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MC
1
Những câu hỏi liên quan
NL
Nguyễn Lê Phước Thịnh
CTVHS
18 tháng 2 2021
1) Vì x=25 thỏa mãn ĐKXĐ nên Thay x=25 vào biểu thức \(A=\dfrac{\sqrt{x}-2}{x+1}\), ta được:
\(A=\dfrac{\sqrt{25}-2}{25+1}=\dfrac{5-2}{25+1}=\dfrac{3}{26}\)
Vậy: Khi x=25 thì \(A=\dfrac{3}{26}\)
2) Ta có: \(B=\dfrac{\sqrt{x}-3}{\sqrt{x}+1}+\dfrac{2x+8\sqrt{x}-6}{x-\sqrt{x}-2}\)
\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}+\dfrac{2x+8\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-5\sqrt{x}+6+2x+8\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3x+3\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3\sqrt{x}}{\sqrt{x}-2}\)
18 tháng 12 2023
55 - 3x = (-5)2
55 - 3x = 25
3x = 55 - 25
3x = 30
x = 30 : 3
x = 10
Vậy x = 10
HN
0
MC
2
DT
Đỗ Thanh Hải
CTVVIP
25 tháng 3 2021
3 careless
4 bleed
V
1 Hung has colllected stamps since 2000
2 The front yard isn't large enough to play soccer in
3 Milk is delivered twice a week by the milkman
4 When did you bought this computer?
5 His car doesn't run as fasst as a race car
6 Two languages can be spoken well by Lan
VI
1 T
2 F
3 T
4 T
VII
1 Mr. Nam is arriving in Hue tonight
2 They are going back to Englang in two months
3 It is not difficult to tránlate this sentence into English
4 Would you mind lending me your dictionary?
31 tháng 10 2021
3 careless
4 bleed
V
1 Hung has colllected stamps since 2000
2 The front yard isn't large enough to play soccer in
3 Milk is delivered twice a week by the milkman
4 When did you bought this computer?
5 His car doesn't run as fasst as a race car
6 Two languages can be spoken well by Lan
VI
1 T
2 F
3 T
4 T
VII
1 Mr. Nam is arriving in Hue tonight
2 They are going back to Englang in two months
3 It is not difficult to tránlate this sentence into English
4 Would you mind lending me your dictionary?
PT
3
NB
0
NL
Nguyễn Lê Phước Thịnh
CTVHS
16 tháng 6
Câu 81:
1: ĐKXĐ: x>=0
\(5\sqrt{12x}+4\cdot\sqrt{3x}+2\cdot\sqrt{48x}=14\)
=>\(5\cdot2\cdot\sqrt{3x}+4\cdot\sqrt{3x}+2\cdot4\sqrt{3x}=14\)
=>\(22\sqrt{3x}=14\)
=>\(\sqrt{3x}=\frac{14}{22}=\frac{7}{11}\)
=>\(3x=\frac{49}{121}\)
=>\(x=\frac{49}{121\cdot3}=\frac{49}{363}\) (nhận)
2: ĐKXĐ: x>=5
\(\sqrt{4x-20}+\sqrt{x-5}-\frac13\cdot\sqrt{9x-45}=4\)
=>\(2\sqrt{x-5}+\sqrt{x-5}-\frac13\cdot3\sqrt{x-5}=4\)
=>\(2\sqrt{x-5}=4\)
=>\(\sqrt{x-5}=2\)
=>x-5=4
=>x=9(nhận)
3: ĐKXĐ: x>=-1
Ta có: \(\sqrt{16x+16}-\sqrt{9x+9}=1\)
=>\(4\cdot\sqrt{x+1}-3\cdot\sqrt{x+1}=1\)
=>\(\sqrt{x+1}=1\)
=>x+1=1
=>x=0(nhận)
4: ĐKXĐ: x>=0
\(3\sqrt{2x}+5\cdot\sqrt{8x}-20-\sqrt{18x}=0\)
=>\(3\sqrt{2x}+5\cdot2\sqrt{2x}-3\sqrt{2x}=20\)
=>\(10\sqrt{2x}=20\)
=>\(\sqrt{2x}=2\)
=>2x=4
=>x=2(nhận)
5: ĐKXĐ: x>=2
\(\sqrt{49x-98}-14\cdot\sqrt{\frac{x-2}{49}}=3\sqrt{x-2}+8\)
=>\(7\sqrt{x-2}-14\cdot\frac{\sqrt{x-2}}{7}-3\sqrt{x-2}=8\)
=>\(2\sqrt{x-2}=8\)
=>\(\sqrt{x-2}=4\)
=>x-2=16
=>x=18(nhận)
6: ĐKXĐ: x>=-3
\(\sqrt{9x+27}+5\sqrt{x+3}-\frac34\cdot\sqrt{16x+48}=5\)
=>\(3\sqrt{x+3}+5\sqrt{x+3}-\frac34\cdot4\sqrt{x+3}=5\)
=>\(5\sqrt{x+3}=5\)
=>\(\sqrt{x+3}=1\)
=>x+3=1
=>x=-2(nhận)
7: ĐKXĐ: x>=2
\(\sqrt{x-2}+\sqrt{9x-18}+\sqrt{81x-162}=4\)
=>\(\sqrt{x-2}+3\sqrt{x-2}+9\sqrt{x-2}=4\)
=>\(13\sqrt{x-2}=4\)
=>\(\sqrt{x-2}=\frac{4}{13}\)
=>\(x-2=\frac{16}{169}\)
=>\(x=2+\frac{16}{169}=\frac{354}{169}\) (nhận)
8: ĐKXĐ: x>=2
\(\sqrt{x-2}+\sqrt{9x-18}+\sqrt{81x-162}=4\)
=>\(\sqrt{x-2}+3\sqrt{x-2}+9\sqrt{x-2}=4\)
=>\(13\sqrt{x-2}=4\)
=>\(\sqrt{x-2}=\frac{4}{13}\)
=>\(x-2=\frac{16}{169}\)
=>\(x=2+\frac{16}{169}=\frac{354}{169}\) (nhận)
9: ĐKXĐ: x>=1
\(\sqrt{16x-16}-\sqrt{9x-9}+\sqrt{4x-4}+\sqrt{x-1}=8\)
=>\(4\sqrt{x-1}-3\sqrt{x-1}+2\sqrt{x-1}+\sqrt{x-1}=8\)
=>\(4\sqrt{x-1}=8\)
=>\(\sqrt{x-1}=2\)
=>x-1=4
=>x=5(nhận)
NQ
4
I
1C
2 D
3A
4 B
II
5 D
6 A
7 D
8 B
9 C
10 A
11 B
12 B
13 B
14 D