5^(\(2x-3\)) + 7.5² = 12.5²

5^{(2\(x\) - 3)}             = 12.5² - 7.5² 

5^{(2\(x\) - 3)}              = 5².(12 - 7)

5^(2\(x-3\))             = 5².5

5^(\(2x-3\))             = 5³ 

2\(x-3\)               = 3 

2\(x\)                      = 3 + 3

2\(x\)                      = 6 

\(x\)                         = 6 : 2 

\(x\)                         = 3

5²ˣ⁻³ + 7.5² = 12.5²

5²ˣ⁻³ = 12.5² - 7.5²

5²ˣ⁻³ = 5².(12 - 7)

5²ˣ⁻³ = 5².5

5²ˣ⁻³ = 5³

2x - 3 = 3

2x = 3 + 3

2x = 6

x = 6 : 2

x = 3

 3/4-2x| 1|=7/8

7.5-3|5-2x|=-4.5

|x +4/15|+ 1/2=3.5       

Ai giúp mình với, Mình đang cần gấp

(Ai làm nhanh mình k nha)

a, 3.(2\(x\) + 4) + 198 = (-3)².10

   3.(2\(x\) + 4) + 198 = 90

   3.(2\(x\) + 4) = 90  - 198

    3.(2\(x\) + 4) = - 108

       2\(x\) + 4 = -108 : 3

       2\(x\) + 4  = -36

       2\(x\)       = - 36 - 4

      2\(x\)       = - 40

       \(x\)       = -40 : 2

        \(x\)      = - 20 

b, 2.(\(x\) + 7) -  6  = 18

    2.(\(x\) + 7) = 18 + 6

    2.(\(x\) + 7)  =24

         \(x\) + 7 = 24 : 2

         \(x\) + 7  = 12

           \(x\)       = 12 - 7

             \(x\)     = 5

c, 5.(\(x\) + 5) - 3.(\(x\) - 2) = 5² + 18

    5\(x\) + 25 - 3\(x\) + 6 = 25 + 18

    2\(x\) + 31 = 43

    2\(x\)          = 43 - 31

   2\(x\)        = 12

     \(x\)        = 12 : 2

     \(x\)        = 6

2)

a) \(2\left|2x-3\right|=1\)

=> \(\left|2x-3\right|=1:2\)

=> \(\left|2x-3\right|=\frac{1}{2}\)

=> \(\left[{}\begin{matrix}2x-3=\frac{1}{2}\\2x-3=-\frac{1}{2}\end{matrix}\right.\) => \(\left[{}\begin{matrix}2x=\frac{1}{2}+3=\frac{7}{2}\\2x=\left(-\frac{1}{2}\right)+3=\frac{5}{2}\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=\frac{7}{2}:2\\x=\frac{5}{2}:2\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{7}{4}\\x=\frac{5}{4}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{7}{4};\frac{5}{4}\right\}.\)

b) \(7,5-3\left|5-2x\right|=-4,5\)

=> \(4,5\left|5x-2\right|=-4,5\)

=> \(\left|5x-2\right|=\left(-4,5\right):4,5\)

=> \(\left|5x-2\right|=-1\)

Ta luôn có: \(\left|x\right|>0\forall x\)

=> \(\left|5x-2\right|>-1\)

=> \(\left|5x-2\right|\ne-1\)

Vậy không tồn tại giá trị nào của \(x\) thỏa mãn yêu cầu đề bài.

c) \(\left|3x-4\right|+\left|3y+5\right|=0\)

Ta có: \(\left|3x-4\right|>\) hoặc \(=0\forall x\)

\(\left|3y+5\right|>\) hoặc \(=0\forall y.\)

=> \(\left|3x-4\right|+\left|3y+5\right|=0\)

=> \(\left[{}\begin{matrix}3x-4=0\\3y+5=0\end{matrix}\right.\) => \(\left[{}\begin{matrix}3x=0+4=4\\3y=0-5=-5\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=4:3\\y=\left(-5\right):3\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{4}{3}\\y=-\frac{5}{3}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{4}{3}\right\};y\in\left\{-\frac{5}{3}\right\}.\)

Chúc bạn học tốt!

