a) Ta có:  ( x + 1 ) 4 = ( 2 x ) 4 nên x +1 = 2x. Do đó x = 1.

b) Ta có:  ( 2 x - l ) 5 = x 5 nên 2x - l = x. Do đó x = l.

1.

\(f\left(x\right)=2x^4+6x^3+8x^2+12x+1\)

2.

\(h\left(x\right)=\left(2x^4+6x^3+8x^2+12x+1\right)-\left(2x^4+6x^3+17x^2+12x-26\right)\)

\(=-9x^2+27\)

3.

\(h\left(x\right)=0\Leftrightarrow-9x^2+27=0\)

\(\Leftrightarrow x^2=3\Rightarrow x=\pm\sqrt{3}\)

a)x³-x²=0

⇔x²(x-1)=0

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

b)3x²-5x=0

⇔ x(3x-5)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{3}\end{matrix}\right.\)

c)x³=x⁵

⇔ x³(1-x²)=0

⇔ x³(1-x)(1+x)=0

⇔\(\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

d)(2x+7)²-4(2x+7)=0

⇔ (2x+7)(2x+3)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-7}{2}\\x=\dfrac{-3}{2}\end{matrix}\right.\)

a) Ta có: \(x^3-x^2=0\)

\(\Leftrightarrow x^2\left(x-1\right)=0\)

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

b) Ta có: \(3x^2-5x=0\)

\(\Leftrightarrow x\left(3x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{3}\end{matrix}\right.\)

c) Ta có: \(x^3=x^5\)

\(\Leftrightarrow x^5-x^3=0\)

\(\Leftrightarrow x^3\left(x^2-1\right)=0\)

\(\Leftrightarrow x^3\left(x-1\right)\left(x+1\right)=0\)

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

d) Ta có: \(\left(2x+7\right)^2-4\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x+7\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-7}{2}\\x=\dfrac{-3}{2}\end{matrix}\right.\)

cho M(x) =0

\(=>x^3-25x=0=>x\left(x^2-25\right)=0\)

\(=>\left[{}\begin{matrix}x=0\\x^2-25=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x^2=25=>\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\end{matrix}\right.\)

M(x) =0

\(=>x^5+27x^2=0=>x^2\left(x^3+27\right)=0\)

\(=>\left[{}\begin{matrix}x^2=0\\x^3=-27\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)

M(x)=x³-25x=x.(x²-25)=x.(x-5)(x+5)

c: Ta có: \(\left|x+\dfrac{5}{6}\right|:\dfrac{4}{5}=\dfrac{3}{8}\)

\(\Leftrightarrow\left|x+\dfrac{5}{6}\right|=\dfrac{3}{8}\cdot\dfrac{4}{5}=\dfrac{3}{10}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{5}{6}=\dfrac{3}{10}\\x+\dfrac{5}{6}=-\dfrac{3}{10}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-8}{15}\\x=-\dfrac{17}{15}\end{matrix}\right.\)

a, y \(\times\) \(\dfrac{4}{3}\) = \(\dfrac{16}{9}\)

    y         =    \(\dfrac{16}{9}\) : \(\dfrac{4}{3}\)

    y         = \(\dfrac{4}{3}\)

b, ( y - \(\dfrac{1}{2}\)) + 0,5 = \(\dfrac{3}{4}\)

    y - 0,5 + 0,5 = \(\dfrac{3}{4}\)

   y                   = \(\dfrac{3}{4}\)

c, \(\dfrac{4}{5}-\dfrac{2}{5}y\) = 0,2

   0,8 - 0,4y = 0,2

           0,4y = 0,8 - 0,2

           0,4y  = 0,6

               y = 1,5

d, (y + \(\dfrac{3}{4}\)) \(\times\) \(\dfrac{5}{7}\) = \(\dfrac{10}{9}\)

    y + \(\dfrac{3}{4}\)           = \(\dfrac{10}{9}\) : \(\dfrac{5}{7}\)

   y + \(\dfrac{3}{4}\)            = \(\dfrac{14}{9}\)

y                    = \(\dfrac{14}{9}\) - \(\dfrac{3}{4}\)

 y                   =   \(\dfrac{29}{36}\)

e, y : \(\dfrac{5}{4}\)         = \(\dfrac{9}{5}\)  + \(\dfrac{1}{2}\)

   y : \(\dfrac{5}{4}\)         =   \(\dfrac{23}{10}\)

  y                =      \(\dfrac{23}{10}\)

  y               =   \(\dfrac{23}{8}\)

f, y \(\times\) \(\dfrac{1}{2}\) + \(\dfrac{3}{2}\) \(\times\) y   = \(\dfrac{4}{5}\)

   y \(\times\) ( \(\dfrac{1}{2}+\dfrac{3}{2}\))      =  \(\dfrac{4}{5}\)

   2y                       = \(\dfrac{4}{5}\)

    y                        = \(\dfrac{2}{5}\)

Bài 1:

a) \(2\cdot3\cdot2\cdot3\cdot2\cdot3=2^3\cdot3^3=6^3\)

b) \(100\cdot100\cdot100=100^3=\left(10^2\right)^3=10^6\)

c) \(2x\cdot2x\cdot2x=\left(2x\right)^3=8x^3\)

d) \(2\cdot2^3\cdot2^5=2^{1+3+5}=2^9\)

e) \(3^{10}\cdot3^5\cdot3^4=3^{10+5+4}=3^{19}\)

Bài 2:

\(40-x=2^6\cdot2^2\)

\(\Rightarrow40-x=2^8\)

\(\Rightarrow40-x=256\)

\(\Rightarrow x=40-256\)

\(\Rightarrow x=-216\)

b) \(3^2\cdot3^x=81\)

\(\Rightarrow3^{2+x}=3^4\)

\(\Rightarrow2+x=4\)

\(\Rightarrow x=4-2=2\)

c) \(2^x=512\)

\(\Rightarrow2^x=2^9\)

\(\Rightarrow x=9\)

d) \(x^5=243\)

\(\Rightarrow x^5=3^5\)

\(\Rightarrow x=3\)

Bài 3:

a) \(3^6=3\cdot3\cdot3\cdot3\cdot3\cdot3=729\)

b) \(8^3=\left(2^3\right)^3=2^9=512\)

c) \(3^3\cdot75+3^3\cdot25=3^3\cdot\left(75+25\right)=3^3\cdot100=27\cdot100=2700\)

d) \(2^3\cdot3-\left(1^{10}+8\right):3=2^3\cdot3-9:3=2^3\cdot3-3\cdot3:3=3\cdot\left(2^3-3:3\right)=3\cdot\left(8-1\right)=21\)

e) \(32-\left[4+\left(5\cdot3^2-42\right)\right]-14=18-\left[4+\left(45-42\right)\right]\)

\(=18-\left(4+3\right)\)

\(=18-7=11\)

2:

a: =>40-x=256

=>x=40-256=-216

b: =>x+2=4

=>x=2

c: =>2^x=2^9

=>x=9

d; =>x^5=3^5

=>x=3

ko dang cau hoi linh tinh

5 mũ 56,nha bạn

kết quả là 5 ^ ⁵⁶

a=10

a=24

câu 1 giống các bn ấy nhưng ko bít cách làm

720 : ( a * 2 + a * 3 ) = 2 * 3 

720 : ( a * 5 ) = 6

720 : 6 : 5 

= 120 : 5 

a = 24