\(4\left(x-5\right)-16=24\cdot3\)

\(\Rightarrow4\left(x-5\right)-16=72\)

\(\Rightarrow4\left(x-5\right)=72+16\)

\(\Rightarrow4\left(x-5\right)=88\)

\(\Rightarrow x-5=\dfrac{88}{4}\)

\(\Rightarrow x-5=22\)

\(\Rightarrow x=22+5\)

\(\Rightarrow x=27\)

\(4.\left(x-5\right)-16=24.3\)

\(4.\left(x-5\right)-16=72\)

\(4.\left(x-5\right)=72+16\)

\(4.x-5=88\)

\(x-5=88:4\)

\(x-5=22\)

\(x=22+5\)

\(x=27\)

$4.\left(x-5\right)-16=24.3$

$4.\left(x-5\right)-16=72$

$4.\left(x-5\right)=72+16$

$4.x-5=88$

$x-5=88:4$

$x-5=22$

$x=22+5$

$x=27$

\(4\left(x-5\right)-2^3=2^4.3\)

\(\Rightarrow4x-20-8=48\)

\(\Rightarrow4x=76\Rightarrow x=19\)

4(x-5) - 2^3 =2^4.3

4(x-5)-2^3=16.3

4(x-5)-2^3=48

4(x-5)=48+2^3

4(x-5)=56

x-5=56:4

x-5=14

x=14+5

x=19

sai mong mn thông cảm nha

\(4\cdot\left(5-x\right)-2^3=2^4\cdot3\)

\(\Rightarrow4\cdot\left(5-x\right)-2^3=16\cdot3\)

\(\Rightarrow4\cdot\left(5-x\right)-2^3=48\)

\(\Rightarrow4\cdot\left(5-x\right)=48+8\)

\(\Rightarrow4\cdot\left(5-x\right)=56\)

\(\Rightarrow5-x=56:4\)

\(\Rightarrow5-x=14\)

\(\Rightarrow x=5-14\)

\(\Rightarrow x=-9\)

\(4\left(x-5\right)-23=24\cdot3\)

\(4\left(x-5\right)-23=72\)

\(4\left(x-5\right)=72+23\)

\(4\left(x-5\right)=95\)

\(x-5=\dfrac{95}{4}\)

\(x=\dfrac{95}{4}+5\)

\(x=\dfrac{115}{4}\)

\(4(x–5)-23=24.3\)

\(4(x–5)-23=72\)

\(4(x–5)=72+23\)

\(4(x-5)=95\)

\(x-5=95:4\)

\(x-5=23,75\)

\(x=23,75+5\)

\(x=28,75\)

4 ( x - 5 ) - 23 = 24 . 3

=> 4(x-5)-23=72

=> 4(x-5)=72+23

=> 4(x-5)=95

=> x-5=95/4

=> x-5 = 23,75

=> x = 23,75 + 5

=>x = 28,75

Đáp án cần chọn là: C

Bất phương trình : \(\Leftrightarrow2^{3x^2}< 2^3.3^{1-x}\Leftrightarrow2^{3x^2-3}< 3^{1-x}\)

                                                       \(\Leftrightarrow\left(3x^2-3\right)\log_32< 1-x\)

                                                       \(\Leftrightarrow\left(x-1\right)\left[3\left(x+1\right)\log_32+1\right]< 0\) 

                                                       \(\Leftrightarrow-\frac{3\log_32+1}{3\log_32}< x< 1\)

Vậy tập nghiệm của bất phương trình là : 

              \(S=\left(-\frac{3\log_32+1}{3\log_32};1\right)\)

ta phân tích thành

\(\frac{1}{1\cdot2}\)x\(\frac{2\cdot2}{2\cdot3}\)x\(\frac{3\cdot3}{3\cdot4}\)x......x\(\frac{5\cdot5}{5\cdot6}\)x\(\frac{6\cdot6}{6\cdot7}\)

Tử nhân tử mẫu nhân mẫu ta có

1x2x2x3x3x.........x5x5x6x6

1x2x2x3x3x4x....x5x6x6x7

Rút gọn ta có

\(\frac{1}{7}\)

Vậy A=\(\frac{1}{7}\)