giai pt vo ti sau
\(\sqrt{2x^2+8x+6}-\sqrt{x^2-1}=2x+2\)
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WN
0
TA
23 tháng 9 2021
ĐK: `{(2x^2+8x+6>=0),(x^2-1>=0),(2x+2>=0):} <=> {(x=-1),(x>=1):}`
`\sqrt(2x^2+8x+6)+\sqrt(x^2-1)=2x+2`
`<=>(2x^2+8x+6)+(x^2-1)+2\sqrt((2x^2+8x+6)(x^2-1))=(2x+2)^2`
`<=>2(x+3)(x+1)+(x-1)(x+2)+2\sqrt((x+1)^2 (x+3)(x-1))=4(x+1)^2`
`<=> (x+1)[2(x+3)+(x-1)+2\sqrt((x+3)(x-1))-4(x+1)]=0`
`<=> [(x=-1\ (TM)),([2(x+3)+(x-1)+2\sqrt((x+3)(x-1))-4(x+1)]=0\ (1)):}`
(1) `<=> x-1=2\sqrt((x+3)(x-1))`
`<=>x^2-2x+1=4(x+3)(x-1)`
`<=>x=1\ `(TM)
Vậy `S={\pm 1}`.
23 tháng 9 2021
\(ĐK:x\le-3;x\ge-1\)
\(PT\Leftrightarrow\sqrt{2\left(x+1\right)\left(x+3\right)}+\sqrt{\left(x-1\right)\left(x+1\right)}-2\left(x+1\right)=0\\ \Leftrightarrow\sqrt{x+1}\left(\sqrt{2\left(x+3\right)}+\sqrt{x-1}-2\sqrt{x+1}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\\sqrt{2\left(x+3\right)}+\sqrt{x-1}-2\sqrt{x+1}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\\sqrt{2\left(x+3\right)}+\sqrt{x-1}=2\sqrt{x+1}\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2\left(x+3\right)+\left(x-1\right)+2\sqrt{2\left(x+3\right)\left(x-1\right)}=4\left(x+1\right)\\ \Leftrightarrow2\sqrt{2\left(x+3\right)\left(x-1\right)}=x-1\\ \Leftrightarrow8\left(x+3\right)\left(x-1\right)-\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(7x+25\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-\dfrac{25}{7}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=1\)
Vậy \(S=\left\{-1;1\right\}\)
NH
1
13 tháng 3 2018
\(x^4-2x^3+3x^2-2x+1=0\)
Chia cả hai vé cho \(x^2\)
\(\Leftrightarrow x^2-2x+3-\dfrac{2}{x}+\dfrac{1}{x^2}\)
\(\Leftrightarrow x^2+2+\dfrac{1}{x^2}-2\left(x+\dfrac{1}{x}\right)+1=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)^2-2\left(x+\dfrac{1}{x}\right)+1=0\)
Đặt x+1/x = a, ta có:
\(a^2-2a+1=0\)
\(\Leftrightarrow\left(a-1\right)^2=0\)
\(\Leftrightarrow a=1\)
\(\Leftrightarrow x+\dfrac{1}{x}=1\)
\(\Leftrightarrow x^2+1=x\)
\(\Leftrightarrow x^2-x+1=0\)
\(\Leftrightarrow x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\)
Do \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+3>0\)
Do đó phương trình vô nghiệm
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 4 2021
a.
ĐKXĐ: \(1\le x\le7\)
\(\Leftrightarrow x-1-2\sqrt{x-1}+2\sqrt{7-x}-\sqrt{\left(x-1\right)\left(7-x\right)}=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-1}-2\right)-\sqrt{7-x}\left(\sqrt{x-1}-2\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-\sqrt{7-x}\right)\left(\sqrt{x-1}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=\sqrt{7-x}\\\sqrt{x-1}=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=7-x\\x-1=4\end{matrix}\right.\)
\(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 4 2021
b. ĐKXĐ: ...
