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\(\frac{7^{x+2}+7^{x+1}+7^x}{57}=\frac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}\)
\(\Rightarrow\frac{7^x.7^2+7^x.7^1+7^x}{57}=\frac{5^{2x}+5^{2x}.5+5^{2x}.5^3}{131}\)
\(\Rightarrow\frac{7^x.\left(7^2+7+1\right)}{57}=\frac{5^{2x}.\left(1+5+5^3\right)}{131}\)
\(\Rightarrow\frac{7^x.57}{57}=\frac{5^{2x}.131}{131}\)
\(\Rightarrow7^x=5^{2x}\)
Bạn tự làm phần còn lại nhé
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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:
\(7^x=5^{2x}\Rightarrow7^x=25^x\Rightarrow\frac{7^x}{25^x}=1\Rightarrow\left(\frac{7}{25}\right)^x=1\Rightarrow x=0\)
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\(\left(2x-3\right)\left(6-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1,5\\x=3\end{matrix}\right.\)
*\(\frac{1}{4}+\frac{1}{3}:\left(2x-1\right)=-5\)
\(\Leftrightarrow\frac{1}{3}\cdot\frac{1}{2x-1}=-5-\frac{1}{4}\)
\(\Leftrightarrow\frac{1}{3\left(2x-1\right)}=\frac{-21}{4}\)
\(\Leftrightarrow-63\left(2x-1\right)=4\)
\(\Leftrightarrow2x-1=-\frac{4}{63}\)
\(\Leftrightarrow2x=\frac{59}{63}\)
\(x=\frac{59}{126}\)
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f) \(\frac{2x-1}{21}=\frac{3}{2x+1}\)( ĐKXĐ : \(x\ne-\frac{1}{2}\))
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=21\cdot3\)
\(\Leftrightarrow4x^2-1=63\)
\(\Leftrightarrow4x^2=64\)
\(\Leftrightarrow x^2=16\)
\(\Leftrightarrow x^2=\left(\pm4\right)^2\)
\(\Leftrightarrow x=\pm4\)(tmđk)
h) \(\frac{10x+5}{6}=\frac{5}{x+1}\)( ĐKXĐ : \(x\ne-1\))
\(\Leftrightarrow\left(10x+5\right)\left(x+1\right)=6\cdot5\)
\(\Leftrightarrow10x^2+15x+5=30\)
\(\Leftrightarrow10x^2+15x+5-30=0\)
\(\Leftrightarrow10x^2+15x-25=0\)
\(\Leftrightarrow5\left(2x^2+3x-5\right)=0\)
\(\Leftrightarrow2x^2+3x-5=0\)
\(\Leftrightarrow2x^2-2x+5x-5=0\)
\(\Leftrightarrow2x\left(x-1\right)+5\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+5\right)\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)(tmđk)
f) \(\frac{2x-1}{21}=\frac{3}{2x+1}\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=21.3\)
\(\Leftrightarrow4x^2-1=63\)
\(\Leftrightarrow4x^2=64\)
\(\Leftrightarrow x^2=16\)\(\Leftrightarrow x^2=4^2\)\(\Leftrightarrow x=4\)
Vậy \(x=4\)
h) \(\frac{10x+5}{6}=\frac{5}{x+1}\)
\(\Leftrightarrow\left(10x+5\right)\left(x+1\right)=5.6\)
\(\Leftrightarrow5\left(2x+1\right)\left(x+1\right)=30\)
\(\Leftrightarrow\left(2x+1\right)\left(x+1\right)=6\)
\(\Leftrightarrow2x^2+3x+1=6\)
\(\Leftrightarrow2x^2+3x-5=0\)
\(\Leftrightarrow\left(2x^2-2x\right)+\left(5x-5\right)=0\)
\(\Leftrightarrow2x\left(x-1\right)+5\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\2x=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{-5}{2}\end{cases}}\)
Vậy \(x\in\left\{\frac{-5}{2};1\right\}\)
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a) 3x - 2 = 0 => 3x = 2 => x = 2/3
b) 2x - 1 = 0 => 2x = 1 => x = 1/2
c) 5 ( 4+2x) = 8+5x
<=> 20 + 10x = 8 + 5x
<=> 10x - 5x = 8 - 20
<=> 5x = -12
x = -12/5
d) \(\frac{1}{2}+\frac{3}{4}x=6-\frac{4}{5}x\)
\(\frac{3}{4}x+\frac{4}{5}x=6-\frac{1}{2}\)
\(\frac{31}{20}x=\frac{11}{2}\)
\(x=\frac{11}{2}:\frac{31}{20}=\frac{110}{31}\)
e) 3 + 2x = 4 - 8x
<=> 2x + 8x = 4 - 3
10 x = 1
x = 1/10
f \(5+\frac{1}{2}\left(x+5\right)=3\)
\(\frac{1}{2}\left(x+5\right)=3-5=-2\)
\(x+5=-2:\frac{1}{2}=-4\)
\(x=-4-5=1\)
Vậy ......
đề bài là gì bạn ơi
Tìm x sao cho các số có dạng sau đều là stn
Bạn lấy luôn cách giải hô mk nhé