\(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-5}+2=0\)
\(2x-x^2+\sqrt{6x^2-12x+7}=0\)
\(\left(x+1\right)\left(x+4\right)-3\sqrt{x^2+5x+2}=6\)
\(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\)
giúp dùm đi mấy pạn
Những câu hỏi liên quan
Giải phương trình sau:
a) \(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
b) \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
c) \(2x-x^2+\sqrt{6x^2-12x+7}=0\)
d) \(\left(x+1\right)\left(x+4\right)-3\sqrt{x^2+5x+2}=6\)
a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow x+5=4\)
hay x=-1
b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\Leftrightarrow\sqrt{x-1}=17\)
\(\Leftrightarrow x-1=289\)
hay x=290
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Giải các phương trình sau
a\(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\)
b \(\sqrt{9x^2+18}+2\sqrt{x^2+2}-\sqrt{25x^2+50}+3=0\)
c\(2x-x^2+\sqrt{6x^2-12+7}=0\)
d\(\left(x+1\right).\left(x+4\right)-3\sqrt{x^2+5x+2}=6\)
a) \(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\)
<=> \(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9\left(x-1\right)}+24\frac{\sqrt{x-1}}{\sqrt{64}}=-17\)
<=>\(\frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
<=>\(\sqrt{x-1}\left(\frac{1}{2}-\frac{9}{2}+\frac{6}{2}\right)=-17\)
<=>\(\sqrt{x-1}=-17\)
<=>x-1=17
<=>x=18
Vậy pt có nghiệm là x=18
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\(a.ĐK:x-1\ge0\Leftrightarrow x\ge1\)
\(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\)
\(\Leftrightarrow\frac{1}{2}\sqrt{x-1}-\frac{27}{2}\sqrt{x-1}+24\sqrt{\frac{x-1}{64}}=-17\)
\(\Leftrightarrow\sqrt{x-1}\left(\frac{1}{2}-\frac{27}{2}+24\sqrt{\frac{1}{64}}\right)=-17\)
\(\Leftrightarrow\sqrt{x-1}.\left(-10\right)=-17\)
\(\Leftrightarrow\sqrt{x-1}=\frac{-17}{-10}=\frac{17}{10}\)
\(\Leftrightarrow x-1=\left(\frac{17}{10}\right)^2\)
\(\Leftrightarrow x=\frac{289}{100}+1=3,89\left(TM\right)\)
Vậy \(S=\left\{3,89\right\}\)
\(b.ĐK:x^2+2\ge0\)
\(\sqrt{9x^2+18}+2\sqrt{x^2+2}-\sqrt{25x^2+50}+3=0\)
\(\Leftrightarrow9\sqrt{x^2+2}+2\sqrt{x^2+2}-25\sqrt{x^2+2}=-3\)
\(\Leftrightarrow\sqrt{x^2+2}\left(9+2-25\right)=-3\)
\(\Leftrightarrow\sqrt{x^2+2}=\frac{-3}{-14}=\frac{3}{14}\)
\(\Leftrightarrow x^2+2=\left(\frac{3}{14}\right)^2\)
\(\Leftrightarrow x=\sqrt{\frac{9}{196}-2}=\sqrt{-\frac{383}{196}}\left(vl\right)\)
Vậy \(S=\varnothing\)
Mấy câu kia làm tương tự
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b)\(\sqrt{9x^2+18}+2\sqrt{x^2+2}-\sqrt{25x^2+50}+3=0\)
<=>\(\sqrt{9\left(x^2+2\right)}+2\sqrt{x^2+2}-\sqrt{25\left(x^2+2\right)}+3=0\)
<=>\(3\sqrt{x^2+2}+2\sqrt{x^2+2}-5\sqrt{x^2+2}+3=0\)
<=>\(\sqrt{x^2+2}\left(3+2-5\right)=-3\)
<=>0x=-3
Vậy pt vô nghiệm
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2. Giải PT:
a) \(\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}.\)
b) \(\sqrt{4x-20}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=4.\)
c) \(2x-x^2+\sqrt{6x^2-12x+7}=0.\)
d) \(\left(x+1\right)\left(x+4\right)-3\sqrt{x^2+5x+2}=6.\)
\(a,\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)\(ĐKXĐ:x\ge-\frac{5}{7}\)
\(\Leftrightarrow9x-7=7x+5\)
\(\Leftrightarrow9x-7x=5+7\)
\(\Leftrightarrow2x=12\)
\(\Leftrightarrow x=6\)
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\(b,\sqrt{4x-20}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow\sqrt{4\left(x-5\right)}+3.\frac{\sqrt{x-5}}{\sqrt{9}}-\frac{1}{3}\sqrt{9\left(x-5\right)}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}\left(2+1-1\right)=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\)
\(\Leftrightarrow x=9\)
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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(...
