1: \(=\dfrac{x-1}{x^2+x+1}+\dfrac{x+1}{x-1}\)

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

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

2: \(=\dfrac{\left(x^2-y^2\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{\left(x-y\right)\left(x+y\right)^2}{x^2+xy+y^2}\)

\(\dfrac{22}{5}\times\dfrac{6}{121}\times\dfrac{11}{4}\times\dfrac{3}{5}\times\dfrac{1}{3}\times\dfrac{5}{4}\)

\(=\left(\dfrac{22}{5}\times\dfrac{5}{4}\right)\times\left(\dfrac{6}{121}\times\dfrac{11}{4}\right)\times\left(\dfrac{3}{5}\times\dfrac{1}{3}\right)\)

\(=\dfrac{11}{2}\times\dfrac{3}{22}\times\dfrac{1}{5}\)

\(=\dfrac{3}{20}\)

\(=\dfrac{22\times6\times11\times3\times1\times5}{5\times121\times4\times5\times3\times4}=\dfrac{11\times2\times6\times11\times1}{11\times11\times4\times5\times4}=\dfrac{2\times6\times1}{4\times5\times4}=\dfrac{18}{100}=\dfrac{9}{50}\)

Ta có:

x³(x+2) – x(x³ + 2³) – 2x(x² – 2²)

= x³ . x + x³ . 2 – (x . x³ + x . 2³) – ( 2x . x² – 2x . 2²)

= x⁴ + 2x³ – (x⁴ + 8x ) – (2x³ – 8x)

= x⁴ + 2x³ – x⁴ – 8x – 2x³ + 8x

= (x⁴ – x⁴) + (2x³ – 2x³) + (-8x + 8x)

= 0

3: =>x(x+1)=0

=>x=0 hoặc x=-1

4: =>(2x-3)(x+2)=0

=>x=3/2 hoặc x=-2

6: =>6x=7 hoặc 6x=-7

=>x=7/6 hoặc x==7/6

\(x^3+\dfrac{3}{4}x+\dfrac{3}{2}x^2+\dfrac{1}{8}=\dfrac{1}{64}\)

\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)

\(\Leftrightarrow x=-\dfrac{1}{4}\)

Chữ số 5 

trả lời 

chữ số 5 tận cùng

 hok tốt

a) \(2^x=16=2^4\Rightarrow x=4\)

b) \(x^3=27=3^3\Rightarrow x=3\)

c) \(x^{50}=x\Rightarrow x\left(x^{49}-1\right)=0\Rightarrow x=0\) hay \(x=1\)

d) \(\left(x-2\right)^2=16=4^2\Rightarrow x-2=4\) hay \(x-2=-4\)

\(\Rightarrow x=6\) hay \(x=-2\)

a) \(2^{300}=2^{3.100}=8^{100}\)

\(3^{200}=3^{2.100}=9^{100}\)

vì \(8^{100}< 9^{100}\)

\(\Rightarrow2^{300}< 3^{200}\)

b) \(3^{500}=3^{5.100}=243^{100}\)

\(7^{300}=7^{3.100}=343^{100}\)

vì \(243^{100}< 343^{100}\)

\(\Rightarrow3^{500}< 7^{300}\)

`@` `\text {Ans}`

`\downarrow`

`1,`

`a,`

`2^x = 16`

`=> 2^x = 2^4`

`=> x = 4`

Vậy, `x = 4`

`b,`

`x^3 = 27`

`=> x^3 = 3^3`

`=> x = 3`

Vậy, `x = 3`

`c,`

\(x^{50}=x\)

`=>`\(x^{50}-x=0\)

`=>`\(x\left(x^{49}-1\right)=0\)

`=>`\(\left[{}\begin{matrix}x=0\\x^{49}-1=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0\\x^{49}=1\end{matrix}\right.\)

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

Vậy, `x \in {0; 1}`

`d,`

`(x-2^2)=16`

`=> x - 2^2 = 16`

`=> x = 16 + 2^2`

`=> x = 20`

Vậy, `x = 20`

`2,`

`a,`

Ta có:

\(2^{300}=\left(2^3\right)^{100}=8^{100}\)

\(3^{200}=\left(3^2\right)^{100}=9^{100}\)

Vì `8 < 9 =>`\(8^{100}< 9^{100}\)

`=>`\(2^{300}< 3^{200}\)

Vậy, \(2^{300}< 3^{200}\)

`b,`

Ta có:

\(3^{500}=\left(3^5\right)^{100}=243^{100}\)

\(7^{300}=\left(7^3\right)^{100}=343^{100}\)

Vì `243 < 343 =>`\(243^{100}< 343^{100}\)

`=>`\(3^{500}< 7^{300}\)

Vậy, \(3^{500}< 7^{300}.\)

Lời giải:

a. $P(x)=x^3+3x^2-2x+2019-(3x^2-2x)=x^3+2019$

b.

$Q(2)=-2^3+2-22=-28$

c.

$P(x)+Q(x)=x^3+2019+(-x^3+x-2022)=x-3$

$P(x)+Q(x)=0$

$x-3=0$

$x=3$ 

Vậy nghiệm của đa thức là $x=3$