Biến đổi pt đầu:
\(x\left(y-1\right)-\left(y-1\right)^2=\sqrt{y-1}-\sqrt{x}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\ge0\\\sqrt{y-1}=b\ge0\end{matrix}\right.\)
\(\Rightarrow a^2b^2-b^4=b-a\)
\(\Leftrightarrow b^2\left(a+b\right)\left(a-b\right)+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(b^2\left(a+b\right)+1\right)=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow\sqrt{x}=\sqrt{y-1}\Rightarrow y=x+1\)
Thế vào pt dưới:
\(3\sqrt{5-x}+3\sqrt{5x-4}=2x+7\)
\(\Leftrightarrow3\left(x-\sqrt{5x-4}\right)+7-x-3\sqrt{5-x}=0\)
\(\Leftrightarrow\dfrac{3\left(x^2-5x+4\right)}{x+\sqrt{5x-4}}+\dfrac{x^2-5x+4}{7-x+3\sqrt{5-x}}=0\)
\(\Leftrightarrow\left(x^2-5x+4\right)\left(\dfrac{3}{x+\sqrt{5x-4}}+\dfrac{1}{7-x+3\sqrt{5-x}}\right)=0\)
\(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 7 2021
c.
ĐKXĐ: \(\left[{}\begin{matrix}x\le-5\\x\ge6\end{matrix}\right.\)
\(\sqrt{\left(x-3\right)\left(x-5\right)}+\sqrt{\left(x-3\right)\left(x+5\right)}=\sqrt{\left(x-3\right)\left(x-6\right)}\)
- Với \(x\ge6\) , do \(x-3>0\) pt trở thành:
\(\sqrt{x-5}+\sqrt{x+5}=\sqrt{x-6}\)
Do \(\left\{{}\begin{matrix}\sqrt{x-5}>\sqrt{x-6}\\\sqrt{x+5}>0\end{matrix}\right.\) \(\Rightarrow\sqrt{x-5}+\sqrt{x+5}>\sqrt{x-6}\) pt vô nghiệm
- Với \(x\le-5\) pt tương đương:
\(\sqrt{\left(3-x\right)\left(5-x\right)}+\sqrt{\left(3-x\right)\left(-x-5\right)}=\sqrt{\left(3-x\right)\left(6-x\right)}\)
Do \(3-x>0\) pt trở thành:
\(\sqrt{5-x}+\sqrt{-x-5}=\sqrt{6-x}\)
\(\Leftrightarrow-2x+2\sqrt{x^2-25}=6-x\)
\(\Leftrightarrow2\sqrt{x^2-25}=x+6\) (\(x\ge-6\))
\(\Leftrightarrow4\left(x^2-25\right)=x^2+12x+36\)
\(\Leftrightarrow3x^2-12x-136=0\Rightarrow x=\dfrac{6-2\sqrt{111}}{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 7 2021
a.
Kiểm tra lại đề, pt này không giải được
b.
