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a.(2x +1). (2x+1)=1
Mà chỉ có 1.1=1
Vậy 2x + 1=1
2x=1-1
2x=0
Suy ra: x= 0
Hoàng Khánh Thi thiếu nha.
a) (2x+1)2 = \(\left(\pm1\right)^2\)
=> 2x + 1 = 1 hoặc 2x + 1 = -1
=> 2x = 0 hoặc 2x = -2
=> x = 0 hoặc x = -1.
a)\(\left(5x+1\right)^2=\frac{36}{49}\\ \left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\\ \Rightarrow\left[{}\begin{matrix}5x+1=\frac{6}{7}\\5x+1=\frac{-6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{-1}{35}\\x=\frac{-13}{35}\end{matrix}\right.\)
vậy...
2.
a) \(\left(5x+1\right)^2=\frac{36}{49}\)
⇒ \(5x+1=\pm\frac{6}{7}\)
⇒ \(\left[{}\begin{matrix}5x+1=\frac{6}{7}\\5x+1=-\frac{6}{7}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}5x=\frac{6}{7}-1=-\frac{1}{7}\\5x=\left(-\frac{6}{7}\right)-1=-\frac{13}{7}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\left(-\frac{1}{7}\right):5\\x=\left(-\frac{13}{7}\right):5\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=-\frac{1}{35}\\x=-\frac{13}{35}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{1}{35};-\frac{13}{35}\right\}.\)
Chúc bạn học tốt!
e)
\(\left(-\frac{3}{4}\right)^{3x-1}=\frac{256}{81}\)
\(\left(-\frac{3}{4}\right)^{3x}=\left(-\frac{4}{3}\right)^4\)
\(\left(-\frac{3}{4}\right)^{3x}=\left(-\frac{3}{4}\right)^{-4}\)
\(3x=-4\)
\(x=-\frac{4}{3}\)
f) \(172x^2-7^9:98^3=2^{-3}\)
\(172x^2-\frac{7^9}{\left(7^2.2\right)^3}=\frac{1}{2^3}\)
\(172x^2-\frac{7^3}{2^3}=\frac{1}{2^3}\)
\(172x^2=\frac{7^3}{2^3}+\frac{1}{2^3}=\frac{344}{8}\)
\(x^2=\frac{344}{8}:172=\frac{1}{4}\)
x=1/2 hoặc x=-1/2
\(a,\frac{1}{2}x+\frac{5}{2}=\frac{7}{2}x-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x+\frac{5}{2}-\frac{7}{2}x=-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{7}{2}x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x=-\frac{13}{4}\)
\(\Leftrightarrow x=-\frac{13}{4}:(-3)=-\frac{13}{4}:\frac{-3}{1}=-\frac{13}{4}\cdot\frac{-1}{3}=\frac{13}{12}\)
\(b,\frac{2}{3}x-\frac{2}{5}=\frac{1}{2}x-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{2}{5}-\frac{1}{2}x=-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{1}{2}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x=\frac{1}{15}\)
\(\Leftrightarrow x=\frac{1}{15}:\frac{1}{6}=\frac{1}{15}\cdot6=\frac{6}{15}=\frac{2}{5}\)
\(c,\frac{1}{3}x+\frac{2}{5}(x+1)=0\)
\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)
\(\Leftrightarrow\frac{11}{15}x=-\frac{2}{5}\)
\(\Leftrightarrow x=-\frac{6}{11}\)
d,e,f Tương tự