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a) \(\left(-\frac{3}{4}\right)^{3x-1}=\frac{-27}{64}\)
\(\Leftrightarrow\left(-\frac{3}{4}\right)^{3x-1}=\left(-\frac{3}{4}\right)^3\)
\(\Leftrightarrow3x-1=3\)
\(\Leftrightarrow3x=4\)
\(\Leftrightarrow x=\frac{4}{3}\)
b) Đề sai ! Sửa :
\(\left(\frac{4}{5}\right)^{2x+5}=\frac{256}{625}\)
\(\Leftrightarrow\left(\frac{4}{5}\right)^{2x+5}=\left(\frac{4}{5}\right)^4\)
\(\Leftrightarrow2x+5=4\)
\(\Leftrightarrow2x=-1\)
\(\Leftrightarrow x=-\frac{1}{2}\)
c) \(\frac{\left(x+3\right)^5}{\left(x+5\right)^2}=\frac{64}{27}\)
\(\Leftrightarrow\left(x+3\right)^3=\left(\frac{4}{3}\right)^3\)
\(\Leftrightarrow x+3=\frac{4}{3}\)
\(\Leftrightarrow x=-\frac{5}{3}\)
d) \(\left(x-\frac{2}{15}\right)^3=\frac{8}{125}\)
\(\Leftrightarrow\left(x-\frac{2}{15}\right)^3=\left(\frac{2}{15}\right)^3\)
\(\Leftrightarrow x-\frac{2}{15}=\frac{2}{15}\)
\(\Leftrightarrow x=\frac{4}{15}\)
Bài 1:
a) \(\frac{1}{5}x^4y^3-3x^4y^3\)
= \(\left(\frac{1}{5}-3\right)x^4y^3\)
= \(-\frac{14}{5}x^4y^3.\)
b) \(5x^2y^5-\frac{1}{4}x^2y^5\)
= \(\left(5-\frac{1}{4}\right)x^2y^5\)
= \(\frac{19}{4}x^2y^5.\)
Mình chỉ làm 2 câu thôi nhé, bạn đăng nhiều quá.
Chúc bạn học tốt!
d,\(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\\ \Leftrightarrow\left(x-\frac{2}{9}\right)^3=\left(\frac{4}{9}\right)^3\\ \Leftrightarrow x-\frac{2}{9}=\frac{4}{9}\\ \Leftrightarrow x=\frac{6}{9}\)
Vậy...
a) \(\left(x-3\right).\left(4-5x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\4-5x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0+3\\5x=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=4:5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=\frac{4}{5}\end{matrix}\right.\)
Vậy \(x\in\left\{3;\frac{4}{5}\right\}.\)
b) \(\left|x+\frac{3}{4}\right|+\frac{1}{3}=0\)
\(\Rightarrow\left|x+\frac{3}{4}\right|=0-\frac{1}{3}\)
\(\Rightarrow\left|x+\frac{3}{4}\right|=-\frac{1}{3}.\)
Ta luôn có: \(\left|x\right|\ge0\) \(\forall x.\)
\(\Rightarrow\left|x+\frac{3}{4}\right|>-\frac{1}{3}\)
\(\Rightarrow\left|x+\frac{3}{4}\right|\ne-\frac{1}{3}.\)
Vậy \(x\in\varnothing.\)
c) \(5^x.\left(5^3\right)^2=625\)
\(\Rightarrow5^x.5^6=5^4\)
\(\Rightarrow5^{x+6}=5^4\)
\(\Rightarrow x+6=4\)
\(\Rightarrow x=4-6\)
\(\Rightarrow x=-2\)
Vậy \(x=-2.\)
Chúc bạn học tốt!
Bài 1:
Ta có: \(x+\left(-\frac{31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x\)
\(\Leftrightarrow2x=\frac{1440}{144}=10\)
\(\Rightarrow x=5\)
Khi đó: \(y^2=\left(\frac{49}{12}\right)^2-5=\frac{1681}{144}\)
=> \(\hept{\begin{cases}y=\frac{41}{12}\\y=-\frac{41}{12}\end{cases}}\)
a) \(\left|x\right|+\frac{1}{4}=\frac{1}{5}\)
\(\left|x\right|=\frac{1}{5}-\frac{1}{4}\)
\(\left|x\right|=\frac{-1}{20}\)(vô lý vì \(\left|x\right|\ge0\)với mọi x . Mà \(\frac{-1}{20}\)>0 )
Vậy không tồn tại x
b)\(\left|x+2\right|-\frac{1}{12}=\frac{1}{4}\)
\(\left|x+2\right|=\frac{1}{4}+\frac{1}{12}\)
\(\left|x+2\right|=\frac{1}{3}\)
\(\Rightarrow x+2\varepsilon\left\{\frac{1}{3};\frac{-1}{3}\right\}\)
+)\(x+2=\frac{1}{3}\Rightarrow x=\frac{-5}{3}\) +)\(x+2=\frac{-1}{3}\Rightarrow x=\frac{-7}{3}\)
Vậy \(x=\frac{-5}{3}\)hoặc \(x=\frac{-7}{3}\)
c)\(\left|x+5\right|=\frac{1}{7}-\left|\frac{4}{3}-\frac{1}{6}\right|\)
\(\left|x+5\right|=\frac{1}{7}-\frac{7}{6}\)
\(\left|x+5\right|=\frac{-43}{42}\)( vô lý vì \(\left|x+5\right|\ge0\)với mọi x , mà \(\frac{-43}{42}< 0\))
Vậy không tồn tại x
d)\(\left|x+\frac{5}{6}\right|=\left|\frac{1}{5}-\frac{2}{3}\right|+\frac{-3}{4}\)
\(\left|x+\frac{5}{6}\right|=\frac{7}{15}+\frac{-3}{4}\)
\(\left|x+\frac{5}{6}\right|=\frac{-17}{60}\)( Vô lý vì \(\left|x+\frac{5}{6}\right|\ge0\)với mọi x mà \(\frac{-17}{60}< 0\))
Vậy không tồn tại x
a.\(3x^2-51=-24\)
\(\Rightarrow3x^2=27\)
\(\Rightarrow x^2=9\)
\(\Rightarrow x=3\)
b.\(5x.\left(5^3\right)^2=625\)
\(\Rightarrow5x=5^5:5^{\left(3x2\right)}\)
\(\Rightarrow5x=5^5:5^6\)
\(\Rightarrow5x=5^{-1}=0,2\)
\(\Rightarrow x=0,2:5\)
\(\Rightarrow x=0,04\)
(1/3)x+1=(1/3)x
=> x+1=x
=> x-x=-1
=> 0=-1 ( vô lí )
(4/5)x=256/625
=> (4/5)x=(4/5)4
=> x=4
Vậy x=4
a) \(\left(\frac{1}{3}\right)^{x+1}=\left(\frac{1}{3}\right)^x\)
\(\left(\frac{1}{3}\right)^{x+1}-\left(\frac{1}{3}\right)^x=0\)
\(\left(\frac{1}{3}\right)^x.\left[\frac{1}{3}-1\right]=0\)
\(\Leftrightarrow\left(\frac{1}{3}\right)^x=0\) (vô lý)
=> Không có giác trị thõa mãn
b) \(\frac{256}{625}=\frac{4^4}{5^4}=\left(\frac{4}{5}\right)^4\)