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\(\left|x+\dfrac{1}{1.5}\right|+\left|x+\dfrac{1}{5.9}\right|+\left|x+\dfrac{1}{9.14}\right|+...+\left|x+\dfrac{1}{397.401}\right|\ge0\)
\(\Rightarrow101x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow x+\dfrac{1}{1.5}+x+\dfrac{1}{5.9}+...+x+\dfrac{1}{397.401}=101x\)
\(\Rightarrow101x+\left(\dfrac{1}{1.5}+\dfrac{1}{5.9}+...+\dfrac{1}{397.401}\right)=x\)
\(\Rightarrow\dfrac{1}{4}\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+...+\dfrac{4}{397.401}\right)=x\)
\(\Rightarrow x=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+....+\dfrac{1}{397}-\dfrac{1}{401}\right)\)
\(\Rightarrow x=\dfrac{1}{4}\left(1-\dfrac{1}{401}\right)\)
\(\Rightarrow x=\dfrac{1}{4}.\dfrac{400}{401}\)
\(\Rightarrow x=\dfrac{100}{401}\)
1) (x−1):0,16=−9:(1−x)
\(\Rightarrow\)(x-1):0,16= 9:(-1):(x-1)
\(\Rightarrow\)(x-1):0,16=9:(x-1)
\(\Rightarrow\)(x-1).(x-1)= 9. 0,16
\(\Rightarrow\)(x-1)\(^2\)= 1,44=1,2\(^2\)=(-1,2)\(^2\)
\(\Rightarrow\)x-1=1,2\(\Rightarrow\)x=2,2
hoặc x-1= -1,2\(\Rightarrow\)x= -0,2
Vậy x =2,2 ; x=0,2
...............................
a. \(\dfrac{-5}{4}\) x4 . \(\dfrac{8}{15}\) x = \(\dfrac{-40}{60}\) x5 = \(\dfrac{-2}{3}\) x5
b. -2x\(\left(\dfrac{3}{4}x^2-x+\dfrac{1}{2}\right)\) = -\(\dfrac{-3}{2}\) x3 + 2x3 - x
c. \(x\left(x-\dfrac{1}{2}\right)\) - (x - 2)(x + 3)
= x2 - \(\dfrac{1}{2}\) x - x2 - 3x - 2x - 6
\(VT=\left|2x+3\right|+\left|2x-1\right|=\left|2x+3\right|+\left|1-2x\right|\ge\left|2x+3+1-2x\right|=\left|4\right|=4\)
\(VP=\frac{8}{3\left(x+1\right)^2+2}\le\frac{8}{2}=4\)
\(VT\ge VP\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}\left(2x+3\right)\left(1-2x\right)\ge0\left(1\right)\\\left(x+1\right)^2=0\left(2\right)\end{cases}}\)
\(\left(2\right)\)\(\Leftrightarrow\)\(x=-1\) ( thỏa mãn\(\left(1\right)\) )
...
\(a,\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\Rightarrow x=\dfrac{5}{6}\\ b,\Rightarrow\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{3}{2}\right)^9=\left(\dfrac{3}{2}\right)^4\\ \Rightarrow2x-1-9=4\\ \Rightarrow2x=14\Rightarrow x=7\\ c,\Rightarrow2^{x-1}+2^{x+2}=9\cdot2^5\\ \Rightarrow2^{x-1}\left(1+2^3\right)=9\cdot2^5\\ \Rightarrow2^{x-1}\cdot9=9\cdot2^5\\ \Rightarrow2^{x-1}=2^5\Rightarrow x-1=5\Rightarrow x=6\\ d,\Rightarrow\left(2x+1\right)^2=12+69=81\\ \Rightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)
\(\Leftrightarrow\left|2x+3\right|+\left|2x-1\right|=\dfrac{8}{3x^2+6x+5}\)
TH1: x<-3/2
Pt sẽ là \(\dfrac{8}{3x^2+6x+5}=-2x-3+1-2x=-4x-2\)
=>(3x^2+6x+5)(-4x-2)=8
=>-12x^3-6x^2-24x^2-12x-20x-10-8=0
=>-12x^3-30x^2-32x-18=0
=>x=-1,35(loại)
TH2: -3/2<=x<1/2
Pt sẽ là \(\dfrac{8}{3x^2+6x+5}=2x+3+1-2x=4\)
=>3x^2+6x+5=2
=>3x^2+6x+3=0
=>x=-1(nhận)
TH3: x>=1/2
=>\(\dfrac{8}{3x^2+6x+5}=2x+3+2x-1=4x+2\)
=>(3x^2+6x+5)(4x+2)=8
=>12x^3+6x^2+24x^2+12x+20x+10-8=0
=>12x^3+30x^2+32x+2=0
=>x=-0,06(loại)
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