\(2016x+\left(\dfrac{7}{12}+\dfrac{4}{21}+\dfrac{2}{24}+\dfrac{11}{30}+\dfrac{3}{41}+\dfrac{15}{56}\right)-\left(\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}\right)=0\)

\(2016x+\left[\left(\dfrac{7}{12}+\dfrac{2}{24}\right)+\left(\dfrac{4}{21}+\dfrac{11}{30}+\dfrac{3}{41}+\dfrac{15}{56}\right)\right]-\dfrac{47}{30}=0\)

\(2016x+\left(\dfrac{2}{3}+\dfrac{1473}{1640}\right)=0+\dfrac{47}{30}\)

\(2016x+\dfrac{2}{3}+\dfrac{1473}{1640}=\dfrac{47}{30}\)

\(2016x+\dfrac{2}{3}=\dfrac{47}{30}-\dfrac{1473}{1640}\)

\(2016x+\dfrac{2}{3}=\dfrac{3289}{4920}\)

\(2016x=\dfrac{3289}{4920}-\dfrac{2}{3}\)

\(2016x=\dfrac{3}{1640}\)

\(x=\dfrac{3}{1640}:2016\)

\(x=\dfrac{1}{1102080}\)

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- Chị nghĩ có thể là em ghi sai đề. Hoặc là cô/thầy của em ra đề như vậy. TvT

- Chúc em học tốt.

X.(X^2014+2016)>0

do đó x>0

do X^2+2016>0 

nên X-3<0

Suy ra:x<3

Bài 1: 

a) 0²⁰⁰² < 0²⁰²³

b) 2022⁰ = 2023⁰

c) 54⁹ < 55¹⁰

d) ( 4 + 5 )³ > 4² + 5²

đ) 9² - 3² > ( 9 - 3 )²

Bài 2:

a) 3² x 4³ - 3² + 333

= 9 x 64 - 9 + 333

= 576 - 9 + 333

= 567 + 333

= 900

b) 5 x 4³ + 24 x 5 + 41⁰

= 5 x 64 + 24 x 5 + 1

= 5 x ( 64 + 24 ) + 1

= 5 x 88 + 1

= 440 + 1

= 441

c) 2³ x 4² + 3² x 5 - 40 x 1²⁰²³

= 8 x 16 + 9 x 5 - 40 x 1

= 128 + 45 - 40

= 133

Bài 1 :

a) \(0^{2002}=0;0^{2023}=0\Rightarrow0^{2002}=0^{2023}\)

b) \(2022^0=1;2023^0=1\Rightarrow2022^0=2023^0\)

c) \(54^9< 55^9;55^9< 55^{10}\Rightarrow54^9< 55^{10}\)

d) \(\left(4+5\right)^3>\left(4+5\right)^2;\left(4+5\right)^2>4^2+5^2\Rightarrow\left(4+5\right)^3>4^2+5^2\)

đ) \(9^2-3^2=81-9=82;\left(9-3\right)^2=6^2=36\Rightarrow9^2-3^2>\left(9-3\right)^2\)

\(\left(x^2+5\right)\left(x^2-49\right)=0\)

\(\left(x^2-49\right)=0\)  vì   \(x^2+5>0\)

\(\Rightarrow x^2=49\)

\(\Rightarrow x=\pm7\)

vậy \(x=\pm7\)

\(\left(x^2+5\right)\left(x^2-7\right)< 0\)

\(\left(x^2-7\right)< 0\)

\(\left(x-\sqrt{7}\right)\left(x+\sqrt{7}\right)< 0\)

\(\hept{\begin{cases}x-\sqrt{7}>0\\x+\sqrt{7}< 0\end{cases}}\)  hoặc \(\hept{\begin{cases}x-\sqrt{7}< 0\\x+\sqrt{7}>0\end{cases}}\)

\(\hept{\begin{cases}x>\sqrt{7}\\x< -\sqrt{7}\end{cases}}\)  hoặc \(\hept{\begin{cases}x< \sqrt{7}\\x>-\sqrt{7}\end{cases}}\)

đến đây tự làm tiếp

\(\left(x^2+5\right)\times\left(x^2-49\right)=0\)

Vậy \(\orbr{\begin{cases}x^2+5=0\\x^2-49=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=-5\\x^2=49\end{cases}}\)

Mà \(x^2\ge0\Rightarrow x^2=-5\)(loại)

Vậy \(x^2=49\)

Nên \(\orbr{\begin{cases}x=7\\x=-7\end{cases}}\)

đeó  bt

 3^x=9

suy ra 3²=9

suy ra x= 2

hello 123

mũ viết thay cái này nè ^

shirt + 6

viết vậy khó hiểu quá

chịu

Chịu 

\(a,\left(-5\right).\left|x\right|=-75\)

\(\left|x\right|=\frac{-75}{-5}=15\)

\(\Rightarrow\orbr{\begin{cases}x=15\\x=-15\end{cases}}\)

Vậy....

\(b,\left(-6\right)^3.x^2=-1944\)

\(-216.x^2=-1944\)

\(x^2=9\)

\(\Rightarrow x=\pm3\)

Vậy....

\(d,\left|9-x\right|=-7+64\)

\(\left|9-x\right|=57\)

\(\Rightarrow\orbr{\begin{cases}9-x=57\\9-x=-57\end{cases}\Rightarrow\orbr{\begin{cases}x=-48\\x=66\end{cases}}}\)

Vậy...

\(e,\left|x+101\right|-\left(-16\right)=\left(-43\right).\left(-5\right)\)

\(\left|x+101\right|+16=215\)

\(\left|x+101\right|=199\)

\(\Rightarrow\orbr{\begin{cases}x+101=199\\x+101=-199\end{cases}\Rightarrow\orbr{\begin{cases}x=98\\x=-300\end{cases}}}\)

Vậy..

hok tốt!!

a,\(\left(-5\right).\left|x\right|=-75\)

\(=>\left|x\right|=-75:\left(-5\right)=15\)

\(=>\orbr{\begin{cases}x=15\\x=-15\end{cases}}\)

b,\(\left(-6\right)^3.x^2=-1944\)

\(=>\frac{1944}{216}=x^2\)

\(=>x=\sqrt{\frac{1944}{216}}=3\)