a: \(125^5:25^3=5^{15}:5^6=5^9\)

b: \(27^6:9^3=3^{18}:3^6=3^{12}\)

c: \(4^{20}:2^{15}=2^{40}:2^{15}=2^{45}\)

d: \(24^n:2^{2n}=24^n:4^n=6^n\)

e: \(64^4\cdot16^5:4^{20}=2^{24}\cdot2^{20}:2^{40}=2^4\)

\(=\left(2^3\cdot3\right)^n:2^{2n}=2^{3n}\cdot3^n:2^{2n}=2^n\cdot3^n=6^n\)

24ⁿ : 2²ⁿ = 24ⁿ : 4ⁿ = (24 : 4)ⁿ = 6ⁿ

nhìn thấy duy nhất một chữ rối!

(x+1/5)^2 =26/25-17/25

<=> (x +1/5)^2 =(3/5)^2

<=> x+1/5=3/5

=> x= 2/5

a) 
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)

\(\left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\)

\(\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)

\(\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(x+\frac{1}{5}=\frac{3}{5}\)

\(x=\frac{3}{5}-\frac{1}{5}\)

\(x=\frac{2}{5}\)
b) 
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)

\(\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1+\frac{x+349}{5}=0\)
\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+4+\frac{x+329}{5}=0\)

\(\left(x+329\right). \left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+4+\frac{1}{5}\right)=0\)
\(\Rightarrow\left(x+329\right)=0\)
\(\Rightarrow x=-329\)

1.

(x + 1/5)² = 26/25 - 17/25 

(x + 1/5)² = 9/25 

Rút căn hai vế : 

|(x + 1/5)| = 3/5 

x = -4/5 
hoặc
x = 2/5

2.

(x + 2) / 327 + (x + 3) / 326 + (x + 4) / 325 + (x + 5) / 324 + (x + 349) / 5 = 0 
<=> (x + 2) / 327 +1+ (x + 3) / 326 +1+ (x + 4) / 325 +1+ (x + 5) / 324 +1+ (x + 349) / 5 -4 = 0 
<=> (x+ 329)/327 + (x+ 329)/326 + (x+ 329)/325 + (x+ 329)//324 + (x+ 329)/5 =0 
<=> (x+ 329).(1/327 + 1/ 326 + 1/325 + 1/324 +1/5) =0 
Do (1/327 + 1/ 326 + 1/325 + 1/324 +1/5) >0 nên x+ 329 =0 => x= -329 

Câu 1 chưa chắc đã đúng ( quên hết kiến thức lớp 6 rùi ) hihi

aaaaaaaa . chết rồi . cho mình sủa câu thứ nhất :

(x+1/5)² + 17/25=26/25

( x + 1/5 ) ² = 26/25 - 17/25

( x + 1/5 ) ² = 3/5²

x + 1/5 = 3/5

x = 2/5.

Lời giải:
Ta thấy \(2^{4n+2}-2=2(2^{4n}-1)=2(16^n-1)\)

$16\equiv 1\pmod 5\Rightarrow 16^n\equiv 1\pmod 5$

$\Rightarrow 16^n-1\equiv 0\pmod 5$

$\Rightarrow 16^n-1\vdots 5$

$\Rightarrow 2(16^n-1)\vdots 10$

Vậy đáp án b.

là b

Chọn B

\(=\left(4\cdot6\right)^n:\left(2^2\right)^n=4^n\cdot6^n:4^n=6^n\)

1) Ta có: \(\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{7}{25}\cdot\dfrac{5}{7}\right)\)

\(=\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{1}{5}\right)\)

=0

2) Ta có: \(\dfrac{8}{17}\cdot\dfrac{4}{15}+\dfrac{8}{17}\cdot\dfrac{22}{15}-\dfrac{8}{15}\cdot\dfrac{9}{17}\)

\(=\dfrac{8}{17}\left(\dfrac{4}{15}+\dfrac{22}{15}-\dfrac{9}{15}\right)\)

\(=\dfrac{8}{17}\cdot\dfrac{15}{15}=\dfrac{8}{17}\)

3) Ta có: \(\dfrac{2021}{2}\cdot\dfrac{1}{3}+\dfrac{4042}{4}\cdot\dfrac{1}{5}+\dfrac{6063}{3}\cdot\dfrac{22}{15}\)

\(=\dfrac{2021}{2}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)+2021\cdot\dfrac{22}{15}\)

\(=\dfrac{2021}{2}\cdot\dfrac{8}{15}+\dfrac{2021}{2}\cdot\dfrac{44}{15}\)

\(=\dfrac{2021}{2}\cdot\dfrac{52}{15}\)

\(=\dfrac{52546}{15}\)

4) Ta có: \(\dfrac{4}{7}\cdot\dfrac{2}{13}+\dfrac{8}{13}:\dfrac{7}{4}+\dfrac{4}{7}:\dfrac{13}{2}+\dfrac{4}{7}\cdot\dfrac{1}{13}\)

\(=\dfrac{4}{7}\left(\dfrac{2}{13}+\dfrac{8}{13}+\dfrac{2}{13}+\dfrac{1}{13}\right)\)

\(=\dfrac{4}{7}\)

1) Tính biểu thức:

  C = 3 - 3² + 3³ - 3⁴ + 3⁵ - 3⁶ +...+ 3²³ - 3²⁴

3C =      3² - 3³ + 3⁴ - 3⁵ + 3⁶-...- 3²³ + 3²⁴ - 3²⁵

3C - C = -3²⁵ - 3

2C      = -3²⁵ - 3

2C = - ( 3²⁵ + 3) = - [(3⁴)⁶. 3 + 3] = - [\(\overline{...1}\)⁶.3+3] = -[ \(\overline{..3}\)  + 3]

2C = - \(\overline{..6}\)

⇒ \(\left[{}\begin{matrix}C=\overline{..3}\\C=\overline{..8}\end{matrix}\right.\) 

⇒ C không thể chia hết cho 420 ( xem lại đề bài em nhé)

b, (\(x+1\))²⁰²² + (\(\sqrt{y-1}\) )²⁰²³ = 0

Vì (\(x+1\))²⁰²² ≥ 0 

\(\sqrt{y-1}\) ≥ 0 ⇒ (\(\sqrt{y-1}\))²⁰²³ ≥ 0

Vậy (\(x\) + 1)²⁰²² + (\(\sqrt{y-1}\))²⁰²³ = 0

⇔ \(\left\{{}\begin{matrix}\left(x+1\right)^{2022}=0\\\sqrt{y-1}=0\end{matrix}\right.\)

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

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

Kết luận: cặp (\(x,y\)) thỏa mãn đề bài là:

(\(x,y\)) = (-1; 1)

Ta thử đáp án là cách nhanh nhất nhé!

Thay vào công thức:

\(F=\sqrt{F_1^2+F_2^2+2\cdot F_1\cdot F_2\cdot cos\alpha}\)

Lần lượt thay ta đc đáp án B thỏa man nhé:

\(F=\sqrt{16^2+12^2+2\cdot16\cdot12\cdot cos90^o}=20N\)

Chọn B