Rút gọn các phân số sau:
a) 2 3 .3 4 2 2 .3 2 .5
b) 2 4 .5 2 .11 2 .7 2 3 .5 3 .7 2 .11
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30 tháng 4 2020
a,\(\frac{2^3.3^4}{2^2.3^2.5}=\frac{2.3^2}{5}=\frac{2.9}{5}=\frac{18}{5}\)
b, \(\frac{2^4.5^2.11^2.7}{2^3.5^3.7^2.11}=\frac{2.11}{5.7}=\frac{22}{35}\)
c,\(\frac{1998.1990+3978}{1992.1991-3984}=\frac{18.111.1990+18.221}{24.83.1991-166.24}\)
\(=\frac{18.\left(111.1990+221\right)}{24.\left(83.1991-166\right)}\)
\(=\frac{3.221111}{4.165087}=\frac{221111}{4.55029}=\frac{221111}{220116}\)
\(\)
HH
2
31 tháng 5 2018
a) \(\frac{2^3.3^4}{2^2.3^2.5}=\frac{2.3^2}{5}\)
b) \(\frac{2^4.5^2.11^2.7}{2^3.5^3.7^2.11}=\frac{2.11}{5.7}\)
c) \(\frac{121.75.130.69}{39.60.11.198}=\frac{11^2.3.5^2.2.5.13.23.3}{13.3.2^3.3.5.11.11.3^2.2}\)
\(=\frac{11^2.3^2.5^3.2.13.23}{13.3^4.2^4.5.11^2}=\frac{5^2.23}{3^2.2^3}\)
31 tháng 5 2018
a,\(\frac{2^3.3^4}{2^2.3^2.5}=\frac{2^2.2.3^2.3^2}{2^2.3^2.5}\)\(=\frac{2.3^2}{5}\)
b,\(\frac{2^4.5^2.11^2.7}{2^3.5^3.7^2.11}=\frac{2^3.2.5^2.11.11.7}{2^3.5^2.5.7.7.11}=\frac{2.11}{5.7}\)
NM
3
27 tháng 8 2021
`a)(2x-1)^2+(x+3)^2-5(x-7)(x+7)`
`=4x^2-4x+1+x^2+6x+9-5(x^2-49)`
`=5x^2-5x^2-4x+6x+1+9+245`
`=2x+255`
`b)(x-2)(x^2+2x+4)-(25+x^3)`
`=x^3-8-x^3-25=-33`
AH
Akai Haruma
Giáo viên
27 tháng 8 2021
Lời giải:
a.
$(2x-1)^2+(x+3)^2-5(x-7)(x+7)$
$=4x^2-4x+1+(x^2+6x+9)-5(x^2-49)$
$=5x^2+2x+10-(5x^2-245)=2x+255$
b.
$(x-2)(x^2+2x+4)-(25+x^3)=(x^3-2^3)-(25+x^3)$
$=-8-25=-33$
26 tháng 5 2022
a: 1/4=2/8
2/5=2/5
b: 2/3=14/21
7/8=14/16
c: 3/4=15/20
5/6=15/18
d: 1/3=7/21
7/9=7/9
e: 3/4=9/12
9/24=9/24
f: 7/10=133/190
19/30=133/210
26 tháng 5 2022
a) 5/20 và 8/20
b) 14/24 và 21/24
c) 18/24 và 20/24
d) 3/9 và 7/9
e) 18/24 và 9/24
f) 21/30 và 19/3
24 tháng 6 2021
a)\(A=\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)
\(=\sqrt[3]{1+3\sqrt{2}+3\sqrt{2^2}+2\sqrt{2}}-\sqrt[3]{2\sqrt{2}-3\sqrt{2^2}+3\sqrt{2}-1}\)
\(=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\sqrt[.3]{\left(\sqrt{2}-1\right)^3}\)
\(=1+\sqrt{2}-\left(\sqrt{2}-1\right)=2\)
b)\(B=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)
\(\Leftrightarrow B^3=5+2\sqrt{13}+3\sqrt[3]{\left(5+2\sqrt{13}\right)\left(5-2\sqrt{13}\right)}\left(\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5+2\sqrt{13}}\right)+5-2\sqrt{13}\)
\(\Leftrightarrow B^3=10+3.\sqrt[3]{-27}.B\)
\(\Leftrightarrow B^3+9B-10=0\)
\(\Leftrightarrow\left(B-1\right)\left(B^2+B+10\right)=0\)
\(\Leftrightarrow B=1\) (vì \(B^2+B+10>0\))
c)\(C=\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}\)
