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3,5 + /x + \(\frac{3}{2}\) / = -1,5(-\(\sqrt{9}\))
=> 3,5 +/ x +\(\frac{3}{2}\) / = -1,5 ( -3 )
=> 3,5 + / x + \(\frac{3}{2}\) / =4,5
=> / x + \(\frac{3}{2}\) / = 4,5 - 3,5
=> / x + \(\frac{3}{2}\) / = 1
=> \(\hept{\begin{cases}x+\frac{3}{2}=1\\x+\frac{3}{2}=-1\end{cases}}\)
=> \(\hept{\begin{cases}x=1-\frac{3}{2}\\x=-1-\frac{3}{2}\end{cases}}\)
=> \(\hept{\begin{cases}x=\frac{-1}{2}\\x=\frac{-5}{2}\end{cases}}\)
vậy x = \(\frac{-1}{2}\)hay x = \(\frac{-5}{2}\)
\(3,5+\left|x+\frac{3}{2}\right|=-1,5.\left(-\sqrt{9}\right)\) \(3,5+\left|x+\frac{3}{2}\right|=-1,5.\left(-3\right)\) \(3,5+\left|x+\frac{3}{2}\right|=4,5\) \(\left|x+\frac{3}{2}\right|=4,5-3,5\) \(\left|x+\frac{3}{2}\right|=1\) \(\Rightarrow\orbr{\begin{cases}x+\frac{3}{2}=1\\x+\frac{3}{2}=-1\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=-\frac{5}{2}\end{cases}}\) Vậy x=\(-\frac{1}{2}\) hoặc x=\(-\frac{5}{2}\)
\(3,5+\left|x+\dfrac{3}{2}\right|=-1,5\cdot\left(-\sqrt{9}\right)\)
\(3,5+\left|x+\dfrac{3}{2}\right|=-1,5\cdot\left(-3\right)\)
\(3,5+\left|x+\dfrac{3}{2}\right|=4,5\)
\(\left|x+\dfrac{3}{2}\right|=4,5-3,5\)
\(\left|x+\dfrac{3}{2}\right|=1\)
\(\Rightarrow x+\dfrac{3}{2}=1\) hoặc \(x+\dfrac{3}{2}=-1\)
\(x=1-\dfrac{3}{2}\) \(x=-1-\dfrac{3}{2}\)
\(x=\dfrac{-1}{2}\) \(x=\dfrac{-5}{2}\)
Vậy \(x=\dfrac{-1}{2}\)hoặc \(x=\dfrac{-5}{2}\)
\(3,5+\left|x+\dfrac{3}{2}\right|=-1,5.\left(-\sqrt{9}\right)\)
\(\Rightarrow3,5+\left|x+\dfrac{3}{2}\right|=4,5\)
\(\Rightarrow\left|x+\dfrac{3}{2}\right|=4,5-3,5=1\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{2}=1\\x+\dfrac{3}{2}=-1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1-\dfrac{3}{2}\\x=-1-\dfrac{3}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=\dfrac{-5}{2}\end{matrix}\right.\)
Vậy..................
a: Trường hợp 1: x>=-2,3
A=2x+4,6-2x+15,4=20
Trường hợp 2: x<-2,3
M=-2x-4,6-2x+15,4=-4x+10,8
b: Trường hợp 1: x<-7,2
M=-x-7,2-(1,2-x)
=-x-7,2-1,2+x=-8,4
Trường hợp 2: -7,2<=x<1,2
B=x+7,2-(1,2-x)
=x+7,2-1,2+x=2x+6
Trường hợp 3: x>=1,2
B=x+7,2-x+1,2=8,4
c: Trường hợp 1: x>=-3
C=8,5x-19,5-(1,5x+4,5)
=8,5x-19,5-1,5x-4,5
=7x-24
Trường hợp 2: x<-3
C=8,5x-19,5-(-1,5x-4,5)
=8,5x-19,5+1,5x+4,5
=10x-15
\(K=5\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{45\cdot46}\right)\)
\(=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{45}-\dfrac{1}{46}\right)\)
=5*45/46=225/46
\(T=\dfrac{1}{5}\cdot\sqrt{6\cdot\dfrac{2}{3}}-\dfrac{3}{2}\cdot\sqrt{\dfrac{2}{3}\cdot\dfrac{8}{75}}+\dfrac{1}{2}\cdot\sqrt{6\cdot\dfrac{8}{75}}\)
\(=\dfrac{1}{5}\cdot2-\dfrac{3}{2}\cdot\dfrac{4}{15}+\dfrac{1}{2}\cdot\dfrac{4}{5}\)
=2/5-12/30+4/10
=2/5
a) \(A=\left|x-3,5\right|+\left|4,1-x\right|\)
do \(3,5\le x\le4,1\) \(\Rightarrow\)\(x-3,5\ge0;\) \(4,1-x\ge0\)
\(\Leftrightarrow\)\(\left|x-3,5\right|=x-3,5;\) \(\left|4,1-x\right|=4,1-x\)
\(A=x-3,5+4,1-x=0,6\)
b) bạn làm tương tự
a: \(=0.5\cdot10-\dfrac{1}{7}+15=20-\dfrac{1}{7}=\dfrac{139}{7}\)
b: \(=6\cdot\dfrac{-2}{3}+12\cdot\dfrac{4}{9}+18\cdot\dfrac{-8}{27}\)
\(=-4+\dfrac{16}{3}-\dfrac{16}{3}=-4\)
c: \(=\left(\dfrac{5}{2}+\dfrac{3}{8}-\dfrac{5}{8}+\dfrac{2}{3}\right):\left(\dfrac{17}{2}+\dfrac{49}{4}-\dfrac{17}{8}+\dfrac{34}{15}\right)\)
\(=\dfrac{35}{12}:\dfrac{2507}{120}=\dfrac{350}{2507}\)
a) \(A=|x-3,5|+|4,1-x|;3,5\le x\le4,1\)
\(=x-3,5+x-4,1\)
\(=2x-7,6\)
b) \(B=|x+1|+|x-3|\)
Nếu x =< -1 =>B=-x-1+3-x=2-2x
Nếu -1<x <3 => B=x+1+x-x=4
Nếu x >= 3 => B=x+1+x-3=2x-2
a) Vì \(3,5\le x\le4,1\)
\(\Rightarrow|x-3,5|\ge0;|4,1-x|\ge0\forall x\)
A= x-3,5+4,1-x = 0,6
b) * với x < -1
B= - (x+1) - (x-3) = -x - 1-x +3 = - 2x + 2
*Với -1 \(\le\)x < 3
thì B = x+1 - (x-3)
B= x+1 - x + 3 = 4
* Với x \(\ge\)3
thì B= x+1+x-3 = 2x - 2
x=144
Ta có: \(\sqrt{x}-3,5=8,5\)
\(\Rightarrow\sqrt{x}=8,5+3,5\)
\(\Rightarrow\sqrt{x}=12\)
\(\Rightarrow x=144\)
Hoc tốt nha^^