqrt{3-2x}-3}-\frac{1}{3-2\sqrt[3]{5+3x}}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\frac{1}{2\sqrt{3-2x}-3}=\frac{1}{3-2\sqrt[3]{5+3x}}\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2\sqrt{3-2x}-3=3-2\sqrt[3]{5+3x}\)
\(\Leftrightarrow\sqrt[3]{5+3x}+\sqrt{3-2x}=3\)
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{5+3x}=a\\\sqrt{3-2x}=b\ge0\end{matrix}\right.\) ta được:
\(\left\{{}\begin{matrix}a+b=3\\2a^3+3b^2=19\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=3-a\\2a^3+3b^2=19\end{matrix}\right.\)
\(\Leftrightarrow2a^3+3\left(3-a\right)^2=19\)
\(\Leftrightarrow2a^3+3a^2-18a+8=0\)
\(\Rightarrow\left[{}\begin{matrix}a=-4\\a=\frac{1}{2}\\a=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt[3]{5+3x}=-4\\\sqrt[3]{5+3x}=\frac{1}{2}\\\sqrt[3]{5+3x}=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}5+3x=-64\\5+3x=\frac{1}{8}\\5+3x=8\end{matrix}\right.\)
\(\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{4x}{x-4}-\frac{2-\sqrt{x}}{2+\sqrt{x}}\right):\frac{4\sqrt{x^3-12\sqrt{x}}}{2x-\sqrt{x^3}}\)
\(\frac{6x-\left(x+6\right)\sqrt{x}-3}{2\left(x-4\sqrt{x}+3\right)\left(2-\sqrt{x}\right)}-\frac{3}{-2x+10\sqrt{x}-12}-\frac{1}{3\sqrt{x}-x-2}\)
1) \(\left(\sqrt{12}-6\sqrt{3}+\sqrt{24}\right)\sqrt{6}-\left(\frac{5}{2}\sqrt{2}+12\right)\)
2) \(\frac{26}{2\sqrt{3}+5}-\frac{4}{\sqrt{3}-2}\)
3) \(\sqrt{x^2-6x+9}=2x\)
4) \(\sqrt{4x^2+1}=2x-1\)
5) \(\sqrt{x^2-4x+4}=\sqrt{x^2-2x+1}\)
1) Ta có: \(\left(\sqrt{12}-6\sqrt{3}+\sqrt{24}\right)\cdot\sqrt{6}-\left(\frac{5}{2}\sqrt{2}+12\right)\)
\(=\left(2\sqrt{3}-6\sqrt{3}+2\sqrt{6}\right)\cdot\sqrt{6}-\left(\sqrt{\frac{25}{4}\cdot2}+12\right)\)
\(=\left(-4\sqrt{3}+2\sqrt{6}\right)\cdot\sqrt{6}-\left(\sqrt{\frac{50}{4}}+12\right)\)
\(=-12\sqrt{2}+12-\frac{5\sqrt{2}}{2}-12\)
\(=\frac{-24\sqrt{2}-5\sqrt{2}}{2}\)
\(=\frac{-29\sqrt{2}}{2}\)
2) Ta có: \(\frac{26}{2\sqrt{3}+5}-\frac{4}{\sqrt{3}-2}\)
\(=\frac{26\left(5-2\sqrt{3}\right)}{\left(5+2\sqrt{3}\right)\left(5-2\sqrt{3}\right)}+\frac{4}{2-\sqrt{3}}\)
\(=\frac{26\left(5-2\sqrt{3}\right)}{25-12}+\frac{4\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=2\left(5-2\sqrt{3}\right)+4\left(2+\sqrt{3}\right)\)
\(=10-4\sqrt{3}+8+4\sqrt{3}\)
\(=18\)
3) ĐK để phương trình có nghiệm là: x≥0
Ta có: \(\sqrt{x^2-6x+9}=2x\)
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=2x\)
\(\Leftrightarrow\left|x-3\right|=2x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x\\x-3=-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-3-2x=0\\x-3+2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x-3=0\\3x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=3\\3x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)
Vậy: S={1}
4) ĐK để phương trình có nghiệm là: \(x\ge\frac{1}{2}\)
Ta có: \(\sqrt{4x^2+1}=2x-1\)
\(\Leftrightarrow\left(\sqrt{4x^2+1}\right)^2=\left(2x-1\right)^2\)
\(\Leftrightarrow4x^2+1=4x^2-4x+1\)
\(\Leftrightarrow4x^2+1-4x^2+4x-1=0\)
\(\Leftrightarrow4x=0\)
hay x=0(loại)
Vậy: S=∅
sqrt{7-frac{1}{5}x} 2) sqrt{frac{4}{3}+2x}
sqrt{frac{2}{x^2+frac{1}{2}}} 4) sqrt{frac{-2}{x^2+frac{1}{3}}}
sqrt{frac{-5}{x^2+4x}} 6) sqrt{frac{5}{x^2+4x}}
sqrt{frac{2}{x^2+3x-4}} 8)sqrt{frac{2}{3x^2+5x-8}}
sqrt{frac{-6}{x^2-25}} 10)sqrt{frac{-7}{4x^2-9}}
sqrt{frac{2}{4x^2+x+3}} Gỉai phương trình
Giúp mk vs các bn mk bí rồi.
