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\(bn\)\(xem\)\(lai\)\(giup\)\(mk\)\(cho\)\(\frac{x+522}{7}\)\(neu\)\(thay\)\(bang\)\(\frac{x+552}{7}\)\(thi\)\(dug\)\(hon\)
thế thì bạn giải thử xem cô t ra đề thế mà ừ thì cứ cho là x + 552 cx đc
\(x+\left(\frac{1}{2}\right)^3=\frac{1}{4}\)
\(x+\frac{1}{8}=\frac{1}{4}\)
\(x=\frac{1}{4}-\frac{1}{8}\)
\(x=\frac{4}{16}-\frac{2}{16}\)
\(x=\frac{1}{8}\)
Vậy \(x=\frac{1}{8}\)
b) \(\left(\frac{2}{3}\right)^3-x=\frac{1}{3}\)
\(\frac{8}{27}-x=\frac{1}{3}\)
\(x=\frac{8}{27}-\frac{1}{3}\)
\(x=\frac{8}{27}-\frac{9}{27}\)
\(x=-\frac{1}{27}\)
Vậy \(x=-\frac{1}{27}\)
c) \(x.\left(-\frac{1}{2}\right)^4=\frac{3}{8}\)
\(x.\frac{1}{16}=\frac{3}{8}\)
\(x=\frac{3}{8}:\frac{1}{16}\)
\(x=\frac{3}{8}.16\)
\(x=6\)
c) \(\left(\frac{1}{2}\right)^3.x=\left(\frac{1}{2}\right)^5\)
\(x=\left(\frac{1}{2}\right)^5:\left(\frac{1}{2}\right)^3\)
\(x=\left(\frac{1}{2}\right)^2\)
\(x=\frac{1}{4}\)
Vậy \(x=\frac{1}{4}\)
Chúc bạn học tốt !!!
a) \(x+\left(\frac{1}{2}\right)^3=\frac{1}{4}\Leftrightarrow x+\frac{1}{8}=\frac{1}{4}\Leftrightarrow x=\frac{1}{4}-\frac{1}{8}\Leftrightarrow x=\frac{1}{8}\)
b) \(\left(\frac{2}{3}\right)^3-x=\frac{1}{3}\Leftrightarrow\frac{8}{27}-x=\frac{1}{3}\Leftrightarrow-x=\frac{1}{3}-\frac{8}{27}\Leftrightarrow-x=\frac{1}{27}\Leftrightarrow x=-\frac{1}{27}\)
c) \(x.\left(\frac{-1}{2}\right)^4=\frac{3}{8}\Leftrightarrow x.\frac{1}{16}=\frac{3}{8}\Leftrightarrow x=\frac{3}{8}:\frac{1}{16}\Leftrightarrow x=6\)
d) \(\left(\frac{1}{2}\right)^2.x=\left(\frac{1}{2}\right)^5\Leftrightarrow\frac{1}{8}.x=\frac{1}{32}\Leftrightarrow x=\frac{1}{32}:\frac{1}{8}\Leftrightarrow x=\frac{1}{4}\)
a)\(\frac{x+3}{x+5}=7\Leftrightarrow x+3=7\left(x+5\right)\)
\(\Leftrightarrow x+3=7x+35\)
\(\Leftrightarrow-6x=32\)
\(\Leftrightarrow x=-\frac{16}{3}\)
b)\(\frac{2x-1}{3x+5}=-\frac{2}{3}\)
\(\Leftrightarrow3\left(2x-1\right)=-2\left(3x+5\right)\)
\(\Leftrightarrow6x-3=-6x-10\)
\(\Leftrightarrow12x=-7\)
\(\Leftrightarrow x=-\frac{7}{12}\)
c)\(\frac{x+1}{4}=\frac{9}{x+1}\Leftrightarrow\left(x+1\right)^2=36\)
\(\Leftrightarrow\left(x+1\right)^2=6^2\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=6\\x+1=-6\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-7\end{cases}}}\)
d)\(\frac{6x-1}{2x+3}=\frac{3x}{x+2}\)
\(\Leftrightarrow\left(6x-1\right)\left(x+2\right)=3x\left(2x+3\right)\)
\(\Leftrightarrow6x^2+12x-x-2=6x^2+9x\)
\(\Leftrightarrow2x=2\Leftrightarrow x=1\)
2.
a) Ta có:
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)=\left(x+1\right)\left(\frac{1}{13}+\frac{1}{14}\right)\)
Vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\ne\frac{1}{13}+\frac{1}{14}\)nên \(x+1=0\Leftrightarrow x=-1\)
Vậy x = -1
b) Ta có:
\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Rightarrow\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)
\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\Rightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}\right)=\left(x+2004\right)\left(\frac{1}{2002}+\frac{1}{2003}\right)\)
Vì \(\frac{1}{2000}+\frac{1}{2001}\ne\frac{1}{2002}+\frac{1}{2003}\)nên \(x+2004=0\Leftrightarrow x=-2004\)
Vậy, x = -2004
(1/2-3/4) x - 7/3 = -5/9
-1/4 x =16/9
x= -64/9