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1) Tính
a) 253 : 52 = (52)3 : 52 = 56 : 52 = 54 = 625
\(b)\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^6=\left(\dfrac{3}{7}\right)^{21}:\left[\left(\dfrac{3}{7}\right)^2\right]^6=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^{12}=\left(\dfrac{3}{7}\right)^9\) d) 9 . 32 . \(\dfrac{1}{81}\) . 32 = 32 . 32 . \(\dfrac{1}{3^4}\) . 32 = 9
2) Tìm x thuộc Q, biết:
a) 3x + 2 = 27
=> 3x + 2 = 33
x + 2 = 3
x = 3 - 2
x = 1
b) \(\left(\dfrac{1}{2}x-3\right)^4=81\)
\(\Rightarrow\left(\dfrac{1}{2}x-3\right)^4=3^4\)
\(\dfrac{1}{2}x-3=3^{ }\)
\(\dfrac{1}{2}x=3+3\)
\(\dfrac{1}{2}x=9\)
\(x=9:\dfrac{1}{2}\)
\(x=18\)
c) \(\left(x-\dfrac{1}{2}\right)^3=-27\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\left(-3\right)^3\)
\(x-\dfrac{1}{2}=-3\)
\(x=-3+\dfrac{1}{2}\)
\(x=\dfrac{-5}{2}\)
d) 5 . 5x + 1 = 125
5x + 1 = 125 : 5
5x + 1 = 25
5x + 1 = 52
x + 1 = 2
x = 2 - 1
x = 1.
Bài 1:
a, \(\dfrac{x+5}{x}=\dfrac{4}{3}\)
\(\Rightarrow3x+15=4x\\ \Rightarrow4x-3x=15\\ \Rightarrow x=15\)
b, \(\dfrac{x-20}{x-10}=\dfrac{x+40}{x+70}\)
\(\Rightarrow\left(x-20\right).\left(x+70\right)=\left(x+40\right)\left(x-10\right)\)
\(\Rightarrow x^2+70x-20x-1400=x^2-10x+40x-400\)
\(\Rightarrow x^2-x^2+70x-20x+10x-40x=-400+1400\)
\(\Rightarrow20x=1000\Rightarrow x=50\)
c, \(4^x=\dfrac{1.2.3.....31}{4.6.8.....64}\)
\(\Rightarrow4^x=\dfrac{1}{2.2.2.2.....2.2.64}\) (có 30 số 2)
\(\Rightarrow4^x=\dfrac{1}{2^{30}.4^3}\Rightarrow4^x=\dfrac{1}{4^{15}.4^3}\)
\(\Rightarrow4^x=\dfrac{1}{4^{18}}\)
\(\Rightarrow4^x=4^{-18}\)
Vì \(4\ne-1;4\ne0;4\ne1\) nên \(x=-18\)
Chúc bạn học tốt!!!
a , \(\dfrac{x+5}{x}=\dfrac{4}{3}\Leftrightarrow3\left(x+5\right)=4x\)
<=> 3x+15=4x
<=> x= 15
b , \(\dfrac{x-20}{x-10}=\dfrac{x+40}{x+70}\)
<=> \(\dfrac{x-10}{x-10}-\dfrac{10}{x-10}=\dfrac{x+70}{x+70}-\dfrac{30}{x+70}\)
<=> \(1-\dfrac{10}{x-10}=1-\dfrac{30}{x+70}\)
<=> \(\dfrac{10}{x-10}=\dfrac{30}{x+70}\Leftrightarrow\dfrac{1}{x-10}=\dfrac{3}{x+70}\)
<=> (x+70)=3(x-10)
<=> x+70 = 3x-30
<=> 100=2x
<=> x= 50
a: \(\dfrac{3x+2}{5x+7}=\dfrac{3x-1}{5x+1}\)
\(\Leftrightarrow\left(3x+2\right)\left(5x+1\right)=\left(3x-1\right)\left(5x+7\right)\)
\(\Leftrightarrow15x^2+3x+10x+2=15x^2+21x-5x-7\)
=>16x-7=13x+2
=>3x=9
hay x=3
b: \(\dfrac{x+1}{2016}+\dfrac{x}{2017}=\dfrac{x+2}{2015}+\dfrac{x+3}{2014}\)
\(\Leftrightarrow\left(\dfrac{x+1}{2016}+1\right)+\left(\dfrac{x}{2017}+1\right)=\left(\dfrac{x+2}{2015}+1\right)+\left(\dfrac{x+3}{2014}+1\right)\)
=>x+2017=0
hay x=-2017
e: \(\left(2x-3\right)^2=144\)
=>2x-3=12 hoặc 2x-3=-12
=>2x=15 hoặc 2x=-9
=>x=15/2 hoặc x=-9/2
a: =>1/6x=-49/60
=>x=-49/60:1/6=-49/60*6=-49/10
b: =>3/2x-1/5=3/2 hoặc 3/2x-1/5=-3/2
=>x=17/15 hoặc x=-13/15
c: =>1,25-4/5x=-5
=>4/5x=1,25+5=6,25
=>x=125/16
d: =>2^x*17=544
=>2^x=32
=>x=5
i: =>1/3x-4=4/5 hoặc 1/3x-4=-4/5
