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6:ĐKXĐ: x>=0; x<>1/25
BPT=>\(\dfrac{3\sqrt{x}}{5\sqrt{x}-1}+3< =0\)
=>\(\dfrac{3\sqrt{x}+15\sqrt{x}-5}{5\sqrt{x}-1}< =0\)
=>\(\dfrac{18\sqrt{x}-5}{5\sqrt{x}-1}< =0\)
=>\(\dfrac{1}{5}< \sqrt{x}< =\dfrac{5}{18}\)
=>\(\dfrac{1}{25}< x< =\dfrac{25}{324}\)
7:
ĐKXĐ: x>=0
BPT \(\Leftrightarrow\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}>\dfrac{8}{3}:\dfrac{8}{3}=1\)
=>\(\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}-1>=0\)
=>\(\dfrac{\sqrt{x}+1-2\sqrt{x}-3}{2\sqrt{x}+3}>=0\)
=>\(-\sqrt{x}-2>=0\)(vô lý)
8:
ĐKXĐ: x>=0; x<>9/4
BPT \(\Leftrightarrow\dfrac{\sqrt{x}-2}{2\sqrt{x}-3}+4< 0\)
=>\(\dfrac{\sqrt{x}-2+8\sqrt{x}-12}{2\sqrt{x}-3}< 0\)
=>\(\dfrac{9\sqrt{x}-14}{2\sqrt{x}-3}< 0\)
TH1: 9căn x-14>0 và 2căn x-3<0
=>căn x>14/9 và căn x<3/2
=>14/9<căn x<3/2
=>196/81<x<9/4
TH2: 9căn x-14<0 và 2căn x-3>0
=>căn x>3/2 hoặc căn x<14/9
mà 3/2<14/9
nên trường hợp này Loại
9:
ĐKXĐ: x>=0
\(BPT\Leftrightarrow\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}< =-\dfrac{1}{3}\)
=>\(\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}+\dfrac{1}{3}< =0\)
=>\(\dfrac{6\sqrt{x}+9+5\sqrt{x}+7}{3\left(5\sqrt{x}+7\right)}< =0\)
=>\(\dfrac{11\sqrt{x}+16}{3\left(5\sqrt{x}+7\right)}< =0\)(vô lý)
10:
ĐKXĐ: x>=0; x<>1/49
\(BPT\Leftrightarrow\dfrac{6\sqrt{x}-2}{7\sqrt{x}-1}+6>0\)
=>\(\dfrac{6\sqrt{x}-2+42\sqrt{x}-6}{7\sqrt{x}-1}>0\)
=>\(\dfrac{48\sqrt{x}-8}{7\sqrt{x}-1}>0\)
=>\(\dfrac{6\sqrt{x}-1}{7\sqrt{x}-1}>0\)
TH1: 6căn x-1>0 và 7căn x-1>0
=>căn x>1/6 và căn x>1/7
=>căn x>1/6
=>x>1/36
TH2: 6căn x-1<0 và 7căn x-1<0
=>căn x<1/6 và căn x<1/7
=>căn x<1/7
=>0<=x<1/49
![](https://rs.olm.vn/images/avt/0.png?1311)
6) ĐKXĐ: \(x\le-6\)
\(\sqrt{\left(x+6\right)^2}=-x-6\Leftrightarrow\left|x+6\right|=-x-6\)
\(\Leftrightarrow x+6=x+6\left(đúng\forall x\right)\)
Vậy \(x\le-6\)
7) ĐKXĐ: \(x\ge\dfrac{2}{3}\)
\(pt\Leftrightarrow\sqrt{\left(3x-2\right)^2}=3x-2\Leftrightarrow\left|3x-2\right|=3x-2\)
\(\Leftrightarrow3x-2=3x-2\left(đúng\forall x\right)\)
Vậy \(x\ge\dfrac{2}{3}\)
8) ĐKXĐ: \(x\ge5\)
\(pt\Leftrightarrow\sqrt{\left(4-3x\right)^2}=2x-10\)\(\Leftrightarrow\left|4-3x\right|=2x-10\)
\(\Leftrightarrow4-3x=10-2x\Leftrightarrow x=-6\left(ktm\right)\Leftrightarrow S=\varnothing\)
9) ĐKXĐ: \(x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow\sqrt{\left(x-3\right)^2}=2x-3\Leftrightarrow\left|x-3\right|=2x-3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x-3\left(x\ge3\right)\\x-3=3-2x\left(\dfrac{3}{2}\le x< 3\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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1)can(2)*(can(2)+1-can(3))
2)-1/(canbậc3của2-1)
3)120
4)1
5)3
6)60
7)chưa làm
8)72
9)47
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(a,A=x-4\sqrt{x+9}=\left(x+9-4\sqrt{x+9}+4\right)-13\\ A=\left(\sqrt{x+9}-2\right)^2-13\ge-13\\ A_{min}=-13\Leftrightarrow x+9=4\Leftrightarrow x=-5\\ b,B=x-3\sqrt{x-10}=\left(x-10-3\sqrt{x-10}+\dfrac{9}{4}\right)+\dfrac{31}{4}\\ B=\left(\sqrt{x-10}+\dfrac{9}{4}\right)^2+\dfrac{31}{4}\ge\dfrac{31}{4}\\ B_{min}=\dfrac{31}{4}\Leftrightarrow x-10=\dfrac{81}{16}\Leftrightarrow x=\dfrac{241}{16}\\ c,C=x-\sqrt{x+1}=\left(x+1-\sqrt{x+1}+\dfrac{1}{4}\right)-\dfrac{5}{4}\\ C=\left(\sqrt{x+1}-\dfrac{1}{2}\right)^2-\dfrac{5}{4}\ge-\dfrac{5}{4}\\ C_{min}=-\dfrac{5}{4}\Leftrightarrow x+1=\dfrac{1}{4}\Leftrightarrow x=-\dfrac{3}{4}\)
\(d,D=x+\sqrt{x+2}=\left(x+2+\sqrt{x+2}+\dfrac{1}{4}\right)-\dfrac{9}{4}\\ D=\left(\sqrt{x+2}+\dfrac{1}{4}\right)^2-\dfrac{9}{4}\ge-\dfrac{9}{4}\\ D_{min}=-\dfrac{9}{4}\Leftrightarrow\sqrt{x+2}=-\dfrac{1}{4}\Leftrightarrow x\in\varnothing\)
Vậy dấu \("="\) ko xảy ra
a: \(A=x-4\sqrt{x}+9\)
\(=\left(\sqrt{x}-2\right)^2+5\ge5\forall x\)
Dấu '=' xảy ra khi x=4
b: \(B=x-3\sqrt{x}-10\)
\(=x-2\cdot\sqrt{x}\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{49}{4}\)
\(=\left(\sqrt{x}-\dfrac{3}{2}\right)^2-\dfrac{49}{4}\ge-\dfrac{49}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{9}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Nếu :
1 x 2 x 3 =5
2 x 3 x 4 = 10
3 x 4 x 5 = 17
4 x 5 x 6 = 26
Thì 6 x 7 x 8= 50
k mình nhé ^_^
10\(\sqrt{8+x^3}=3\left(x^2-x+6\right)\) dk tu xd
\(10\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}=3\left(x+2\right)\left(x+3\right)\)
\(\Leftrightarrow\sqrt{x+2}\left(10\sqrt{x^2-2x+4}-3\left(x+3\right)\sqrt{x+2}\right)=0\)
đến đây tự giải quyết nhé