`@` `\text {Ans}`

`\downarrow`

`a)`

\(\left(\dfrac{x}{2}-1\right)^3+2=-\dfrac{11}{8}\) phải k bạn nhỉ? `11/8` k có bậc lũy thừa nào `=5` á.

`=>`\(\left(\dfrac{x}{2}-1\right)^3=-\dfrac{11}{8}-2\)

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

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

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

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

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

`=> x=1`

Vậy, `x=1`

`b)`

\(\left(\dfrac{x}{3}+\dfrac{1}{2}\right)\left(75\%-1\dfrac{1}{2}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{x}{3}+\dfrac{1}{2}=0\\0,75-1\dfrac{1}{2}x=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}\dfrac{x}{3}=-\dfrac{1}{2}\\-\dfrac{3}{2}x=\dfrac{75}{100}\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}2x=-3\\-3x\cdot100=2\cdot75\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=-\dfrac{3}{2}\\-3x\cdot100=150\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=-\dfrac{3}{2}\\-3x=1,5\end{matrix}\right.\)

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

Vậy, `x={-3/2; -1/2}.`

cảm ơn xong chẳng có ai :)))

Bài làm :

\(a,115-4x=7\)

\(4x=115-7\)

\(4x=108\)

\(x=27\)

b, Sai đề bài rồi cậu ơi

\(c,25+3.\left(x-8\right)=106\)

\(3\left(x-8\right)=106-25\)

\(3\left(x-8\right)=81\)

\(x-8=27\)

\(x=27+8\)

\(x=35\)

\(d,3^x=81\)

\(3^x=3^4\)

\(\Rightarrow x=4\)

Học tốt

a. 115-4x=7                                                    c.  3^x=81=3⁴   

Suy ra:4x=115-7=108                                   Suy ra:x=4

             x=108:4=27

  Vậy x=27

c. 25+3.(x-8)=106

          3.(x-8)=106-25=81

            x-8=81:3=27

           x     =27+8=35

115-4x=7,4x=115-7,4x=108,x=108:4,x=27

Bài 2:

\(a,\Leftrightarrow x^5-x^3+5x+a=\left(x+1\right)\cdot a\left(x\right)\)

Thay \(x=-1\Leftrightarrow-1+1-5+a=0\Leftrightarrow a=5\)

\(b,\Leftrightarrow x^4+x^3+ax-2=\left(x-2\right)\cdot b\left(x\right)\)

Thay \(x=2\Leftrightarrow16+8+2a-2=0\Leftrightarrow2a=-22\Leftrightarrow a=-11\)

Bài 1:

\(x^{19}-x-3=\left(x+1\right)\cdot a\left(x\right)+R\) với R là hằng số (do x+1 bậc 1)

Thay \(x=-1\Leftrightarrow-1+1-3=R\Leftrightarrow R=-3\)

Vậy phép chia dư -3

Gấp lắm rùi. Các bạn giải nhanh lên nhé.

a) nếu \(5x-3\ge0\)hay \(x\ge\frac{3}{5}\) ta có \(\left|5x-3\right|=5x-3\)

nếu \(5x-3< 0\) hay \(x< \frac{3}{5}\) ta có \(\left|5x-3\right|=3-5x\)

với \(x\ge\frac{3}{5}\) ta có

\(\left|5x-3\right|=x+7\) \(< =>5x-3=x+7\)

                                           \(< =>5x-x=7+3\)

                                            \(< =>4x=10\)

                                             \(< =>x=\frac{10}{4}=\frac{5}{2}\) (thoả mãn khoảng xét: \(\frac{5}{2}>\frac{3}{5}\))

với \(x< \frac{3}{5}\)ta được

\(\left|5x-3\right|=x+7\) \(< =>3-5x=x+7\)

                                          \(< =>-5x-x=7-3\)

                                          \(< =>-6x=4\)

