bai 3 . Rút Gọn
a=4^2x25^2x125
__________________
3^2x5^2
Rút gọn
A = \(\sqrt{\dfrac{3\sqrt{3}-4}{2\sqrt{3}+1}}+\sqrt{\dfrac{\sqrt{3}+4}{5-2\sqrt{3}}}\)
bài 4: rút gọn
A= 1+5+5^2+5^3+5^4 + ........ +5^99 + 5^100
B= 1-5+5^2-5^3 + ...... - 3^99 + 5^100
bài 1: rút gọn bthuc
a.\(\dfrac{a+\sqrt{a}}{\sqrt{a}}\) b.\(\dfrac{\sqrt{\left(x-3\right)^2}}{3-x}\)
b2: rút gọn
a.\(\dfrac{\sqrt{9x^2-6x+1}}{9x^2-1}\) b.4-x-\(\sqrt{4-4x+x^2}\) c.\(\sqrt{4x^2-4x\text{x^2 +2*x-3 >0}}-\sqrt{4x^2+4x+1}\)
Bài 1:
a) \(\dfrac{a+\sqrt{a}}{\sqrt{a}}=\sqrt{a}+1\)
b) \(\dfrac{\sqrt{\left(x-3\right)^2}}{3-x}=\dfrac{\left|x-3\right|}{3-x}=\pm1\)
Bài 2:
a) \(\dfrac{\sqrt{9x^2-6x+1}}{9x^2-1}=\dfrac{\left|3x-1\right|}{\left(3x-1\right)\left(3x+1\right)}=\pm\dfrac{1}{3x+1}\)
b) \(4-x-\sqrt{x^2-4x+4}=4-x-\left|x-2\right|=\left[{}\begin{matrix}6-2x\left(x\ge2\right)\\2\left(x< 2\right)\end{matrix}\right.\)
tính rồi rút gọn
a)2-1/4
b)7/9-4/9
c)15/21-4/7
a: =8/4-1/4=7/4
b: =3/9=1/3
c: =5/7-4/7=1/7
a,\(\dfrac{7}{4}\)
b,\(\dfrac{1}{3}\)
c,\(\dfrac{1}{7}\)
có ai biết giải bài này k hộ mình vs ( chi tiết hộ mình nhé )
bài 1: trục căn thức ở mẫu và rút gọn
a, \(\dfrac{1}{2\sqrt{2}-3\sqrt{3}}\)
b, \(\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}\)
bài 2: trục căn thức ở mẫu và rút gọn
a, \(\dfrac{\sqrt{8}}{\sqrt{5}-\sqrt{3}}\)
b, \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)
bài 3: trục căn thức và thực hiện phép tính
a, M=\(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right).\left(\sqrt{6}+11\right)\)
b, N= \(\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right).\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
Bài 1:
a.
\(\frac{1}{2\sqrt{2}-3\sqrt{3}}=\frac{2\sqrt{2}+3\sqrt{3}}{(2\sqrt{2}-3\sqrt{3})(2\sqrt{2}+3\sqrt{3})}=\frac{2\sqrt{2}+3\sqrt{3}}{(2\sqrt{2})^2-(3\sqrt{3})^2}=\frac{2\sqrt{2}+3\sqrt{3}}{-19}\)
b.
\(=\sqrt{\frac{(3-\sqrt{5})^2}{(3-\sqrt{5})(3+\sqrt{5})}}=\sqrt{\frac{(3-\sqrt{5})^2}{3^2-5}}=\sqrt{\frac{(3-\sqrt{5})^2}{4}}=\sqrt{(\frac{3-\sqrt{5}}{2})^2}=|\frac{3-\sqrt{5}}{2}|=\frac{3-\sqrt{5}}{2}\)
Bài 2.
a.
\(=\frac{\sqrt{8}(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}=\frac{2\sqrt{2}(\sqrt{5}+\sqrt{3})}{5-3}=\sqrt{2}(\sqrt{5}+\sqrt{3})=\sqrt{10}+\sqrt{6}\)
b.
\(=\sqrt{\frac{(2-\sqrt{3})^2}{(2-\sqrt{3})(2+\sqrt{3})}}=\sqrt{\frac{(2-\sqrt{3})^2}{2^2-3}}=\sqrt{(2-\sqrt{3})^2}=|2-\sqrt{3}|=2-\sqrt{3}\)
Bài 3:
a.
