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

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

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

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

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

________

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

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

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

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

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

_____________

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

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

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

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

\(x=2\)

________

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

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

ĐKXĐ: \(x\notin\left\{2;3;\dfrac{1}{2}\right\}\)

Ta có: \(\dfrac{x+4}{2x^2-5x+2}+\dfrac{x+1}{2x^2-7x+3}=\dfrac{2x+5}{2x^2-7x+3}\)

\(\Leftrightarrow\dfrac{x+4}{\left(x-2\right)\left(2x-1\right)}+\dfrac{x+1}{\left(x-3\right)\left(2x-1\right)}=\dfrac{2x+5}{\left(2x-1\right)\left(x-3\right)}\)

\(\Leftrightarrow\dfrac{\left(x+4\right)\left(x-3\right)}{\left(x-2\right)\left(2x-1\right)\left(x-3\right)}+\dfrac{\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)\left(2x-1\right)}=\dfrac{\left(2x+5\right)\left(x-2\right)}{\left(2x-1\right)\left(x-3\right)\left(x-2\right)}\)

Suy ra: \(x^2-3x+4x-12+x^2-2x+x-2=2x^2-4x+5x-10\)

\(\Leftrightarrow2x^2-14=2x^2+x-10\)

\(\Leftrightarrow2x^2-14-2x^2-x+10=0\)

\(\Leftrightarrow-x-4=0\)

\(\Leftrightarrow-x=4\)

hay x=-4(nhận)

Vậy: S={-4}

\(\dfrac{3}{x-5}-\dfrac{x+1}{x\left(x-5\right)}\left(dkxd:x\ne0,x\ne5\right)\\ =\dfrac{3x-x-1}{x\left(x-5\right)}=\dfrac{2x-1}{x^2-5x}\)

----------------------------------------

\(\dfrac{8\left(y+2\right)}{3x^2}.\dfrac{15x^5}{4\left(y+2\right)^2}\left(dkxd:x\ne0,y\ne-2\right)\\ =\dfrac{8}{4}.\dfrac{15x^2.x^3}{3x^2}=10x^3\)

------------------------------------------

\(\dfrac{8\left(y-1\right)}{3x^2-3}:\dfrac{4\left(y-1\right)^3}{x^2-2x+1}\left(dkxd:x\ne1,x\ne-1\right)\\ =\dfrac{8\left(y-1\right)}{3\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)^2}{4\left(y-1\right)^3}\\ =\dfrac{2\left(x-1\right)}{3\left(x+1\right)\left(y-1\right)^2}\)

1a.

ĐKXĐ: \(x\ne\left\{1;3\right\}\)

\(\Leftrightarrow\dfrac{6}{x-1}=\dfrac{4}{x-3}+\dfrac{4}{x-3}\)

\(\Leftrightarrow\dfrac{3}{x-1}=\dfrac{4}{x-3}\Leftrightarrow3\left(x-3\right)=4\left(x-1\right)\)

\(\Leftrightarrow3x-9=4x-4\Rightarrow x=-5\)

b.

ĐKXĐ: \(x\ne\left\{-1;2\right\}\)

\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{3}{2-x}+\dfrac{1}{2-x}\)

\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{4}{2-x}\Leftrightarrow5\left(2-x\right)=4\left(x+1\right)\)

\(\Leftrightarrow10-2x=4x+4\Leftrightarrow6x=6\Rightarrow x=1\)

1c.

ĐKXĐ: \(x\ne\left\{2;5\right\}\)

\(\Leftrightarrow\dfrac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}=\dfrac{-3x}{\left(x-2\right)\left(x-5\right)}\)

\(\Leftrightarrow3x\left(x-5\right)-x\left(x-2\right)=-3x\)

\(\Leftrightarrow2x^2-10x=0\Leftrightarrow2x\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=5\left(loại\right)\end{matrix}\right.\)

2a.

\(\Leftrightarrow-4x^2-5x+6=x^2+4x+4\)

\(\Leftrightarrow5x^2+9x-2=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{5}\end{matrix}\right.\)

2b.

