Phương trình không có nghiệm nguyên \(x\), nhưng có nghiệm duy nhất trong khoảng:

\(6 < x < 7.\)

Nếu muốn nghiệm gần đúng:

\(x \approx \left(log ⁡\right)_{3} \left(\right. 850.5 \left.\right) \approx 6.1.\)

\(\frac{21}{x}=\frac{7}{-4}\Leftrightarrow7x=21.\left(-4\right)\Leftrightarrow7x=-84\Leftrightarrow x=-84:7\Leftrightarrow x=-12\)

\(\frac{114}{2x}=-\frac{8}{12}\Leftrightarrow\frac{57}{x}=-\frac{2}{3}\Leftrightarrow-2x=57.3\Leftrightarrow2x=171\Leftrightarrow x=\frac{171}{2}\)

Bài 3:

a) \(\left(x-\frac{1}{2}\right)^2=0\)

\(\Rightarrow x-\frac{1}{2}=0\)

\(\Rightarrow x=\frac{1}{2}\)

Vậy \(x=\frac{1}{2}\)

b) \(\left(x-2\right)^2=1\)

\(\Rightarrow x-2=\pm1\)

+) \(x-2=1\Rightarrow x=3\)

+) \(x-2=-1\Rightarrow x=1\)

Vậy \(x=3\) hoặc \(x=1\)

c) \(\left(2x-1\right)^3=-8\)

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

\(\Rightarrow2x-1=-2\)

\(\Rightarrow2x=-1\)

\(\Rightarrow x=\frac{-1}{2}\)

Vạy \(x=\frac{-1}{2}\)

d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)

\(\Rightarrow\left(x+\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2\)

\(\Rightarrow x+\frac{1}{2}=\frac{1}{4}\)

\(\Rightarrow x=\frac{-1}{4}\)

Vậy \(x=\frac{-1}{4}\)

 A=5-3(2x+1)^2

Ta có : (2x+1)^2\(\ge\)0

\(\Rightarrow\)-3(2x-1)^2\(\le\)0

\(\Rightarrow\)5+(-3(2x-1)^2)\(\le\)5

Dấu = xảy ra khi : (2x-1)^2=0

=> 2x-1=0 =>x=\(\frac{1}{2}\)

Vậy : A=5 tại x=\(\frac{1}{2}\)

Ta có : (x-1)^2 \(\ge\)0

=> 2(x-1)^2\(\ge\)0

=>2(x-1)^2+3 \(\ge\)3

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

Dấu = xảy ra khi : (x-1)^2 =0

=> x = 1

Vậy : B = \(\frac{1}{3}\)khi x = 1

\(\frac{x^2+8}{x^2+2}\)= \(\frac{x^2+2+6}{x^2+2}=1+\frac{6}{x^2+2}\)

Làm như câu B                   GTNN = 4 khi x =0 

k vs nha

bài 12 :

a,\(\left(x-\frac{1}{2}\right)^2=0\)

Mà: 0²=0

=> \(\left(x-\frac{1}{2}\right)^2=0^2\)

\(\Rightarrow x-\frac{1}{2}=0\)

\(\Rightarrow x=\frac{1}{2}\)

b,  \(\left(x-2\right)^2=1\)

Mà : 1=1²

\(\Rightarrow\left(x-2\right)^2=1^2\)

=> x - 2 = 1

=> x = 3

c, \(\left(2x-1\right)^3=-8\)

\(\Rightarrow\left(2x-1\right)=-2\)

Vì -8 =-2³

nên ...

=> 2x =-1

=> x=0.5

d.\(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)

cái này cũng như mấy cái trên thôi

Bài 12:

a) \(\left(x-\frac{1}{2}\right)^2=0\)

\(\Rightarrow x-\frac{1}{2}=0\)

\(x=\frac{1}{2}\)

b) \(\left(x-2\right)^2=1\)

\(x-2=\pm1\)

• Nếu \(x-2=1\)

\(x=3\)

• Nếu \(x-2=-1\)

\(x=1\)

c) \(\left(2x-1\right)^3=-8\)

\(\Rightarrow2x-1=-2\)

\(2x=-1\)

\(x=-\frac{1}{2}\)

d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)

\(x+\frac{1}{12}=\pm\frac{1}{4}\)

• Nếu \(x+\frac{1}{12}=\frac{1}{4}\)

\(x=\frac{1}{6}\)

• Nếu \(x+\frac{1}{12}=-\frac{1}{4}\)

\(x=-\frac{1}{3}\)

Bài 13: có người làm rồi

Bài 14:

a) \(25^3\div5^2\)

\(=\left(5^2\right)^3\div5^2\)

\(=5^6\div5^2=5^4\)

b) \(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6\)

\(=\left(\frac{3}{7}\right)^{21}:\left[\left(\frac{3}{7}\right)^2\right]^6\)

\(=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{12}=\left(\frac{3}{7}\right)^9\)

c) \(3-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2\)

\(=3-1+\frac{1}{4}:2\)

\(=2+\frac{1}{8}=2\frac{1}{8}\)

a) \(\left|3-2x\right|+\frac{3}{4}=\left|-2\frac{3}{4}\right|\)

