1) Tìm số nguyên x, biết : 

a) 3^{x + 2} + 4 . 3^{x + 1} = 7.3⁶

3^{x + 1} ( 3 + 4) = 7.3⁶

3^{x + 1} .7 = 7.3⁶

=> x + 1 = 6

=> x = 5

b) 4^{x + 3} - 3.4^{x + 1} = 13 . 4¹¹

4^{x + 1} ( 4² - 3) = 13 . 4¹¹

4^{x + 1} . 13 = 13 . 4¹¹

=> x + 1 = 11

=> x = 10

mk nghĩ là đúng

Bài 1: 

a: \(\left(2x-1\right)^4=16\)

=>2x-1=2 hoặc 2x-1=-2

=>2x=3 hoặc 2x=-1

=>x=3/2 hoặc x=-1/2

b: \(\left(2x-y+7\right)^{2012}+\left|x-3\right|^{2013}< =0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-y+7=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=2x+7=y=2\cdot3+7=13\end{matrix}\right.\)

c: \(10800=2^4\cdot3^3\cdot5^2\)

mà \(2^{x+2}\cdot3^{x+1}\cdot5^x=10800\)

nên \(\left\{{}\begin{matrix}x+2=4\\x+1=3\\x=2\end{matrix}\right.\Leftrightarrow x=2\)

Tìm số ự nhiên x biết.

a, 2^x=4^6.16³

2^x=(2²)².(2⁴)³

2^x=2⁴.2¹²

2^x=2¹⁶

=>x=16

Vậy x=16

học tốt

a, \(2^x=4^6\cdot16^3\)

\(2^x=\left(2^2\right)^6\cdot\left(2^4\right)^3\)

\(2^x=2^{12}\cdot2^{12}\)

\(2^x=2^{24}\)

\(\Rightarrow\text{ }x=24\)

a: (x-3)²=49

=>x-3=7 hoặc x-3=-7

=>x=10 hoặc x=-4

b: \(\left(x^4\right)^2=\dfrac{x^{12}}{x^5}\)

\(\Leftrightarrow x^8-x^7=0\)

\(\Leftrightarrow x^7\left(x-1\right)=0\)

=>x=0 hoặc x=1

c: \(\Leftrightarrow x^{10}-25x^8=0\)

\(\Leftrightarrow x^8\left(x^2-25\right)=0\)

\(\Leftrightarrow x^8\left(x-5\right)\left(x+5\right)=0\)

hay \(x\in\left\{0;5;-5\right\}\)

a) \(\left(3,5\right)^3=42,875\)

b) \(\left(-\dfrac{4}{11}\right)^2=\dfrac{16}{121}\)

c) \(\left(0,5\right)^4\cdot6^4=\left(0,5\cdot6\right)^4=3^4=81\)

d) \(\left(\dfrac{-1}{3}\right)^5:\left(\dfrac{1}{6}\right)^5=\left(\dfrac{-1}{3}:\dfrac{1}{6}\right)^5=\left(-2\right)^5=-32\)

\(a)\left(3,5\right)^3=42,875\)

\(b)\left(-\dfrac{4}{11}\right)^2=\dfrac{16}{121}\)

\(c)\left(0,5\right)^4.6^4=\left(0,5.6\right)^4=3^4=81\)

\(d)\left(-\dfrac{1}{3}\right)^5:\left(\dfrac{1}{6}\right)^5=\left(\dfrac{-1}{3}:\dfrac{1}{6}\right)^5=\left(-2\right)^5=-32\)

Chúc bn học tốt !

Bài 1 :

A + B = 4x² - 5xy + 3y² + 3x² + 2xy - y²

= ( 4x² + 3x² ) - ( 5xy - 2xy ) + ( 3y² - y² )

= 7x² - 3xy + 2y²

A - B = 4x² - 5xy + 3y² - ( 3x² + 2xy - y² )

= 4x² - 5xy + 3y² - 3x² - 2xy + y²

= ( 4x² - 3x² ) - ( 5xy + 2xy ) + ( 3y² + y² )

= x² - 7xy + 4y²

Bài 2 :

a) M + (5x² - 2xy) = 6x² + 9xy - y²

M = 6x² + 9xy - y² - (5x² - 2xy)

M = 6x² + 9xy - y² - 5x² + 2xy

M = ( 6x² - 5x² ) + ( 9xy + 2xy ) - y²

M = x² + 11xy - y²

Vậy M = x² + 11xy - y²

b) (3xy - 4y²) - N = x² - 7xy + 8y²

N = 3xy - 4y² - x² - 7xy + 8y²

N = ( 3xy - 7xy ) - ( 4y² - 8y² ) - x²

N = -4xy + 4y² - x²

Vậy N = -4xy + 4y² - x²

3, Cho đa thức

A(x)+B(x) = (3x⁴-\(\dfrac{3}{4}\)x³+2x²-3)+(8x⁴+\(\dfrac{1}{5}\)x³-9x+\(\dfrac{2}{5}\))

= 3x⁴-\(\dfrac{3}{4}\)x³+2x²-3+8x⁴+\(\dfrac{1}{5}\)x³-9x+\(\dfrac{2}{5}\)

= (3x⁴+8x⁴⁾+(-3/4x³+1/5x³)+(-3+2/5)+2x²-9x

= 11x⁴ -0.55x³-2.6+2x²-9x

A(x)-B(x)=(3x⁴-\(\dfrac{3}{4}\)x³+2x²-3)-(8x⁴+\(\dfrac{1}{5}\)x³-9x+\(\dfrac{2}{5}\))

= 3x⁴-\(\dfrac{3}{4}\)x³+2x²-3-8x⁴-\(\dfrac{1}{5}\)x³+9x-\(\dfrac{2}{5}\)

= (3x⁴-8x⁴⁾+(-3/4x³-1/5x³)+(-3-2/5)+2x²+9x

= -5x⁴-0.95x³-3.4+2x²+9x

B(x)-A(x)=(8x⁴+\(\dfrac{1}{5}\)x³-9x+\(\dfrac{2}{5}\))-(3x⁴-\(\dfrac{3}{4}\)x³+2x²-3)

=8x⁴+\(\dfrac{1}{5}\)x³-9x+\(\dfrac{2}{5}\)-3x⁴+\(\dfrac{3}{4}\)x³-2x²+3

=(8x⁴-3x⁴)+(1/5x³+3/4x³)+(2/5+3)-9x-2x²

⁼ 5x⁴+0.95x³+2.6-9x-2x²