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P = 3, 7 + | 4, 3 - x |
Ta có : | 4, 3 - x | ≥ 0 ∀ x
=> 3, 7 + | 4, 3 - x | ≥ 3, 7 ∀ x
Dấu "=" xảy ra <=> 4, 3 - x = 0 => x = 4, 3
=> MinP = 3, 7 <=> x = 4, 3
Q = 5,5 - | 2x - 1, 5 |
Ta có : - | 2x -1, 5 | ≤ 0 ∀ x
=> 5, 5 - | 2x - 1, 5 | ≤ 5, 5 ∀ x
Dấu "=" xảy ra <=> 2x - 1, 5 = 0 => x = 3/4
=> MaxQ = 5, 5 <=> x = 3/4
|\(x\)| = 1 ⇒ (|\(x\)|)2 = 1 ⇒ \(x^2\) = 1
Thay \(x^2\) = 1 vào biểu thức: M = (\(x^{2^{ }}\) + a)(\(x^2\) + b)(\(x^2\) + c) ta có:
M = (1 + a)(1 + b)(1 + c)
M = (1 + b + a + ab)(1 + c)
M = 1 + b + a + ab + c + bc + ac + abc
M = 1 + ( a + b + c) + (ab + bc + ac) + abc
M = 1 + 2 + (-5) + 3
M = (1+2+3) - 5
M = 1
Sửa đề :
\(A=\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^{2005}}\)
\(2A=2+1+\frac{1}{2}+...+\frac{1}{2^{2004}}\)
\(2A-A=\left(1+2+\frac{1}{2}+...+\frac{1}{2^{2004}}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^{2005}}\right)\)
\(A=2-\frac{1}{2^{2005}}\)
\(a,\dfrac{a+b}{a-b}=\dfrac{c+a}{c-a}\Leftrightarrow\left(a+b\right)\left(c-a\right)=\left(c+a\right)\left(a-b\right)\\ \Leftrightarrow ac-a^2+bc-ab=ac-bc+a^2-ab\\ \Leftrightarrow2bc=2a^2\Leftrightarrow a^2=bc\Leftrightarrow m=a^2-bc=0\)
\(b,\Leftrightarrow\dfrac{abz-acy}{a^2}=\dfrac{bcx-abz}{b^2}=\dfrac{acy-bcx}{c^2}=\dfrac{abz-acy+bcx-abz+acy-bcx}{a^2+b^2+c^2}=0\\ \Leftrightarrow\left\{{}\begin{matrix}abz-acy=0\\bcx-abz=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}bz=cy\\cx=az\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{z}{c}=\dfrac{y}{b}\\\dfrac{x}{a}=\dfrac{z}{c}\end{matrix}\right.\\ \Leftrightarrow\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\)
cách 2:
a=\(6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)
a=(6-5-3)-(2/3+5/3-7/3)+(1/2+3/2-5/2)
a=-2-1/2
a=-5/2
Bài 1:
a, \(\dfrac{-x-2}{3}\) = - \(\dfrac{6}{7}\)
- \(x\) - 2 = - \(\dfrac{18}{7}\)
\(x\) = - 2 + \(\dfrac{18}{7}\)
\(x\) = - \(\dfrac{4}{7}\)
Bài b, \(\dfrac{4}{7-x}\) = \(\dfrac{1}{3}\)
12 = 7 - \(x\)
\(x\) = 7 - 12
\(x\) = -5
\(=\frac{\left(3\cdot3\cdot5\right)^{10}\cdot5^{20}}{\left(3\cdot5\cdot5\right)^{15}}\)
\(=\frac{3^{10}\cdot3^{10}\cdot5^{10}\cdot5^{20}}{3^{15}\cdot5^{15}\cdot5^{15}}\)
\(=\frac{3^{20}\cdot5^{30}}{3^{15}\cdot5^{30}}\)
\(=3^5=243\)
nhớ nha