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\(x=\dfrac{1}{2}-\dfrac{2}{3}=\dfrac{3-4}{6}=-\dfrac{1}{6}\) là phương án c
a) Vì A=\(\dfrac{15^{16}+1}{15^{17}+1}\) < 1
\(\Rightarrow\dfrac{15^{16}+1}{15^{17}+1}< \dfrac{15^{16}+1+14}{15^{17}+1+14}=\dfrac{15^{16}+15}{15^{17}+15}\) \(=\dfrac{15\left(15^{15}+1\right)}{15\left(15^{16}+1\right)}\) \(=\dfrac{15^{15}+1}{15^{16}+1}\)
Vậy A<B
b) A=\(\dfrac{2006^{2007}+1}{2006^{2006}+1}>1\)
\(\Rightarrow\dfrac{2006^{2007}+1+2005}{2006^{2006}+1+2005}\)
= \(\dfrac{2006^{2007}+2006}{2006^{2006}+2006}\)
= \(\dfrac{2006\left(2006^{2006}+1\right)}{2006\left(2006^{2005}+1\right)}\)
= \(\dfrac{2006^{2006+1}}{2006^{2005}+1}\)
Vậy A>B
a) Ta có: \(\left|2x-\dfrac{1}{3}\right|\ge0\forall x\)
\(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|-\dfrac{7}{4}\ge-\dfrac{7}{4}\forall x\)
Dấu '=' xảy ra khi \(2x=\dfrac{1}{3}\)
hay \(x=\dfrac{1}{6}\)
Vậy: \(A_{min}=-\dfrac{7}{4}\) khi \(x=\dfrac{1}{6}\)
b) Ta có: \(\dfrac{1}{3}\left|x-2\right|\ge0\forall x\)
\(\left|3-\dfrac{1}{2}y\right|\ge0\forall y\)
Do đó: \(\dfrac{1}{3}\left|x-2\right|+\left|3-\dfrac{1}{2}y\right|\ge0\forall x,y\)
\(\Leftrightarrow\dfrac{1}{3}\left|x-2\right|+\left|3-\dfrac{1}{2}y\right|+4\ge4\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-2=0\\3-\dfrac{1}{2}y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\end{matrix}\right.\)
Vậy: \(B_{min}=4\) khi x=2 và y=6
a) 7.28=x.x
=> 196=x2
=> \(\left(\pm14\right)^2=x^2\)
=> x=\(\pm14\)
b) DK: x≠-17
pt<=> 4.(10+2)=6.(17+x)
=> 4.12=17.6+6x
=> 48-102=6x
=>-66=6x
=>x=-11
c) 7.(x+40)=6.(17+x)
=> 7x+280=102+6x
=> 7x-6x=102-280
=> x=-178
Giải:
a) \(\dfrac{7}{x}=\dfrac{x}{28}\)
\(\Leftrightarrow x^2=196\)
\(\Leftrightarrow x=\pm\sqrt{196}=\pm14\)
Vậy ...
b) \(\dfrac{10+2}{17+x}=\dfrac{3}{4}\)
\(\Leftrightarrow40+8=51+3x\)
\(\Leftrightarrow3x=40+8-51=-3\)
\(\Leftrightarrow x=-\dfrac{3}{3}=-1\)
Vậy ...
c) \(\dfrac{40+x}{17+x}=\dfrac{6}{7}\)
\(\Leftrightarrow280+7x=102+6x\)
\(\Leftrightarrow7x-6x=102-280\)
\(\Leftrightarrow x=-178\)
Vậy ...
\(A=\dfrac{636363\cdot37-373737\cdot63}{1+2+3+...+2006}\)
\(=\dfrac{37^2\cdot3^3\cdot7^2\cdot13-37^2\cdot3^3\cdot7^2\cdot13}{\left(2006+1\right)\cdot1003}\)
=0
A=\(x.\dfrac{1}{5}+x.\dfrac{2}{3}-x.\dfrac{1}{4}\)
=\(x.\left(\dfrac{1}{5}+\dfrac{2}{3}-\dfrac{1}{4}\right)\)
=\(x.\dfrac{37}{60}\)
Thay x=\(\dfrac{1}{2}\) vào A ta được
A=\(\dfrac{1}{2}.\dfrac{37}{60}=\dfrac{37}{120}\)
\(\dfrac{2006.125+1000}{126.2006-1006}\)
\(=\)\(\dfrac{2006.125+1000}{(125+1).2006-1006}\)
\(=\)\(\dfrac{2006.125+1000}{125.2006+1.2006-1006}\)
\(=\)\(\dfrac{2006.125+1000}{125.2006+2006-1006}\)
\(=\)\(\dfrac{2006.125+1000}{125.2006+1000}\)
\(=\) 1
a)A=3\(\dfrac{1}{11}\) x \(\dfrac{27}{46}\) x 1\(\dfrac{6}{17}\) x 2\(\dfrac{4}{9}\)
A=\(\dfrac{34}{11}\) x \(\dfrac{27}{46}\) x \(\dfrac{23}{17}\) x \(\dfrac{22}{9}\)
A=\(\dfrac{34\times27\times23\times22}{11\times46\times17\times9}\)
A=\(\dfrac{2\times3}{1}\)
A=6
mk cảm ơn bạn nhìu nha . bk có thể giải cho mk câu B đc ko