a: \(\left(\sqrt{3}\right)^x=243\)

=>\(3^{\dfrac{1}{2}\cdot x}=3^5\)

=>\(\dfrac{1}{2}\cdot x=5\)

=>x=10

b: \(0,1^x=1000\)

=>\(\left(\dfrac{1}{10}\right)^x=1000\)

=>\(10^{-x}=10^3\)

=>-x=3

=>x=-3

c: \(\left(0,2\right)^{x+3}< \dfrac{1}{5}\)

=>\(\left(0,2\right)^{x+3}< 0,2\)

=>x+3>1

=>x>-2

d: \(\left(\dfrac{3}{5}\right)^{2x+1}>\left(\dfrac{5}{3}\right)^2\)

=>\(\left(\dfrac{3}{5}\right)^{2x+1}>\left(\dfrac{3}{5}\right)^{-2}\)

=>2x+1<-2

=>2x<-3

=>\(x< -\dfrac{3}{2}\)

e: \(5^{x-1}+5^{x+2}=3\)

=>\(5^x\cdot\dfrac{1}{5}+5^x\cdot25=3\)

=>\(5^x=\dfrac{3}{25,2}=\dfrac{1}{8,4}=\dfrac{10}{84}=\dfrac{5}{42}\)

=>\(x=log_5\left(\dfrac{5}{42}\right)=1-log_542\)

a: =>x-1/2=1/3

=>x=5/6

b: =>|2x-1|=x+1

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-1\\\left(2x-1-x-1\right)\left(2x-1+x+1\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-1\\\left(x-2\right)\left(3x\right)=0\end{matrix}\right.\)

hay \(x\in\left\{2;0\right\}\)

c: \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{3}{5}>\dfrac{2}{5}\\\dfrac{1}{2}x-\dfrac{3}{5}< -\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x>1\\\dfrac{1}{2}x< \dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\x< \dfrac{2}{5}\end{matrix}\right.\)

Lời giải:

a.

$(\frac{-1}{3})^3.x=\frac{1}{81}=(\frac{-1}{3})^4$
$\Rightarrow x=(\frac{-1}{3})^4: (\frac{-1}{3})^3=\frac{-1}{3}$
b.

$2^2.16> 2^x> 4^2$
$\Rightarrow 2^2.2^4> 2^x> (2^2)^2$

$\Rightarrow 2^6> 2^x> 2^4$

$\Rightarrow 6> x> 4$

$\Rightarrow x=5$ (với điều kiện $x$ là số tự nhiên nhé)

c.

$9.27< 3^x< 243$

$3.3^3< 3^x< 3^5$

$\Rightarrow 3^4< 3^x< 3^5$

$\Rightarrow 4< x< 5$

Với $x$ là stn thì không có số nào thỏa mãn.

a: \(2^{x^2-1}=256\)

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

=>\(x^2-1=8\)

=>\(x^2=9\)

=>\(x\in\left\{3;-3\right\}\)

b: \(3^{x^2+3x}=81\)

=>\(3^{x^2+3x}=3^4\)

=>\(x^2+3x=4\)

=>\(x^2+3x-4=0\)

=>(x+4)(x-1)=0

=>\(\left[{}\begin{matrix}x+4=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=1\end{matrix}\right.\)

c: \(2^{x^2-5x}=64\)

=>\(2^{x^2-5x}=2^6\)

=>\(x^2-5x=6\)

=>\(x^2-5x-6=0\)

=>(x-6)(x+1)=0

=>\(\left[{}\begin{matrix}x-6=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-1\end{matrix}\right.\)

d: \(\left(\dfrac{1}{3}\right)^x=243\)

=>\(\left(\dfrac{1}{3}\right)^x=3^5=\left(\dfrac{1}{3}\right)^{-5}\)

=>x=-5

e: \(\left(\dfrac{1}{3}\right)^{x+5}=3^{2x+1}\)

=>\(3^{-x-5}=3^{2x+1}\)

=>-x-5=2x+1

=>-3x=6

=>x=-2

a: =>x=(-2/3)^5:(-2/3)^2=(-2/3)^3=-8/27

b: =>x*(-1/3)^3=(-1/3)^4

=>x=-1/3

d: =>3x-2=-3

=>3x=-1

=>x=-1/3

a.\(3^{x-1}=243\)

\(3^x:3^1=243\)

\(3^x=729\)

\(\Leftrightarrow3^6=729\)

\(\Leftrightarrow x=6\)

b.\(\left(\dfrac{2}{3}\right)^{x+1}=\dfrac{8}{4}\)

\(\left(\dfrac{2}{3}\right)^x.\left(\dfrac{2}{3}\right)=\dfrac{8}{4}\)

\(\left(\dfrac{2}{3}\right)^x=3\)

Câu b tính đến đây rồi không mò đc x nữa.

