Rút gon biểu thức :` 7^3 + 8^3 + ... + 2010^3`
rút gon biểu thức :
(2x - 3)(x + 5) - x(2x + 7)
\(\left(2x-3\right)\left(x+5\right)-x\left(2x+7\right)\\ =2x^2-3x+10x-15-2x^2-7x\\ =\left(2x^2-2x^2\right)+\left(10x-3x-7x\right)-15\\ =-15\)
\(\left(2x-3\right)\left(x+5\right)-x\left(2x+7\right)\\ =2x^2+10x-3x-15-2x^2-7x\\ =-15\)
bài 1 rút gon biểu thức
\(6\sqrt{\dfrac{2}{3}}-\sqrt{24}+2\sqrt{\dfrac{3}{8}}+2\sqrt{\dfrac{1}{6}}\)
`6\sqrt(2/3)-\sqrt(24)+2\sqrt(3/8)+2\sqrt(1/6)`
`=6. \sqrt6/3 - \sqrt(2^2 .6) + 2. \sqrt(24)/8 + 2. \sqrt6/6`
`=2\sqrt6-2\sqrt6+ \sqrt6/2 + \sqrt6/3`
`=\sqrt6/2+\sqrt6/3`
`=(3\sqrt6+2\sqrt6)/6`
`=(5\sqrt6)/6`
Ta có: \(6\sqrt{\dfrac{2}{3}}-\sqrt{24}+2\sqrt{\dfrac{3}{8}}+2\sqrt{\dfrac{1}{6}}\)
\(=\dfrac{6\sqrt{2}}{\sqrt{3}}-2\sqrt{6}+2\cdot\dfrac{\sqrt{3}}{2\sqrt{2}}+\dfrac{2}{\sqrt{6}}\)
\(=2\sqrt{6}-2\sqrt{6}+\dfrac{\sqrt{3}}{\sqrt{2}}+\dfrac{\sqrt{2}}{\sqrt{3}}\)
\(=\dfrac{\sqrt{6}}{2}+\dfrac{\sqrt{6}}{3}=\dfrac{5\sqrt{6}}{6}\)
rút gon biểu thức sau : A= 4 + 2^2+ 2^3 +...+ 2^2021
\(A=2^2+2^2+2^3+...+2^{2021}\\ 2A=2^3+2^3+2^4+...+2^{2022}\\ A=2^{2022}+2^3-2^2-2^2\)
\(2A=8+2^3+...+2^{2022}\)
\(\Leftrightarrow A=2^{2022}+8-4-2^2=2^{2022}\)
7/24+13/24=
3/8+6/15
2+3/4+7/12
tính rồi rút gon giải chi tiết
a)\(=\dfrac{20}{24}=\dfrac{5}{6}\)
b)\(=\dfrac{31}{40}\)
c)\(=\dfrac{24}{12}+\dfrac{9}{12}+\dfrac{7}{12}=\dfrac{40}{12}=\dfrac{10}{3}\)
Bài 1 rút gon biểu thức
a,\(\sqrt{72}-3\sqrt{48}-5\sqrt{8}+4\sqrt{27}\)
b,\(\sqrt{2+\sqrt{3}}\cdot\sqrt{2-\sqrt{3}}\)
a: \(=6\sqrt{2}-12\sqrt{3}-10\sqrt{2}+12\sqrt{3}=-4\sqrt{2}\)
b: \(=\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=\sqrt{4-3}=1\)
rút gon biểu thức
(2x+5)^2+3(x-2)(x+2)
= 4x^2 + 20x + 25 + 3x^2 - 12
= x^2 + 20x + 13
rút gon biểu thức
M=|2x-3|+|x-1| với x>1,5
rút gọn biểu thức: P=\(\frac{3^{2010}-6^{2010}+9^{2010}-12^{2010}+15^{2010}-18^{2010}}{-1+2^{2010}-3^{2010}+4^{2010}-5^{2010}+6^{2010}}\)
Rút gọn biểu thức:
\(P=\frac{3^{2010}-6^{2010}+9^{2010}-12^{2010}+15^{2010}-18^{2010}}{-1+2^{2010}-3^{2010}+4^{2010}-5^{2010}+6^{2010}}\)
\(P=\frac{3^{2010}-6^{2010}+9^{2010}-12^{2010}+15^{2010}-18^{2010}}{-1+2^{2010}-3^{2010}+4^{2010}-5^{2010}+6^{2010}}\)
\(P=\frac{-3^{2010}.\left(-1+2^{2010}-3^{2010}+4^{2010}-5^{2010}+6^{2010}\right)}{-1+2^{2010}-3^{2010}+4^{2010}-5^{2010}+6^{2010}}\)
\(P=-3^{2010}\)
rút gon biểu thức
E=\(\sqrt[3]{182+\sqrt{33125}}+\sqrt[3]{182-\sqrt{33125}}\)
\(E^3=182+\sqrt{33125}+182-\sqrt{33125}+3\sqrt[3]{182^2-33125}\left(E\right)\)
=\(364-3E\)
\(\Rightarrow E^3+3E-364=0\)
\(\Leftrightarrow E^3-7E^2+7E^2-49E+52E-364=0\)
\(\Leftrightarrow\left(E-7\right)\left(E^2+7E+52\right)=0\)
\(\Rightarrow E=7\)
ta có \(E^3=\left(\sqrt[3]{182+\sqrt{33125}}+\sqrt[3]{182-\sqrt{33125}}\right)^3\)
\(E^3=\left(182+\sqrt{33125}\right)+\left(182-\sqrt{33125}\right)+3\cdot E\cdot\sqrt[3]{33124-33125}\)
\(E^3=364-3E\)
giải phương trình \(E^3+3E-364=0\)
suy ra E= 7