\(2x^2-6x+1=0\Rightarrow x=\dfrac{3\pm\sqrt{7}}{2}\)

Bài 1:

a) \(3^7:3^5-\left(\dfrac{5}{17}\right)^0=3^{7-5}-1=3^2-1=9-1=8\)

b) \(\left(\dfrac{5}{2}\right)^{13}:\left(\dfrac{1}{2}+2\right)^3\)

\(=\left(\dfrac{5}{2}\right)^{13}:\left(\dfrac{5}{2}\right)^3\)

\(=\left(\dfrac{5}{2}\right)^{10}\)

c) \(8.\left(\dfrac{1}{4}\right)^3+\left(\dfrac{2}{27}\right)^0-\dfrac{1}{8}\)

\(=8.\dfrac{1}{64}+1-\dfrac{1}{8}\)

\(=\dfrac{1}{8}+1-\dfrac{1}{8}\)

\(=1\)

Bài 2:

a) \(\dfrac{3^4.4^4}{6^4}=\dfrac{3^4.\left(2^2\right)^4}{\left(2.3\right)^4}=\dfrac{3^4.2^8}{2^4.3^4}=\dfrac{2^8}{2^4}=2^4=16\)

b) \(\dfrac{15^3}{10^3}=\dfrac{\left(3.5\right)^3}{ \left(2.5\right)^3}=\dfrac{3^3.5^3}{2^3.5^3}=3^3:2^3=\dfrac{27}{8}\)

c) \(\dfrac{4^2.12^5}{9^2.2^{10}}=\dfrac{\left(2^2\right)^2.\left[3.\left(2^2\right)\right]^5}{\left(3^2\right)^2.2^{10}}=\dfrac{2^4.3^5.2^{10}}{3^4.2^{10}}=2^4.3=16.3=48\)

d) \(\dfrac{6^2+5.2^2+4}{15}=\dfrac{\left(2.3\right)^2+5.2^2+2^2}{15}=\dfrac{2^2.3^2+5.2^2+2^2}{15}=\dfrac{2^2\left(3^2+5+1\right)}{15}=\dfrac{2^2.15}{15}=2^2=4\)

Bài 3:

a) \(\dfrac{\left(\dfrac{2}{3}\right)^3.\left(\dfrac{-3}{4}\right)^2.\left(-1\right)^5}{\left(\dfrac{2}{5}\right)^2.\left(\dfrac{-5}{12}\right)^2}\)

\(=\dfrac{\left(\dfrac{2}{3}\right)^3.\left(\dfrac{-3}{4}\right)^2.-1}{\left[\dfrac{2}{5}.\left(\dfrac{-5}{12}\right)\right]^2}\)

\(=\dfrac{\left(\dfrac{2}{3}\right)^3. \left(\dfrac{-3}{4}\right)^2.-1}{\left(\dfrac{-1}{6}\right)^2}\)

\(=\left(\dfrac{2}{3}\right)^3.\left[\left(\dfrac{-3}{4}\right).-6\right]^2.-1\)

\(=\left(\dfrac{2}{3}\right)^3.\left(\dfrac{9}{2}\right)^2.-1\)

\(=\left(\dfrac{2}{3}\right)^2.\dfrac{2}{3}.\left(\dfrac{9}{2}\right)^2.-1\)

\(=\left(\dfrac{2}{3}.\dfrac{9}{2}\right)^2.\dfrac{2}{3}.-1\)

\(=9.\dfrac{2}{3}.-1\)

\(=6.-1=-6\)

b) \(\dfrac{6^6+6^3.3^3+3^6}{-73}=\dfrac{\left(2.3\right)^6+\left(2.3\right)^3.3^3+3^6}{-73}=\dfrac{2^6.3^6+2^3.3^3.3^3+3^6}{-73}=\dfrac{2^6.3^6+2^3.3^6+3^6}{-73}=\dfrac{3^6\left(2^6+2^3+1\right)}{-73}=\dfrac{3^6.73}{-73}=\dfrac{3^6}{-1}=\left(-3\right)^6\)

\(#Wendy.Dang\)

Lần sau bnn gửi từng bài thôi nha, chứ như vầy nhiều quá thì làm không nổi mất. đánh máy nãy giờ lú luôn gòi nè :))

