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\(a,\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\Rightarrow x=\dfrac{5}{6}\\ b,\Rightarrow\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{3}{2}\right)^9=\left(\dfrac{3}{2}\right)^4\\ \Rightarrow2x-1-9=4\\ \Rightarrow2x=14\Rightarrow x=7\\ c,\Rightarrow2^{x-1}+2^{x+2}=9\cdot2^5\\ \Rightarrow2^{x-1}\left(1+2^3\right)=9\cdot2^5\\ \Rightarrow2^{x-1}\cdot9=9\cdot2^5\\ \Rightarrow2^{x-1}=2^5\Rightarrow x-1=5\Rightarrow x=6\\ d,\Rightarrow\left(2x+1\right)^2=12+69=81\\ \Rightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)
(Lần sau, "bé" nhớ sử dụng công cụ công thức trực quan nhé, dịch của "bé" mình mệt ghê :V)
a,
\(\frac{x+1}{55}+\frac{x+3}{53}+\frac{x+5}{51}=\frac{x+7}{49}+\frac{x+9}{47}+\frac{x+11}{45}\\ \Leftrightarrow\frac{x+1}{55}+1+\frac{x+3}{53}+1+\frac{x+5}{51}+1=\frac{x+7}{49}+1+\frac{x+9}{47}+1+\frac{x+11}{45}+1\\ \Leftrightarrow\frac{x+56}{55}+\frac{x+56}{53}+\frac{x+56}{51}=\frac{x+56}{49}+\frac{x+56}{47}+\frac{x+56}{45}\\ \Leftrightarrow\left(x+56\right)\left(\frac{1}{55}+\frac{1}{53}+\frac{1}{51}-\frac{1}{49}-\frac{1}{47}-\frac{1}{45}\right)=0\)
And... u know dat right? ( ͡~ ͜ʖ ͡°)
b,
\(\frac{x+29}{31}-\frac{x+27}{33}=\frac{x+17}{43}-\frac{x+15}{45}\\ \Leftrightarrow\frac{x+29}{31}+1-\frac{x+27}{33}+1=\frac{x+17}{43}+1-\frac{x+15}{45}+1\\ \Leftrightarrow\frac{x+60}{31}-\frac{x+60}{33}=\frac{x+60}{43}-\frac{x+60}{45}\\ \Leftrightarrow\left(x+60\right)\left(\frac{1}{31}-\frac{1}{33}-\frac{1}{43}+\frac{1}{45}\right)=0\)
Ok, phần còn lại là của bạn nhé. :)
Chúc bạn học tốt nha.
Ta có :
\(\frac{x-3}{97}+\frac{x-27}{73}+\frac{x-67}{33}+\frac{x-73}{27}=4\)
\(\Leftrightarrow\left(\frac{x-3}{97}-1\right)+\left(\frac{x-27}{73}-1\right)+\left(\frac{x-67}{33}-1\right)+\left(\frac{x-73}{27}-1\right)=0\)
\(\Leftrightarrow\frac{x-100}{97}+\frac{x-100}{73}+\frac{x-100}{33}+\frac{x-100}{27}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}\right)=0\)
Vì \(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}>0\) Nên \(x-100=0\)
\(\Leftrightarrow x=100\)
Vậy \(x=100\)
\(\Leftrightarrow\frac{x-3}{87}+\frac{x-27}{79}+\frac{x-67}{33}+\frac{x-73}{27}-4=0\)
\(\Leftrightarrow\left(\frac{x-3}{97}-1\right)+\left(\frac{x-27}{73}-1\right)+\left(\frac{x-67}{33}-1\right)+\left(\frac{x-73}{27}-1\right)=0\)
\(\Leftrightarrow\left(\frac{x-3-97}{97}\right)+\left(\frac{x-27-73}{73}\right)+\left(\frac{x-67-33}{33}\right)+\left(\frac{x-73-27}{27}\right)=0\)
\(\Leftrightarrow\frac{x-100}{97}+\frac{x-100}{73}+\frac{x-100}{33}+\frac{x-100}{27}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}\right)=0\)
Vì \(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}\ne0\)
\(\Rightarrow x-100=0\Leftrightarrow x=100\)
a. \(\dfrac{x}{-27}=\dfrac{-3}{x}\)
\(\Leftrightarrow xx=\left(-27\right)\left(-3\right)\)
\(\Leftrightarrow x^2=81\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=9^2\\x^2=\left(-9\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-9\end{matrix}\right.\)
Vậy ...
\(\dfrac{x}{-27}=\dfrac{-3}{x}\)
\(\Rightarrow x^2=81\)
\(\Rightarrow\left[{}\begin{matrix}x=9\\x=-9\end{matrix}\right.\)
\(\dfrac{-9}{x}=\dfrac{-x}{\dfrac{4}{49}}\)
\(\Rightarrow-x^2=\dfrac{-36}{49}\)
\(-x^2\ge0;\dfrac{-36}{49}< 0\)
Vậy đa thức vô nghiệm
Dấu " / " là phân số nhé :
a) 5 5/27 + 7/23 + 0,5 - 5/27 + 16/23
= 140/27 + 7/23 + 1/2 - 5/27 + 16/23
= ( 140/27 - 5/27 ) + ( 7/23 + 16/23 ) + 1/2
= 5 + 1 + 1/2
= 6 + 1/2
= 13/2
b) 3/8 . 27 1/5 - 51 .1/5 . 3/8 + 19
= 3/8 . 136/5 - 51 . 1/5 . 3/8 + 19
= 3/8 . ( 136/5 - 1/5 ) - 51 + 19
= 3/8 . 27 - 51 + 19
= 81/8 - 52 + 19
= -183/8
Sorry nha mik sửa lại con B
b) 3/8 . 27 1/5 - 51 . 1/5 . 3/8 + 19
= 3/8 . 136/5 - 51/5 . 3/8 + 19
= 3/8 . ( 136/5 - 51/5 ) + 19
= 3/8 . 17 + 19
= 51/8 + 19
= 203/8