\(\sqrt{x+4\sqrt{x-4}}\)+\(\sqrt{x-4\sqrt{x-4}}\)=4
Giải phương trình
K
Khách
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NT
1
AH
Akai Haruma
Giáo viên
6 tháng 2
Lời giải:
ĐKXĐ: $x\geq 1$
Đặt $\sqrt{x+1}=a; \sqrt{x-1}=b$ (ĐK: $a,b\geq 0$)
PT đã cho trở thành:
$\frac{a^2+b^2}{2}+ab=a+b+4$
$\Leftrightarrow a^2+b^2+2ab=2(a+b)+8$
$\Leftrightarrow (a+b)^2-2(a+b)-8=0$
$\Leftrightarrow (a+b-4)(a+b+2)=0$
Với $a\geq 0; b\geq 0$ thì $a+b+2\geq 2>0$
$\Rightarrow a+b-4=0$
$\Leftrightarrow a+b=4$
$\Leftrightarrow \sqrt{x+1}+\sqrt{x-1}=4$
$\Leftrightarrow \sqrt{x+1}=4-\sqrt{x-1}$
$\Rightarrow x+1=15+x-8\sqrt{x-1}$ (bp 2 vế)
$\Leftrightarrow 14=8\sqrt{x-1}$
$\Leftrightarrow x-1=(\frac{7}{4})^2=\frac{49}{16}$
$\Leftrightarrow x=\frac{65}{16}$ (tm)
EO
2
SA
26 tháng 2 2021
a) đk: \(1\le x\le5\)
\(\sqrt[4]{5-x}+\sqrt[4]{x-1}=\sqrt{2}\)
<=> \(\left(\sqrt[4]{5-x}+\sqrt[4]{x-1}\right)^4=\sqrt{2}^4\)
<=> \(5-x+x-1+4\sqrt[4]{5-x}^3.\sqrt[4]{x-1}+6\sqrt[4]{5-x}^2.\sqrt[4]{x-1}^2+4\sqrt[4]{5-x}.\sqrt[4]{x-1}^3=4\)
<=> \(\sqrt[4]{\left(5-x\right)\left(x-1\right)}.\left(2\sqrt[4]{5-x}^2+3\sqrt[4]{5-x}.\sqrt[4]{x-1}+2\sqrt[4]{x-1}^2\right)=0\)
<=> \(\left[{}\begin{matrix}\sqrt[4]{\left(5-x\right)\left(x-1\right)}=0\left(2\right)\\2\sqrt[4]{5-x}^2+3\sqrt[4]{\left(5-x\right)\left(x-1\right)}+2\sqrt[4]{x-1}^2=0\left(1\right)\end{matrix}\right.\)
Giải (2) <=> \(\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\left(tm\right)\)
Giải (1) : Đặt \(\sqrt[4]{5-x}=a;\sqrt[4]{x-1}=b\)(đk : a, b \(\ge\)0)
Khi đó, ta có: \(2a^2+3ab+2b^2=0\)
<=> 2(a2 + 3/2ab + 9/16b2) + \(\dfrac{7}{8}b^2=0\)
<=> \(2\left(a+\dfrac{3}{4}b\right)^2+\dfrac{7}{8}b^2=0\)
<=> \(\left\{{}\begin{matrix}a+\dfrac{3}{4}b=0\\b=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}a=0\\b=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}\sqrt[4]{x-1}=0\\\sqrt[4]{5-x}=0\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)(vô lí)
PP
3 tháng 12 2021
sao cách này rắc rối quá vậy , có cách nào đơn giản hơn không? mà pt này rõ ràng có nghiệm chứ có phải vô nghiệm đâu
23 tháng 6 2021
a) Áp dụng bđt AM-GM có:
\(\sqrt[3]{\left(9-x\right).8.8}\le\dfrac{9-x+8+8}{3}=\dfrac{25-x}{3}\)\(\Leftrightarrow\sqrt[3]{9-x}\le\dfrac{25-x}{12}\)
\(\sqrt[3]{\left(7+x\right).8.8}\le\dfrac{7+x+8+8}{3}=\dfrac{23+x}{3}\)\(\Leftrightarrow\sqrt[3]{7+x}\le\dfrac{23+x}{12}\)
Cộng vế với vế \(\Rightarrow\sqrt[3]{9-x}+\sqrt[3]{7+x}\le4\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}9-x=8\\7+x=8\end{matrix}\right.\)\(\Rightarrow x=1\)
Vậy...