                                          \(< =>x=-\frac{4}{6}=-\frac{2}{3}\) (thoả mãn khoảng xét : \(-\frac{2}{3}< \frac{3}{5}\))

b) bạn lập bảng xét dấu rồi xét từng trường hợp là được

\(\left(x-3\right)\left(x-12\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-12=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=12\end{cases}}\)

\(\Rightarrow x\in\left\{3;12\right\}\)

\(\left(x^2-81\right)\left(x^2+9\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^2-81=0\\x^2+9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=9\\x\in\varnothing\end{cases}}\Leftrightarrow x=9\)

\(\Rightarrow x=9\)

\(\left(x-4\right)\left(x+2\right)< 0\)

\(\Rightarrow\hept{\begin{cases}x-4\\x+2\end{cases}}\)trái dấu

\(TH1:\hept{\begin{cases}x-4>0\\x+2< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>4\\x< -2\end{cases}}\Leftrightarrow x\in\varnothing\)

\(TH2:\hept{\begin{cases}x-4< 0\\x+2>0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x< 4\\x>-2\end{cases}}\Leftrightarrow x\in\left\{-1;0;1;2;3\right\}\)

Vậy \(x\in\left\{-1;0;1;2;3\right\}\)

a) \(\frac{1}{81}\): 3^x = \(\frac{1}{729}\)

     3^x = \(\frac{1}{81}\): \(\frac{1}{729}\)

     3^x = 9 

=> x = 2 ( vì 3² = 9 ) 

a) -3x - 8 = -7

-3x = -7 + 8

-3x = 1

x = -1/3

b) (3 - x)(x + 7) = 0

3 - x = 0 hoặc x + 7 = 0

x = 3 hoặc x = -7

a) -3x-8=-7

<=> -3x=1

<=> x=\(-\frac{1}{3}\)

Vậy x=\(-\frac{1}{3}\)

b) (3-x)(x+7)=0

<=> 3-x=0 hoặc x+7=0

<=> x=3 hoặc x=-7

Vậy x=3; x=-7

\(a,\frac{1}{2}x+\frac{5}{2}=\frac{7}{2}x-\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{2}x+\frac{5}{2}-\frac{7}{2}x=-\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{2}x-\frac{7}{2}x+\frac{5}{2}=-\frac{3}{4}\)

\(\Leftrightarrow-3x+\frac{5}{2}=-\frac{3}{4}\)

\(\Leftrightarrow-3x=-\frac{13}{4}\)

\(\Leftrightarrow x=-\frac{13}{4}:(-3)=-\frac{13}{4}:\frac{-3}{1}=-\frac{13}{4}\cdot\frac{-1}{3}=\frac{13}{12}\)

\(b,\frac{2}{3}x-\frac{2}{5}=\frac{1}{2}x-\frac{1}{3}\)

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

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

\(\Leftrightarrow\frac{1}{6}x-\frac{2}{5}=-\frac{1}{3}\)

\(\Leftrightarrow\frac{1}{6}x=\frac{1}{15}\)

\(\Leftrightarrow x=\frac{1}{15}:\frac{1}{6}=\frac{1}{15}\cdot6=\frac{6}{15}=\frac{2}{5}\)

\(c,\frac{1}{3}x+\frac{2}{5}(x+1)=0\)

\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)

\(\Leftrightarrow\frac{11}{15}x=-\frac{2}{5}\)

\(\Leftrightarrow x=-\frac{6}{11}\)

d,e,f Tương tự

a; \(\dfrac{1}{2}x\) + \(\dfrac{5}{2}\) = \(\dfrac{7}{2}\)\(x\) - \(\dfrac{3}{4}\)

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

    \(\dfrac{3}{4}\) + \(\dfrac{10}{4}\) = 3\(x\)

    3\(x\)          = \(\dfrac{13}{4}\)

     \(x\)          = \(\dfrac{13}{4}\) : 3

     \(x=\dfrac{13}{12}\) 

Vậy \(x=\dfrac{13}{12}\)