=>1/3x=4,8 hoặc 1/3x=-0,8+4=3,2
=>x=14,4 hoặc x=9,6
j: =>(2x-1)(2x+1)=0
=>x=1/2 hoặc x=-1/2
a) Ta có:
\(\left|x-2017\right|\ge0\) với \(\forall x\)
\(\left|y-2018\right|\ge0\) với \(\forall x\)
\(\Rightarrow\left|x-2017\right|+\left|y-2018\right|\ge0\) với \(\forall x\)
\(\Rightarrow\) Không có giá trị của x; y thỏa mãn yêu cầu
Vậy \(x;y\in\varnothing\)
b) Ta có:
\(3.\left|x-y\right|^5\ge0\)
\(10.\left|y+\dfrac{2}{3}\right|^7\ge0\)
\(3.\left|x-y\right|^5+10.\left|y+\dfrac{2}{3}\right|^7\ge0\left(1\right)\)
Theo bài ra ta có: \(3.\left|x-y\right|^5+10.\left|y+\dfrac{2}{3}\right|^7\le0\left(2\right)\)
Từ (1) và (2)
\(\Rightarrow3.\left|x-y\right|^5+10.\left|y+\dfrac{2}{3}\right|^7=0\)
\(\Rightarrow\left\{{}\begin{matrix}3.\left|x-y\right|^5=0\\10.\left|y+\dfrac{2}{3}\right|^7=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|x-y\right|^5=0\\\left|y+\dfrac{2}{3}\right|^7=0\end{matrix}\right.\Rightarrow}\left\{{}\begin{matrix}x-y=0\\y+\dfrac{2}{3}=0\end{matrix}\right.\Rightarrow}\left\{{}\begin{matrix}x=y\\y=\dfrac{-2}{3}\end{matrix}\right.\Rightarrow}\left\{{}\begin{matrix}x=\dfrac{-2}{3}\\y=\dfrac{-2}{3}\end{matrix}\right.\)\(\)
a) \(7-\sqrt{x}=0\)
\(\Rightarrow\sqrt{x}=7\)
\(\Rightarrow x=\left(\sqrt{7}\right)^2\)
b) \(5\sqrt{x}+1=40\)
\(\Rightarrow5\sqrt{x}=39\)
\(\Rightarrow\sqrt{x}=7,8\)
\(\Rightarrow x=\left(\sqrt{7,8}\right)^2\)
c) \(\dfrac{5}{12}\sqrt{x}-\dfrac{1}{6}=\dfrac{1}{3}\)
\(\Rightarrow\dfrac{5}{12}\sqrt{x}=\dfrac{1}{2}\)
\(\Rightarrow\sqrt{x}=1,2\)
\(\Rightarrow x=\left(\sqrt{1,2}\right)^2\)
d) \(4x^2-1=0\)
\(\Rightarrow\left(2x-1\right)\left(2x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=0\Rightarrow x=0,5\\2x+1=0\Rightarrow x=-0,5\end{matrix}\right.\)
e) \(\sqrt{x+1}-2=0\)
\(\Rightarrow\sqrt{x+1}=2\)
\(\Rightarrow x+1=1,414\)
\(\Rightarrow x=0,414\)
f) \(2x^2+0,82=1\)
\(\Rightarrow2x^2=0,18\)
\(\Rightarrow x^2=0,09\)
\(\Rightarrow x=\pm0,3\)
g) Không có kết quả
a) \(-5\cdot\left(x+\dfrac{1}{5}\right)-\dfrac{1}{2}\cdot\left(x-\dfrac{2}{3}\right)=\dfrac{3}{2}\cdot x-\dfrac{5}{6}\\ -5\cdot x+1-\dfrac{1}{2}\cdot x-\dfrac{1}{3}=\dfrac{3}{2}\cdot x-\dfrac{5}{6}\\ x\cdot\left(-5-\dfrac{1}{2}\right)+\dfrac{1}{3}+\dfrac{5}{6}=\dfrac{3}{2}\cdot x\\ x\cdot\dfrac{-11}{2}+\dfrac{7}{6}=\dfrac{3}{2}\cdot x\\ \dfrac{3}{2}\cdot x-\dfrac{-11}{2}\cdot x=\dfrac{7}{6}\\ x\cdot\left(\dfrac{3}{2}-\dfrac{-11}{2}\right)=\dfrac{7}{6}\\ x\cdot7=\dfrac{7}{6}\\ x=\dfrac{7}{6}:7\\ x=\dfrac{1}{6}\)
Vậy x = \(\dfrac{1}{6}\)
b, \(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}=2^x\\ \dfrac{1\cdot2\cdot3\cdot4\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot30\cdot31}{2^{30}\cdot\left(1\cdot2\cdot3\cdot4\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot30\cdot31\right)\cdot64}=2^x\\ \dfrac{1}{2^{30}\cdot2^6}=2^x\\ \dfrac{1}{2^{36}}=2^x\\ 2^{-36}=2^x\\ \Rightarrow x=-36\)
5a.
