3¹⁴ : ( 3⁹.27 ) + 5².2⁵ - 4³

= 3¹⁴ : ( 3⁹.3³) + 5².2⁵ - ( 2²)³

= 3¹⁴ : 3¹² + 5².2⁵ - 2⁶

= 3² + 2⁵.(25 - 2)

= 9 + 32.23

= 9 + 736

=745

2³.5 - 3².4 + 4.6

= 2³.5 - 3².2² + 2².2.3

= 2².( 2.5 - 9 + 2.3)

= 4 .( 10 - 9 + 6 )

= 4 . 7

= 28

kq  = 28

a: =5-78*32

=5-2496

=-2491

b: \(=6\left(9-6\right)=6\cdot3=18\)

c: \(=46\cdot\dfrac{\left(123-42\right)}{81}=46\)

d: \(=181+3-84+8\cdot25\)

=100+200

=300

e: \(=64\cdot35+140\cdot84-1=2240-1+11760\)

=14000-1

=13999

f: \(=3^3+25\cdot8-1=26+200=226\)

g: \(=3+2^4+1=16+4=20\)

h: \(=36:4\cdot3+2\cdot25-1=27+50-1=27+49=76\)

1.  25 : 5,7 = 250/57

2.  30:2.8.4 = 480

3.  20:2^2.14= 70

4.  125:5^3.170= 170

5.  64:2^5.30.4=240

6.  (25:5^2.30): 15.7=14

       bạn à! Nhiều quá mình ko làm hết được. sorry nha.^-^                     

dù dài quá nhưng cố  gắng giúp mình làm hết nha

a: \(=\dfrac{-4\cdot13\cdot9\cdot5}{3\cdot4\cdot5\cdot2\cdot13}=\dfrac{3}{2}\)

b: \(=\dfrac{1}{2}\cdot\dfrac{1}{3}\cdot5=\dfrac{5}{6}\)

\(\frac{4^2.2^5+32.125}{2^3.5^2}\) \(=\)\(\frac{4^2.2^5+2^5.5^3}{2^3.5^2}\)

                                     \(=\frac{\left(4^2+5^3\right).2^5}{2^3.5^2}\)

                                     \(=\)  \(\frac{\left(4^2+5^3\right).2^2}{5^2}\)

                                    \(=\)    \(\frac{564}{25}\)

\(\frac{4^2.2^5+32.125}{2^3.5^2}=\frac{4^2.2^5+2^5.5^3}{2^3.2^5}\)

                                   \(=\frac{(4^2+5^3).2^5}{2^3.2^5}\)

                                     \(=\frac{(4^2+5^3).2^2}{5^2}\)

                                       \(=\frac{564}{25}\)

\(6^2.2^2.5:\left[3.12-\left(2x-6\right)\right]=2^3.5\)

\(\Rightarrow36.4.5:\left[36-\left(2x-6\right)\right]=8.5\)

\(\Rightarrow720:\left[36-\left(2x-6\right)\right]=40\)

\(\Rightarrow36-\left(2x-6\right)=720:40=18\)

\(\Rightarrow2x-6=36-18=18\)

\(\Rightarrow2x=18+6=24\)

\(\Rightarrow x=24:2=12\)

1. 

$(3^2-2^3)x+3^2.2^2=4^2.3$

$\Leftrightarrow x+36=48$

$\Leftrightarrow x=48-36=12$

2.

$x^5-x^3=0$

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

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

$\Leftrightarrow x^3=0$ hoặc $x-1=0$ hoặc $x+1=0$

$\Leftrightarrow x=0$ hoặc $x=\pm 1$
3.

$(x-1)^2+(-3)^2=5^2(-1)^{100}$

$\Leftrightarrow (x-1)^2+9=25$

$\Leftrightarrow (x-1)^2=25-9=16=4^2=(-4)^2$

$\Rightarrow x-1=4$ hoặc $x-1=-4$

$\Leftrightarrow x=5$ hoặc $x=-3$

4.

$(2x-1)^2-(2x-1)=0$

$\Leftrightarrow (2x-1)(2x-1-1)=0$

$\Leftrightarrow (2x-1)(2x-2)=0$

$\Leftrightarrow 2x-1=0$ hoặc $2x-2=0$

$\Leftrightarrow x=\frac{1}{2}$ hoặc $x=1$

$\Lef

`@` `\text {Ans}`

`\downarrow`

\((3^2-2^3)x+3^2.2^2=4^2.3\)

`=> x + (3*2)^2 = 48`

`=> x+6^2 = 48`

`=> x + 36 = 48`

`=> x = 48 - 36`

`=> x=12`

Vậy, `x=12`

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

`=> x^3(x^2 - 1)=0`

`=>`\(\left[{}\begin{matrix}x^3=0\\x^2-1=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0\\x^2=1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)

Vậy, `x \in {0; +- 1 }`

\(\left(x-1\right)^2+\left(-3\right)^2=5^2\cdot\left(-1\right)^{100}\)

`=> (x-1)^2 + 9 = 25*1`

`=> (x-1)^2 + 9 = 25`

`=> (x-1)^2 = 25 - 9`

`=> (x-1)^2 = 16`

`=> (x-1)^2 = (+-4)^2`

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

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

`=>`\(\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

Vậy, `x \in {5; -3}`

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

`=> (2x-1)(2x-1) - (2x-1)=0`

`=> (2x-1)(2x-1-1)=0`

`=>`\(\left[{}\begin{matrix}2x-1=0\\2x-2=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}2x=1\\2x=2\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)

Vậy, `x \in {1; 1/2}`

Ta có: \(5^2\cdot x-3-2\cdot2^5=5^2\cdot3\)

\(\Leftrightarrow25\cdot x-3-64=75\)

\(\Leftrightarrow25\cdot x=122\)

hay \(x=\dfrac{122}{25}\)

Vậy: \(x=\dfrac{122}{25}\)

\(3^2.2^4-2^3.4^2-2^3.5^0+4^2.2^2\)

\(=9.16-8.16-8.1+16.4\)

\(=16.\left(9-8\right)-8+64\)

\(=16.1-8+64\)

\(=8+64\)

\(=72\)

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