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a) \(\left(x-\frac{1}{2}\right)^4=\frac{1}{81}\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^4=\left(\frac{1}{3}\right)^4\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{2}=\frac{1}{3}\\x-\frac{1}{2}=\frac{-1}{3}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{5}{6}\\x=\frac{1}{6}\end{cases}}\)
Vậy ...
a.
\(-\frac{2}{3}-\frac{1}{3}\times\left(2x-5\right)=\frac{3}{2}\)
\(-\frac{2}{3}-\frac{2}{3}x+\frac{5}{3}=\frac{3}{2}\)
\(\left(-\frac{2}{3}+\frac{5}{3}\right)-\frac{2}{3}x=\frac{3}{2}\)
\(\frac{3}{3}-\frac{2}{3}x=\frac{3}{2}\)
\(1-\frac{2}{3}x=\frac{3}{2}\)
\(\frac{2}{3}x=1-\frac{3}{2}\)
\(\frac{2}{3}x=\frac{2}{2}-\frac{3}{2}\)
\(\frac{2}{3}x=-\frac{1}{2}\)
\(x=-\frac{1}{2}\div\frac{2}{3}\)
\(x=-\frac{1}{2}\times\frac{3}{2}\)
\(x=-\frac{3}{4}\)
b.
\(\frac{1}{3}x+\frac{2}{5}\times\left(x-1\right)=0\)
\(\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0\)
\(x\times\left(\frac{1}{3}+\frac{2}{5}\right)=\frac{2}{5}\)
\(x\times\left(\frac{5}{15}+\frac{6}{15}\right)=\frac{2}{5}\)
\(x\times\frac{11}{15}=\frac{2}{5}\)
\(x=\frac{2}{5}\div\frac{11}{15}\)
\(x=\frac{2}{5}\times\frac{15}{11}\)
\(x=\frac{6}{11}\)
Chúc bạn học tốt
a ) \(-\frac{2}{3}-\frac{1}{3}\left(2x-5\right)=\frac{3}{2}\)
\(\frac{1}{3}\left(2x-5\right)=-\frac{2}{3}-\frac{3}{2}\)
\(\frac{1}{3}\left(2x-5\right)=\frac{-13}{6}\)
\(\left(2x-5\right)=-\frac{13}{6}:\frac{1}{3}\)
\(\left(2x-5\right)=-\frac{13}{6}.\frac{3}{1}\)
\(\left(2x-5\right)=-\frac{13}{2}\)
\(2x=-\frac{13}{2}+5\)
\(2x=-\frac{3}{2}\)
\(\Rightarrow x=-\frac{3}{2}:2\)
\(\Rightarrow x=-\frac{3}{2}.\frac{1}{2}\)
\(\Rightarrow x=-\frac{3}{4}\)
\(\left|x^2-3x\right|+\left|\left(x+1\right)\left(x-3\right)\right|=0\)
\(\Leftrightarrow\hept{\begin{cases}\left|x^2-3x\right|=0\\\left|\left(x+1\right)\left(x-3\right)\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x^2-3x=0\\\left(x+1\right)\left(x-3\right)=0\end{cases}}\)
Xét \(x^2-3x=0\)
\(\Rightarrow x\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
Xét \(\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)
Vì xét 2 trị biểu thức , một cái có 2 giá trị (0 or 3) , một cái (-1 or 3)
Nên ta lấy cái chung
=> x = 3
Ta có:(3x-y)\(^2\)\(\ge\) 0 \(\forall\) x
|x+y|\(\ge\) 0 \(\forall\)i x,y
=>(3x-y)\(^2\)+|x+y|\(\ge\)0 \(\forall\) x,y
=>(3x-y)\(^2\)+|x+y|-3\(\ge\)-3 \(\forall\)x,y
Vậy GTNN của biểu thức B là -3
Dấu "=" xảy ra khi (3x-y)\(^2\)=|x+y|=0
Với (3x-y)\(^2\)=0=>3x-y=0=>3x=y=>x=y=0
Với |x+y|=0=>x+y=0=>x=x=0
Vậy biểu thức B đạt GTNN là -3 khi x=y=0
Ta có : \(\left|x+\frac{13}{14}\right|=-\left|x-\frac{3}{7}\right|\)
\(\Rightarrow\left|x+\frac{13}{14}\right|+\left|x-\frac{3}{7}\right|=0\)
Mà : \(\left|x+\frac{13}{14}\right|\ge0\forall x\)
\(\left|x-\frac{3}{7}\right|\ge0\forall x\)
Nên : \(\orbr{\begin{cases}\left|x+\frac{13}{14}\right|=0\\\left|x-\frac{3}{7}\right|=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{13}{14}=0\\x-\frac{3}{7}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{13}{14}\\x=\frac{3}{7}\end{cases}}\)
\(\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|=0\)
vì \(\left|\frac{3}{2}x+\frac{1}{9}\right|\ge0;\left|\frac{1}{5}y-\frac{1}{2}\right|\ge0=>\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|\ge0\) (với mọi x,y)
Mà \(\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|=0\) (theo đề)
Nên \(\left|\frac{3}{2}x+\frac{1}{9}\right|=0=>\frac{3}{2}x=-\frac{1}{9}=>x=-\frac{2}{27}\)
\(\left|\frac{1}{5}y-\frac{1}{2}\right|=0=>\frac{1}{5}y=\frac{1}{2}=>y=\frac{5}{2}\)
Vậy...........
\(a^{\left(2n+6\right)\left(3n+9\right)}=1\)
=>(2n+6)(3n+9)=0
=>n=-3