b)Đk:\(x\ge2\)
Pt \(\Leftrightarrow\left(x-1\right)^2.\left(x^2-4\right)=\left(x-2\right)^2.\left(x^2-1\right)\)
\(\Leftrightarrow\left(x-1\right)^2\left(x-2\right)\left(x+2\right)=\left(x-2\right)^2\left(x+1\right)\left(x-1\right)\)
Do \(x\ge2\Rightarrow x-1>0\)
Chia cả hai vế của pt cho x-1 ta được:
\(\left(x-1\right)\left(x-2\right)\left(x+2\right)=\left(x-2\right)^2\left(x+1\right)\)
\(\Leftrightarrow\left(x-2\right)\left[\left(x-1\right)\left(x+2\right)-\left(x-2\right)\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left[x^2+x-2-x^2+3x-2\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(4x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=1\left(ktm\right)\end{matrix}\right.\)
Vậy S={2}
c)Đk:\(\left\{{}\begin{matrix}9-x^2\ge0\\x^2-1\ge0\\x-3\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}-3\le x\le3\\\left[{}\begin{matrix}x\ge1\\x\le-1\end{matrix}\right.\\x\ge3\end{matrix}\right.\)\(\Rightarrow x=3\)
Thay x=3 vào pt thấy thỏa mãn
Vậy S={3}
23 tháng 6 2021
a) Quên mất, ko áp dụng đc AM-GM, xin lỗi
Pt \(\Leftrightarrow\sqrt[3]{9-x}-2=2-\sqrt[3]{7+x}\)
\(\Leftrightarrow\dfrac{9-x-8}{\sqrt[3]{\left(9-x\right)^2}+2\sqrt[3]{9-x}+4}=\dfrac{8-\left(7-x\right)}{4+2\sqrt[3]{7+x}+\sqrt[3]{\left(7+x\right)^2}}\)
\(\Leftrightarrow\dfrac{1-x}{\sqrt[3]{\left(9-x\right)^2}+2\sqrt[3]{9-x}+4}=\dfrac{1-x}{4+2\sqrt[3]{7+x}+\sqrt[3]{\left(7+x\right)^2}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\dfrac{1}{\sqrt[3]{\left(9-x\right)^2}+2\sqrt[3]{9-x}+4}=\dfrac{1}{4+2\sqrt[3]{7+x}+\sqrt[3]{\left(7+x\right)^2}}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\sqrt[3]{\left(9-x\right)^2}+2\sqrt[3]{9-x}+4=4+2\sqrt[3]{7+x}+\sqrt[3]{\left(7+x\right)^2}\left(1\right)\end{matrix}\right.\)
Từ (1) \(\Leftrightarrow\sqrt[3]{\left(9-x\right)^2}-\sqrt[3]{\left(7+x\right)^2}+2\left(\sqrt[3]{9-x}-\sqrt[3]{7+x}\right)=0\)
\(\Leftrightarrow\left(\sqrt[3]{9-x}-\sqrt[3]{7+x}\right)\left(\sqrt[3]{9-x}+\sqrt[3]{7+x}\right)+2\left(\sqrt[3]{9-x}-\sqrt[3]{7+x}\right)=0\)
\(\Leftrightarrow\left(\sqrt[3]{9-x}-\sqrt[3]{7+x}\right).4+2\left(\sqrt[3]{9-x}-\sqrt[3]{7+x}\right)=0\)
\(\Leftrightarrow\sqrt[3]{9-x}-\sqrt[3]{7+x}=0\)
\(\Leftrightarrow\sqrt[3]{9-x}=\sqrt[3]{7+x}\)\(\Leftrightarrow9-x=7+x\)
\(\Leftrightarrow x=1\)
Vậy S={1}
15 tháng 9 2021
\(1,\sqrt{x+2+4\sqrt{x-2}}=5\left(x\ge2\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x-2}+4\right)^2}=5\\ \Leftrightarrow\sqrt{x-2}+4=5\\ \Leftrightarrow\sqrt{x-2}=1\\ \Leftrightarrow x-2=1\Leftrightarrow x=3\\ 2,\sqrt{x+3+4\sqrt{x-1}}=2\left(x\ge1\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x-1}+4\right)^2}=2\\ \Leftrightarrow\sqrt{x-1}+4=2\\ \Leftrightarrow\sqrt{x-1}=-2\\ \Leftrightarrow x\in\varnothing\left(\sqrt{x-1}\ge0\right)\)
\(3,\sqrt{x+\sqrt{2x-1}}=\sqrt{2}\left(x\ge\dfrac{1}{2};x\ne1\right)\\ \Leftrightarrow x+\sqrt{2x-1}=2\\ \Leftrightarrow x-2=-\sqrt{2x-1}\\ \Leftrightarrow x^2-4x+4=2x-1\\ \Leftrightarrow x^2-6x+5=0\\ \Leftrightarrow\left(x-5\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=1\left(loại\right)\end{matrix}\right.\)
