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AH
Akai Haruma
Giáo viên
14 tháng 7 2019

a)

\(\sqrt[3]{(\sqrt{2}+1)(3+2\sqrt{2})}=\sqrt[3]{(\sqrt{2}+1)(2+2\sqrt{2}+1)}\)

\(=\sqrt[3]{(\sqrt{2}+1)(\sqrt{2}+1)^2}=\sqrt[3]{(\sqrt{2}+1)^3}=\sqrt{2}+1\)

b)

\(\sqrt[3]{(4-2\sqrt{3})(\sqrt{3}-1)}=\sqrt[3]{(3-2\sqrt{3}+1)(\sqrt{3}-1)}\)

\(=\sqrt[3]{(\sqrt{3}-1)^2(\sqrt{3}-1)}=\sqrt[3]{(\sqrt{3}-1)^3}=\sqrt{3}-1\)

c)

\((\sqrt[3]{4}+1)^3-(\sqrt[3]{4}-1)^3=[(\sqrt[3]{4}+1-(\sqrt[3]{4}-1)][(\sqrt[3]{4}+1)^2+(\sqrt[3]{4}+1)(\sqrt[3]{4}-1)+(\sqrt[3]{4}-1)^2]\)

\(=2[\sqrt[3]{16}+1+2\sqrt[3]{4}+\sqrt[3]{16}-1+\sqrt[3]{16}+1-2\sqrt[3]{4}]\)

\(=2(3\sqrt[3]{16}+1)\)

AH
Akai Haruma
Giáo viên
14 tháng 7 2019

d)

\((\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4})(\sqrt[3]{3}+\sqrt[3]{2})=[(\sqrt[3]{3})^2-\sqrt[3]{3}.\sqrt[3]{2}+(\sqrt[3]{2})^2](\sqrt[3]{3}+\sqrt[3]{2})\)

\(=(\sqrt[3]{3})^3+(\sqrt[3]{2})^3=3+2=5\)

e)

\(E=\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\)

Áp dụng công thức $(a+b)^3=a^3+b^3+3ab(a+b)$ ta có:

\(E^3=20+14\sqrt{2}+20-14\sqrt{2}+3\sqrt[3]{(20+14\sqrt{2})(20-14\sqrt{2})}.E\)

\(E^3=40+3\sqrt[3]{20^2-(14\sqrt{2})^2}.E\)

\(E^3=40+3\sqrt[3]{8}.E=40+6E\)

\(\Leftrightarrow E^2(E-4)+4E(E-4)+10(E-4)=0\)

\(\Leftrightarrow (E-4)(E^2+4E+10)=0\)

Dễ thấy $E^2+4E+10=(E+2)^2+6\neq 0$ nên $E-4=0$ hay $E=4$

26 tháng 6 2021

\(A=2.\left|\left(-3\right)\right|^3+2.\left(-2\right)^2-4\left|\left(-2\right)^3\right|\)

\(=54+8-32=30\)

\(B=\left|\sqrt{2}-2\right|+\left|\sqrt{2}-3\right|=2-\sqrt{2}+3-\sqrt{2}\)

\(=5-2\sqrt{2}\)

\(C=\left|3-\sqrt{3}\right|-\left|1+\sqrt{3}\right|=3-\sqrt{3}-1-\sqrt{3}\)

\(=2-2\sqrt{3}\)

\(D=\left|5+\sqrt{6}\right|-\left|\sqrt{6}-5\right|=5+\sqrt{6}-5+\sqrt{6}\)

\(=2\sqrt{6}\)

\(E=\sqrt{15^2}-\sqrt{5^2}=15-5=10\)

26 tháng 6 2021

`A=2sqrt{(-3)^6}+2sqrt{(-2)^4}-4sqrt{(-2)^6}=2|(-3)^3|+2|(-2)^2|-4|(-2)^3|=54+8-32=30` $\\$ `B=sqrt{(sqrt2-2)^2}+sqrt{(sqrt2-3)^2}=2-sqrt2+3-sqrt2=5-2sqrt2` $\\$ `C=sqrt{(3-sqrt3)^2}-sqrt{(1+sqrt3)^2}=3-sqrt3-sqrt3-1=2-2sqrt3` $\\$ `D=sqrt{(5+sqrt6)^2}-sqrt{(sqrt6-sqrt5)^2}=5+sqrt6-5+sqrt6=2sqrt6` $\\$ `E=sqrt{17^2-8^2}-sqrt{3^2+4^2}=sqrt{289-64}-sqrt{9+16}=sqrt(225)-sqrt{25}=15-5=10`

26 tháng 8 2021

a, \(\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(3\sqrt{2}-2\right)^2}\)

\(=\left|\sqrt{2}-1\right|+\left|3\sqrt{2}-2\right|\)

\(=\sqrt{2}-1+3\sqrt{2}-2=4\sqrt{2}-3\)

b, \(2\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(2\sqrt{3}+1\right)^2}\)