\(M=\left[\frac{15(\sqrt{6}-1)}{(\sqrt{6}+1)(\sqrt{6}-1)}+\frac{4(\sqrt{6}+2)}{(\sqrt{6}-2)(\sqrt{6}+2)}-\frac{12(3+\sqrt{6})}{(3-\sqrt{6})(3+\sqrt{6})}\right](\sqrt{6}+11)\)
\(=\left[\frac{15(\sqrt{6}-1)}{6-1}+\frac{4(\sqrt{6}+2)}{6-2^2}-\frac{12(3+\sqrt{6})}{3^2-6}\right](\sqrt{6}+11)\)
\(=[3(\sqrt{6}-1)+2(\sqrt{6}+2)-4(3+\sqrt{6})](\sqrt{6}+11)=(\sqrt{6}-11)(\sqrt{6}+11)=6-11^2=-115\)
b.
\(N=\left[1-\frac{\sqrt{5}(\sqrt{5}+1)}{\sqrt{5}+1}\right].\left[\frac{\sqrt{5}(\sqrt{5}-1)}{1-\sqrt{5}}-1\right]\)
\(=(1-\sqrt{5})(-\sqrt{5}-1)=(\sqrt{5}-1)(\sqrt{5}+1)=5-1=4\)
Rút gọn
a) \(\dfrac{x^5-2x^4+2x^3-4x^2-3x+6}{x+4}\)
b) \(\dfrac{x^4-4x^2+3}{x^4+6x^2-7}\)
c) \(\dfrac{x^4+x^3-x-1}{x^4+x^3+2x^2+x+1}\)
\(a,=\dfrac{x^4\left(x-2\right)+2x^2\left(x-2\right)-3\left(x-2\right)}{x+4}\\ =\dfrac{\left(x-2\right)\left(x^4+2x^2-3\right)}{x+4}\\ =\dfrac{\left(x-2\right)\left(x^4-x^2+3x^2-3\right)}{x+4}\\ =\dfrac{\left(x-2\right)\left(x-1\right)\left(x^2+3\right)}{x+4}\)
\(b,=\dfrac{x^4-3x^2-x^2+3}{x^4-x^2+7x^2-7}=\dfrac{\left(x^2-3\right)\left(x^2-1\right)}{\left(x^2+7\right)\left(x^2-1\right)}=\dfrac{x^2-3}{x^2+7}\\ c,=\dfrac{\left(x^3-1\right)\left(x+1\right)}{x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)}\\ =\dfrac{\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)}{\left(x^2+1\right)\left(x^2+x+1\right)}=\dfrac{x^2-1}{x^2+1}\)
1.Rút gọn
a) C = 1/2 + 1/2^2 + 1/2^3 + ....+ 1/2^2020
\(C=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}\\ 2C=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2019}}\\ 2C-C=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2019}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}\right)\\ C=1-\dfrac{1}{2^{2020}}=\dfrac{2^{2020}-1}{2^{2020}}\)
1.Rút gọn
a) C = 1/2 + 1/2^2 + 1/2^3 + ....+ 1/2^2020
Giải:
C=1/2 + 1/2^2 + 1/2^3 + ... + 1/2^2020
2C=1 + 1/2 + 1/2^2 + ... +1/2^2019
2C-C=(1+1/2+1/2^2+...+1/2^2019)-(1/2+1/2^2+1/2^3+...+1/2^2020)
C=1-1/2^2020
Chúc bạn học tốt!
Rút gọn
a, √21+3√48 - √21-3√48
b, √-2√10 - √7 + 2√10
c.ơn các bạn trước :3
`a)Đặt \, A=sqrt{21+3sqrt{48}}-sqrt{21-3sqrt{48}}`
Vì `21+3sqrt{48}>21-3sqrt{48}`
`=>sqrt{21+3sqrt{48}}-sqrt{21-3sqrt{48}}>0`
Hay `A>0`
`<=>A^2=21+3sqrt{48}+21-3sqrt{48}-2sqrt{21^2-9.48}`
`<=>A^2=42-2sqrt{9}=32-2.3=26`
`<=>A=sqrt{26}(do \ A>0)`
b)Chắc đề là như này:
`sqrt{7-2sqrt{10}}-sqrt{7+2sqrt{10}}`
`=sqrt{5-2sqrt{5}.sqrt2+2}-sqrt{5+2sqrt{5}.sqrt2+2}`
`=sqrt{(sqrt5-sqrt2)^2}-sqrt{(sqrt5+sqrt2)^2}`
`=sqrt5-sqrt2-sqrt5-sqrt2=-2sqrt2`
a) Ta có: \(\sqrt{21+3\sqrt{48}}-\sqrt{21-3\sqrt{48}}\)
\(=\sqrt{12+2\cdot2\sqrt{3}\cdot3+9}-\sqrt{12-2\cdot2\sqrt{3}\cdot3+9}\)
\(=2\sqrt{3}+3-2\sqrt{3}+3\)
=6
b) Ta có: \(\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)
\(=\sqrt{5}-\sqrt{2}-\sqrt{5}-\sqrt{2}\)
\(=2\sqrt{2}\)