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a) \(A=x^3+2x^2+7x-4-x-x^3-2x^2+1\)
\(A=\left(x^3-x^3\right)+\left(2x^2-2x^2\right)+\left(7x-x\right)+\left(-4+1\right)\)
\(A=6x-3\)
b) Thay x = (-5)
\(\Rightarrow A=6.\left(-5\right)-3\)
\(\Rightarrow A=-30-3\)
\(\Rightarrow A=-33\)
c) \(A=6x-3\)
\(10=6x-3\)
\(13=6x\)
\(x=\frac{13}{6}\)
a)\(x+\frac{1}{3}=\frac{3}{4}\)
\(\Rightarrow x=\frac{3}{4}-\frac{1}{3}\)
\(\Rightarrow x=\frac{5}{12}\)
b)\(x-\frac{2}{5}=\frac{5}{7}\)
\(\Rightarrow x=\frac{5}{7}+\frac{2}{5}\)
\(\Rightarrow x=1\frac{4}{35}\)
c)\(-x-\frac{2}{3}=-\frac{6}{7}\)
\(\Rightarrow-x=-\frac{6}{7}+\frac{2}{3}\)
\(\Rightarrow-x=-\frac{4}{21}\)
\(\Rightarrow x=\frac{4}{21}\)
d)\(\frac{4}{7}-x=\frac{1}{3}\)
\(x=\frac{4}{7}-\frac{1}{3}\)
\(\Rightarrow x=\frac{5}{21}\)
nhận giá trị âm tức là giá trị của biểu thức nhỏ hơn 0 và ngược lại!
a) \(15-3x< 0\)
\(\Leftrightarrow-3x< -15\)
\(\Leftrightarrow3x>5\)
b) \(27x+9< 0\)
\(\Leftrightarrow27x< -9\)
\(\Leftrightarrow x< -\frac{1}{3}\)
c) \(2y^2-4x< 0\)
\(\Leftrightarrow2\cdot\left(y^2-2x\right)< 0\)
\(\Leftrightarrow y^2-2x< 0\)
......
Bài 2 :
a,\(\frac{x-1}{3}=2-\frac{x}{-2}\)
\(\Leftrightarrow\frac{x-1}{3}=\frac{-4-x}{-2}\Leftrightarrow-2x+2=-12-3x\Leftrightarrow x=-14\)
b, \(\frac{x-1}{x+5}=\frac{6}{7}\Leftrightarrow7x-7=6x+30\Leftrightarrow x=37\)
c, \(\frac{2x-1}{4}=\frac{4}{2x-1}\Leftrightarrow\left(2x-1\right)^2=16\)
\(\Leftrightarrow\left(2x-1\right)^2-4^2=0\Leftrightarrow\left(2x-5\right)\left(2x+3\right)=0\Leftrightarrow x=\frac{5}{2};-\frac{3}{2}\)
1)
a) \(\dfrac{2}{5}+\dfrac{3}{10}=\dfrac{4}{10}+\dfrac{3}{10}=\dfrac{7}{10}=0,7\)
b)\(\dfrac{-5}{12}+\dfrac{4}{15}=\dfrac{-25}{60}+\dfrac{16}{60}=\dfrac{-9}{60}=-0,15\)
2)
a) \(2x-\dfrac{3}{2}=\dfrac{5}{2}\)
\(\Leftrightarrow2x=\dfrac{5}{2}+\dfrac{3}{2}\)
\(\Leftrightarrow2x=4\)
\(\Leftrightarrow x=2\)
Vậy x=2
b) \(\left|x+3,8\right|=5,8\)
\(\Leftrightarrow x+3,8=\pm5,8\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3,8=5,8\\x+3,8=-5,8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-9,6\end{matrix}\right.\)
Vậy x=2 ; x=-9,6
3)
\(2^{255}=2^{3.85}=\left(2^3\right)^{85}=8^{85}\)
\(3^{170}=3^{2.85}=\left(3^2\right)^{85}=9^{85}\)
Vì \(8< 9\)
Nên \(8^{85}< 9^{85}\)
Vậy \(2^{255}< 3^{170}\)
1)
a, \(\dfrac{2}{5}+\dfrac{3}{10}=\dfrac{4}{10}+\dfrac{3}{10}=\dfrac{7}{10}=0,7\)
b)\(\dfrac{-5}{12}+\dfrac{4}{15}=\dfrac{-25}{60}+\dfrac{16}{60}=\dfrac{-9}{60}=-0,15\)
2)
a) \(2x-\dfrac{3}{2}=\dfrac{5}{2}\)
\(\Leftrightarrow2x=\dfrac{5}{2}+\dfrac{3}{2}\)
\(\Leftrightarrow2x=4\)
Vậy x=2
b) \(\left|x+3,8\right|=5,8\)
\(\Leftrightarrow x+3,8=\pm5,8\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3,8=5,8\\x+3,8=-5,8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-9,6\end{matrix}\right.\)
Vậy x=2 ; x=-9,6
3)
\(2^{255}=2^{3.85}=\left(2^3\right)^{85}=8^{85}\)
\(3^{170}=3^{2.85}=\left(3^2\right)^{85}=9^{85}\)
Vì \(8< 9\)
Nên \(8^{85}< 9^{85}\)
Vậy \(2^{255}< 3^{170}\)