\(\frac{7^{x+2}+7^{x+1}+7x}{57}=\frac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}\)

\(\Rightarrow\frac{7x\left(7^2+7^1+1\right)}{57}=\frac{5^{2x}\left(1+5^1+5^3\right)}{131}\)

\(\Rightarrow\frac{7x\left(49+7+1\right)}{57}=\frac{5^{2x}\left(1+5+125\right)}{131}\)

\(\Rightarrow\frac{7x.57}{57}=\frac{5^{2x}.131}{131}\)

\(\Rightarrow7x=25x\)

\(\Rightarrow x=0\)

\(\left(4x-3\right)^4=\left(4x-3\right)^2\)

\(\Rightarrow\left(4x-3\right)^4-\left(4x-3\right)^2=0\)

\(\Rightarrow\left(4x-3\right)^2\left[\left(4x-3\right)^2-1\right]=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(4x-3\right)^2=0\\\left(4x-3\right)^2=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}4x-3=0\\4x-3=-1\\4x-3=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{4}\\x=\frac{1}{2}\\x=1\end{cases}}\)

Bài mình làm sai à hay sao mà t]ick sai thế???

a/

\(\left(1-3x\right)^3=\left(-4\right)^3\Leftrightarrow1-3x=-4\Leftrightarrow x=\frac{5}{3}\)

b/

\(\left(4-3x\right)^4=\left(4-3x\right)^2\Leftrightarrow\left(4-3x\right)^2\left[\left(4-3x\right)^2-1\right]=0\)

\(\Leftrightarrow\left(4-3x\right)^2\left(5-3x\right)\left(3-3x\right)=0\)

\(\Leftrightarrow3-3x=0\) hoặc \(4-3x=0\) hoặc \(5-3x=0\)

\(\Leftrightarrow x=1\) hoặc \(x=\frac{4}{3}\) hoặc \(x=\frac{5}{3}\)

c/

\(\frac{7^{x+2}+7^{x+1}+7^x}{57}=\frac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}\)\(\Leftrightarrow\frac{49.7^x+7.7^x+7^x}{57}=\frac{5^{2x}+5.5^{2x}+125.5^{2x}}{131}\)

\(\Leftrightarrow7^x=5^{2x}\Leftrightarrow7^x=25^x\Leftrightarrow\left(\frac{7}{25}\right)^x=1=\left(\frac{7}{25}\right)^0\)

\(\Rightarrow x=0\)

\(\left(1-3x\right)^3=-64\)

=> \(1-3x=-4\)

=> \(-3x=-4+1\) (chuyển vế)

=> \(-3x=-3\Rightarrow x=-3:\left(-3\right)=1\)

Tìm x biết
a) (7^x-2+7^x-1+7)/57=(5^x+5^2x+1+5^2x+3)/131
b) (4x-3)^4=(4x-3)^2

Bài 3:

a) \(\left(x-6\right).\left(2x-5\right).\left(3x+9\right)=0\)

\(\Leftrightarrow\left(x-6\right).\left(2x-5\right).3.\left(x+3\right)=0\)

Vì \(3\ne0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\2x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\2x=5\\x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\frac{5}{2}\\x=-3\end{matrix}\right.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{6;\frac{5}{2};-3\right\}.\)

b) \(2x.\left(x-3\right)+5.\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right).\left(2x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\frac{5}{2}\end{matrix}\right.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{3;-\frac{5}{2}\right\}.\)

c) \(\left(x^2-4\right)-\left(x-2\right).\left(3-2x\right)=0\)

\(\Leftrightarrow\left(x^2-2^2\right)-\left(x-2\right).\left(3-2x\right)=0\)

\(\Leftrightarrow\left(x-2\right).\left(x+2\right)-\left(x-2\right).\left(3-2x\right)=0\)

\(\Leftrightarrow\left(x-2\right).\left(x+2-3+2x\right)=0\)

\(\Leftrightarrow\left(x-2\right).\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{3}\end{matrix}\right.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2;\frac{1}{3}\right\}.\)

Chúc bạn học tốt!

Bài 4 xem lại đề nhé bác

1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)

=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)

b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)

=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)

c) TT

a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)

\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)

=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)

=> \(\left|50x-140\right|=\left|25x+24\right|\)

=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)

c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)

=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)

Bài 2 : a. |2x - 5| = x + 1

 TH1 : 2x - 5 = x + 1

    => 2x - 5 - x = 1

    => 2x - x - 5 = 1

    => 2x - x = 6

    => x = 6

TH2 : -2x + 5 = x + 1

   => -2x + 5 - x = 1

   => -2x - x + 5 = 1

   => -3x = -4

   => x = 4/3

Ba bài còn lại tương tự

cho mình

d, \(=>\left(\frac{4}{5}\right)^{2x+7}=\left(\frac{4}{5}\right)^4.\)

=> \(2x+7=4\) 

=> 2x= -3

=> x=-3/2     . Vậy x=-3/2

e, => \(\frac{7^x.7^2+7^x.7+7^x}{57}=\frac{5^{2x}+5^{2x}.5+5^{2x}.5^2}{131}.\)

=> \(\frac{7^x\left(7^2+7+1\right)}{57}=\frac{5^{2x}\left(1+5+5^2\right)}{131}\)

= > \(\frac{7^x.57}{57}=\frac{5^{2x}.131}{131}\)

=> \(7^x=5^{2x}\)

Đến đoạn này là mik nghĩ không ra nhé

Cô làm tiếp giúp Linh Đan:

\(7^x=5^{2x}\Rightarrow7^x=25^x\Rightarrow\frac{7^x}{25^x}=1\Rightarrow\left(\frac{7}{25}\right)^x=1\Rightarrow x=0\)

a)\(\left(\frac{4}{5}\right)^{2x+7}=\left(\frac{4}{5}\right)^4\)

=> 2x + 7 = 4 

     2x        = 4 - 7 

     2x        = -3

       x        = -3 : 2

       x         = -1,5

   Vậy x = -1,5

Đây là lớp 8 nha các b giúp mk với

Do mk viết nhầm