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a) ( x - 3)4 + ( x - 5)4 = 82
Đặt : x - 4 = a , ta có :
( a + 1)4 + ( a - 1)4 = 82
⇔ a4 + 4a3 + 6a2 + 4a + 1 + a4 - 4a3 + 6a2 - 4a + 1 = 82
⇔ 2a4 + 12a2 - 80 = 0
⇔ 2( a4 + 6a2 - 40) = 0
⇔ a4 - 4a2 + 10a2 - 40 = 0
⇔ a2( a2 - 4) + 10( a2 - 4) = 0
⇔ ( a2 - 4)( a2 + 10) = 0
Do : a2 + 10 > 0
⇒ a2 - 4 = 0
⇔ a = + - 2
+) Với : a = 2 , ta có :
x - 4 = 2
⇔ x = 6
+) Với : a = -2 , ta có :
x - 4 = -2
⇔ x = 2
KL.....
b) ( n - 6)( n - 5)( n - 4)( n - 3) = 5.6.7.8
⇔ ( n - 6)( n - 3)( n - 5)( n - 4) = 1680
⇔ ( n2 - 9n + 18)( n2 - 9n + 20) = 1680
Đặt : n2 - 9n + 19 = t , ta có :
( t - 1)( t + 1) = 1680
⇔ t2 - 1 = 1680
⇔ t2 - 412 = 0
⇔ ( t - 41)( t + 41) = 0
⇔ t = 41 hoặc t = - 41
+) Với : t = 41 , ta có :
n2 - 9n + 19 = 41
⇔ n2 - 9n - 22 = 0
⇔ n2 + 2n - 11n - 22 = 0
⇔ n( n + 2) - 11( n + 2) = 0
⇔ ( n + 2)( n - 11) = 0
⇔ n = - 2 hoặc n = 11
+) Với : t = -41 ( giải tương tự )
@Giáo Viên Hoc24.vn
@Giáo Viên Hoc24h
@Giáo Viên
@giáo viên chuyên
@Akai Haruma
14dm5cm=14,5dm;3dm7cm=3,7dm
chu vi hình chữ nhật đó là:
(14,5+3,7)x2=36,4(dm)
ĐS:36,4dm
14 dm 5 cm = 14,5 dm
3 dm 7 cm = 3,7 dm
Chiều rộng HCN là :
14,5 - 3,7 = 10,8 ( dm )
chu vi HCN là :
( 14,5 + 10,8 ) x 2 = 50,6 ( dm )
ĐS:..
\(\left(1+1+1\right)!=6.\)
\(2+2+2=6\)
\(3.3-3=6\)
\(\sqrt{4}+\sqrt{4}.\sqrt{4}=6\)
\(5+5\div5=6\)
\(6.6\div6=6\)
\(7-7\div7=6\)
\(\sqrt{8+8\div8}!=6\)
\(\sqrt{9}.\sqrt{9}-\sqrt{9}=6\)
\(\sqrt{10-10\div10}!\)
a) \(6\sqrt{x-1}-\dfrac{1}{3}\cdot\sqrt{9x-9}+\dfrac{7}{2}\sqrt{4x-4}=24\) (ĐK: \(x\ge1\))
\(\Leftrightarrow6\sqrt{x-1}-\dfrac{1}{3}\cdot\sqrt{9\left(x-1\right)}+\dfrac{7}{2}\sqrt{4\left(x-1\right)}=24\)
\(\Leftrightarrow6\sqrt{x-1}-\dfrac{1}{3}\cdot3\sqrt{x-1}+\dfrac{7}{2}\cdot2\sqrt{x-1}=24\)
\(\Leftrightarrow6\sqrt{x-1}-\sqrt{x-1}+7\sqrt{x-1}=24\)
\(\Leftrightarrow12\sqrt{x-1}=24\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{24}{12}\)
\(\Leftrightarrow\sqrt{x-1}=2\)
\(\Leftrightarrow x-1=4\)
\(\Leftrightarrow x=4+1\)
\(\Leftrightarrow x=5\left(tm\right)\)
b) \(\dfrac{1}{2}\sqrt{4x+8}-2\sqrt{x+2}-\dfrac{3}{7}\sqrt{49x+98}=-8\) (ĐK: \(x\ge-2\))
\(\Leftrightarrow\dfrac{1}{2}\cdot2\sqrt{x+2}-2\sqrt{x+2}-\dfrac{3}{7}\cdot7\sqrt{x+2}=-8\)
\(\Leftrightarrow\sqrt{x+2}-2\sqrt{x+2}-3\sqrt{x+2}=-8\)
\(\Leftrightarrow-4\sqrt{x+2}=-8\)
\(\Leftrightarrow\sqrt{x+2}=\dfrac{-8}{-4}\)
