a)\(2\sqrt{\dfrac{16}{3}}-3\sqrt{\dfrac{1}{27}}-6\sqrt{\dfrac{4}{75}}\)

\(=2.\sqrt{\dfrac{4^2}{3}}-3.\sqrt{\dfrac{1}{3.3^2}}-6\sqrt{\dfrac{2^2}{3.5^2}}\)

\(=2.\dfrac{4}{\sqrt{3}}-3.\dfrac{1}{3\sqrt{3}}-6.\dfrac{2}{5\sqrt{3}}=\dfrac{8}{\sqrt{3}}-\dfrac{1}{\sqrt{3}}-\dfrac{12}{5\sqrt{3}}\)\(=\dfrac{23}{5\sqrt{3}}=\dfrac{23\sqrt{3}}{15}\)

b)\(\left(6\sqrt{\dfrac{8}{9}}-5\sqrt{\dfrac{32}{25}}+14\sqrt{\dfrac{18}{49}}\right).\sqrt{\dfrac{1}{2}}\)

\(=6\sqrt{\dfrac{8}{9}.\dfrac{1}{2}}-5\sqrt{\dfrac{32}{25}.\dfrac{1}{2}}+14\sqrt{\dfrac{18}{49}.\dfrac{1}{2}}\)

\(=6\sqrt{\dfrac{4}{9}}-5\sqrt{\dfrac{16}{25}}+14\sqrt{\dfrac{9}{49}}\)\(=6.\dfrac{2}{3}-5.\dfrac{4}{5}+14.\dfrac{3}{7}=6\)

c)\(\sqrt{\left(\sqrt{2}-2\right)^2}-\sqrt{6+4\sqrt{2}}=\left|\sqrt{2}-2\right|-\sqrt{4+2.2\sqrt{2}+2}=2-\sqrt{2}-\sqrt{\left(2+\sqrt{2}\right)^2}\)

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

a: \(A=\left(1-\sqrt{7}\right)\cdot\left(1+\sqrt{7}\right)=1-7=-6\)

b: \(B=3\sqrt{3}+8\sqrt{3}-15\sqrt{3}=-4\sqrt{3}\)

c: \(C=4\sqrt{2}-5\sqrt{2}+3\sqrt{2}=2\sqrt{2}\)

a) \(2\sqrt{98}-3\sqrt{18}+\dfrac{1}{2}\sqrt{32}=14\sqrt{2}-9\sqrt{2}+2\sqrt{2}=7\sqrt{2}\)

b) \(\left(5\sqrt{2}+2\sqrt{5}\right).\sqrt{5}-\sqrt{250}=5\sqrt{10}+10-5\sqrt{10}=10\)

c) \(\left(2\sqrt{3}-5\sqrt{2}\right).\sqrt{3}-\sqrt{36}=6-5\sqrt{6}-6=5\sqrt{6}\)

d) \(3\sqrt{48}+2\sqrt{27}-\dfrac{1}{3}\sqrt{243}=12\sqrt{3}+6\sqrt{3}-3\sqrt{3}=15\sqrt{3}\)

e) \(6\sqrt{\dfrac{1}{3}}+\dfrac{9}{\sqrt{3}}-\dfrac{2}{\sqrt{3}-1}=2\sqrt{3}+3\sqrt{3}=\left(\sqrt{3}+1\right)=4\sqrt{3}-1\)

f) \(4\sqrt{\dfrac{1}{2}}-\dfrac{6}{\sqrt{2}}.\dfrac{2}{\sqrt{2}+1}=2\sqrt{2}-\left(12-6\sqrt{2}\right)=8\sqrt{2}-12\)

a: \(A=3^{\dfrac{2}{5}}\cdot3^{\dfrac{1}{5}}\cdot3^{\dfrac{1}{5}}=3^{\dfrac{2}{5}+\dfrac{1}{5}+\dfrac{1}{5}}=3^{\dfrac{4}{5}}\)

b: \(B=\left(-27\right)^{\dfrac{1}{3}}=\left[\left(-3\right)^3\right]^{\dfrac{1}{3}}=\left(-3\right)^{\dfrac{1}{3}\cdot3}=\left(-3\right)^1=-3\)

c: \(C=\sqrt[3]{-64}\cdot\left(\dfrac{1}{2}\right)^3\)

\(=\sqrt[3]{\left(-4\right)^3}\cdot\dfrac{1}{2^3}=-4\cdot\dfrac{1}{8}=-\dfrac{4}{8}=-\dfrac{1}{2}\)

d: \(D=\left(-27\right)^{\dfrac{1}{3}}\cdot\left(\dfrac{1}{3}\right)^4\)