\(\Leftrightarrow2C=\sqrt[3]{8\sqrt{5}+16}-\sqrt[3]{8\sqrt{5}-16}=\sqrt[3]{1+3\sqrt{5}+3\sqrt{5^2}+5\sqrt{5}}-\sqrt[3]{5\sqrt{5}-3\sqrt{5^2}+3\sqrt{5}-1}\)
\(=\sqrt[3]{\left(1+\sqrt{5}\right)^3}-\sqrt[3]{\left(\sqrt{5}-1\right)^3}\)
\(=1+\sqrt{5}-\left(\sqrt{5}-1\right)\)
\(\Rightarrow C=1\)
d) \(D=\dfrac{10}{\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}}\left(\dfrac{1+\sqrt{2}}{\sqrt{4-2\sqrt{3}}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\right)\)
\(=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{\left(\sqrt[3]{3}+\sqrt[3]{2}\right)\left(\sqrt[3]{9^2}-\sqrt[3]{6}+\sqrt[3]{2^2}\right)}\left(\dfrac{1+\sqrt{2}}{\sqrt{\left(1-\sqrt{3}\right)^2}}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\right)\)
\(=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{5}.\dfrac{1+\sqrt{2}}{\left|1-\sqrt{3}\right|}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)
\(=2\left(\sqrt[3]{3}+\sqrt[3]{2}\right).\dfrac{\left(1+\sqrt{2}\right)\left(\sqrt{2}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=2\left(\sqrt[3]{3}+\sqrt[3]{2}\right).\dfrac{\left(\sqrt{2}\right)^2-1}{\left(\sqrt{3}\right)^2-1}\)
\(=\sqrt[3]{3}+\sqrt[3]{2}\)
Vậy...
19 tháng 8 2021
a: Ta có: \(\dfrac{8}{\left(\sqrt{5}+\sqrt{3}\right)^2}-\dfrac{8}{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\dfrac{8}{8+2\sqrt{15}}-\dfrac{8}{8-2\sqrt{15}}\)
\(=\dfrac{64-16\sqrt{15}-64-16\sqrt{15}}{4}\)
\(=\dfrac{-32\sqrt{15}}{4}=-8\sqrt{15}\)
b: Ta có: \(\dfrac{1}{4-3\sqrt{2}}-\dfrac{1}{4+3\sqrt{2}}\)
\(=\dfrac{4+3\sqrt{2}-4+3\sqrt{2}}{-2}\)
\(=-\dfrac{6\sqrt{2}}{2}=-3\sqrt{2}\)
19 tháng 8 2021
b) \(\dfrac{1}{4-3\sqrt{2}}-\dfrac{1}{4+3\sqrt{2}}=\dfrac{4+3\sqrt{2}-4+3\sqrt{2}}{\left(4-3\sqrt{2}\right)\left(4+3\sqrt{2}\right)}=\dfrac{6\sqrt{2}}{-2}=-3\sqrt{2}\)
c) \(\left(\dfrac{\sqrt{7}+3}{\sqrt{7}-3}-\dfrac{\sqrt{7}-3}{\sqrt{7}+3}\right):\sqrt{28}=\dfrac{\left(\sqrt{7}+3\right)^2-\left(\sqrt{7}-3\right)^2}{\left(\sqrt{7}-3\right)\left(\sqrt{7}+3\right)}:\sqrt{28}=\dfrac{16+6\sqrt{7}-16+6\sqrt{7}}{7-9}=\dfrac{12\sqrt{7}}{-2}=-6\sqrt{7}\)
12 tháng 2 2018
a) \(\frac{2^3.3^4}{2^2.3^2.5}=\frac{2.3^2}{1.1.5}=\frac{18}{5}\)
b) \(\frac{2^4.5^2.11^2.7}{2^3.5^3.7^2.11}=\frac{2.1.11.1}{1.5.7.1}=\frac{22}{35}\)
5 tháng 2 2022
a: \(=6+2\sqrt{11}-4+\sqrt{11}=2+3\sqrt{11}\)
b: \(=\dfrac{3x+9\sqrt{x}-2x+4\sqrt{x}}{\left(\sqrt{x}+3\right)\left(x-2\sqrt{x}\right)}\cdot\dfrac{\left(\sqrt{x}+3\right)^2}{\sqrt{x}+13}=\dfrac{\sqrt{x}+3}{x-2\sqrt{x}}\)
a) 2 3 .3 4 2 2 .3 2 .5 = 2 3 − 2 .3 4 − 2 5 = 18 5
b) 2 4 .5 2 .11 2 .7 2 3 .5 3 .7 2 .11 = 22 35