\(\sqrt{2x-5}+2\sqrt{7-x}=\sqrt{3}x^2-8\sqrt{3x}+19\sqrt{3}\)
Những câu hỏi liên quan
\(\left(5\right)\sqrt{x+3-4\sqrt{x-1}}\sqrt{x+8+6\sqrt{x-1}}=5\)
\(\left(6\right)2x^2+3x+\sqrt{2x^2+3x+9}=33\)
\(\left(7\right)\sqrt{3x^2+6x+12}+\sqrt{5x^4-10x^2+30}=8\)
\(\left(8\right)x+y+z+8=2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}\)
6: \(\Leftrightarrow2x^2+3x+9+\sqrt{2x^2+3x+9}-42=0\)
Đặt \(\sqrt{2x^2+3x+9}=a\left(a>=0\right)\)
Phương trình sẽ trở thành là: a^2+a-42=0
=>(a+7)(a-6)=0
=>a=-7(loại) hoặc a=6(nhận)
=>2x^2+3x+9=36
=>2x^2+3x-27=0
=>2x^2+9x-6x-27=0
=>(2x+9)(x-3)=0
=>x=3 hoặc x=-9/2
8: \(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)
=>\(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)
=>\(\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=4\\z-3=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\\z=12\end{matrix}\right.\)
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Giải phương trình:
1) \(3x^2+4x-3=4x\sqrt{4x-3}\)
2) \(\sqrt{x^2+2x}+\sqrt{x+2}=\sqrt{x}+\sqrt{x^2+2x-2}\)
3) \(\sqrt{3x+8}-\sqrt{3x+5}=\sqrt{5x-4}-\sqrt{5x-7}\)
4) \(\sqrt{7-x^2+x\sqrt{x+5}}=\sqrt{3-2x-x^2}\)
1/ \(3x^2+4x-3=4x\sqrt{4x-3}\)
\(\Leftrightarrow\left(4x^2-4x\sqrt{4x-3}+4x-3\right)-x^2=0\)
\(\Leftrightarrow\left(2x-\sqrt{4x-3}\right)^2-x^2=0\)
\(\Leftrightarrow\left(3x-\sqrt{4x-3}\right)\left(x-\sqrt{4x-3}\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}3x=\sqrt{4x-3}\\x=\sqrt{4x-3}\end{matrix}\right.\)
\(\Leftrightarrow\left[\begin{matrix}9x^2-4x+3=0\\x^2-4x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[\begin{matrix}x=1\\x=3\end{matrix}\right.\)
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3.\(pt\Leftrightarrow\sqrt{3x+8}-\sqrt{3x+5}=\sqrt{5x-4}-\sqrt{5x-7}\)
\(\Leftrightarrow\frac{3x+8-5x+4}{\sqrt{3x+8}+\sqrt{5x+4}}-\frac{3x+5-5x+7}{\sqrt{3x+5}+\sqrt{5x+7}}=0\)
\(\Leftrightarrow\left(12-2x\right)\left(\frac{1}{\sqrt{3x+8}+\sqrt{5x+4}}+\frac{1}{\sqrt{3x+5}+\sqrt{5x+7}}\right)=0\)
\(\Rightarrow x=6\)
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Giải phương trình:
1, \(4\sqrt{x+3}+\sqrt{19-3x}=x^2+2x+9\)
2, \(\sqrt{3x-8}-\sqrt{x+1}=\dfrac{2x-11}{5}\)
3, \(\sqrt{x+\dfrac{3}{x}}=\dfrac{x^2+7}{2\left(x+1\right)}\)
a) \(\sqrt{3x^2-5x+7}\)+\(\sqrt{3x^2+x+1}\) = 12x-12
b) \(\sqrt{x^2+33}\)+3 = 2x+\(\sqrt{x^2-12}\)
c) 3x-\(8\sqrt{x+14}\) = \(2\sqrt{2x-3}\) - 28
d) \(x^2\)+\(\sqrt{x+7}\) = 7
Giải các phương trình sau:1) sqrt{3x^2+5x+8}-sqrt{3x^2+5x+1}12) x^2-2x-12+4sqrt{left(4-xright)left(2+xright)}03) 3sqrt{x}+dfrac{3}{2sqrt{x}}2x+dfrac{1}{2x}-74) sqrt{x}-dfrac{4}{sqrt{x+2}}+sqrt{x+2}05)left(x-7right)sqrt{dfrac{x+3}{x-7}}x+46) 2sqrt{x-4}+sqrt{x-1}sqrt{2x-3}+sqrt{4x-16}7) sqrt{x+2sqrt{x-1}}+sqrt{x-2sqrt{x-1}}dfrac{x+3}{2}Giúp mình với ajk, mink đang cần gấp
