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a,\(\left(x-\frac{7}{9}\right)^3=\left(\left(\frac{2}{3}\right)^2\right)^3\)
\(x-\frac{7}{9}=\frac{4}{9}\)
\(x=\frac{4}{9}+\frac{7}{9}\)
\(x=\frac{11}{9}\)
Vậy x=\(\frac{11}{9}\)
bạn ơi trả lời được câu này kông
( x + 1 ) + ( x - 3 ) + ( x + 5 ) + ............ + ( x +9) = 35
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
a) \(2x\left(x-\frac{1}{7}\right)=0\)
\(x\left(x-\frac{1}{7}\right)=0\)
\(\Rightarrow2x-2.\frac{1}{7}=0\)
\(2x-\frac{2}{7}=0\)
=> \(2x=\frac{2}{7}\)
=> x=\(\frac{1}{7}\)
b) (x-9)(\(x+\frac{3}{5}\))=0
\(\Rightarrow\orbr{\begin{cases}x-9=0\\x+\frac{3}{5}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{-3}{5}\end{cases}}\)
Vậy x=0 hoặc x=-3/5
c) \(\left(\frac{-4}{7}-2x\right)\left(x-\frac{5}{4}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{-4}{7}-2x=0\\x-\frac{5}{4}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-2}{7}\\x=\frac{5}{4}\end{cases}}\)
Vậy x=-2/7 hoặc x=5/4
a, => x.(x-1/7) = 0:2 = 0
=> x=0 hoặc x-1/7=0
=> x=0 hoặc x=1/7
Vậy x thuộc {0;1/7}
b, => x-9=0 hoặc x+3/5=0
=> x=9 hoặc x=-3/5
Vậy x thuộc {-3/5;9}
c, => -4/7-2x=0 hoặc x-5/4=0
=> x=-2/7 hoặc x=5/4
Vậy x thuộc {-2/7;5/4}
Tk mk nha
a)
\((3x-7)^5=0\Rightarrow 3x-7=0\Rightarrow x=\frac{7}{3}\)
b)
\(\frac{1}{4}-(2x-1)^2=0\)
\(\Leftrightarrow (2x-1)^2=\frac{1}{4}=(\frac{1}{2})^2=(-\frac{1}{2})^2\)
\(\Rightarrow \left[\begin{matrix} 2x-1=\frac{1}{2}\\ 2x-1=\frac{-1}{2}\end{matrix}\right.\Rightarrow \Rightarrow \left[\begin{matrix} x=\frac{3}{4}\\ x=\frac{1}{4}\end{matrix}\right.\)
c)
\(\frac{1}{16}-(5-x)^3=\frac{31}{64}\)
\(\Leftrightarrow (5-x)^3=\frac{1}{16}-\frac{31}{64}=\frac{-27}{64}=(\frac{-3}{4})^3\)
\(\Leftrightarrow 5-x=\frac{-3}{4}\)
\(\Leftrightarrow x=\frac{23}{4}\)
d)
\(2x=(3,8)^3:(-3,8)^2=(3,8)^3:(3,8)^2=3,8\)
\(\Rightarrow x=3,8:2=1,9\)
e)
\((\frac{27}{64})^9.x=(\frac{-3}{4})^{32}\)
\(\Leftrightarrow [(\frac{3}{4})^3]^9.x=(\frac{3}{4})^{32}\)
\(\Leftrightarrow (\frac{3}{4})^{27}.x=(\frac{3}{4})^{32}\)
\(\Leftrightarrow x=(\frac{3}{4})^{32}:(\frac{3}{4})^{27}=(\frac{3}{4})^5\)
f)
\(5^{(x+5)(x^2-4)}=1\)
\(\Leftrightarrow (x+5)(x^2-4)=0\)
\(\Leftrightarrow \left[\begin{matrix} x+5=0\\ x^2-4=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x+5=0\\ x^2=4=2^2=(-2)^2\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=-5\\ x=\pm 2\end{matrix}\right.\)
g)
\((x-2,5)^2=\frac{4}{9}=(\frac{2}{3})^2=(\frac{-2}{3})^2\)
\(\Rightarrow \left[\begin{matrix} x-2,5=\frac{2}{3}\\ x-2,5=\frac{-2}{3}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{19}{6}\\ x=\frac{11}{6}\end{matrix}\right.\)
h)
\((2x+\frac{1}{3})^3=\frac{8}{27}=(\frac{2}{3})^3\)
\(\Rightarrow 2x+\frac{1}{3}=\frac{2}{3}\Rightarrow x=\frac{1}{6}\)
\(\frac{32-x}{7}=\frac{x-42}{9}\)
=\(\frac{\left(32-x\right)9}{63}=\frac{\left(x-42\right)7}{63}\)
\(\Rightarrow\)\(\left(32-x\right)9=\left(x-42\right)7\)
=\(288-x9=x7-294\)
=\(288+294=x9+x7\)
=\(x=-36\frac{6}{16}\)
=\(x\times16=-582\)
\(x=-582\div16\)
a,\(\frac{32-x}{7}=\frac{x-42}{9}\)
\(\Leftrightarrow9\left(32-x\right)=7\left(x-42\right)\)
\(\Leftrightarrow288-9x-7x-294=0\)
\(\Leftrightarrow9x+7x=288-294\)
\(\Leftrightarrow2x=-6\)
\(\Leftrightarrow x=-3\)
b. \(\left(2x-1\right)^2+\left|x+3\right|=0\)
\(\Leftrightarrow\left|x+3\right|=-4x^2+4x-1\)
\(\left|x+3\right|=x+3\)khi \(x+3\ge0\)hay \(x\ge-3\)
\(\left|x+3\right|=-\left(x+3\right)\)khi \(x+3< 0\)hay \(x< -3\)
với \(x\ge-3\Rightarrow x+3=-4x^2+4x-1\)
\(\Leftrightarrow4x^2-4x+1+x+3=0\)
\(\Leftrightarrow4x^2-3x+4=0\)\(\Leftrightarrow\)vô nghiệm
với \(x< -3\)\(\Rightarrow-x-3=-4x+4-1\)
\(\Leftrightarrow4x^2-4x+1-x-3=0\)
\(\Leftrightarrow4x^2-5x-2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5+\sqrt{57}}{8}\left(tm\right)\\x=\frac{5-\sqrt{57}}{8}\left(L\right)\end{cases}}\)