Bài 1:

a) \(-15,5.20,8+3,5.9,2-15,5.9,2+3,5.20,8\)

\(=20,8.\left(-15,5+3,5\right)+9,2.\left(-15,5+3,5\right)\)

\(=\left(-15,5+3,5\right).\left(20,8+9,2\right)\)

\(=\left(-12\right).30=-360\)

b) \(\left[\left(-19,95\right)+\left(-45,75\right)\right]+\left[4,95+5,75\right]\)

\(=\left[\left(-19,95\right)+4,95\right]+\left[\left(-45,75\right)+5,75\right]\)

\(=-15+\left(-40\right)=-55\)

Bài 2 :

\(a,2.\left|2x-3\right|=1\)

\(\Leftrightarrow\left|2x-3\right|=\frac{1}{2}\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{7}{2}\\2x=\frac{5}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{4}\\x=\frac{5}{4}\end{matrix}\right.\)

Vậy : \(x\in\left\{\frac{7}{4},\frac{5}{4}\right\}\)

\(b,7.5-3\left|5-2x\right|=-4.5\)

\(\Leftrightarrow3.\left|5-2x\right|=7.5-\left(-4.5\right)=12\)

\(\Leftrightarrow\left|5-2x\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}5-2x=4\\5-2x=-4\end{matrix}\right.\)

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

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

Vậy : \(x\in\left\{\frac{1}{2},\frac{9}{2}\right\}\)

\(c,\left|3x-4\right|+\left|3y+5\right|=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x-4=0\\3y+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x=4\\3y=-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{4}{3}\\y=-\frac{5}{3}\end{matrix}\right.\)

Vậy : \(\left(x,y\right)=\left(\frac{4}{3},-\frac{5}{3}\right)\)

Bài 3 :

a) \(2^{300}\) và \(3^{200}\)

Ta có : \(2^{300}=\left(2^3\right)^{100}=8^{100}\)

\(3^{200}=\left(3^2\right)^{100}=9^{100}\)

mà : \(9^{100}>8^{100}\Rightarrow3^{200}>2^{300}\)

Vậy : \(3^{200}>2^{300}\)

b) \(2^{30}+3^{30}+4^{30}\) và \(3.2.4^{10}\)

Ta có : \(3.2.4^{10}=6.\left(2^2\right)^{10}=6.2^{20}=3.2^{21}\)

Ta thấy : \(2^{30}>3.2^{21}\Rightarrow2^{30}+3^{30}+4^{30}>3.2^{21}\)

hay : \(2^{30}+3^{30}+4^{30}>3.2.4^{10}\)

Vậy : \(2^{30}+3^{30}+4^{30}>3.2.4^{10}\)

Chúc bạn học tốt !

Bài 4 là:

\(\dfrac{1}{2.5}\) + \(\dfrac{1}{5.8}\) +....+\(\dfrac{1}{92.95}\) + \(\dfrac{1}{95.98}\)

                  Đúng không bạn

Bạn chỉ nên đưa câu khó lên vì làm cả đề thì ko làm kịp đc

2a) \(\frac{3^6+45^4-15^3.4^5}{27^4.25^3+45^6}\)

= \(\frac{3^6+\left(3^2.5\right)^4-\left(3.5\right)^3.\left(2^2\right)^5}{\left(3^3\right)^4.\left(5^2\right)^3+\left(3^2.5\right)^6}\)

= \(\frac{3^6+3^8.5^4-3^3.5^3.4^{10}}{3^{12}.5^6-3^{12}.5^6}=\frac{3^3.\left(3^3+3^5.5^4-5^3.4^{10}\right)}{0}\)(xem lại đề)

b)  \(\frac{\left(\frac{2}{5}\right)^7.5^7+\left(\frac{16}{3}\right)^3:\left(\frac{4}{9}\right)^3}{2^7.5^2+512}\)

= \(\frac{\left(\frac{2}{5}.5\right)^7+\left(\frac{16}{3}:\frac{4}{9}\right)^3}{2^7.5^2+2^9}\)

= \(\frac{2^7+12^3}{2^7\left(5^2+2^2\right)}\)

= \(\frac{2^7+\left(2^2.3\right)^3}{2^7.29}\)

= \(\frac{2^7+2^6.3^3}{2^7.29}\)

= \(\frac{2^6\left(1+27\right)}{2^7.29}=\frac{28}{2.29}=\frac{14}{29}\)

mk xin lỗi bn nhé...mk vt nhầm đề...mk vt lại nha:

2a,\(\frac{3^6+45^4-15^3.9^5}{27^4.25^3+45^6}\)

a: Ta có: \(x\left(2x-3\right)-\left(2x-1\right)\left(x+5\right)=17\)

\(\Leftrightarrow2x^2-3x-2x^2-10x+x+5=17\)

\(\Leftrightarrow-12x=12\)

hay x=-1