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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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28 tháng 11 2019 lúc 23:30
Giải các pt sau:
a) sqrt{x+8}+frac{9x}{sqrt{x+8}}-6sqrt{x}0
b) x^4-2x^3+sqrt{2x^3+x^2+2}-20
c) 3xsqrt[3]{x+7}left(x+sqrt[3]{x+7}right)7x^3+12x^2+5x-6
d) 4x^2+left(8x-4right)sqrt{x}-13x+2sqrt{2x^2+5x-3}
e) 16x^2+19x+7+4sqrt{-3x^2+5x+2}left(8x+2right)left(sqrt{2-x}+2sqrt{3x+1}right)
f) left(5x+8right)sqrt{2x-1}+7xsqrt{x+3}9x+8-left(x+26right)sqrt{x-1}
g) sqrt[3]{3x+1}+sqrt[3]{5-x}+sqrt[3]{2x-9}-sqrt[3]{4x-3}0
Đọc tiếp
Giải các pt sau:
a) \(\sqrt{x+8}+\frac{9x}{\sqrt{x+8}}-6\sqrt{x}=0\)
b) \(x^4-2x^3+\sqrt{2x^3+x^2+2}-2=0\)
c) \(3x\sqrt[3]{x+7}\left(x+\sqrt[3]{x+7}\right)=7x^3+12x^2+5x-6\)
d) \(4x^2+\left(8x-4\right)\sqrt{x}-1=3x+2\sqrt{2x^2+5x-3}\)
e) \(16x^2+19x+7+4\sqrt{-3x^2+5x+2}=\left(8x+2\right)\left(\sqrt{2-x}+2\sqrt{3x+1}\right)\)
f) \(\left(5x+8\right)\sqrt{2x-1}+7x\sqrt{x+3}=9x+8-\left(x+26\right)\sqrt{x-1}\)
g) \(\sqrt[3]{3x+1}+\sqrt[3]{5-x}+\sqrt[3]{2x-9}-\sqrt[3]{4x-3}=0\)
Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen
help me, pleaseee
Cần gấp lắm ạ!
1)7sqrt{3x-7}+left(4x-7right)sqrt{7-x}322)4x^2-11x+6left(x-1right)sqrt{2x^2-6x+6}3)9+3sqrt{xleft(3-2xright)}7sqrt{x}+5sqrt{3-2x}4)sqrt{2x^2+4x+7}x^4+4x^3+3x^2-2x-75)frac{6-2x}{sqrt{5-x}}+frac{6+2x}{sqrt{5+x}}frac{8}{3}6)2left(5x-3right)sqrt{x+1}+left(x+1right)sqrt{3-x}3left(5x+1right)7)sqrt{7x+7}+sqrt{7x-6}+2sqrt{49x^2+7x-42}181-14x
Đọc tiếp
1)\(7\sqrt{3x-7}+\left(4x-7\right)\sqrt{7-x}=32\)
2)\(4x^2-11x+6=\left(x-1\right)\sqrt{2x^2-6x+6}\)
3)\(9+3\sqrt{x\left(3-2x\right)}=7\sqrt{x}+5\sqrt{3-2x}\)
4)\(\sqrt{2x^2+4x+7}=x^4+4x^3+3x^2-2x-7\)
5)\(\frac{6-2x}{\sqrt{5-x}}+\frac{6+2x}{\sqrt{5+x}}=\frac{8}{3}\)
6)\(2\left(5x-3\right)\sqrt{x+1}+\left(x+1\right)\sqrt{3-x}=3\left(5x+1\right)\)
7)\(\sqrt{7x+7}+\sqrt{7x-6}+2\sqrt{49x^2+7x-42}=181-14x\)
a)Giải các phương trình sau bằng phương pháp đặt ẩn phụ:1) x^2-3x-3frac{3left(sqrt[3]{x^3-4x^2+4}-1right)}{1-x} ;2)1+frac{2}{3}sqrt{x-x^2}sqrt{x}+sqrt{1-x}b) Giải các phương trình sau(không giới hạn phương pháp):1)2left(1-xright)sqrt{x^2+2x-1}x^2-2x-1 ; 2)sqrt{2x+4}-2sqrt{2-x}frac{12x-8}{sqrt{9x^2+16}}3)frac{3x^2+3x-1}{3x+1}sqrt{x^2+2x-1} ; 4) frac{2x^3+3x^2+11x-8}{3x^2+4x+1}sqrt{frac{10x-8}{x+1}}5)13x-17+4sqrt{x+1}6sqrt{x-2}left(1+2sqrt{x+1}right);6)x^2+8x+2left(x+1right)sqrt{x+6}6sqrt{x+...
