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a, 24-x=32=25
=> 4-x=5
<=> x=-1
b, (x+1,5)2+(y-2,5)10=0
Vì (x+1,5)2\(\ge\)0, (y-2,5)10\(\ge\)0
\(\Rightarrow\hept{\begin{cases}x+1,5=0\\y-2,5=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-1,5\\y=2,5\end{cases}}}\)
a)\(2^{4-x}\)=32
=>\(2^{4-x}\)=32=\(2^5\)
=>4-x=5
=>x=4-5=-1
=>x=-1
Có \(\left|x-1,5\right|\) \(\ge\) 0
\(\left|2,5-x\right|\) \(\ge\) 0
Suy ra \(\left|x-1,5\right|\) + \(\left|2,5-x\right|\) \(\ge\) 0
Mà \(\left|x-1,5\right|\) + \(\left|2,5-x\right|\) =0
Suy ra \(\left|x-1,5\right|\) = 0
\(\left|2,5-x\right|\) =0
Suy ra x-1,5=0
2,5-x=0
Suy ra x=1,5
x=2,5 (Vô lý)
Vậy k có giá trị của x thỏa mãn
\(\left|x-1,5\right|+\left|2,5-x\right|=0\)
\(\left\{{}\begin{matrix}\left|x-1,5\right|\ge0\\\left|2,5-x\right|\ge0\end{matrix}\right.\) \(\Rightarrow\left|x-1,5\right|+\left|2,5-x\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|x-1,5\right|=0\\\left|2,5-x\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1,5\\x=2,5\end{matrix}\right.\)
a: =>x/3=-5/2
hay x=-15/2
b: \(\Leftrightarrow\dfrac{7}{3}:x=\dfrac{1}{5}-\dfrac{4}{9}=\dfrac{9-20}{45}=\dfrac{-11}{45}\)
\(\Leftrightarrow x=\dfrac{7}{3}:\dfrac{-11}{45}=\dfrac{7}{3}\cdot\dfrac{-45}{11}=\dfrac{-105}{11}\)
c: \(\Leftrightarrow x=\dfrac{-7}{2}\cdot2=-7\)
d: =>x/27=-1/3+2/9=2/9-3/9=-1/9=-3/27
=>x=-3
\(\left|x-1,5\right|=2\\ \Rightarrow\left[{}\begin{matrix}x-1,5=2\\x-1,5=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3,5\\x=-0,5\end{matrix}\right.\)
Vậy \(x\in\left\{3,5;-0,5\right\}\)
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\(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\\ \Rightarrow\left|x+\frac{3}{4}\right|=\frac{1}{2}\\ \Rightarrow\left[{}\begin{matrix}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{2}\\x=-\frac{5}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{1}{2};-\frac{5}{4}\right\}\)
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\(\left|x-2\right|=x\left(ĐK:x\ge0\right)\\ \Rightarrow\left[{}\begin{matrix}x-2=x\\x-2=-x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-x=2\\x+x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}0=2\left(\text{vô lý}\right)\\2x=2\end{matrix}\right.\\ \Rightarrow x=1\left(tmđk\right)\)
Vậy \(x=1\)
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\(\left|x-3,4\right|+\left|2,6-x\right|=0\\ \Rightarrow\left|x-3,4\right|=-\left|2,6-x\right|\)
Mà \(\left|2,6-x\right|\ge0\forall x\Rightarrow-\left|2,6-x\right|\le0\forall x\)
\(\Rightarrow\left|x-3,4\right|\le0\forall x\left(\text{vô lý}\right)\)
Vậy \(x\in\varnothing\)
a/ \(\left|x-1,5\right|=2\)
\(\Rightarrow\left[{}\begin{matrix}x-1,5=2\\x-1,5=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2+1,5=3,5\\x=-2+1,5=-0,5\end{matrix}\right.\)
