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24 tháng 10 2021
1) ĐKXĐ: \(x\ge-2\)
\(pt\Leftrightarrow\dfrac{\sqrt{x+2}}{2}+5\sqrt{x+2}-2\sqrt{x+2}=14\)
\(\Leftrightarrow\dfrac{\sqrt{x+2}+6\sqrt{x+2}}{2}=14\Leftrightarrow7\sqrt{x+2}=28\)
\(\Leftrightarrow\sqrt{x+2}=4\Leftrightarrow x+2=16\Leftrightarrow x=14\left(tm\right)\)
2) ĐKXĐ: \(x\ge0\)
\(pt\Leftrightarrow2x+3=x^2\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)
3) \(pt\Leftrightarrow\sqrt{\left(5x+2\right)^2}=1\Leftrightarrow\left|5x+2\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+2=1\\5x+2=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
4) ĐKXĐ: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1\ge0\\2x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1\le0\\2x-1< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{1}{2}\\x\le-1\end{matrix}\right.\)
\(pt\Leftrightarrow\dfrac{x+1}{2x-1}=4\Leftrightarrow x+1=8x-4\)
\(\Leftrightarrow7x=5\Leftrightarrow x=\dfrac{5}{7}\left(tm\right)\)
5) ĐKXĐ: \(x\ge2\)
\(pt\Leftrightarrow\dfrac{x-2}{3x+1}=36\)
\(\Leftrightarrow x-2=108x+36\Leftrightarrow107x=-38\Leftrightarrow x=-\dfrac{38}{107}\left(ktm\right)\)
Vậy \(S=\varnothing\)
8 tháng 8 2021
a: Ta có: \(A=\dfrac{x\sqrt{x}-3}{x-2\sqrt{x}-3}-\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{3-\sqrt{x}}\)
\(=\dfrac{x\sqrt{x}-3-2\left(x-6\sqrt{x}+9\right)-x-4\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x\sqrt{x}-x-4\sqrt{x}-6-2x+12\sqrt{x}-18}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x\sqrt{x}-3x+8\sqrt{x}-24}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x\left(\sqrt{x}-3\right)+8\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x+8}{\sqrt{x}+1}\)
AH
Akai Haruma
Giáo viên
6 tháng 8 2021
Lời giải:
ĐKXĐ: $x\geq 0; x\neq 9$
a. \(A=\frac{x\sqrt{x}-3}{(\sqrt{x}+1)(\sqrt{x}-3)}-\frac{2(\sqrt{x}-3)^2}{(\sqrt{x}+1)(\sqrt{x}-3)}-\frac{(\sqrt{x}+3)(\sqrt{x}+1)}{(\sqrt{x}+1)(\sqrt{x}-3)}\)
\(=\frac{x\sqrt{x}-3x+8\sqrt{x}-24}{(\sqrt{x}+1)(\sqrt{x}-3)}=\frac{(\sqrt{x}-3)(x+8)}{(\sqrt{x}+1)(\sqrt{x}-3)}=\frac{x+8}{\sqrt{x}+1}\)
b.
\(14-6\sqrt{5}=(3-\sqrt{5})^2\Rightarrow \sqrt{x}=3-\sqrt{5}\)
\(A=\frac{14-6\sqrt{5}+8}{3-\sqrt{5}+1}=\frac{22-6\sqrt{5}}{4-\sqrt{5}}=\frac{58-2\sqrt{5}}{11}\)
c.
Áp dụng BĐT Cô-si:
$x+4\geq 4\sqrt{x}\Rightarrow x+8\geq 4(\sqrt{x}+1)$
$\Rightarrow A=\frac{x+8}{\sqrt{x}+1}\geq 4$
Vậy $A_{\min}=4$. Giá trị này đạt tại $x=4$
BT
10 tháng 8 2017
\(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right]:\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
a/ \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt[]{x-3}\right)}\right]:\left(\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
=> \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3}{\sqrt[]{x-3}}\right]:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
=> \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}-3}+1\right]:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
=> \(R=\left[\frac{2\sqrt{x}+\sqrt{x}-3}{\sqrt{x}-3}\right].\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
=> \(R=\frac{3\sqrt{x}-3}{\sqrt{x}-3}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
b/ Để R<-1 => \(\frac{3\left(\sqrt{x}-1\right)}{\sqrt{x}+1}< -1\)
<=> \(3\sqrt{x}-3< -\sqrt{x}-1\)
<=> \(4\sqrt{x}< 2\)=> \(\sqrt{x}< \frac{1}{2}\) => \(-\frac{1}{4}< x< \frac{1}{4}\)
10 tháng 8 2017
Chỗ => R = \(\left(\frac{2\sqrt{x}}{\sqrt{x}-3}+1\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\) là sao vậy ạ?