Đọc tiếp
\(\sqrt{7-\frac{1}{5}x}\) 2) \(\sqrt{\frac{4}{3}+2x}\)
\(\sqrt{\frac{2}{x^2+\frac{1}{2}}}\) 4) \(\sqrt{\frac{-2}{x^2+\frac{1}{3}}}\)
\(\sqrt{\frac{-5}{x^2+4x}}\) 6) \(\sqrt{\frac{5}{x^2+4x}}\)
\(\sqrt{\frac{2}{x^2+3x-4}}\) 8)\(\sqrt{\frac{2}{3x^2+5x-8}}\)
\(\sqrt{\frac{-6}{x^2-25}}\) 10)\(\sqrt{\frac{-7}{4x^2-9}}\)
\(\sqrt{\frac{2}{4x^2+x+3}}\) Gỉai phương trình
Giúp mk vs các bn mk bí rồi.
1. x^3-x^2+12xsqrt{x-1}+200
2. x^3+sqrt{left(x-1right)^3}9x+8
3. sqrt{2x^2+x+1}+sqrt{x^2-x+1}3x
4. x^6+left(x^3-3right)^33x^5-9x^2-1
5. x^2-6left(x+3right)sqrt{x+1}+14x+3sqrt{x+1}+130
6. x^2-4x+left(x-3right)sqrt{x^2-x+1}-1
7. sqrt{2x-1}+sqrt{5-x}x-2+2sqrt{-2x^2+11x-5}
8. sqrt{5x+11}-sqrt{6-x}+5x^2-14x-600
9. x^2+6x+83sqrt{x+2}
10. 2x^2+3x-2left(2x-1right)sqrt{2x^2+x-3}
11. sqrt{x+1}+sqrt{4-x}-sqrt{left(x+1right)left(4-xright)}1
12. x^2-sqrt{x^2-4x}4left(...
