1)     \(-1,6:\left(1+\frac{8}{12}\right)\)

  \(=-1,6:\left(1+\frac{2}{3}\right)\)

     \(=-1,6:1-1,6:\frac{2}{3}\)

       \(=-1,6+\frac{-1,6.3}{2}\)

          \(=-1,6-2,4\)

         \(=4\)

2)    \(\frac{3}{8}-\frac{4}{5}-\frac{-17}{40}\)

      \(=\frac{15-32+17}{40}\)

          =0

3)   \(\frac{3}{4}-\frac{16}{32}+\frac{4}{-3}\)

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

\(=\frac{9-6-16}{12}\)

\(=\frac{-13}{12}\)

4)  \(\frac{-4}{7}+\frac{2}{3}.\frac{-9}{14}\)  

\(=\frac{-4}{7}+\frac{-3}{7}=-1\)

5)   \(2\frac{3}{4}-\left(\frac{1}{7}+1\frac{3}{4}\right)\)

\(=\frac{11}{4}-\left(\frac{1}{7}+\frac{7}{4}\right)\)

\(=\frac{11}{4}-\frac{4+49}{28}\)

\(=\frac{11}{4}-\frac{53}{28}\)

\(=\frac{77-53}{28}=\frac{24}{28}=\frac{6}{7}\)

6)   \(8\frac{2}{7}-\left(3\frac{4}{9}+4\frac{2}{7}\right)\)

\(=\frac{58}{7}-\left(\frac{31}{9}+\frac{30}{7}\right)=\frac{522-217+270}{63}\)

\(=\frac{575}{63}\)

Ta có \(2^{x+2}-2^x=96\)

\(\Rightarrow2^x.\left(2^2-1\right)=96\)

\(\Rightarrow2^x.3=96\)

\(\Rightarrow2^x=32\)

\(\Rightarrow2^x=2^5\)

\(\Rightarrow x=5\)

b, Ta có \(5^x+5^{x+2}=650\)

\(\Rightarrow5^x.\left(1+5^2\right)=650\)

\(\Rightarrow5^x.26=650\)

\(\Rightarrow5^x=25\)

\(\Rightarrow5^x=5^2\)

\(\Rightarrow x=2\)

a:

2^x+2-2^x=96
=> 2^x.2²-2^x=99
=>2^x.4-2^x=96
=>2^x(4-1)=96
=>2^x.3=96
=>2^x=96:3=32=2^5
=>x=5

^a:

5^x+5^x+2=650
=>5^x+5^x.5²=650
=>5^x+5^x.25=650
=>5^x(1+25)=650
=>5^x=650:26=25
=>x=5
----HỌC TỐT NHA----

\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)

Bài này mình làm bài KT ở lớp rồi bạn ạ,ko thể tính được mà so sánh với -1/2

\(A=-\left(\frac{3\cdot8\cdot15}{4\cdot9\cdot16}....\frac{9999}{10000}\right)\)Vì A có 99 số hạng (số lẻ)

\(A=-\left(\frac{1\cdot3\cdot2\cdot4\cdot3\cdot5}{2\cdot2\cdot3\cdot3\cdot4\cdot4}...\frac{99\cdot101}{100\cdot100}\right)\)

\(A=-\left(\frac{1}{2}\cdot\left(\frac{3\cdot2\cdot4\cdot3}{2\cdot3\cdot3\cdot4}...\frac{99}{100}\right)\cdot\frac{101}{100}\right)\)

\(A=-\left(\frac{1}{2}\cdot\frac{101}{100}\right)< \left(-\frac{1}{2}\cdot\frac{100}{100}\right)\Leftrightarrow-\frac{101}{200}< \frac{-100}{200}=-\frac{1}{2}\)

\(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{99^2}-1\right)\)

\(=-\frac{3}{2^2}.\frac{-8}{3^2}...\frac{-9999}{100^2}\)

\(=\frac{-\left(3.8...9999\right)}{\left(2.3.4...100\right)^2}=\frac{-\left(1.3.2.4....99.101\right)}{\left(2.3.4...100\right)^2}\)

\(=\frac{-\left[\left(1.2.3..99\right).\left(3.4.5...101\right)\right]}{\left(2.3..4...100\right).\left(2.3.4...100\right)}=\frac{-101}{100.2}=\frac{-101}{200}\)

LẠM DỤNG QUÁ NHIỀU

\(3^{2x+2}=9^{x+3}\)

\(\Rightarrow3^{2x+2}=3^{2x+6}\)

\(\Rightarrow2x+6=2x+2\)

\(\Rightarrow\left(2x-2x\right)+\left(6-2\right)=0\)

\(\Rightarrow0x=-4\left(loại\right)\)

\(b,\left(x-3\right)^4=\left(x-3\right)^6\)

\(\Rightarrow\left(x-3\right)^4-\left(x-3\right)^4.\left(x-3\right)^2=0\)

\(\Rightarrow\left(x-3\right)^4.\left[1-\left(x-3\right)^2\right]=0\)

\(\Rightarrow\hept{\begin{cases}\left(x-3\right)^4=0\\1-\left(x-3\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=3\\x-3\in\left\{\pm1\right\}\end{cases}\Rightarrow}\hept{\begin{cases}x=3\\x\in\left\{4;2\right\}\end{cases}\Rightarrow}x\in\left\{2;3;4\right\}}\)

a, => 3^2x+2  =3^2.(x+3)

        2^x+2=2.(x+3)

        2(x+1)=2(x+3)

           x+1=x+3

=> x= rỗng

vậy............

c, x¹⁵-x²=0

           x²(x¹³-1)=0

\(\orbr{\begin{cases}x^2=0\\x^{13}=0\end{cases}}< =>\orbr{\begin{cases}x=0\\x^{13}=1^{13}\end{cases}}< =>\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

vậy.........

Ở câu a) số mũ lúc nào cug dương mà bạn ( 45^-10 = 45¹⁰). Nếu số mũ là dương thì:

            a)\(\frac{45^{10}.5^{20}}{75^{15}}\)

        = \(\frac{\left(3^2.5\right)^{10}.5^{20}}{\left(3.5^2\right)^{15}}\)

        = \(\frac{3^{20}.5^{10}.5^{20}}{3^{15}.5^{30}}\)

         = \(\frac{3^{20}.5^{30}}{3^{15}.5^{30}}\)

         = \(\frac{3^5.1}{1.1}\)

         = \(\frac{243}{1}\)

         = 243

      b)\(\frac{2^{15}.9^4}{6^6.8^2}\)

          = \(\frac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^2}\)

          = \(\frac{2^{15}.3^8}{2^6.3^6.2^6}\)

          = \(\frac{2^{15}.3^8}{2^{12}.3^6}\)

          = \(\frac{2^3.3^2}{1.1}\)

          = \(\frac{8.9}{1}\)

          = \(\frac{72}{1}\)

          = 72

cảm ơn bn

dễ mà tự mà làm đi

gọi số đó là a

suy ra a+1 chia het cho 4,5,6

muốn a nhỏ nhất thì a+1 nhỏ nhất

số nhỏ nhất chia hết cho 4,5,6 là 60

vậy a=59

số cần tìm là 59

lxl=\(\hept{\begin{cases}-x\\x\end{cases}}\)

x o hai chu i ngan la ma ahihihi k cho 10 diem day