24 tháng 9 2021
\(1,\\ a,ĐK:x-2\ge0\Leftrightarrow x\ge2\\ b,ĐK:2-3x\ge0\Leftrightarrow x\le\dfrac{2}{3}\\ 2,\\ a,=\sqrt{16}-3\sqrt{4}=4-6=-2\\ b,=\dfrac{-\sqrt{7}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=-\sqrt{7}\\ c,=\sqrt{4}\cdot\sqrt{36}=2\cdot6=12\\ d,=\sqrt{\dfrac{25}{81}}\cdot\sqrt{\dfrac{16}{49}}=\dfrac{5}{9}\cdot\dfrac{4}{7}=\dfrac{20}{63}\\ 3,\\ a,=\sqrt{19+2\sqrt{34}}-\sqrt{19-2\sqrt{34}}\\ =\sqrt{\left(\sqrt{17}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{17}-\sqrt{2}\right)^2}=\sqrt{17}+\sqrt{2}-\sqrt{17}+\sqrt{2}=2\sqrt{2}\\ b,=3-4+2\cdot5=9\)
\(4,ĐK:x\ge-5\\ PT\Leftrightarrow2\sqrt{x+5}-2\sqrt{x+5}+3\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}=2\\ \Leftrightarrow x+5=4\Leftrightarrow x=-1\left(tm\right)\\ 5,\\ a,B=\dfrac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\left(\sqrt{x}+2\right)^2}{1-\sqrt{x}}=\dfrac{\sqrt{x}+2}{\sqrt{x}}\\ b,B=\dfrac{5}{2}\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}}=\dfrac{5}{2}\\ \Leftrightarrow2\sqrt{x}+4=5\sqrt{x}\\ \Leftrightarrow3\sqrt{x}=4\Leftrightarrow\sqrt{x}=\dfrac{4}{3}\Leftrightarrow x=\dfrac{16}{9}\)
DK
2 tháng 6 2021
\(M=3\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}+4\right)^2+14\)
\(=3\left(x+2\sqrt{x}+1\right)-\left(x+8\sqrt{x}+16\right)+14\)
\(=3x+6\sqrt{x}+3-x-8\sqrt{x}-16+14\)
\(=2x-2\sqrt{x}+1\)
\(=2\left(x-4\sqrt{x}+4\right)+6\sqrt{x}-7\)
\(=2\left(\sqrt{x}-2\right)^2+6\sqrt{x}-7\ge2.0+6.\sqrt{4}-7=5\)
Dấu "=" \(x=4\)
Vậy GTNN của M là 4 <=> x = 4
DK
4 tháng 6 2021
\(\left\{{}\begin{matrix}xz=x+4\left(1\right)\\2y^2=7xz-3x-14\\x^2+y^2=35-z^2\left(3\right)\end{matrix}\right.\left(2\right)\)
Nhận thấy \(x=0\) không là nghiệm của (1) .
\(\rightarrow z=\dfrac{x+4}{x}\)(4)
Thế (1) vào (2) .
\(2y^2=7\left(x+4\right)-3x-14=4x+14\leftrightarrow y^2=2x+7\)(\(x\ge-\dfrac{7}{2}\)) (5)
Thế (4)(5) vào (3)
\(x^2+2x+7=35-\left(\dfrac{x+4}{x}\right)^2\)
\(\Leftrightarrow x^4+2x^3-27x^2+8x+16=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-1\right)\left(x^2+7x+4\right)=0\)\(\)
TH1 : \(x-4=0\Leftrightarrow x=4\Leftrightarrow\left\{{}\begin{matrix}y=\pm\sqrt{15}\\z=2\end{matrix}\right.\)
TH2 : \(x-1=0\Leftrightarrow x=1\Leftrightarrow\left\{{}\begin{matrix}y=\pm3\\z=5\end{matrix}\right.\)
TH3 : \(x^2+7x+4=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-7+\sqrt{33}}{2}\left(TM\right)\\x=\dfrac{-7-\sqrt{33}}{2}\left(KTM\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{-7+\sqrt{33}}{2}\Leftrightarrow\left\{{}\begin{matrix}y=\pm\sqrt[4]{33}\\z=-\dfrac{5+\sqrt{33}}{2}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 12 2020
a.
ĐKXĐ: \(x\ge1\)
\(\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+\sqrt{\left(x-1\right)\left(x^3+x^2+x+1\right)}\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x^3+x^2+x+1}-1\right)-\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x^3+x^2+x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x^3+x^2+x=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 12 2020
b.
ĐKXĐ: \(x\ge-1\)
\(x^2-6x+9+x+1-4\sqrt{x+1}+4=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(\sqrt{x+1}-2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\\sqrt{x+1}-2=0\end{matrix}\right.\)
\(\Leftrightarrow x=3\)
c.
ĐKXĐ: \(-2\le x\le\dfrac{4}{5}\)
\(VT=2x+3\sqrt{4-5x}+1.\sqrt{x+2}\)
\(VT\le2x+\dfrac{1}{2}\left(9+4-5x\right)+\dfrac{1}{2}\left(1+x+2\right)=8\)
Dấu "=" xảy ra khi và chỉ khi \(x=-1\)
16 tháng 6 2021
a) ĐK: \(x\ge3\)
PT \(\Leftrightarrow\sqrt{\left(x-3\right)\left(x-2\right)}-\sqrt{x-2}+\sqrt{x+1}-\sqrt{\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-3}-1\right)+\sqrt{x+1}\left(1-\sqrt{x-3}\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2}-\sqrt{x+1}\right)\left(\sqrt{x-3}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=\sqrt{x+1}\\\sqrt{x-3}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x-2=x+1\\x-3=1\end{matrix}\right.\) \(\Leftrightarrow x=4\) (Thỏa mãn)
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