Đọc tiếp
a)Giải các phương trình sau bằng phương pháp đặt ẩn phụ:
1) \(x^2-3x-3=\frac{3\left(\sqrt[3]{x^3-4x^2+4}-1\right)}{1-x}\) ;2)\(1+\frac{2}{3}\sqrt{x-x^2}=\sqrt{x}+\sqrt{1-x}\)
b) Giải các phương trình sau(không giới hạn phương pháp):
1)\(2\left(1-x\right)\sqrt{x^2+2x-1}=x^2-2x-1\) ; 2)\(\sqrt{2x+4}-2\sqrt{2-x}=\frac{12x-8}{\sqrt{9x^2+16}}\)
3)\(\frac{3x^2+3x-1}{3x+1}=\sqrt{x^2+2x-1}\) ; 4) \(\frac{2x^3+3x^2+11x-8}{3x^2+4x+1}=\sqrt{\frac{10x-8}{x+1}}\)
5)\(13x-17+4\sqrt{x+1}=6\sqrt{x-2}\left(1+2\sqrt{x+1}\right)\);
6)\(x^2+8x+2\left(x+1\right)\sqrt{x+6}=6\sqrt{x+1}\left(\sqrt{x+6}+1\right)+9\)
7)\(x^2+9x+2+4\left(x+1\right)\sqrt{x+4}=\frac{5}{2}\sqrt{x+1}\left(2+\sqrt{x+4}\right)\)
8)\(8x^2-26x-2+5\sqrt{2x^4+5x^3+2x^2+7}\)
À do nãy máy lag sr :) Chứ bài đặt ẩn phụ mệt lắm :)
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Xem thêm câu trả lời
giải phương trình :1, sqrt{4-x^2}+2sqrt[3]{x^4-4x^3+4x^2}left(x-1right)^2+1-left|xright|2, 2x^3+9x^2-6xleft(1+2sqrt{6x-1}right)+2sqrt{6x-1}+803, x^3-3x+1sqrt{8-3x^2}4, left(4x^2+x-1right)sqrt{x^2+x+2}left(4x^2+3x+5right)sqrt{x^2-1}5, sqrt[3]{3-x^3}2x^3+x-36, sqrt[3]{x^2+3x+3}+sqrt[3]{2x^2+3x+2}6x^2+12x+87, frac{x^2+2x-8}{x^2-2x+3}left(x+1right)left(sqrt{x+2}-2right)8, frac{4x-1}{sqrt{4x-3}}+frac{11-2x}{sqrt{5-x}}frac{15}{2}9, ...
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giải phương trình :
1, \(\sqrt{4-x^2}+2\sqrt[3]{x^4-4x^3+4x^2}=\left(x-1\right)^2+1-\left|x\right|\)
2, \(2x^3+9x^2-6x\left(1+2\sqrt{6x-1}\right)+2\sqrt{6x-1}+8=0\)
3, \(x^3-3x+1=\sqrt{8-3x^2}\)
4, \(\left(4x^2+x-1\right)\sqrt{x^2+x+2}=\left(4x^2+3x+5\right)\sqrt{x^2-1}\)
5, \(\sqrt[3]{3-x^3}=2x^3+x-3\)
6, \(\sqrt[3]{x^2+3x+3}+\sqrt[3]{2x^2+3x+2}=6x^2+12x+8\)
7, \(\frac{x^2+2x-8}{x^2-2x+3}=\left(x+1\right)\left(\sqrt{x+2}-2\right)\)
8, \(\frac{4x-1}{\sqrt{4x-3}}+\frac{11-2x}{\sqrt{5-x}}=\frac{15}{2}\)
9, \(x^2-4x+14+\sqrt{x+4}=2\sqrt{1+12x}+\sqrt{1+\sqrt{1+12x}}\)
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vậy đến bố tổ conf biết
k thì 2 nha
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giải pt:
a,\(\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{-4x^2+16x-15}\)
b,\(\left(9x-2\right)\sqrt{3x-1}+\left(10-9x\right)\sqrt{3-3x}-4\sqrt{-9x^2+12x-3}=4\)
c, \(\left(6x-5\right)\sqrt{x+1}-\left(6x+2\right)\sqrt{x-1}+4\sqrt{x^2-1}=4x-3\)