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+....+\dfrac{1}{19.21}\\ =\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+....+\dfrac{1}{19}-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}.\dfrac{20}{21}=\dfrac{10}{21}\)
b.
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\\ =\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+....+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)< \dfrac{1}{2}.1=\dfrac{1}{2}\)
a, Ta có: \(\left(2x+\dfrac{1}{4}\right)^4\ge0\rightarrow\left(2x+\dfrac{1}{4}\right)^4+6\ge6\)
Dấu ''=" xảy ra khi \(2x+\dfrac{1}{4}=0\rightarrow2x=\dfrac{-1}{4}\rightarrow x=\dfrac{-1}{8}\)
Vậy MinE=6\(\Leftrightarrow x=\dfrac{-1}{8}\)
b, Ta có: \(\left(5-3x\right)^2\ge0\rightarrow\left(5-3x\right)^2-2013\ge-2013\)
Dấu ''='' xảy ra khi \(5-3x=0\rightarrow3x=5\rightarrow x=\dfrac{5}{3}\)
Vậy MinE=-2013\(\Leftrightarrow x=\dfrac{5}{3}\)
a) \(E=\left(2x+\dfrac{1}{4}\right)^4+6\)
Vì \(\left(2x+\dfrac{1}{4}\right)^4\ge0\)
Nên \(\left(2x+\dfrac{1}{4}\right)^4+6\ge6\)
Vậy GTNN của \(E=6\) khi \(2x+\dfrac{1}{4}=0\Leftrightarrow x=\dfrac{-1}{8}\)
b) \(E=\left(5-3x\right)^2-2013\)
Vì \(\left(5-3x\right)^2\ge0\)
Nên \(\left(5-3x\right)^2-2013\ge-2013\)
Vậy GTNN của \(E=-2013\) khi \(5-3x=0\Leftrightarrow x=\dfrac{5}{3}\)
c) \(A=2013+\left|2x-3\right|\)
Vì \(\left|2x-3\right|\ge0\)
Nên \(2013+\left|2x-3\right|\ge2013\)
Vậy GTNN của \(A=2013\) khi \(2x-3=0\Leftrightarrow x=\dfrac{3}{2}\)
d) \(B=-1+\left|\dfrac{1}{2}x-3\right|\)
Vì \(\left|\dfrac{1}{2}x-3\right|\ge0\)
Nên \(-1+\left|\dfrac{1}{2}x-3\right|\ge-1\)
Vậy GTNN của \(B=-1\) khi \(\dfrac{1}{2}x-3=0\Leftrightarrow x=6\)
a,\(\dfrac{2}{7}x-\dfrac{1}{2}=\dfrac{3}{4}:\sqrt{\dfrac{49}{64}}\)
\(\Leftrightarrow\dfrac{2}{7}x-\dfrac{1}{2}=\dfrac{6}{7}\)
\(\Leftrightarrow\dfrac{2}{7}x=\dfrac{19}{14}\)
\(\Leftrightarrow x=\dfrac{19}{4}\)
Với mọi \(x\in R\)
\(\left|x+2016\right|+\left|x+2017\right|+\left|x+2018\right|\ge0\Leftrightarrow6x\ge0\Leftrightarrow x\ge0\)
với \(x\ge0\) ta được: \(\left\{{}\begin{matrix}\left|x+2016\right|=x+2016\\\left|x+2017\right|=x+2017\\\left|x+2018\right|=x+2018\end{matrix}\right.\)
\(pt\Leftrightarrow3x+6051=6x\Leftrightarrow3x=6051\Leftrightarrow x=2017\)