b/ \(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)
\(\Rightarrow\left|x+\frac{3}{4}\right|=0+\frac{1}{2}=\frac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}-\frac{3}{4}=\frac{2}{4}-\frac{3}{4}=-\frac{1}{4}\\x=-\frac{1}{2}-\frac{3}{4}=\left(-\frac{2}{4}\right)+\left(-\frac{3}{4}\right)=-\frac{5}{4}\end{matrix}\right.\)
c/ \(\left|x-2\right|=x\)
\(\Rightarrow\left[{}\begin{matrix}x-2=x\\x-2=-x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-x=2\\x+x=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}0=2\left(vô-lý\right)\\2x=2\end{matrix}\right.\)
=> 2x = 2
=> x = 2 : 2 = 1
d/ \(\left|x-3,4\right|+\left|2,6-x\right|=0\)
Ta có: \(\left\{{}\begin{matrix}\left|x-3,4\right|\ge0\\\left|2,6-x\right|\ge0\end{matrix}\right.\)
=> Để \(\left|x-3,4\right|+\left|2,6-x\right|=0\) thì \(\left\{{}\begin{matrix}\left|x-3,4\right|=0\\\left|2,6-x\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-3,4=0\\2,6-x=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=0+3,4=3,4\\x=2,6-0=2,6\end{matrix}\right.\)
a) \(\frac{21}{47}+\frac{9}{45}+\frac{26}{47}+\frac{4}{5}\)
\(=\left(\frac{21}{47}+\frac{26}{47}\right)+\left(\frac{9}{45}+\frac{4}{5}\right)\)
\(=\frac{47}{47}+\left(\frac{1}{5}+\frac{4}{5}\right)\)
\(=1+1=2\)
b) \(12.\left(-\frac{2}{3}\right)^2+\frac{4}{3}\)
\(=12.\frac{4}{9}+\frac{4}{3}\)
\(=\frac{16}{3}+\frac{4}{3}\)
\(=\frac{20}{3}\)
c) \(12,5.\left(-\frac{5}{7}\right)+15.\left(-\frac{5}{7}\right)\)
\(=\left(-\frac{5}{7}\right).\left(12,5+15\right)\)
\(=\left(-\frac{5}{7}\right).27,5\)
\(=\left(-\frac{5}{7}\right).\frac{55}{2}\)
\(=-\frac{275}{14}\)
d) \(\frac{4}{5}.\left(\frac{7}{2}+\frac{1}{4}\right)^2\)
\(=\frac{4}{5}.\left(\frac{14}{4}+\frac{1}{4}\right)^2\)
\(=\frac{4}{5}.\left(\frac{15}{4}\right)^2\)
\(=\frac{4}{5}.\frac{225}{16}\)
\(=\frac{45}{4}\)
a)\(\frac{21}{47}+\frac{9}{45}+\frac{26}{47}+\frac{4}{5}\)
=\(\frac{21}{47}+\frac{1}{5}+\frac{26}{47}+\frac{4}{5}\)
=\(\left(\frac{21}{47}+\frac{26}{47}\right)+\left(\frac{1}{5}+\frac{4}{5}\right)\)
=\(\frac{47}{47}+\frac{5}{5}=1+1=2\)
b)\(12.\left(-\frac{2}{3}\right)^2+\frac{4}{3}\)
=\(12.\frac{4}{9}+\frac{4}{3}\)
=\(\frac{12}{1}.\frac{4}{9}+\frac{4}{3}=\frac{48}{9}+\frac{4}{3}\)
=\(\frac{16}{3}+\frac{4}{3}=\frac{20}{3}\)
c)\(12,5.\left(-\frac{5}{7}\right)+1,5.\left(-\frac{5}{7}\right)\)
=\(\left(-\frac{5}{7}\right).\left(12,5+1,5\right)\)
=\(\left(-\frac{5}{7}\right).14=\left(-\frac{5}{7}\right).\frac{14}{1}=-10\)
d)\(\frac{4}{5}.\left(\frac{7}{2}+\frac{1}{4}\right)^2\)
=\(\frac{4}{5}.\left(\frac{14}{4}+\frac{1}{4}\right)^2\)
=\(\frac{4}{5}.\left(\frac{15}{4}\right)^2\)
=\(\frac{4}{5}.\frac{225}{16}\)
=\(\frac{900}{80}=\frac{45}{4}\)
Nhớ tick cho mình nha!
