-1\3+5\2
a,1/3 .(x-2/5)=3/4 b, 7/3:(x-2/3)=4/5 c,1/3.(x-2/5)=4/5 d, 2/3.(x-1/2)-1/4.(x-2/5)=7/3 e,3/7 .(x-2/3)+1/2=5/4.(x-2) f,1/2.(x-3)+1/3.(x-4)+1/4.(x-5)=1/5 g,[2/3.(x-1/2)-4/5]:(x-1/3)=21/5 h, {x-[1/2.(x-3)+11/5]}:(x-1/2)=3/5 i,x.(x-2/5)-(x+2).x+11/4=4/3
a: =>x-2/5=3/4:1/3=3/4*3=9/4
=>x=9/4+2/5=45/20+8/20=53/20
b: =>x-2/3=7/3:4/5=7/3*5/4=35/12
=>x=35/12+2/3=43/12
c: 1/3(x-2/5)=4/5
=>x-2/5=4/5*3=12/5
=>x=12/5+2/5=14/5
d: =>2/3x-1/3-1/4x+1/10=7/3
=>5/12x-7/30=7/3
=>5/12x=7/3+7/30=77/30
=>x=77/30:5/12=154/25
e: \(\Leftrightarrow x\cdot\dfrac{3}{7}-\dfrac{2}{7}+\dfrac{1}{2}-\dfrac{5}{4}x+\dfrac{5}{2}=0\)
=>\(x\cdot\dfrac{-23}{28}=\dfrac{2}{7}-3=\dfrac{-19}{7}\)
=>x=19/7:23/28=76/23
f: =>1/2x-3/2+1/3x-4/3+1/4x-5/4=1/5
=>13/12x=1/5+3/2+4/3+5/4=257/60
=>x=257/65
i: =>x^2-2/5x-x^2-2x+11/4=4/3
=>-12/5x=4/3-11/4=-17/12
=>x=17/12:12/5=85/144
Rút gọn : a . P = 3+2√3 / √3 + 2+√2 / √2+1 - ( √2 + √3 ) ; b. N = ( 1 - 5 + √5 / 1 + √5 ) ( 5 - √5 / 1- √5 - 1 ) ; c. Q = ( 5 - 2√5 / 2 - √5 - 2 ) ( 3+3 √5 / 3 + √5 - 2 ). Giúp mik vs ạ
a: \(P=\dfrac{\sqrt{3}\left(2+\sqrt{3}\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}-1\right)}{1}-\sqrt{3}-\sqrt{2}\)
\(=2+\sqrt{3}+2-\sqrt{2}-\sqrt{3}-\sqrt{2}\)
\(=4-2\sqrt{2}\)
b: \(N=\left(1-\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)\left(\dfrac{-\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}-1\right)\)
\(=\left(1-\sqrt{5}\right)\left(-\sqrt{5}-1\right)\)
\(=\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)=5-1=4\)
a) \(P=\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+\sqrt{3}\right)\)
\(P=\dfrac{\sqrt{3}\cdot\left(\sqrt{3}+2\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\cdot\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\sqrt{2}-\sqrt{3}\)
\(P=\sqrt{3}+2+\sqrt{2}-\sqrt{2}-\sqrt{3}\)
\(P=2\)
b) \(N=\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
\(N=\left[1-\dfrac{\sqrt{5}\left(1+\sqrt{5}\right)}{1+\sqrt{5}}\right]\left[1+\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right]\)
\(N=\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)\)
\(N=1^2-\left(\sqrt{5}\right)^2\)
\(N=-4\)
c) \(Q=\left(\dfrac{5+2\sqrt{5}}{2-\sqrt{5}}-2\right)\left(\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}-2\right)\)
\(Q=\left[\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}+2\right]\left[\dfrac{\sqrt{5}\left(\sqrt{5}+3\right)}{\sqrt{5}+3}-2\right]\)
\(Q=\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)\)
\(Q=\left(\sqrt{5}\right)^2-2^2\)
\(Q=1\)
bài 1:
1/4 + 2/3 2/7 + 2/3 2/5 + 1/3 2/3 + 1/2 1/3 + 3/5 4/5 + 1/3
1/8 + 3/4 1/36 + 5/12 1/3 + 1/6 + 1/18.
bài 2:
15/16 - 3/16 17/18 - 5/6 3/4 - 4/9 1/2 - 2/5 5/6 - 3/10 3-1/3
4/5 - 1/10 5/2 - 1 5/8 - 2/5.
Tính:
5 – 1 = … | 4 – 1 = … | 3 – 1 = … | 2 + 3 = …. |
5 – 2 = … | 4 – 2 = … | 3 – 2 = … | 3 + 2 = …. |
5 – 3 = …. | 4 – 3 = … | 2 – 1 = … | 5 – 2 = …. |
5 – 4 = …. | 5 – 3 = …. |
Lời giải chi tiết:
5 – 1 = 4 | 4 – 1 = 3 | 3 – 1 = 2 | 2 + 3 = 5 |
5 – 2 = 3 | 4 – 2 = 2 | 3 – 2 = 1 | 3 + 2 = 5 |
5 – 3 = 2 | 4 – 3 = 1 | 2 – 1 = 1 | 5 – 2 = 3 |
5 – 4 = 1 | 5 – 3 = 2 |
5-1=4 4-1=3 3-1=2 2+3=5
5-2=3 4-2=2 3-2=1 3+2=5
5-3=2 4-3=1 2-1=1 5-2=3
5-4=1 5-3=2
Tính:
5 – 1 = 4 | 4 – 1 = 3 | 3 – 1 = 2 | 2 + 3 = 5 |
5 – 2 = 3 | 4 – 2 = 2 | 3 – 2 = 1 | 3 + 2 = 5 |
5 – 3 = 2 | 4 – 3 = 1 | 2 – 1 = 1 | 5 – 2 = 3 |
5 – 4 = 1 | 5 – 3 = 2 |
Bài 1
a, Tính P=1+1/2(1+2)+1/3(1+2+3)+1/4(1+2+3+4)+....+1/2012(1+2+3+...+2012)
b,Tìm x thỏa mãn 4^5+4^5+4^5+4^5/3^5+3^5+3^5.6^5+6^5+6^5+6^5+6^5+6^5/2^5+2^5=2^x
Kho..................wa.....................troi.....................thi......................lanh.................ret.......................ai........................tich..........................ung.....................ho........................minh.....................cho....................do....................lanh
Kho..................wa.....................troi.....................thi......................lanh.................ret.......................ai........................tich..........................ung.....................ho........................minh.....................cho....................do....................lanh
>, <, = ?