\(\Leftrightarrow\sqrt{x+2}=2\)
\(\Leftrightarrow x+2=4\)
\(\Leftrightarrow x=4-2\)
\(\Leftrightarrow x=2\left(tm\right)\)
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
a) \(15\sqrt{\dfrac{4}{3}}-5\sqrt{48}+2\sqrt{12}-6\sqrt{\dfrac{1}{3}}\)
\(=\sqrt{15^2\cdot\dfrac{4}{3}}-5\cdot4\sqrt{3}+2\cdot2\sqrt{3}-\sqrt{6^2\cdot\dfrac{1}{3}}\)
\(=\sqrt{\dfrac{225\cdot4}{3}}-20\sqrt{3}+4\sqrt{3}-\sqrt{\dfrac{36}{3}}\)
\(=\sqrt{75\cdot4}-16\sqrt{3}-\sqrt{12}\)
\(=10\sqrt{3}-16\sqrt{3}-2\sqrt{3}\)
\(=-8\sqrt{3}\)
b) \(\dfrac{15}{\sqrt{6}+1}-\dfrac{3}{\sqrt{7}-\sqrt{2}}-15\sqrt{6}+3\sqrt{7}\)
\(=\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}-\dfrac{3\left(\sqrt{7}+\sqrt{2}\right)}{\left(\sqrt{7}-\sqrt{2}\right)\left(\sqrt{7}+\sqrt{2}\right)}-15\sqrt{6}+3\sqrt{7}\)
\(=\dfrac{15\left(\sqrt{6}-1\right)}{6-1}-\dfrac{3\sqrt{7}+3\sqrt{2}}{7-2}-15\sqrt{6}+3\sqrt{7}\)
\(=3\left(\sqrt{6}-1\right)-\dfrac{3\sqrt{7}+3\sqrt{2}}{5}-15\sqrt{6}+3\sqrt{7}\)
\(=3\sqrt{6}-3-\dfrac{3\sqrt{7}+3\sqrt{2}}{5}-15\sqrt{6}+3\sqrt{7}\)
\(=-12\sqrt{6}-3+3\sqrt{7}-\dfrac{3\sqrt{7}+3\sqrt{2}}{5}\)
\(=\dfrac{-60\sqrt{6}-15+15\sqrt{7}-3\sqrt{7}-3\sqrt{2}}{5}\)
\(=\dfrac{-60\sqrt{6}-15+12\sqrt{7}-3\sqrt{2}}{5}\)
`sqrt{5+2sqrt6}`
`=sqrt{3+2sqrt3sqrt2+2}`
`=sqrt{(sqrt3+sqrt2)^2}`
`=|sqrt3+sqrt2|=sqrt3+sqrt2`
`7. sqrt(4+2sqrt3)`
`=sqrt{3+2sqrt3+1}`
`=sqrt{(sqrt3+1)^2}`
`=sqrt3+1`
`8. sqrt(4-2sqrt3)`
`=sqrt{3-2sqrt3+1}`
`=sqrt{(sqrt3-1)^2}`
`=sqrt3-1`
`9. sqrt(11-2sqrt(30))`
`=sqrt{6-2sqrt5sqrt6+5}`
`=sqrt{(sqrt6-sqrt5)^2}`
`=sqrt6-sqrt5`
`10. sqrt(21-4sqrt(17))`
`=sqrt{17-2.2.sqrt{17}+4}`
`=sqrt{(sqrt{17}-2)^2}`
`=sqrt{17}-2`
Bài 1:
1) Ta có: \(3-2\sqrt{2}\)
\(=2-2\cdot\sqrt{2}\cdot1+1\)
\(=\left(\sqrt{2}-1\right)^2\)
2) Ta có: \(8+2\sqrt{7}\)
\(=7+2\cdot\sqrt{7}\cdot1+1\)
\(=\left(\sqrt{7}+1\right)^2\)
3) Ta có: \(x-2\sqrt{x-1}\)
\(=x-1-2\cdot\sqrt{x-1}\cdot1+1\)
\(=\left(\sqrt{x-1}-1\right)^2\)
4) Ta có: \(6-4\sqrt{2}\)
\(=4-2\cdot2\cdot\sqrt{2}+2\)
\(=\left(2-\sqrt{2}\right)^2\)
5) Ta có: \(7+4\sqrt{3}\)
\(=4+2\cdot2\cdot\sqrt{3}+3\)
\(=\left(2+\sqrt{3}\right)^2\)
6) Ta có: \(9-4\sqrt{5}\)
\(=5-2\cdot\sqrt{5}\cdot2+4\)
\(=\left(\sqrt{5}-2\right)^2\)
7) Ta có: \(10+2\sqrt{21}\)
\(=7+2\cdot\sqrt{7}\cdot\sqrt{3}+3\)
\(=\left(\sqrt{7}+\sqrt{3}\right)^2\)
8) Ta có: \(49+20\sqrt{6}\)
\(=25+2\cdot5\cdot2\sqrt{6}+24\)
\(=\left(5+2\sqrt{6}\right)^2\)