a) \(\left(\frac{2}{3}\right)^x=\left(\frac{4}{9}\right)^{50}\)

\(\Rightarrow\left(\frac{2}{3}\right)^x=\left(\frac{2^2}{3^2}\right)^{50}\)

\(\Rightarrow\left(\frac{2}{3}\right)^x=\left(\frac{2}{3}\right)^{100}\)

\(\Rightarrow x=100\)

Vậy x = 100

b) \(\left(\frac{2}{3}-x\right)^2=\frac{1}{36}\)

\(\Rightarrow\left(\frac{2}{3}-x\right)^2=\left(\frac{1}{6}\right)^2\)

\(\Rightarrow\frac{2}{3}-x=\frac{1}{6}\)

\(\Rightarrow x=\frac{2}{3}-\frac{1}{6}\)

\(\Rightarrow x=\frac{1}{2}\)

Vậy \(x=\frac{1}{2}\)

2) 

Ta có:

\(74^{m+1}+74^m=74^m.74^1+74^m=74^m.\left(74+1\right)=74^m.75⋮25\)

( vì \(75⋮25\) )

\(\Rightarrowđpcm\)

giup minh vs mai minh di hoc rui

\(F\left(x\right)=x^2-3x^3-\sqrt{25}+\frac{1}{2}x-\left(-3x^3+\frac{5x}{2}-\sqrt{36}\right)\)

=> \(F\left(x\right)=x^2-3x^3-5+\frac{1}{2}x+3x^3-\frac{5x}{2}+6\)

=> \(F\left(x\right)=x^2+\left(3x^3-3x^3\right)+\left(6-5\right)+\left(\frac{x}{2}-\frac{5x}{2}\right)\)

=> \(F\left(x\right)=x^2+1-2x\)

mk sắp phải đi học rồi các bạn giúp mình với có đc ko mk nhớ sẽ đền đáp công ơn của bạn 

a) (5 - x) +12 = -25

<-> 5 - x + 12 = -25

<-> 17 - x = - 25

<-> x = 42

b) 12 - 4(x - 2) = -4

<-> 12 - 4x + 8 = -4

<-> 20 - 4x = -4

<-> 4x = 24

<-> x = 6

a) (5 - x) + 12 = -25

<=> -x = -25 - 12 - 5

<=> -x = -42

<=> x = 42

b) 12 - 4(x - 2) = -4

<=> 12 - 4x + 8 = -4

<=> -4x = -4 - 8 - 12

<=> -4x = -24

<=> x = 6

c) -15 - |3 - x| = -19

<=> -|3 - x| = -4

<=> 3 - x = 4 hoặc 3 - x = -4

<=> x = -1 hoặc x = 7

Đặt \(a=24-x,b=x-25\)

Khi đó pt ban đầu trở thành :

\(\frac{a^2+ab+b^2}{a^2-ab+b^2}=\frac{19}{49}\)

\(\Leftrightarrow49\left(a^2+ab+b^2\right)=19\left(a^2-ab+b^2\right)\)

\(\Leftrightarrow30a^2+68ab+30b^2=0\)

\(\Leftrightarrow15a^2+34ab+15b^2=0\)

\(\Leftrightarrow\left(3a+5b\right)\left(5a+3b\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3a=-5b\\5a=-3b\end{cases}}\)

Đến đây bạn thay vào là dễ rồi nhé ! Chúc bạn học tốt !

mai bn đăng lại nhé mk lm cho h đi ngủ

(12/25)^x=(5/3)^-2-(-3/5)⁴

(12/25)^x=144/625

=> (12/25)^x=(12/25)²

=> x=2

Vậy x=2

\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{9}{25}\\ \left|\left(x+\frac{1}{5}\right)\right|=\frac{3}{5}\)

 TH1:   \(x=\frac{3}{5}-\frac{1}{5}\\ x=\frac{2}{5}\)

TH2: \(\left|\left(x+\frac{1}{5}\right)\right|=-\frac{3}{5}\\ x=-\frac{3}{5}-\frac{1}{5}\\ x=-\frac{4}{5}\)

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

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

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

\(\Rightarrow x+\frac{1}{5}=\frac{3}{5}\)

\(\Rightarrow x=\frac{2}{5}\)

\(b,-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)

\(\Rightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)

\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}+\frac{24}{27}\)

\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)

\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)

\(\Rightarrow3x-\frac{7}{9}=-\frac{2}{3}\)

\(\Rightarrow3x=-\frac{2}{3}+\frac{7}{9}\)

\(\Rightarrow3x=\frac{1}{9}\)

\(\Rightarrow x=\frac{1}{27}\)

\(c,\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)

\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\)  \(\Rightarrow\)  \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\)  \(\Rightarrow\)  \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)

Bổ sung câu a: \(\Rightarrow\) \(\left[\begin{array}{nghiempt}\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\\\left(x+\frac{1}{5}\right)^2=\left(-\frac{3}{5}\right)^2\end{array}\right.\)\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{5}=\frac{3}{5}\\x+\frac{1}{5}=-\frac{3}{5}\end{array}\right.\) \(\Rightarrow\)  \(\left[\begin{array}{nghiempt}x=\frac{2}{5}\\x=-\frac{4}{5}\end{array}\right.\)