a) \(\left(x-1\right)^3=8=2^3\)

\(x-1=2\)

\(x=2+1=3\)

b) \(7^{2x-6}=49=7^2\)

\(2x-6=2\)

\(2x=6+2=8\)

\(x=8:2=4\)

c) \(\left(2x-14\right)^7=128=2^7\)

\(2x-14=2\)

\(2x=14+2=16\)

\(x=16:2=8\)

d) \(x^4\cdot x^5=5^3\cdot5^6=5^4\cdot5^5\)

\(x=5\)

e) \(3\cdot\left(x+2\right):7\cdot4=120\)

\(x+2=120:3\cdot7:4\)

\(x+2=70\)

\(x=70-2=68\)

Lời giải:

a. $(x-1)^3=8=2^3$
$\Rightarrow x-1=2$

$\Rightarrow x=3$

b. $7^{2x-6}=49=7^2$
$\Rightarrow 2x-6=2$

$\Rightarrow 2x=8$

$\Rightarrow x=4$

c. $(2x-14)^7=128=2^7$

$\Rightarrow 2x-14=2$

$\Rightarrow 2x=16$

$\Rightarrow x=18$

d.

$x^4.x^5=5^3.5^6$

$x^9=5^9$

$\Rightarrow x=5$

e. 

$3(x+2):7=120:4=30$

$3(x+2)=30.7=210$

$x+2=210:3=70$

$x=70-2=68$

Ta có : \(\hept{\begin{cases}\left(x-3,5\right)^2\ge0\forall x\\\left(y-\frac{1}{10}\right)^4\ge0\forall y\end{cases}}\Rightarrow\left(x-3,5\right)^2+\left(y-\frac{1}{10}\right)^4\ge0\forall x,y\)(1)

mà đề bài cho \(\left(x-3,5\right)^2+\left(y-\frac{1}{10}\right)^4\le0\)(2)

Từ (1) và (2) => \(\left(x-3,5\right)^2+\left(y-\frac{1}{10}\right)^4=0\)

=> \(\hept{\begin{cases}x-3,5=0\\y-\frac{1}{10}=0\end{cases}}\Rightarrow\hept{\begin{cases}x=3,5\\y=\frac{1}{10}\end{cases}}\)

Vậy ...

(x-3,5)mux2+(y-1 phần 10) mũ 4

=(x+y) mũ 2 nhân (3,5-1 phần 10)mũ 4

=xy mũ 2 nhân 3,4 mũ 4

= 3,4xy mũ 6

\(\Leftrightarrow\dfrac{1}{2}\left[\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}\right]=\dfrac{49}{99}\\ \Leftrightarrow1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow1-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow\dfrac{1}{2x+1}=\dfrac{1}{99}\\ \Leftrightarrow2x+1=99\Leftrightarrow x=49\)

Ta có: \(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{x\left(x+2\right)}=\dfrac{20}{41}\)

\(\Leftrightarrow\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{40}{41}\)

\(\Leftrightarrow1-\dfrac{2}{x+2}=\dfrac{40}{41}\)

\(\Leftrightarrow\dfrac{2}{x+2}=\dfrac{1}{41}\)

Suy ra: x+2=82

hay x=80

#)Giải :

\(x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+4,5\right)=3,5\)

\(\Leftrightarrow\left(2x^2-x\right)\left(x+5\right)-2x^3-9x^2-x-4,5=3,5\)

\(\Leftrightarrow2x^3+10x^2-x^2-5x-2x^3-9x^2-x-4,5=3,5\)

\(\Leftrightarrow-5x-4,5=3,5\)

\(\Leftrightarrow-5x=8\)

\(\Leftrightarrow x=-\frac{8}{5}\)

\(x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+4,5\right)=3,5\)

\(\Leftrightarrow\left(2x^2-x\right)\left(x+5\right)-\left(2x^3+9x^2+x+4,5\right)=3,5\)

\(\Leftrightarrow2x^3+10x^2-x^2-5x-2x^3-9x^2-x-4,5=3,5\)

\(\Leftrightarrow-6x=8\)

\(\Leftrightarrow x=\frac{-8}{6}=\frac{-4}{3}\)

Cách làm 

Bài 1:     25 + 3(x - 8) = 106

                      3(x - 8) = 106 - 25

                      3(x - 8) = 81

                        (x - 8) = 81 : 3

                        (x - 8) = 27

                                x = 27 + 8

                                x = 25

Bài 2:       720 : [41 - (2x - 5)] = 2³ . 5

                720 : [41 - (2x - 5)] = 8 . 5

                720 : [41 - (2x - 5)] = 40

                         [41 - (2x - 5)] = 720 : 40

                         [41 - (2x - 5)] = 18

                                  (2x - 5) = 41 - 18

                                  (2x - 5) = 23

                                          2x = 23 + 5

                                          2x = 28

                                            x = 28 : 2

                                            x = 14

Xinloi, t ghi thiếu đề

\(\left|x+\frac{1}{1.3}\right|+\left|x+\frac{1}{3.5}\right|+...+\left|x+\frac{1}{97.99}\right|=50x\)

\(\left|x+\frac{1}{1.3}\right|+\left|x+\frac{1}{3.5}\right|+...+\left|x+\frac{1}{97.99}\right|=50x\)

Vì \(\left|x+\frac{1}{1.3}\right|\ge0\forall x\)

     \(\left|x+\frac{1}{3.5}\right|\ge0\forall x\)

        ................

       \(\left|x+\frac{1}{97.99}\right|\ge0\forall x\)

(VT: Vế trái; VP: Vế phải)

\(\Rightarrow VT\ge0\Rightarrow VP=50x\ge0\)mà \(50>0\)

\(\Rightarrow x>0\)

\(\Rightarrow x+\frac{1}{1.3}>0\forall x\)

        ..............

      \(x+\frac{1}{97.99}>0\forall x\)(1)

(1) \(\Leftrightarrow x+\frac{1}{1.3}+x+\frac{1}{3.5}+...+x+\frac{1}{97.99}=50x\)

\(\Leftrightarrow49x+\left(\frac{1}{1.3}+...+\frac{1}{97.99}\right)=50x\)

\(\Leftrightarrow50x-49x=\frac{1}{2}\left(\frac{2}{1.3}+...+\frac{2}{97.99}\right)\)

\(\Leftrightarrow x=\frac{1}{2}\left(1-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(\Leftrightarrow x=\frac{1}{2}\left(1-\frac{1}{99}\right)\)

\(\Leftrightarrow x=\frac{1}{2}\cdot\frac{98}{99}=\frac{49}{99}\)

Vậy....

P/s: Làm bừa :) Ko chắc đúng nhé