Ta chứng minh 2 bất đẳng thức phụ sau: với x, y, z dương thì:

\(x^4+y^4+z^4\ge xyz\left(x+y+z\right)\left(1\right)\)

\(\left(1+x\right)\left(1+y\right)\left(1+z\right)\ge\left(1+\sqrt[3]{xyz}\right)^3\left(2\right)\)

+ Chứng minh BĐT (1), sử dụng BĐT AM - GM:

\(x^4+x^4+y^4+z^4\ge4x^2yz\)

\(y^4+y^4+x^4+z^4\ge4xy^2z\)

\(z^4+z^4+x^4+y^4\ge4xyz^2\)

Cộng dồn lại ta có: \(x^4+y^4+z^4\ge xyz\left(x+y+z\right)\)

+ Chứng minh BĐT (2). Ta có:

\(\left(1+x\right)\left(1+y\right)\left(1+z\right)=1+x+y+z+xy+yz+xyz\ge1+3\sqrt[3]{xyz}+3\sqrt[3]{x^2y^2z^2}+xyz=\left(1+\sqrt[3]{xyz}\right)^3\)

Bây giờ ta quay lại chứng minh BĐT ở đề.

BĐT cần chứng minh tương đương với BĐT sau:

\(\sqrt[4]{\left(1+\dfrac{1}{a}\right)^4+\left(1+\dfrac{1}{b}\right)^4+\left(1+\dfrac{1}{c}\right)^4}\ge\sqrt[4]{3}+\dfrac{\sqrt[4]{243}}{2+abc}\)

\(\Leftrightarrow\left(1+\dfrac{1}{a}\right)^4+\left(1+\dfrac{1}{b}\right)^4+\left(1+\dfrac{1}{c}\right)^4\ge3\left(1+\dfrac{3}{2+abc}\right)^4\)

Sử dụng BĐT (1) ta có:

\(\left(1+\dfrac{1}{a}\right)^4+\left(1+\dfrac{1}{b}\right)^4+\left(1+\dfrac{1}{c}\right)^4\ge\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\left(3+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

Sử dụng BĐT (2) và BĐT AM - GM ta có:

\(\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\left(3+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\left(1+\dfrac{1}{\sqrt[3]{abc}}\right)^3\left(3+\dfrac{3}{\sqrt[3]{abc}}\right)\)

\(\Rightarrow\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\left(3+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\left(1+\dfrac{1}{\sqrt[3]{abc.1.1}}\right)^4\ge3\left(1+\dfrac{3}{2+abc}\right)^4\)

Vậy BĐT đã được chứng minh. Đẳng thức xảy ra <=> a = b = c.

h) \(5^x+5^{x+2}=650\)

\(\Leftrightarrow5^x+5^x.5^2=650\)

\(\Leftrightarrow5^x\left(1+25\right)=650\)

\(\Leftrightarrow5^x.26=650\)

\(\Leftrightarrow5^x=25\)

\(\Leftrightarrow x=2\)

haizzz,đăng ít thôi,chứ nhìn hoa mắt quá =.=

4, \(\Leftrightarrow4x+4+9\left(2x+1\right)=4x+6\left(x+1\right)+7+12x\)

\(\Leftrightarrow22x+13=22x+13\)vậy pt có vô số nghiệm 

5, \(\dfrac{2x}{3}+\dfrac{2x-1}{6}=4-\dfrac{x}{3}\Rightarrow4x+2x-1=24-2x\)

\(\Leftrightarrow8x=25\Leftrightarrow x=\dfrac{25}{8}\)

6, \(\dfrac{x-1}{2}+\dfrac{x-1}{4}=1-\dfrac{2\left(x-1\right)}{3}\Rightarrow6x-6+3x-3=12-8\left(x-1\right)\)

\(\Leftrightarrow9x-9=20-8x\Leftrightarrow17x=29\Leftrightarrow x=\dfrac{29}{17}\)