Đọc tiếp
1. \(x^3-x^2+12x\sqrt{x-1}+20=0\)
2. \(x^3+\sqrt{\left(x-1\right)^3}=9x+8\)
3. \(\sqrt{2x^2+x+1}+\sqrt{x^2-x+1}=3x\)
4. \(x^6+\left(x^3-3\right)^3=3x^5-9x^2-1\)
5. \(x^2-6\left(x+3\right)\sqrt{x+1}+14x+3\sqrt{x+1}+13=0\)
6. \(x^2-4x+\left(x-3\right)\sqrt{x^2-x+1}=-1\)
7. \(\sqrt{2x-1}+\sqrt{5-x}=x-2+2\sqrt{-2x^2+11x-5}\)
8. \(\sqrt{5x+11}-\sqrt{6-x}+5x^2-14x-60=0\)
9. \(x^2+6x+8=3\sqrt{x+2}\)
10. \(2x^2+3x-2=\left(2x-1\right)\sqrt{2x^2+x-3}\)
11. \(\sqrt{x+1}+\sqrt{4-x}-\sqrt{\left(x+1\right)\left(4-x\right)}=1\)
12. \(x^2-\sqrt{x^2-4x}=4\left(x+3\right)\)
13. \(x^2-x-4=2\sqrt{x-1}\left(1-x\right)\)
14. \(\frac{1}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}=1\)
15. \(\sqrt{2x^2+3x+2}+\sqrt{4x^2+6x+21}=11\)
16. \(\sqrt{x+3+3\sqrt{2x-3}}+\sqrt{x-1+\sqrt{2x-1}}=2\sqrt{2}\)
17. \(\left(x-2\right)^2\left(x-1\right)\left(x-3\right)=12\)
18. \(2x^2+\sqrt{x^2-2x-19}=4x+74\)
19. \(x^4+x^2-20=0\)
20. \(x+\sqrt{4-x^2}=2+3x\sqrt{4-x^2}\)
21. \(\left(x^2+x+1\right)\left(\sqrt[3]{\left(3x-2\right)^2}+\sqrt[3]{3x-2}+1\right)=9\)
22. \(\sqrt{x^2-3x+5}+x^2=3x+7\)
23. \(x^2+6x+5=\sqrt{x+7}\)
24. \(\frac{2x^2-3x+10}{x+2}=3\sqrt{\frac{x^2-2x+4}{x+2}}\)
25. \(5\sqrt{x-1}-\sqrt{x+7}=3x-4\)
26. \(2\left(x^2+2\right)=5\sqrt{x^3+1}\)
27. \(\sqrt{x-1}+\sqrt{5-x}-2=2\sqrt{\left(x-1\right)\left(5-x\right)}\)
28. \(x^2+\frac{9x^2}{\left(x-3\right)^2}=40\)
29. \(\frac{26x+5}{\sqrt{x^2+30}}+2\sqrt{26x+5}=3\sqrt{x^2+30}\)
30. \(\frac{\sqrt{27+x^2+x}}{2+\sqrt{5-\left(x^2+x\right)}}=\frac{\sqrt{27+2x}}{2+\sqrt{5-2x}}\)
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\(\Leftrightarrow\frac{\sqrt{x}-1+\sqrt{x}+1}{x-1}=1\Leftrightarrow2\sqrt{x}=x-1\)\(\Leftrightarrow x-2\sqrt{x}+1=2\Leftrightarrow\left(\sqrt{x}-1\right)^2=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3+2\sqrt{2}\\x=3-2\sqrt{2}\end{matrix}\right.\)
Bạn ơi lần sau bạn đăng bài thì cố gắng đăng giãn giãn bớt bớt/ chia nhỏ bài ra chứ một cục bài như thế này nhìn rất đáng sợ và gây tâm lý ngại đọc nhé.
1. ĐKXĐ: $x\geq 1$
Đặt $x\sqrt{x-1}=a\Rightarrow x^3-x^2=x^2(x-1)=a^2$. PT đã cho trở thành:
$a^2+12a+20=0(*)$
Lại thấy rằng vì $x\geq 1$ nên $a\geq 0$
$\Rightarrow a^2+12a+20\geq 20>0$. Do đó $(*)$ vô nghiệm. Kéo theo PT ban đầu vô nghiệm.
2. ĐK: $x\geq -1$
$x^3+\sqrt{(x+1)^3}=9x+8$
$\Leftrightarrow x^3-9x-8+\sqrt{(x+1)^3}=0$
$\Leftrightarrow (x^2-x-8)(x+1)+(x+1)\sqrt{x+1}=0$
$\Leftrightarrow (x+1)(x^2-x-8+\sqrt{x+1})=0$
Nếu $x+1=0\Rightarrow x=-1$ (thỏa mãn)
Nếu $x^2-x-8+\sqrt{x+1}=0$
$\Leftrightarrow (x^2-9)-(x-3)+(\sqrt{x+1}-2)=0$
$\Leftrightarrow (x-3)\left(x+3+\frac{1}{\sqrt{x+1}+2}}-1\right)=0$
Dễ thấy với $x\geq -1$ thì biểu thức trong ngoặc lớn luôn lớn hơn $0$
Do đó $x-3=0\Rightarrow x=3$
Vậy $x=-1$ hoặc $x=3$
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