\(4,\sqrt{x-2+\sqrt{2x-5}}=3\sqrt{2}\left(x\ge\dfrac{5}{2}\right)\\ \Leftrightarrow\sqrt{2x-4+2\sqrt{2x-5}}=6\\ \Leftrightarrow\sqrt{\left(\sqrt{2x-5}+1\right)^2}=6\\ \Leftrightarrow\sqrt{2x-5}+1=6\\ \Leftrightarrow\sqrt{2x-5}=5\\ \Leftrightarrow2x-5=25\Leftrightarrow x=15\left(TM\right)\)
DL
0
HP
7 tháng 8 2021
a, ĐK: \(x\le-1,x\ge3\)
\(pt\Leftrightarrow2\left(x^2-2x-3\right)+\sqrt{x^2-2x-3}-3=0\)
\(\Leftrightarrow\left(2\sqrt{x^2-2x-3}+3\right).\left(\sqrt{x^2-2x-3}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-2x-3}=-\dfrac{3}{2}\left(l\right)\\\sqrt{x^2-2x-3}=1\end{matrix}\right.\)
\(\Leftrightarrow x^2-2x-3=1\)
\(\Leftrightarrow x^2-2x-4=0\)
\(\Leftrightarrow x=1\pm\sqrt{5}\left(tm\right)\)
HP
7 tháng 8 2021
b, ĐK: \(-2\le x\le2\)
Đặt \(\sqrt{2+x}-2\sqrt{2-x}=t\Rightarrow t^2=10-3x-4\sqrt{4-x^2}\)
Khi đó phương trình tương đương:
\(3t-t^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=0\\t=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2+x}-2\sqrt{2-x}=0\\\sqrt{2+x}-2\sqrt{2-x}=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2+x=8-4x\\2+x=17-4x+12\sqrt{2-x}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\left(tm\right)\\5x-15=12\sqrt{2-x}\left(1\right)\end{matrix}\right.\)
Vì \(-2\le x\le2\Rightarrow5x-15< 0\Rightarrow\left(1\right)\) vô nghiệm
Vậy phương trình đã cho có nghiệm \(x=\dfrac{6}{5}\)
30 tháng 7 2021
a) ĐKXĐ: \(x\ge0\)
Ta có: \(\left(x+3\sqrt{x}+2\right)\left(x+9\sqrt{x}+18\right)=168x\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+6\right)=168x\)
\(\Leftrightarrow\left(x+6\right)^2+12\sqrt{x}\left(x+6\right)-133=0\)
\(\Leftrightarrow\left(x+6\right)^2+19\sqrt{x}\left(x+6\right)-7\sqrt{x}\left(x+6\right)-133=0\)
\(\Leftrightarrow\left(x+6\right)\left(x+19\sqrt{x}+6\right)-7\sqrt{x}\left(x+19\sqrt{x}+6\right)=0\)
\(\Leftrightarrow\left(x-7\sqrt{x}+6\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=36\end{matrix}\right.\)
NT
30 tháng 7 2021
Dòng thứ 2 qua dòng thứ 3 anh làm chậm lại được không ạ, tại tắt quá e không hiểu
27 tháng 7 2021
4) Ta có: \(\left(x+3\right)\cdot\sqrt{10-x^2}=x^2-x-12\)
\(\Leftrightarrow\left(x+3\right)\cdot\sqrt{10-x^2}-\left(x-4\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(\sqrt{10-x^2}-x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\\sqrt{10-x^2}=x-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\10-x^2=x^2-8x+16\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x^2-8x+16-10+x^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\2x^2-8x+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\2\left(x^2-4x+3\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\\left(x-1\right)\left(x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\\x=3\end{matrix}\right.\)
\(\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}=4.\)
\(\Rightarrow\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}=4\)
\(\Rightarrow\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}=4\)
\(\Rightarrow\sqrt{x-4}+2+\sqrt{x-4}-2=0\)
\(\Rightarrow2\sqrt{x-2}=0\)
\(\Rightarrow\sqrt{x-2}=0\Rightarrow x-2=0\Rightarrow x=2\)