a: \(\left|x-1.5\right|+\left|2.5-x\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1.5=0\\2.5-x=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
b: \(\left(x-\dfrac{1}{2}\right)^2=0\)
=>x-1/2=0
hay x=1/2
c: \(2^x=16\)
nên \(2^x=2^4\)
=>x=4
d: \(3^{x+1}=9^x\)
\(\Leftrightarrow3^{2x}=3^{x+1}\)
=>2x=x+1
=>x=1
e: \(2^{3x+2}=4^{x+5}\)
=>3x+2=2x+10
=>x=8
bài 1 :
b) (x-1/2 )2 = 0
<=> x - 1/2 = 0
<=> x = 0+ 1/2
<=> x = 1/2
c) ( x - 2 ) 2 = 1
<=> x -2 = 1
<=> x = 1 +2 = 3
d) ( 2x -1 )3 = -8
<=> ( 2x - 1) 3 = ( -2 ) 3
<=> 2x - 1 = -2
<=> 2x = -2+1 = -1
<=> x = -1/2
Bài 2 :
c) 32x-1=243
<=> 32x-1= 35
<=> 2x-1 = 5
<=> 2x = 6
<=> x = 6:2 = 3
Mk chỉ giải đc như vậy thôi
bạn thông cảm nhé !
a, \(\dfrac{5}{6}-\left|2-x\right|=\dfrac{1}{3}\Rightarrow\dfrac{5}{6}-\dfrac{1}{3}=\left|2-x\right|\)
<=> \(\dfrac{1}{2}=\left|2-x\right|\) \(\Leftrightarrow\left[{}\begin{matrix}2-x=\dfrac{1}{2}\\2-x=\dfrac{-1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
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Mấy câu sau tương tự thôi
a)\(\dfrac{3}{2}hay\dfrac{-3}{2}\)
b)\(\dfrac{13}{20}hay\dfrac{-13}{20}\)
c)\(\dfrac{11}{6}hay\dfrac{-11}{6}\)
d)\(\dfrac{4}{3}hay\dfrac{-4}{3}\)
e)\(\dfrac{1}{5}hay\dfrac{-1}{5}\)
Đây là câu trả lời của mình
Hay có nghĩa là hoặc
|2,5-x|=1,3
\(\orbr{\begin{cases}2,5-x=1,3\\2,5-x=-1,3\end{cases}}\Rightarrow\orbr{\begin{cases}x=1,2\\x=3,8\end{cases}}\)
Vậy x=1,2 hoặc x=3,8
|x-1,5|+|2,5-x|=0
\(\Rightarrow\hept{\begin{cases}VT:x-1,5=0\\VP:2,5-x=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1,5\\x=2,5\end{cases}}\)
Vậy x của VT là 1,5 và x của VP là 2,5
\(\left(x-\frac{1}{2}\right)^2=0\)
\(\Rightarrow x-\frac{1}{2}=0\)
x=\(0+\frac{1}{2}\)
x=\(\frac{1}{2}\)
(x-2)2=1
=> x-2=1
x=1+2
x=3
=> x-2=-1
x=(-1)+2
x=1
a, / 2,5 - x / = 1,3
Với 2,5 - x > hoặc = 0 => 2, 5 - x = 1,3
=> x = 1, 2
Với 2,5 - x < hoặc = 0 => - ( 2,5 - x ) = 1,3
=> - 2,5 + x = 1,3
=> x = 3,8
Vậy x thuộc tập hợp 1,2 ; 3,8
p/s: > hoặc = 0, < hoặc = 0 , thuộc tập hợp bạn ghi kí hiệu nha
bn ơi bài này làm 2 trường hợp
\(\left|x-1,5\right|=2\\ x-1,5=2\\ x=0,5\)