\(=\left[\left(-3\right)^3\right]^{\dfrac{1}{3}}\cdot\dfrac{1}{3^4}\)

\(=\left(-3\right)^{3\cdot\dfrac{1}{3}}\cdot\dfrac{1}{81}=\dfrac{-3}{81}=\dfrac{-1}{27}\)

e: \(E=\left(\sqrt{3}+1\right)^{106}\cdot\left(\sqrt{3}-1\right)^{106}\)

\(=\left[\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\right]^{106}\)

\(=\left(3-1\right)^{106}=2^{106}\)

f: \(F=360^{\sqrt{5}+1}\cdot20^{3-\sqrt{5}}\cdot18^{3-\sqrt{5}}\)

\(=360^{\sqrt{5}+1}\cdot\left(20\cdot18\right)^{3-\sqrt{5}}\)

\(=360^{\sqrt{5}+1}\cdot360^{3-\sqrt{5}}=360^{\sqrt{5}+1+3-\sqrt{5}}=360^4\)

g: \(G=2023^{3+2\sqrt{2}}\cdot2023^{2\sqrt{2}-3}\)

\(=2023^{3+2\sqrt{2}+2\sqrt{2}-3}\)

\(=2023^{4\sqrt{2}}\)

a) Để tính giá trị của biểu thức P=(x^3+12x−9)^{2005}=(√3+12√−9)^{2005} với x=3√4(√5+1)−3√4(√5−1). Đầu tiên, ta thay x bằng giá trị đã cho vào biểu thức P: P=(3√4(√5+1)−3√4(√5−1))^3+12(3√4(√5+1)−3√4(√5−1))−9)^{2005} Tiếp theo, ta thực hiện các phép tính để đơn giản hóa biểu thức: P=(4(5+1)^{1/2}−4(5−1)^{1/2})^3+12(4(5+1)^{1/2}−4(5−1)^{1/2})−9)^{2005} =(4√6−4√4)^3+12(4√6−4√4)−9)^{2005} =(4√6−8)^3+12(4√6−8)−9)^{2005} =(64√6−192+96√6−96−9)^{2005} =(160√6−297)^{2005} ≈ 1.332 × 10^3975

b) Để tính giá trị của biểu thức Q=x^3+ax+b=√3+√a+√b^2+√a^3+√3+√a−√b^2+√a^3 với x=3√−b^2+√b^2/4+a^3/(27+3√−b^2−√b^2/4+a^3/27). Tương tự như trên, ta thay x bằng giá trị đã cho vào biểu thức Q: Q=(3√−b^2+√b^2/4+a^3/(27+3√−b^2−√b^2/4+a^3/27))^3+a(3√−b^2+√b^2/4+a^3/(27+3√−b^2−√b^2/4+a^3/27))+b Tiếp theo, ta thực hiện các phép tính để đơn giản hóa biểu thức: Q=(−b+√b^2/4+a^3/(27−b+√b^2/4+a^3/27))^3+a(−b+√b^2/4+a^3/(27−b+√b^2/4+a^3/27))+b =−b^3+3√b^2/4+a^3/(27−3b√b^2/4+a^3/(27))+a(−b+√b^2/4+a^3/(27−b+√b^2/4+a^3/27))+b =−b^3+3√b^2/4+a^3/(27−3b√b^2/4+a^3/(27))+a(−b+√b^2/4+a^3/(27−b+√b^2/4+a^3/27))+b =−b^3+3√b^2/4+a^3/(27−3b√b^2/4+a^3/(27))+a(−b+√b^2/4+a^3/(27−b+√b^2/4+a^3/27))+b =−b^3+3√b^2/4+a^3/(27−3b√b^2/4+a^3/(27))+a(−b+√b^2/4+a^3/(27−b+√b^2/4+a^3/27))+b ≈ −b^3+3√b^2/4+a^3/(27−3b√b^2/4+a^3/(27))+a(−b+√b^2/4+a^3/(27−b+√b^2/4+a^3/27))+b

a) Ta có: \(4\sqrt{28}+3\sqrt{63}-3\sqrt{112}-2\sqrt{175}\)

\(=8\sqrt{7}+9\sqrt{7}-12\sqrt{7}-10\sqrt{7}\)

\(=-5\sqrt{7}\)

b) Ta có: \(\sqrt{5}\left(\sqrt{5}-3\sqrt{20}+2\sqrt{80}\right)\)

\(=\sqrt{5}\left(\sqrt{5}-6\sqrt{5}+8\sqrt{5}\right)\)