\(=2\left|\sqrt{3}-1\right|-\left|2\sqrt{3}+1\right|\)

\(=2\sqrt{3}-2-2\sqrt{3}-1=-3\)

26 tháng 8 2021

c, \(4\sqrt{\left(2-\dfrac{\sqrt{3}}{2}\right)^2}-3\sqrt{\left(\sqrt{3}-1\right)^2}\)

\(=4\left|2-\dfrac{\sqrt{3}}{2}\right|-3\left|\sqrt{3}-1\right|\)

\(=8-2\sqrt{3}-3\sqrt{3}+3=11-5\sqrt{3}\)

8 tháng 6 2021

a,\(\left(\sqrt{1\dfrac{9}{16}}-\sqrt{\dfrac{9}{16}}\right):5=\left(\sqrt{\dfrac{25}{16}}-\dfrac{3}{4}\right):5=\left(\dfrac{5}{4}-\dfrac{3}{4}\right):5\)

\(=\dfrac{1}{2}:5=\dfrac{1}{10}\)

b,\(\left(\sqrt{3}-2\right)^2\left(\sqrt{3}+2\right)^2=\left[\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)\right]^2\)

\(=\left[3-4\right]^2=1\)

c,\(\left(11-4\sqrt{3}\right)\left(11+4\sqrt{3}\right)=11^2-\left(4\sqrt{3}\right)^2\)

\(=121-48=73\)

d,\(\left(\sqrt{2}-1\right)^2-\dfrac{3}{2}\sqrt{\left(-2\right)^2}+\dfrac{4\sqrt{2}}{5}+\sqrt{1\dfrac{11}{25}}.\sqrt{2}\)

\(=2-2\sqrt{2}+1-3+\dfrac{4\sqrt{2}}{5}+\sqrt{\dfrac{36}{25}.2}\)

\(=-2\sqrt{2}+\dfrac{4\sqrt{2}+6\sqrt{2}}{5}\)

\(=-2\sqrt{2}+\dfrac{10\sqrt{2}}{5}=-2\sqrt{2}+2\sqrt{2}=0\)

e,\(\left(1+\sqrt{2021}\right)\sqrt{2022-2\sqrt{2021}}\)

\(=\left(1+\sqrt{2021}\right)\sqrt{2021-2\sqrt{2021}.1+1}\)

\(=\left(1+\sqrt{2021}\right)\sqrt{\left(\sqrt{2021}-1\right)^2}\)

\(=\left(1+\sqrt{2021}\right)\left(\sqrt{2021}-1\right)\)

\(=\sqrt{2021}-1+\sqrt{2021^2}-\sqrt{2021}=2020\)

 

17 tháng 7 2018

a,( √6+2)(√3-√2)

<=> ( √2√3+2)(√3-√2)

<=> √2(√3+√2)(√3-√2)

<=> √2( (√3)2-(√2)2) = √2

b, (√3+1)2-2√3+4

<=> (√3)+2√3 +1 -2√3+4 =8

c, (1+√2-√3)(√2+√3)

<=>√2+√3+(√2)2+√6-√6-(√3)2

<=> √2+√3-1

d, √3(√2-√3)2-(√3+√2)

<=> √3( 2-2√6+3)-√3-√2

<=> 5√3-2√18-√3-√2

<=> 4√3-√2(√36-1)

<=> 4√3 - 3√2

e, (1+2√3-√2)(1+2√3+√2)

<=> (1+2√3)2-(√2)2

<=> (1+4√3+(2√3)2)-2

<=> 1+4√3+12-2= 11+4√3

g, (1-√3)2(1+2√3)2

<=>(1-2√3+3)(1+4√3+12)

<=>( 4-2√3)(13+4√3)

<=> 52+16√3-26√3-24

<=> -10√3+28

27 tháng 6 2021

\(A=\left|2-\sqrt{3}\right|+\left|1+\sqrt{3}\right|=2-\sqrt{3}+1+\sqrt{3}=3\)

\(B=\left|4-\sqrt{5}\right|-\left|2-\sqrt{5}\right|=4-\sqrt{5}-\sqrt{5}+2=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\)

\(C=\left|1-\sqrt{5}\right|-\left|2-\sqrt{5}\right|=\sqrt{5}-1-\sqrt{5}+2=1\)

27 tháng 6 2021

\(A=\left|2-\sqrt{3}\right|+\left|1+\sqrt{3}\right|=2-\sqrt{3}+1+\sqrt{3}=3\)

\(B=\left|4-\sqrt{5}\right|-\left|2-\sqrt{5}\right|=4-\sqrt{5}-\sqrt{5}+2=6-2\sqrt{5}\)

C=\(\left|1-\sqrt{5}\right|-\left|2-\sqrt{5}\right|=\sqrt{5}-1-\sqrt{5}+2=1\)