4…5 | 1…4 | 2…3 | 1…1 |
2…2 | 5…2 | 2…4 | 5…1 |
3…1 | 3…3 | 2…5 | 3…5 |
Lời giải chi tiết:
4 < 5 | 1 < 4 | 2 < 3 | 1 = 1 |
2 = 2 | 5 > 2 | 2 < 4 | 5 > 1 |
3 > 1 | 3 = 3 | 2 < 5 | 3 < 5 |
a)(5+1/5-2/9)-(2-1/23-3/35+5/6)-(8+2/7-1/18)
b) 1/3-3/4(-3/5+1/64- -2/9-1/36+1/15
c) -5/7-(-5/67)+13/10+1/2+(-1/6)+1 3/14-(-2/5)
d)3/5:(-1/15-1/6)+3/5:(-1/3-1 1/15)
a,1/1*2+1/2*3+1/3*4+1/4*5+......+1/9*10
b,2/1*3+2/3*5+2/5*7+2/7*9+2/9*11
c,3/1*3+3/3*5+3/5*7+3/7*9+3/9*11
d,5/1*3+5/3*5+5/5*7+5/7*9+5/9*11
Giúp mình với
a)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{3}{4}+...+\frac{1}{9}-\frac{1}{10}\)
= \(1+\left(\frac{-1}{2}+\frac{1}{2}\right)+\left(\frac{-1}{3}+\frac{1}{3}\right)+...+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{10}\)
= \(1-\frac{1}{10}\)
=\(\frac{9}{10}\)
b)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
=\(1+\left(\frac{-1}{3}+\frac{1}{3}\right)+\left(\frac{-1}{5}+\frac{1}{5}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{11}\)
=\(1-\frac{1}{11}\)
= \(\frac{10}{11}\)
c) đặt A=\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}\)
\(\frac{1}{3}A\)=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(\frac{2}{3}A\)=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
\(\frac{2}{3}A\)=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(\frac{2}{3}A\)=\(1+\left(\frac{-1}{3}+\frac{1}{3}\right)+\left(\frac{-1}{5}+\frac{1}{5}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{11}\)
\(\frac{2}{3}A\)=\(\frac{10}{11}\)
A= \(\frac{10}{11}:\frac{2}{3}\)
A= \(\frac{10}{11}.\frac{3}{2}\)=\(\frac{15}{11}\)
d) giả tương tự câu c kết quả \(\frac{25}{11}\)
tổng đặc biệt đó bạn
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(1-\frac{1}{10}=\frac{9}{10}\)
những câu sau cũng áp dụng như vậy nhé
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{9}-\frac{1}{11}\)
\(=1-\frac{1}{11}=\frac{10}{11}\)
\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}\)
\(=\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{3}{2}.\left(1-\frac{1}{11}\right)=\frac{3}{2}.\frac{10}{11}=\frac{15}{11}\)
Trục căn thức ở mẫu.
1) 5/√5 ; 3/2√3 ; 5/√7 ; 2√3/5√7 ; 5/2√3
2) 1/√3 ; 2/√3 + 1 ; 3/√5 - 1 ; 12/√5 - √3 ; 4√3 - 2/7 × √2
1)
\(\dfrac{5}{\sqrt{5}}=\dfrac{5\sqrt{5}}{5}\sqrt{5}\)
\(\dfrac{3}{2\sqrt{3}}=\dfrac{3\sqrt{3}}{2\sqrt{3}}=\sqrt{\dfrac{3}{2}}\)
\(\dfrac{5}{\sqrt{7}}=\dfrac{5\sqrt{7}}{\sqrt{49}}=\left(\dfrac{5}{7}\right)\sqrt{7}\)
1:
\(\dfrac{2\sqrt{3}}{5\sqrt{7}}=\dfrac{2\sqrt{21}}{35}\)
\(\dfrac{5}{2\sqrt{3}}=\dfrac{5\sqrt{3}}{6}\)
2: \(\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{3}\)
\(\dfrac{2}{\sqrt{3}+1}=\sqrt{3}-1\)
\(\dfrac{3}{\sqrt{5}-1}=\dfrac{3+3\sqrt{5}}{4}\)
\(\dfrac{12}{\sqrt{5}-\sqrt{3}}=6\left(\sqrt{5}+\sqrt{3}\right)=6\sqrt{5}+6\sqrt{3}\)
Các phân số 1/3: 1/6: 5/2: 3/2 xếp theo thứ tự tăng dần
a. 1/3; 1/6; 5/2; 3/2 b. 1/3; 5/2; 1/6; 3/2 c. 1/6; 1/3; 3/2; 5/2 d. 1/6; 3/2; 3/2; 5/2