\(=\sqrt{5}\cdot3\sqrt{5}=15\)

c) Ta có: \(\left(\sqrt{\dfrac{16}{3}}-\sqrt{\dfrac{25}{3}}\right)\cdot\sqrt{3}\)

\(=\dfrac{-1}{\sqrt{3}}\cdot\sqrt{3}\)

=-1

e) Ta có: \(\left(\sqrt{\dfrac{32}{3}}-\sqrt{54}+\sqrt{\dfrac{50}{3}}\right)\cdot\sqrt{6}\)

\(=\left(\dfrac{4\sqrt{2}}{\sqrt{3}}+\dfrac{5\sqrt{2}}{\sqrt{3}}-3\sqrt{6}\right)\cdot\sqrt{6}\)

\(=\dfrac{9\sqrt{12}}{\sqrt{3}}-18\)

\(=0\)

f) Ta có:  \(\left(\sqrt{6}-2\right)\left(\sqrt{3}+\sqrt{2}\right)\)

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

\(=\sqrt{2}\)

Bài 1:

a: \(5\sqrt{8}-4\sqrt{27}-2\sqrt{75}+\sqrt{108}\)

\(=5\cdot2\sqrt{2}-4\cdot3\sqrt{3}-2\cdot5\sqrt{3}+6\sqrt{3}\)

\(=10\sqrt{2}-12\sqrt{3}-10\sqrt{3}+6\sqrt{3}\)

\(=10\sqrt{2}-16\sqrt{3}\)

b: \(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(1-\sqrt{6}\right)^2}\)

\(=\left|3-\sqrt{6}\right|+\left|1-\sqrt{6}\right|\)

\(=3-\sqrt{6}+\sqrt{6}-1\)

=3-1=2

c: \(\dfrac{5\sqrt{3}-3\sqrt{5}}{\sqrt{5}-\sqrt{3}}+\dfrac{1}{4+\sqrt{15}}\)

\(=\dfrac{\sqrt{15}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}+\dfrac{1\left(4-\sqrt{15}\right)}{16-15}\)

\(=\sqrt{15}+4-\sqrt{15}=4\)

d: \(\dfrac{2\sqrt{3-\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{\sqrt{10}-\sqrt{2}}-\dfrac{\sqrt{15}+\sqrt{5}}{\sqrt{12}+2}\)

\(=\dfrac{\sqrt{3-\sqrt{5}}\cdot\sqrt{2}\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}\left(\sqrt{3}+1\right)}{2\left(\sqrt{3}+1\right)}\)

\(=\dfrac{\sqrt{6-2\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}}{2}\)

\(=\sqrt{\left(\sqrt{5}-1\right)^2}\cdot\dfrac{\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}}{2}\)

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

Bài 2:

Vẽ đồ thị:

Phương trình hoành độ giao điểm là:

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

=>\(\dfrac{1}{2}x+3x=3+4\)

=>\(\dfrac{7}{2}x=7\)

=>x=2

Thay x=2 vào y=-3x+3, ta được:

\(y=-3\cdot2+3=-3\)

Vậy: (d1) cắt (d2) tại A(2;-3)

a, \(x^2\)  - 19 = 5.9

     \(x^2\) - 19 = 45

     \(x^2\)         = 45 + 19

     \(x^2\)         = 64

      \(x^2\)        = 8²

      \(x\)         = 8 

b, (2\(x\) + 1)³ = -0,001

    (2\(x\) + 1)³ = (-0,1)³

     2\(x\) + 1   = -0,1

     2\(x\)        = -0,1 - 1

     2\(x\)       = - 1,1

       \(x\)      = -1,1: 2

       \(x\)      = -  0,55

\(x^2-19=5\cdot9\\\Rightarrow x^2-19=45\\\Rightarrow x^2=45+19\\\Rightarrow x^2=64\\\Rightarrow x^2=(\pm8)^2\)

\(\Rightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)

\(---\)

\((2x+1)^3=-0,001\\\Rightarrow (2x+1)^3=(-0,1)^3\\\Rightarrow2x+1=-0,1\\\Rightarrow2x=-0,1-1\\\Rightarrow2x=-1,1\\\Rightarrow x=-1,1:2\\\Rightarrow x=\dfrac{-11}{20}\\---\)

\(\bigg(\dfrac56\bigg)^{2x-1}=\bigg(\dfrac56\bigg)^5\\\Rightarrow 2x-1=5\\\Rightarrow2x=5+1\\\Rightarrow2x=6\\\Rightarrow x=6:2\\\Rightarrow x=3\\---\)