Võ Ngọc Phương

Bài  3b, kết quả -(3)⁶ = - 729 em nhá chứ không phải (-3)⁶

\(b,=-\dfrac{40}{30}-\dfrac{12}{30}-\dfrac{45}{30}=-\dfrac{97}{30}\\ c,=\left(\dfrac{4}{5}+\dfrac{7}{10}\right)+\dfrac{2}{7}=\dfrac{3}{2}+\dfrac{2}{7}=\dfrac{25}{14}\\ d,=\dfrac{2}{3}+\dfrac{7}{4}+\dfrac{1}{2}+\dfrac{3}{8}\\ =\left(\dfrac{2}{3}+\dfrac{1}{2}\right)+\left(\dfrac{7}{4}+\dfrac{3}{8}\right)=\dfrac{7}{6}+\dfrac{17}{8}=\dfrac{79}{24}\)

c: \(\dfrac{4}{5}-\dfrac{-2}{7}-\dfrac{-7}{10}\)

\(=\dfrac{56}{70}+\dfrac{20}{70}+\dfrac{49}{70}\)

\(=\dfrac{125}{70}=\dfrac{25}{14}\)

1: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^6\)

\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{18}\)

=>4x=18

hay x=9/2

2: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^{36}\)

\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{108}\)

=>4x=108

hay x=27

3: \(\left(\dfrac{1}{81}\right)^x=\left(\dfrac{1}{27}\right)^4\)

\(\Leftrightarrow\left(\dfrac{1}{3}\right)^{4x}=\left(\dfrac{1}{3}\right)^{12}\)

=>4x=12

hay x=3

\(a,3-x=x+1,8\)

\(\Rightarrow-x-x=1,8-3\)

\(\Rightarrow-2x=-1,2\)

\(\Rightarrow x=0,6\)

\(b,2x-5=7x+35\)

\(\Rightarrow2x-7x=35+5\)

\(\Rightarrow-5x=40\)

\(\Rightarrow x=-8\)

\(c,2\left(x+10\right)=3\left(x-6\right)\)

\(\Rightarrow2x+20=3x-18\)

\(\Rightarrow2x-3x=-18-20\)

\(\Rightarrow-x=-38\)

\(\Rightarrow x=38\)

\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)

\(\Rightarrow8x-3+1=1+6x+x\)

\(\Rightarrow8x-3=7x\)

\(\Rightarrow8x-7x=3\)

\(\Rightarrow x=3\)

\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)

\(\Rightarrow-3x+x=\dfrac{4}{3}-\dfrac{2}{9}\)

\(\Rightarrow-2x=\dfrac{10}{9}\)

\(\Rightarrow x=-\dfrac{5}{9}\)

\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)

\(\Rightarrow\dfrac{1}{2}x-\dfrac{3}{4}x=-\dfrac{1}{2}-\dfrac{5}{6}\)

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

\(\Rightarrow x=\dfrac{16}{3}\)

\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)

\(\Rightarrow x-4=5-x\)

\(\Rightarrow x+x=5+4\)

\(\Rightarrow2x=9\)

\(\Rightarrow x=\dfrac{9}{2}\)

\(k,7x^2-11=6x^2-2\)

\(\Rightarrow7x^2-6x^2=-2+11\)

\(\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

\(m,5\left(x+3\cdot2^3\right)=10^2\)

\(\Rightarrow5\left(x+24\right)=100\)

\(\Rightarrow x+24=20\)

\(\Rightarrow x=-4\)

\(n,\dfrac{4}{9}-\left(\dfrac{1}{6^2}\right)=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)

\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{4}{9}-\dfrac{1}{36}\)

\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{5}{12}\)

\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2=0\)

\(\Rightarrow x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)

#\(Urushi\text{☕}\)

a: 3-x=x+1,8

=>-2x=-1,2

=>x=0,6

b: 2x-5=7x+35

=>-5x=40

=>x=-8

c: 2(x+10)=3(x-6)

=>3x-18=2x+20

=>x=38

d; 8(x-3/8)+1=6(1/6+x)+x

=>8x-3+1=1+6x+x

=>8x-2=7x+1

=>x=3

e: =>-3x+x=4/3-2/9

=>-2x=12/9-2/9=10/9

=>x=-5/9

g: =>3/4x-1/2x=5/6+1/2

=>1/4x=5/6+3/6=8/6=4/3

=>x=4/3*4=16/3

h: =>x-4=-x+5

=>2x=9

=>x=9/2