a) \(4.8^6.2.8^3\)

\(=2^2.\left(2^3\right)^6.2.\left(2^3\right)^3\)

\(=2^2.2^{18}.2.2^9\)

\(=2^{2+18+1+9}\)

\(=2^{30}\)

______

b) \(12^2.2.12^3.6\)

\(=12^2.12^3.2.6\)

\(=12^2.12^3.12\)

\(=12^{2+3+1}\)

\(=12^6\)

c) \(6^3.2.6^4.3\)

\(=6^3.6^4.2.3\)

\(=6^3.6^4.6\)

\(=6^{3+4+1}\)

\(6^8\)

a) \(4\cdot8^6\cdot2\cdot8^3\)

\(=2^2\cdot\left(2^3\right)^6\cdot2\cdot\left(2^3\right)^3\)

\(=2^2\cdot2^{18}\cdot2\cdot2^9\)

\(=2^{30}\)

b) \(12^2\cdot2\cdot12^3\cdot6\)

\(=12^2\cdot12\cdot12^3\)

\(=12^6\)

c) \(6^3\cdot2\cdot6^4\cdot3\)

\(=6^3\cdot6\cdot6^4\)

\(=6^8\)

Ta có: \(A=\frac{7x-8}{2x-3}=\frac{1}{2}.\frac{14x-16}{2x-3}=\frac{1}{2}.\frac{14x-21+5}{2x-3}=\frac{1}{2}.\frac{7\left(2x-3\right)+5}{2x-3}\)\(=\frac{1}{2}\left(7+\frac{5}{2x-3}\right)\)

Để A đạt GTLN thì \(\frac{1}{2}\left(7+\frac{5}{2x-3}\right)\) lớn nhất

\(\Rightarrow7+\frac{5}{2x-3}\) lớn nhất

\(\Rightarrow\frac{5}{2x-3}\) lớn nhất

\(\Rightarrow2x-3\) nhỏ nhất hay x nhỏ nhất và x > 0

Vì \(x\inℤ\) nên \(2x-3\inƯ\left(5\right)=\left\{1;5\right\}\)

\(\Rightarrow2x\in\left\{4;8\right\}\)

\(\Rightarrow x\in\left\{2;4\right\}\)

Mà x nhỏ nhất và x > 0 nên x = 2

Thay x = 2 vào A ta được: \(A=\frac{1}{2}.\left(7+\frac{5}{2.2-3}\right)=\frac{1}{2}.12=6\)

Vậy MaxA = 6 tại x = 2.

1.

a) \(2^x=128\)

\(2^x=2^7\)

\(=>x=7\)

b) \(8^{x-1}=64\)

\(8^{x-1}=8^2\)

\(=>x-1=2\)

\(x=2+1\)

\(=>x=3\)

c) \(3+3^x=30\)

\(3^x=30-3\)

\(3^x=27=3^3\)

\(=>x=3\)

d) \(\left(x+2\right)=64\) -> đề có thiếu không vậy?

e) \(3^2.x=3^5\)

\(x=3^5:3^2\)

\(=>x=3^3=27\)

f) \(\left(2x-1\right)^3=343\)

\(\left(2x-1\right)^3=7^3\)

\(=>2x-1=7\)

\(2x=7+1\)

\(2x=8\)

\(x=8:2\)

\(=>x=4\)

\(#Wendy.Dang\) 

a,\(2^x\)=128           b,\(8^{x-1}\)=64              c,3+\(3^x\)=30           d,x+2=64

\(2^7\)=128               \(8^{x-1}\)=\(8^2\)                 \(3^x\)=30-3                  x=64-2

=>x=7              =>x-1=2                  \(3^x\)=27                      x=62

                         x=2+1=3                \(3^x\)=\(3^3\)

                                                     =>x=3

e,\(3^2\).x=\(3^5\)                             f,(2x-\(1^3\))=343

x=\(3^5\):\(3^2\)                                 2x=1+343

x=27                                     2x=344

                                               x=344:2

                                               x=172

\(\Leftrightarrow x^2+x-2=x^2+7x-\frac{1}{2}\)

giả sử căn(x) = a thì x = a²?

1: \(-1\le\sin2x\le1\)

=>\(2\ge-2\cdot\sin2x\ge-2\)

=>2+3>=-2sin2x+3>=-2+3

=>5>=y>=1

y max=5 khi sin 2x=-1

=>\(2x=-\frac{\pi}{2}+k2\pi\)

=>\(x=-\frac{\pi}{4}+k\pi\)

y min=1 khi sin 2x=1

=>\(2x=\frac{\pi}{2}+k2\pi\)

=>\(x=\frac{\pi}{4}+k\pi\)

2: \(-1\le cos7x\le1\)

=>-4<4cos7x<=4

=>-4-2<=4*cos7x-2<=4-2

=>-6<=y<=2

y min=-6 khi cos7x=-1

=>\(7x=\pi+k2\pi\)

=>\(x=\frac{\pi+k2\pi}{7}\)

y max=2 khi cos7x=1

=>\(7x=k2\pi\)

=>\(x=\frac{k2\pi}{7}\)

3: \(0\le\sin^24x\le1\)

=>\(0\ge-\sin^24x\ge-1\)

=>\(0+1\ge-\sin^24x+1\ge-1+1\)

=>\(1\ge-\sin^24x+1\ge0\)

=>\(\frac15\ge\frac{-\sin^24x+1}{5}\ge0\)

y max=1/5 khi \(\sin^24x=0\)

=>sin 4x=0

=>\(4x=k\pi\)

=>\(x=\frac{k\pi}{4}\)

y min=0 khi \(\sin^24x=1\)

=>\(cos^24x=0\)

=>cos4x=0

=>\(4x=\frac{\pi}{2}+k\pi\)

=>\(x=\frac{\pi}{8}+\frac{k\pi}{4}\)

6: \(y=\sin x+cosx+2\)

\(=\sqrt2\cdot\sin\left(x+\frac{\pi}{4}\right)+2\)

Ta có: \(-1\le\sin\left(x+\frac{\pi}{4}\right)\le1\)

=>\(-\sqrt2\le\sqrt2\cdot\sin\left(x+\frac{\pi}{4}\right)\le\sqrt2\)

=>\(-\sqrt2+2\le\sqrt2\cdot\sin\left(x+\frac{\pi}{4}\right)+2\le\sqrt2+2\)

\(y_{\min}=-\sqrt2+2\) khi \(\sin\left(x+\frac{\pi}{4}\right)=-1\)

=>\(x+\frac{\pi}{4}=-\frac{\pi}{2}+k2\pi\)

=>\(x=-\frac34\pi+k2\pi\)

y max=\(\sqrt2+2\) khi \(\sin\left(x+\frac{\pi}{4}\right)=1\)

=>\(x+\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\)

=>\(x=\frac{\pi}{4}+k2\pi\)

khó hiểu vcl

đúng lun ko hiểu một chút nào