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

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

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

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

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

________

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

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

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

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

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

_____________

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

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

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

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

\(x=2\)

________

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

\(\left(2x-1\right)^2=\left(\dfrac{1}{2}\right)^2\)           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=\(\dfrac{15}{4}\)}

b) Ta có: \(\left(5-x\right)^3+27=0\)

\(\Leftrightarrow\left(5-x\right)^3=-27\)

\(\Leftrightarrow5-x=-3\)

hay x=8

d) Ta có: \(\left(x^2-1\right)\left(x+7\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=-7\end{matrix}\right.\)

f) Ta có: \(\left(x^2+81\right)\left(x-7\right)\left(x^2-2\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)

a: \(=\dfrac{\left(x^4-y^4\right)^2}{x^2+y^2}=\left(x^2-y^2\right)^2\cdot\left(x^2+y^2\right)\)

b: \(=\dfrac{\left(4x+3\right)\left(16x^2-12x+9\right)}{16x^2-12x+9}=4x+3\)

Bài 1:

\(A=26^2-24^2=\left(26-24\right)\left(26+24\right)=2\cdot50=100\)

\(B=27^2-25^2=\left(27-25\right)\left(27+25\right)=2\cdot52=104\)

=>A<B

Bài 2:

\(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)=11\)

=>\(4\left(x^2+2x+1\right)+4x^2-4x+1-8\left(x^2-1\right)=11\)

=>\(4x^2+8x+4+4x^2-4x+1-8x^2+8=11\)

=>4x+13=11

=>4x=-2

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

a) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=27\)

\(\Rightarrow x^3+3^3-x\left(x^2-4\right)=27\)

\(\Rightarrow x^3+27-x^3+4x=27\)

\(\Rightarrow27+4x=27\)

\(\Rightarrow4x=0\)

\(\Rightarrow x=0\)

b) \(2x^2+7x+3=0\)

\(\Rightarrow2x^2+x+6x+3=0\)

\(\Rightarrow x\left(2x+1\right)+3\left(2x+1\right)=0\)

\(\Rightarrow\left(2x+1\right)\left(x+3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x+1=0\\x+3=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=-1\\x=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-3\end{matrix}\right.\)

c) Trùng đề bài a

d) \(2x^2+11x+9=0\)

\(\Rightarrow2x^2+2x+9x+9=0\)

\(\Rightarrow2x\left(x+1\right)+9\left(x+1\right)=0\)

\(\Rightarrow\left(x+1\right)\left(2x+9\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\2x+9=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\2x=-9\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{9}{2}\end{matrix}\right.\)

a) \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x+5\right)=x+2-\left(x-5\right)\)

\(\Rightarrow\left(6x^2+21x-2x-7\right)-\left(6x^2+5x+6x+5\right)=x+2-x+5\)

\(\Rightarrow6x^2+19x-7-6x^2-11x-5=7\)

\(\Rightarrow8x-12=7\)

\(\Rightarrow8x=7+12\)

\(\Rightarrow8x=19\)

\(\Rightarrow x=\dfrac{19}{8}\)

b) \(\left(x+1\right)\left(x+2\right)\left(x+5\right)-x^3-8x^2=27\)

\(\Rightarrow\left(x^2+3x+2\right)\left(x+5\right)-x^3-8x^2=27\)

\(\Rightarrow x^3+3x^2+2x+5x^2+15x+10-x^3-8x^2=27\)

\(\Rightarrow17x+10=27\)

\(\Rightarrow17x=17\)

\(\Rightarrow x=1\)

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

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

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

\(---\)

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

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

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

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

a: \(x=\left(-\dfrac{2}{3}\right)^5:\left(-\dfrac{2}{3}\right)^2=\left(-\dfrac{2}{3}\right)^3=-\dfrac{8}{27}\)

b: =>x-1/2=1/3

=>x=5/6

c: =>2/3x-1=0 hoặc 3/4x+1/2=0

=>x=3/2 hoặc x=-1/2:3/4=-1/2*4/3=-4/6=-2/3

d =>4/9:x=10/3:9/4=10/3*4/9=40/27

=>x=4/9:40/27=4/9*27/40=108/360=3/10

a) \(\left(x-10\right)^2-x\left(x+8\right)=-12x+100=-11,76+100=88,24\)

b) \(x^3-9x^2+27x-27=\left(x-3\right)^3=\left(5-3\right)^3=8\)

c) \(6x\left(2x-7\right)-\left(3x-5\right)\left(4x+7\right)=-43x+35=121\)

\(a)\) \(\left(x-10\right)^{^2}-x.\left(x+8\right)\) \(với\) \(x=0,98\)

\(=-12x+100\)

\(=-11,76+100\)

\(=88,24\)

\(b)\) \(x^3-9x^2+27.x-27\) \(với\) \(x=5\)

\(=\left(x-3\right)^3\)

\(=\left(5-3\right)^3\)

\(=8\)

\(c)\)\(6x.\left(2x-7\right)-\left(3x-5\right).\left(4x+7\right)\) \(tại\) \(x=-2\)

\(=-43+35\)

\(=121\)

Chúc bạn hôc tốt nha ❤