\(\bigg(\dfrac13x-\dfrac23\bigg)^3=27\\\Rightarrow\bigg(\dfrac13x-\dfrac23\bigg)^3=3^3\\\Rightarrow\dfrac13x-\dfrac23=3\\\Rightarrow\dfrac13x=3+\dfrac23\\\Rightarrow\dfrac13x=\dfrac{11}{3}\\\Rightarrow x=\dfrac{11}{3}:\dfrac13\\\Rightarrow x=11\\---\)

\(\bigg(\dfrac{1}{32}\bigg)^x=\bigg(\dfrac12\bigg)^{15}\\\Rightarrow\bigg(\dfrac{1}{32}\bigg)^x=\bigg[\bigg(\dfrac{1}{2}\bigg)^5\bigg]^3\\\Rightarrow\bigg(\dfrac{1}{32}\bigg)^x=\bigg(\dfrac{1^5}{2^5}\bigg)^3\\\Rightarrow\bigg(\dfrac{1}{32}\bigg)^x=\bigg(\dfrac{1}{32}\bigg)^3\\\Rightarrow x=3\\Toru\)

\(\left(3-x\right)^3=-\dfrac{27}{64}\)

\(\left(3-x\right)^3=\left(\dfrac{-3}{4}\right)^3\)

\(=>3-x=\dfrac{-3}{4}\)

\(x=3-\dfrac{-3}{4}=\dfrac{12}{4}+\dfrac{3}{4}\)

\(x=\dfrac{15}{4}\)

________

\(\left(x-5\right)^3=\dfrac{1}{-27}\)

\(\left(x-5\right)^3=\left(\dfrac{-1}{3}\right)^3\)

\(=>x-5=\dfrac{-1}{3}\)

\(x=\dfrac{-1}{3}+5=\dfrac{-1}{3}+\dfrac{15}{3}\)

\(x=\dfrac{14}{3}\)

_____________

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{27}{8}\)

\(\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{3}{2}\right)^3\)

\(=>x-\dfrac{1}{2}=\dfrac{3}{2}\)

\(x=\dfrac{3}{2}+\dfrac{1}{2}\)

\(x=2\)

________

\(\left(2x-1\right)^2=\dfrac{1}{4}\)            

\(\left(2x-1\right)^2=\left(\dfrac{1}{2}\right)^2\)           hoặc              \(\left(2x-1\right)^2=\left(\dfrac{-1}{2}\right)^2\)

\(=>2x-1=\dfrac{1}{2}\)                                       \(2x-1=\dfrac{-1}{2}\)

\(2x=\dfrac{1}{2}+1=\dfrac{1}{2}+\dfrac{2}{2}\)                               \(2x=\dfrac{-1}{2}+1=\dfrac{-1}{2}+\dfrac{2}{2}\)

\(2x=\dfrac{3}{2}\)                                                     \(2x=\dfrac{1}{2}\)

\(x=\dfrac{3}{2}:2=\dfrac{3}{2}.\dfrac{1}{2}\)                                     \(x=\dfrac{1}{2}:2=\dfrac{1}{2}.\dfrac{1}{2}\)

\(x=\dfrac{3}{4}\)                                                       \(x=\dfrac{1}{4}\)

____________

\(\left(2-3x\right)^2=\dfrac{9}{4}\)

\(\left(2-3x\right)^2=\left(\dfrac{3}{2}\right)^2\)                hoặc                  \(\left(2-3x\right)^2=\left(\dfrac{-3}{2}\right)^2\)

\(=>2-3x=\dfrac{3}{2}\)                                               \(2-3x=\dfrac{-3}{2}\)

\(3x=2-\dfrac{3}{2}=\dfrac{4}{2}-\dfrac{3}{2}\)                                      \(3x=2-\dfrac{-3}{2}=\dfrac{4}{2}+\dfrac{3}{2}\)

\(3x=\dfrac{1}{2}\)                                                            \(3x=\dfrac{7}{2}\)

\(x=\dfrac{1}{2}.\dfrac{1}{3}\)                                                          \(x=\dfrac{7}{2}.\dfrac{1}{3}\)

\(x=\dfrac{1}{6}\)                                                               \(x=\dfrac{7}{6}\)

______________

\(\left(1-\dfrac{2}{3}\right)^2=\dfrac{4}{9}\) -> Kiểm tra đề câu này

(3-x)^{3_{=(-\(\dfrac{3}{4}\))}3}

^{3-x=-\(\dfrac{3}{4}\)}

  ^{x=3-(-\(\dfrac{3}{4}\))}

  ^{x=\(\dfrac{15}{4}\)}