`a)`

Quy đồng mẫu số ta có :

`4/15 = 8/30`

`1/3 = 10/30`

`=> 8 < x < 10`

`=> x ∈{ 9}`

____________________________________________

`(-5)/12 < x/9 < 2/(-9)`

Quy đồng mẫu số :

`(-5)/12 = (-15)/36`

`2/(-9) = 8/(-36)`

`=>(-5) < x < (-36)`

`=> x ∈ ∅`

a) 9/30

3,3(x - 1,2) = 9,9

(x - 1,2) = 9,9 : 3,3

(x - 1,2) = 3

x = 3 + 1,2

x = 4,2.

\(3,3x-3,96=9,9\Rightarrow x=4,2\)

A = -1 -2 - 3 -4 -......-2010

A = - (1+2+3+4+......2010)

A = - (2010 +1)[ (2010  -1):1 +1] : 2

A = - 2011.2010:2

A = -2021055

-x^2-1=10
-x^2   = 10+1

-x^2   = 11

=> 11 không là tích của hai số nào nên 11 là hợp số

=> Không có giá trị thỏa mãn

   Vậy x không có giá trị thỏa mãn.

Đúng ko??

-x^2-1 = 10

=> -x¹ = 10

=> -x = 10

=> x = -10.

Lời giải:
\(B=\frac{4.\frac{17}{4}}{\frac{119}{36}.\frac{16}{7}}=\frac{17}{\frac{68}{9}}=17.\frac{9}{68}=\frac{9}{4}\)

a)2.3^x+1=10.3¹²+8.3¹²

=>2.3^x+1=3¹².18

=>3^x+1=3¹².9

=>3^x+1=3¹².3²

=>3^x+1=3¹⁴

=>x+1=14

=>x=13

b)2^x+2^x+3=144

=>2^x+2^x.2³=144

=>2^x(1+8)=144

=>2^x.9=144

=>2^x=16

=>2^x=2⁴

=>x=4

anh giải câu b nhá, câu a anh không biết đề em viết kiểu sao

\(2^x+2^{x+3}=144\)

\(2^x+2^x.8=144\)

\(2^x\left(1+8\right)=144\)

\(2^x=16=2^4\)

\(\Rightarrow x=4\)

\(PA=2PB\Rightarrow\dfrac{PA}{AB}=\dfrac{2}{3};QA=2QC\Rightarrow\dfrac{QA}{AC}=\dfrac{2}{3}\)

Hai tg APK và tg ABK có chung đường cao từ K->AB nên

\(\dfrac{S_{APK}}{S_{ABK}}=\dfrac{PA}{AB}=\dfrac{2}{3}\Rightarrow S_{APK}=\dfrac{2}{3}xS_{ABK}\)

Hai tg AQK và tg ACK có chung đường cao từ K->AC nên

\(\dfrac{S_{AQK}}{S_{ACK}}=\dfrac{QA}{AC}=\dfrac{2}{3}\Rightarrow S_{AQK}=\dfrac{2}{3}xS_{ACK}\)

\(\Rightarrow S_{APKQ}=S_{APK}+S_{AQK}=\dfrac{2}{3}x\left(S_{ABK}+S_{ACK}\right)=\dfrac{2}{3}xS_{ABC}=\dfrac{2}{3}x360=240cm^2\)

Ta có:

\(S_{APM}=\dfrac{2}{3}\times S_{ABM}\) (chung đường cao hạ từ \(M\), \(AP=\dfrac{2}{3}\times AB\))

\(S_{APK}=\dfrac{1}{2}\times S_{APM}\) (chung đường cao hạ từ \(P\), \(AK=\dfrac{1}{2}\times AM\))

\(=\dfrac{1}{2}\times\dfrac{2}{3}\times S_{ABM}=\dfrac{1}{3}\times S_{ABM}\)

Tương tự \(S_{AQK}=\dfrac{1}{3}\times S_{ACM}\)

suy ra \(S_{APKQ}=S_{APK}+S_{AQK}=\dfrac{1}{3}\times S_{ABM}+\dfrac{1}{3}\times S_{ACM}\)

\(=\dfrac{1}{3}\times\left(S_{ABM}+S_{ACM}\right)=\dfrac{1}{3}\times S_{ABC}=\dfrac{1}{3}\times360=120\left(cm^2\right)\)

\(\left(5x-4\right)^{10}=\left(5x-4\right)^7\) 

\(\left(5x-4\right)^{10}-\left(5x-4\right)^7=0\)

\(\left(5x-4\right)^7\left\{\left(5x-4\right)^3-1\right\}=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(5x-4\right)^7=0\\\left(5x-4\right)^3-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x-4=0\\\left(5x-4\right)^3=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\5x-4=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=1\end{matrix}\right.\)

4/5 và 1 nha

`4.(x-3)-7.7=1`

`4.(x-3)-49=1`

`4.(x-3)=1+49=50`

`x-3=50:4=25/2`

`x=25/2+3=31/2`

_____________________________________________________

`9.(x+4)-25=5.4`

`9.(x+4)-25=20`

`9.(x+4)=20+25=45`

`x+4=45:9=5`

`x=5-4=1`

\(4.\left(x-3\right)-7.7=1\)

\(4.\left(x-3\right)-49=1\)

\(4.\left(x-3\right)=1+49\)

\(4.\left(x-3\right)=50\)

\(\left(x-3\right)=50\div4\)

\(x-3=12,5\)

\(x=12,5+3\)

\(x=15,5\)

\(9.\left(x+4\right)-25=5.4\)

\(9.\left(x+4\right)-25=20\)

\(9.\left(x+4\right)=20+25\)

\(9.\left(x+4\right)=45\)

\(\left(x+4\right)=45\div9\)

\(x+4=5\)

\(x=5-4\)

\(x=1\)

A = \(\dfrac{1}{9}\) + \(\dfrac{1}{18}\) + \(\dfrac{1}{30}\) + ......+ \(\dfrac{1}{135}\)

A = \(\dfrac{1}{3}\) ( \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) +........+ \(\dfrac{1}{45}\))

A = \(\dfrac{2}{3}\) ( \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\) +.....+\(\dfrac{1}{90}\))

A = \(\dfrac{2}{3}\) ( \(\dfrac{1}{2\times3}\) + \(\dfrac{1}{3\times4}\) + \(\dfrac{1}{4\times5}\)+......+\(\dfrac{1}{9\times10}\))

A = \(\dfrac{2}{3}\)( \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)+  \(\dfrac{1}{3}\)- \(\dfrac{1}{4}\)+  \(\dfrac{1}{4}\)-  \(\dfrac{1}{5}\)+........+ \(\dfrac{1}{9}\)- \(\dfrac{1}{10}\))

A = \(\dfrac{2}{3}\)( \(\dfrac{1}{2}\)- \(\dfrac{1}{10}\))

A = \(\dfrac{2}{3}\) x( \(\dfrac{5}{10}\) - \(\dfrac{1}{10}\))

A = \(\dfrac{2}{3}\)x \(\dfrac{4}{10}\)

A = \(\dfrac{4}{15}\)

ai giúp mình nhanh đi mà