ĐKXĐ: \(x\ge0\)
\(\sqrt{x\left(x+1\right)}-\sqrt{x}+1-\sqrt{x+1}=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x+1}-1\right)-\left(\sqrt{x+1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x+1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow...\)
NA
2
BH
29 tháng 7 2019
\(pt\Leftrightarrow\sqrt{2x^2+8x+6}-4+\sqrt{x^2-1}-2x+2=0\)
\(\Leftrightarrow\frac{2\left(x-1\right)\left(x+5\right)}{\sqrt{2x^2+8x+6}+4}+\sqrt{x^2-1}-2\left(x-1\right)=0\)
Giải nốt nhá
29 tháng 7 2019
\(\sqrt{2x^2+8x+6}+\sqrt{x^2-1}=2x+2\)
\(\Leftrightarrow\sqrt{2\left(x^2+4x+3\right)}+\sqrt{x^2-1}=2x+2\)
\(\Leftrightarrow\sqrt{2\left(x+1\right)\left(x-3\right)}+\sqrt{x^2-1}=2x+2\)
\(\Leftrightarrow\sqrt{2\left(x+1\right)\left(x+3\right)}+\sqrt{x^2-1^2}=2x+2\)
\(\Leftrightarrow\sqrt{2\left(x+1\right)\left(x+3\right)}+\sqrt{\left(x+1\right)\left(x-1\right)}=2x+2\)
\(\Leftrightarrow2x^2+8x+6+\left(2x+2\right)\sqrt{2\left(x+3\right)\left(x-1\right)}+\left(x+1\right)\left(x-1\right)=4\left(x+1\right)^2\)
\(\Leftrightarrow\left(2x+2\right)\sqrt{2\left(x+3\right)\left(x-1\right)}=4\left(x+1\right)^2-2x^2-8x-6-\left(x+1\right)\left(x-1\right)\)
\(\Leftrightarrow8\left(x+1\right)^3.\left(x+3\right)\left(x-1\right)=\left(x+1\right)^2.\left(x-1\right)^2\)
\(\Leftrightarrow8x^4-8x^3+24x^3-24x^2+16x^3-16x^2+48x^2-48x+8x^2-8x+24x-24\)\(=x^4-2x^3+x^2+2x^3-4x^2+2x+x-2x+1\)
\(\Leftrightarrow8x^4+32x^3+16x^3-32x=x^4-2x^3+x^2+2x^3-4x^2+2x+x^2-2x+1\)
\(\Leftrightarrow8x^4+32x^3+16x^2-32x-24=x^4-2x^2+1\)
\(\Leftrightarrow8x^4+32x^2+16x^2-32x-24-x^4+2x^2-1=0\)
\(\Leftrightarrow7x^4+32x^3+18x^2-32x-25=0\)
\(\Leftrightarrow\left(7x^3+39x^2+57x+25\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(7x^2+25x+7x+25\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[x\left(7x+25\right)+\left(7x+25\right)\right]\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(7x+25\right)\left(x+1\right)\left(x-1\right)=0\)
Nhưng \(7x+25\ne0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=1\end{cases}}\)
Vậy: nghiệm phương trình là x = 1; x = -1
DT
1
SA
18 tháng 11 2018
Đặt \(\sqrt[3]{x+2}=a;\sqrt[3]{3x+2}=2\)
Ta có: \(\left\{{}\begin{matrix}a-b=2\\3a^3-b^3=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b+2\left(1\right)\\3a^3-b^3=4\end{matrix}\right.\)
Thay (1) vào (2) ta có:
3(b + 2)3 - b3 = 4
<=> 3(b3 + 6b2 + 12b + 8) - b3 = 4
<=> 2b3 + 6b2 + 12b + 4 = 0
<=> b3 + 3b2 + 6b + 2 = 0
Đến đây chắc phải dùng công thức nghiệm tổng quát, vô lý @@
CN
27 tháng 11 2018
ban giai sai roi, bài này ra no là (\(-46-18\sqrt{6}\);\(-46+18\sqrt{6}\);-1)
ĐKXĐ: \(x\ge1;x\le-3;x=-1\)
\(\sqrt{2\left(x+1\right)\left(x+3\right)}-\sqrt{\left(x-1\right)\left(x+1\right)}=2\left(x+1\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=0\left(1\right)\\\sqrt{2\left(x+3\right)}-\sqrt{x-1}=2\sqrt{x+1}\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x+1=0\Rightarrow x=-1\)
\(\left(2\right)\Leftrightarrow\sqrt{2x+6}=\sqrt{x-1}+2\sqrt{x+1}\)
\(\Leftrightarrow2x+6=x-1+4\sqrt{\left(x-1\right)\left(x+1\right)}+4x+4\)
\(\Leftrightarrow4\sqrt{x^2-1}=3-3x\) \(\Leftrightarrow\left\{{}\begin{matrix}3-3x\ge0\\16\left(x^2-1\right)=\left(3-3x\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\7x^2+18x-25=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-25}{7}\end{matrix}\right.\)
Vậy pt có 3 nghiệm: \(x=-1;1;\dfrac{-25}{7}\)
thank