Đọc tiếp
Giải các phương trình sau:
1) \(\sqrt{3x^2+5x+8}-\sqrt{3x^2+5x+1}=1\)
2) \(x^2-2x-12+4\sqrt{\left(4-x\right)\left(2+x\right)}=0\)
3) \(3\sqrt{x}+\dfrac{3}{2\sqrt{x}}=2x+\dfrac{1}{2x}-7\)
4) \(\sqrt{x}-\dfrac{4}{\sqrt{x+2}}+\sqrt{x+2}=0\)
5)\(\left(x-7\right)\sqrt{\dfrac{x+3}{x-7}}=x+4\)
6) \(2\sqrt{x-4}+\sqrt{x-1}=\sqrt{2x-3}+\sqrt{4x-16}\)
7) \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\dfrac{x+3}{2}\)
Giúp mình với ajk, mink đang cần gấp
\(x^2+5x+6=3x+3\cdot4+2x-9\)
\(2\sqrt{x}+8x+5=5x-4+3x+19\)
\(5\sqrt{x}+2x-8=5x+4-3x-19\)
\(2x^2+5z+8+\sqrt{x}=x^2+3x+35+x^2+2x-7\)
\(3\sqrt{x}+7x+5=\sqrt{x}+4x-6+3x+18\)
\(2\sqrt{3x}+11x-18=5x+2+6\cdot\sqrt{3x}+6x-21\)
Giai phuong trinh
1/ sqrt{x^2+4x+5}+sqrt{x^2-6x+13}3
2/ sqrt{3x^2-18x+28}+sqrt{4x^2-24x+45}6x-x^2-5
3/ sqrt{2x^2-4x+27}+sqrt{3x^2-6x+12}4x^2+8x+4
4/ sqrt{x^2+x+7}+sqrt{x^2+x+2}sqrt{3x^2+3x+19}
5/ left(x+2right)left(x+3right)-sqrt{x^2+5x+1}9
6/ left(x+4right)left(x+1right)-3sqrt{x^2+5x+2}6
7/ sqrt{2x^2+3x+5}+sqrt{2x^2-3x+5}3sqrt{x}
Đọc tiếp
Giai phuong trinh
1/ \(\sqrt{x^2+4x+5}+\sqrt{x^2-6x+13}=3\)
2/ \(\sqrt{3x^2-18x+28}+\sqrt{4x^2-24x+45}=6x-x^2-5\)
3/ \(\sqrt{2x^2-4x+27}+\sqrt{3x^2-6x+12}=4x^2+8x+4\)
4/ \(\sqrt{x^2+x+7}+\sqrt{x^2+x+2}=\sqrt{3x^2+3x+19}\)
5/ \(\left(x+2\right)\left(x+3\right)-\sqrt{x^2+5x+1}=9\)
6/ \(\left(x+4\right)\left(x+1\right)-3\sqrt{x^2+5x+2}=6\)
7/ \(\sqrt{2x^2+3x+5}+\sqrt{2x^2-3x+5}=3\sqrt{x}\)
Em xin phép làm bài EZ nhất :)
4,ĐK :\(\forall x\in R\)
Đặt \(x^2+x+2=t\) (\(t\ge\dfrac{7}{4}\))
\(PT\Leftrightarrow\sqrt{t+5}+\sqrt{t}=\sqrt{3t+13}\)
\(\Leftrightarrow2t+5+2\sqrt{t\left(t+5\right)}=3t+13\)
\(\Leftrightarrow t+8=2\sqrt{t^2+5t}\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge-8\\\left(t+8\right)^2=4t^2+20t\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\3t^2+4t-64=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left(t-4\right)\left(3t+16\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left[{}\begin{matrix}t=4\left(tm\right)\\t=-\dfrac{16}{3}\left(l\right)\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow x^2+x+2=4\)\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy ....
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Giải Phương Trình
\(\sqrt{2x-5}+2\sqrt{7-x}=\sqrt{3}x^2-8\sqrt{3}x+19\sqrt{3}\)
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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