Tìm x : (2x-1/5)2+1=29/25
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Em lớp 5 nên giải đc câu đầu thôi ạ :))
\(\frac{x-25}{1990}+\frac{x-29}{1986}+\frac{x+1}{1008}=4\)
\(\Leftrightarrow\left(\frac{x-25}{1990}-1\right)+\left(\frac{x-29}{1986}-1\right)+\left(\frac{x+1}{1008}-2\right)=0\)
\(\Leftrightarrow\frac{x-2015}{1990}+\frac{x-2015}{1986}+\frac{x-2015}{1008}=0\)
\(\Leftrightarrow\left(x-2015\right)\left(\frac{1}{1990}+\frac{1}{1986}+\frac{1}{1008}\right)=0\)
\(\Leftrightarrow x-2015=0\)
\(\Leftrightarrow x=2015\)
\(\left(2x-5\right)^3-\left(x-2\right)^3=\left(x-3\right)^3\)
\(\Leftrightarrow\left(2x-5-x+2\right)^3+3\left(2x-5\right)\left(x-2\right)\left(2x-5-x+2\right)=\left(x-3\right)^3\)
\(\Leftrightarrow\left(x-3\right)^2+3\left(2x-5\right)\left(x-2\right)\left(x-3\right)=\left(x-3\right)^3\)
\(\Leftrightarrow\left(x-3\right)^2+3\left(2x-5\right)\left(x-2\right)\left(x-3\right)-\left(x-3\right)^3=0\)
\(\Leftrightarrow\left(x-3\right)\left(\left(x-3\right)+3\left(2x-5\right)\left(x-2\right)-\left(x-3\right)^2\right)=0\)
Đến đây bắt buộc nhân đa thức với đa thức rồi tính thôi :v , tự tính nha
a: ta có: \(7-\left(2x-\frac13\right)^2=3\)
=>\(\left(2x-\frac13\right)^2=7-3=4\)
=>\(\left[\begin{array}{l}2x-\frac13=2\\ 2x-\frac13=-2\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=2+\frac13=\frac73\\ 2x=-2+\frac13=-\frac53\end{array}\right.\)
=>\(\left[\begin{array}{l}x=\frac73:2=\frac76\\ x=-\frac53:2=-\frac56\end{array}\right.\)
b: \(\left(2x+\frac13\right)^2-\frac38=\frac18\)
=>\(\left(2x+\frac13\right)^2=\frac38+\frac18=\frac48=\frac12\)
=>\(2x+\frac13=\sqrt{\frac12}=\frac{\sqrt2}{2}\)
=>\(2x=\frac{\sqrt2}{2}-\frac13=\frac{3\sqrt2-2}{6}\)
=>\(x=\frac{3\sqrt2-2}{12}\)
c: \(12:\left\lbrack29-\left(x-\frac23\right)^2\right\rbrack=3\)
=>\(29-\left(x-\frac23\right)^2=12:3=4\)
=>\(\left(x-\frac23\right)^2=29-4=25\)
=>\(\left[\begin{array}{l}x-\frac23=5\\ x-\frac23=-5\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5+\frac23=\frac{17}{3}\\ x=-5+\frac23=-\frac{13}{3}\end{array}\right.\)
d: \(\left(3x-\frac12\right)^3+\frac83=\frac{29}{9}-\frac{14}{27}\)
=>\(\left(3x-\frac12\right)^3+\frac83=\frac{87}{27}-\frac{14}{27}=\frac{73}{27}\)
=>\(\left(3x-\frac12\right)^3=\frac{73}{27}-\frac83=\frac{73}{27}-\frac{72}{27}=\frac{1}{27}=\left(\frac13\right)^3\)
=>\(3x-\frac12=\frac13\)
=>\(3x=\frac12+\frac13=\frac56\)
=>\(x=\frac{5}{18}\)
e: \(2\left(2x-\frac13\right)^2+\frac43=\frac56+\frac{13}{18}\)
=>\(2\left(2x-\frac13\right)^2=\frac{15}{18}+\frac{13}{18}-\frac43=\frac{28}{18}-\frac{24}{18}=\frac{4}{18}=\frac29\)
=>\(\left(2x-\frac13\right)^2=\frac19\)
=>\(\left[\begin{array}{l}2x-\frac13=\frac13\\ 2x-\frac13=-\frac13\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=\frac23\\ 2x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac13\\ x=0\end{array}\right.\)
=> [2x -1/5]^2 = 4/25 => 2x -1/5 = -2/5 hoặc 2/5
=> 2x = -1/5 hoặc 3/5 => x= -1/10 hoặc x= 3/10