Bài 1: viết thành căn bậc hai
1) 3-2√2
2) 8+2√7
3) x-2√x-1
4) 6-4√2
5) 7+4√3
6) 9-4√5
7) 10+2√21
8) 49+20√6
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)
\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)
\(=\dfrac{-1621}{126}\)
b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)
\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)
\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)
\(=-\dfrac{49}{20}\)
`4/7+4`
`=4/7+4/1`
`=4/7+28/7`
`=32/7`
__
`3+6/11`
`=33/11+6/11`
`=39/11`
__
`3-5/7`
`=3/1-5/7`
`=21/7-5/7`
`=16/7`
__
`21/9-2`
`=21/9-18/9`
`=3/9`
`=1/3`
__
`15/24+2`
`=15/24+48/24`
`=63/24`
`=21/16`
__
`63/45-20/25`
`=63/45-4/5`
`=63/45-36/45`
`=27/45`
`=9/15`
__
`3/4-2/8`
`=3/4-1/4`
`=2/4`
__
`6/7-5/8`
`=48/56-35/56`
`=13/56`
__
`37/45-5/9`
`=37/45-25/45`
`=12/45`
`=4/15`
__
`46/39-11/13`
`=46/39-33/39`
`=13/39`
`=1/2`
__
`5/12+3/4+1/3`
`=5/12+9/12+4/12`
`=14/12+4/12`
`=18/12`
`=3/2`
__
`1/2+3/7+11/14`
`=7/14+6/14+11/14`
`=13/14+11/14`
`=24/14`
`=12/7`
__
`7/10-(1/5+1/4)`
`=7/10-(4/20+5/20)`
`=7/10-9/20`
`=14/20-9/20`
`=5/20`
`=1/4`
__
`15/4-2/3-3/4`
`=(15/4-3/4)-2/3`
`=12/4-2/3`
`=3-2/3`
`=9/3-2/3`
`=7/3`
1) 1/3 x 1/2 x 3/7 = 3/42 = 1/14
2) 5/4 x 1/3 +1/7 = 5/12 + 1/7 = 35/84 + 12/84 = 47/84
3) 8 x ( 8/9 - 2/3 ) = 8 x 2/9 = 16/9
4) 5/6 x 48/20 x 1/2 = 240/240 = 1
5) ( 2/5 + 3/4 ) + 8 = 23/20 + 8 = 23//20 + 160/20 = 183/20
6) 10 x ( 1/2 - 1/5 ) = 10 x 3/10 = 10/1 x 3/10 = 30/10 = 3
2/3 + 3/4=17/12
9/4 + 3/5=57/20
5/24 + 1/4=11/24
3/15 - 5/35=2/35 18/27 - 2/6=1/3 37/12 - 3=1/12
11/9 x 3/22=1/6 7/13 x 13/7=1 4 x6/7=24/7
2/5 : 3/10=4/3 3/8 : 9/4=1/6 8/21 : 4/7=2/3
Lời giải chi tiết:
2 = 1 + 1 |
6 = 2 + 4 |
8 = 5 + 3 |
10 = 8 + 2 |
3 = 1 + 2 |
6 = 3 + 3 |
8 = 4 + 4 |
10 = 7 + 3 |
4 = 3 + 1 |
7 = 6 + 1 |
9 = 8 + 1 |
10 = 6 + 4 |
4 = 2 + 2 |
7 = 5 + 2 |
9 = 7 + 2 |
10 = 5 + 5 |
5 = 4 + 1 |
7 = 4 + 3 |
9 = 6 + 3 |
10 = 10 + 0 |
5 = 3 + 2 |
8 = 7 + 1 |
9 = 5+ 4 |
10 = 0 + 10 |
6 = 5 + 1 |
8 = 6 + 2 |
10 = 9 + 1 |
1 = 0 + 1 |
2=1+1 6=2+4 8=5+3 10=8+2
3=1+2 6=3+3 8=4+4 10=7+3
4=3+1 7=6+1 9=8+1 10=6+4
4=2+2 7=5+2 9=7=2 10=5+5
5=4+1 7=4+3 9=6+3 10=10+0
5=3+2 8=7+1 9=5=4 10=0+10
6=5+1 8=6=2 10=9+1 1=0+1
1) 1/3 x 1/2 x 3/7 = 1/6 x 3/7 = 1/14
2) 5/4 x 1/3 + 1/7 = 5/12 + 1/7 = 47/84
3) 8 x (8/9 - 2/3) = 8 x 2/9 = 16/9
4) 5/6 x 48/20 x 1/2 = 2 x 1/2 = 1
5) (2/5 + 3/4) x 8 = 23/20 x 8 = 46/5
6) 10 x (1/2 - 1/5) = 10 x 3/10 = 3
có rút gọn phân số 35/6 về phân số đc ko ? Nếu có thì mn rút gọn cho mik nhé !
Ai xong tr tui k cho
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
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: \(49+20\sqrt{6}\)
\(=25+2\cdot5\cdot2\sqrt{6}+24\)
\(=\left(5+2\sqrt{6}\right)^2\)