⇔ | 3 - 2x | + 3/4 = 11/4

⇔ | 3 - 2x | = 8/4 = 2

⇔ \(\orbr{\begin{cases}3-2x=2\\3-2x=-2\end{cases}}\text{⇔}\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{5}{2}\end{cases}}\)

b) 2^x+2 - 2^x = 96

⇔ 2^x( 2² - 1 ) = 96

⇔ 2^x.3 = 96

⇔ 2^x = 32

⇔ 2^x = 2⁵

⇔ x = 5

c) ( 2x + 5 )³ = -27

⇔ ( 2x + 5 )³ = (-3)³

⇔ 2x + 5 = -3

⇔ 2x = -8

⇔ x = -4

a. \(\left|3-2x\right|+\frac{3}{4}=\left|-2\frac{3}{4}\right|\)

\(\Rightarrow\left|3-2x\right|+\frac{3}{4}=\left|-\frac{11}{4}\right|\)

\(\Rightarrow\left|3-2x\right|+\frac{3}{4}=\frac{11}{4}\)

\(\Rightarrow\left|3-2x\right|=2\)

\(\Leftrightarrow\orbr{\begin{cases}3-2x=2\\3-2x=-2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{5}{2}\end{cases}}\)

b. 2^x+2 - 2^x = 96

<=> 2^x . 2² - 2^x = 96

<=> 2^x ( 2² - 1 ) = 96

<=> 2^x . 3 = 96

<=> 2^x = 32 = 2⁵

<=> x = 5

c. ( 2x + 5 )³ = - 27

<=> ( 2x + 5 )³ = ( - 3 )³

<=> 2x + 5 = - 3

<=> 2x = - 8

<=> x = - 4

e)

\(\left(x+3\right)^3=\left(x+3\right)^5\)

\(\Rightarrow\)\(x+3=1;0\)

TH1:                                                                   TH2

\(x+3=0\)                                                 \(x+3=1\)

\(x=-3\)                                                      \(x=-2\)

\(x\in\left\{-3;-2\right\}\)

a) /2x/-/2,5/=/-7,5/

/2x/-(-2,5)=7,5

/2x/        =7,5+(-2,5)

/2x/        =5

2x=5     hoặc       2x= -5

  x=5:2                  x= -5:2

  x=2,5                  x= -2,5

Vậy x=2,5 hoặc x= -2,5

a) 3(x-4)+2,6=5,9

\(3\left(x-4\right)=3,3\)

\(x-4=\frac{11}{10}\)

\(x=\frac{51}{10}\)

b)|1+x|+3=3

\(\Rightarrow|1+x|=0\)

\(1+x=0\)

\(x=-1\)

c) \(\left(\frac{-2}{3}\right)^x=\frac{-8}{27}\)

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

\(x=3\)

chúc bạn học tốt

a)\(\left(-3\right)^{x+3}=-\frac{1}{27}\)

\(\left(-3\right)^{x+3}=\left(-\frac{1}{3}\right)^3\)

\(\left(-3\right)^{x+3}=\left(-\frac{3^0}{3^1}\right)^3\)

\(\left(-3\right)^{x+3}=\left(-3^{-1}\right)^3\)

\(\left(-3\right)^{x+3}=\left(-3\right)^{-3}\)

\(\Rightarrow x+3=-3\)

\(\Rightarrow x=-6\)

b)\(\left(-6\right)^{2x+2}=\frac{1}{36}\)

\(\left(-6\right)^{2x+2}=\left(-\frac{1}{6}\right)^2\)

\(\left(-6\right)^{2x+2}=\left(-\frac{6^0}{6^1}\right)^2\)

\(\left(-6\right)^{2x+2}=\left(-6^{-1}\right)^2\)

\(\left(-6\right)^{2x+2}=\left(-6\right)^{-2}\)

\(\Rightarrow2x+2=-2\)

\(\Rightarrow2x=-4\)

\(\Rightarrow x=-2\)

c)\(\left(-3\right)^{x+5}=\frac{1}{81}\)

\(\left(-3\right)^{x+5}=\left(-\frac{1}{3}\right)^4\)

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

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

\(\left(-3\right)^{x+5}=\left(-3\right)^{-4}\)

\(\Rightarrow x+5=-4\)

\(\Rightarrow x=-9\)

d)\(\left(\frac{1}{9}\right)^x=\left(\frac{1}{27}\right)^6\)

\(\left[\left(\frac{1}{3}\right)^2\right]^x=\left[\left(\frac{1}{3}\right)^3\right]^6\)

\(\left(\frac{1}{3}\right)^{2x}=\left(\frac{1}{3}\right)^{18}\)

\(\Rightarrow2x=18\)

\(\Rightarrow x=9\)

e)\(\left(\frac{4}{9}\right)^x=\left(\frac{8}{27}\right)^6\)

\(\left[\left(\frac{2}{3}\right)^2\right]^x=\left[\left(\frac{2}{3}\right)^3\right]^6\)

\(\left(\frac{2}{3}\right)^{2x}=\left(\frac{2}{3}\right)^{18}\)

\(\Rightarrow2x=18\)

\(\Rightarrow x=9\)