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a, \(\left|2x-3\right|-\dfrac{1}{3}=0\Leftrightarrow\left|2x-3\right|=\dfrac{1}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=\dfrac{1}{3}\\2x-3=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)
b, tương tự
c, \(\left|2x-1\right|-\left|x+\dfrac{1}{3}\right|=0\Leftrightarrow\left|2x-1\right|=\left|x+\dfrac{1}{3}\right|\)
TH1 : \(2x-1=x+\dfrac{1}{3}\Leftrightarrow x=\dfrac{4}{3}\)
TH2 : \(2x-1=-x-\dfrac{1}{3}\Leftrightarrow3x=\dfrac{2}{3}\Leftrightarrow x=\dfrac{2}{9}\)
d, \(3x-\left|x+15\right|=\dfrac{5}{4}\Leftrightarrow\left|x+15\right|=3x-\dfrac{5}{4}\)ĐK : x >= 5/12
TH1 : \(x+15=3x-\dfrac{5}{4}\Leftrightarrow-2x=-\dfrac{65}{4}\Leftrightarrow x=\dfrac{65}{8}\)( tm )
TH2 : \(x+15=\dfrac{5}{3}-3x\Leftrightarrow4x=-\dfrac{40}{3}\Leftrightarrow x=-\dfrac{10}{3}\)
a, \(\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{2}{5}\right)>0\)
\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{1}{3}>0\\x+\dfrac{2}{5}>0\end{matrix}\right.\) hay \(\left\{{}\begin{matrix}x-\dfrac{1}{3}< 0\\x+\dfrac{2}{5}< 0\end{matrix}\right.\)
+,Xét \(\left\{{}\begin{matrix}x-\dfrac{1}{3}>0\\x+\dfrac{2}{5}>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>\dfrac{1}{3}\\x>-\dfrac{2}{5}\end{matrix}\right.\)
\(\Rightarrow x>\dfrac{1}{3}\)
+, Xét \(\left\{{}\begin{matrix}x-\dfrac{1}{3}< 0\\x+\dfrac{2}{5}< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x< \dfrac{1}{3}\\x< -\dfrac{2}{5}\end{matrix}\right.\)
\(\Rightarrow x< -\dfrac{2}{5}\)
Vậy...........
b, \(\left(x+\dfrac{3}{5}\right)\left(x+1\right)< 0\)
Vì \(x+\dfrac{3}{5}< x+1\) với mọi \(x\in R\)
\(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{3}{5}< 0\\x+1>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x< -\dfrac{3}{5}\\x>-1\end{matrix}\right.\)
Vậy...........
c, \(\dfrac{3}{7}x-\dfrac{2}{5}x=\dfrac{-17}{35}\)
\(\Rightarrow\dfrac{1}{35}x=\dfrac{-17}{35}\)
\(\Rightarrow x=-17\)
d, \(\left(\dfrac{3}{4}x-\dfrac{9}{10}\right)\left(\dfrac{1}{3}+\dfrac{-3}{5}x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x-\dfrac{9}{10}=0\\\dfrac{1}{3}+\dfrac{-3}{5}x=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{9}{10}\\-\dfrac{3}{5}x=-\dfrac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\\x=\dfrac{5}{9}\end{matrix}\right.\)
Vậy.........
Chúc bạn học tốt!!!
a/ \(\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{2}{5}\right)>0\)
TH1:\(\left\{{}\begin{matrix}x-\dfrac{1}{3}>0\\x+\dfrac{2}{5}>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>\dfrac{1}{3}\\x>-\dfrac{2}{5}\end{matrix}\right.\)\(\Rightarrow x>\dfrac{1}{3}\)
TH2:\(\left\{{}\begin{matrix}x-\dfrac{1}{3}< 0\\x+\dfrac{2}{5}< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x< \dfrac{1}{3}\\x< -\dfrac{2}{5}\end{matrix}\right.\)\(\Rightarrow x< -\dfrac{2}{5}\)
Vậy \(x>\dfrac{1}{3}\) hoặc \(x< -\dfrac{2}{5}\) thì tm
b/ \(\left(x+\dfrac{3}{5}\right)\left(x+1\right)< 0\)
TH1:\(\left\{{}\begin{matrix}x+\dfrac{3}{5}< 0\\x+1>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x< -\dfrac{3}{5}\\x>-1\end{matrix}\right.\) \(\Rightarrow-1< x< -\dfrac{3}{5}\)
TH2:\(\left\{{}\begin{matrix}x+\dfrac{3}{5}>0\\x+1< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>-\dfrac{3}{5}\\x< -1\end{matrix}\right.\)(vô lý)
Vậy....................
c/ \(\dfrac{3}{7}x-\dfrac{2}{5}x=-\dfrac{17}{35}\)
\(\Rightarrow\left(\dfrac{3}{7}-\dfrac{2}{5}\right)x=-\dfrac{17}{35}\)
\(\Rightarrow\dfrac{1}{35}x=-\dfrac{17}{35}\)
\(\Rightarrow x=-\dfrac{17}{35}:\dfrac{1}{35}=-17\)
Vậy.............
d/ \(\left(\dfrac{3}{4}x-\dfrac{9}{10}\right)\left(\dfrac{1}{3}+\dfrac{-3}{5}x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x-\dfrac{9}{10}=0\\\dfrac{1}{3}-\dfrac{3}{5}x=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{9}{10}\\\dfrac{3}{5}x=\dfrac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\\x=\dfrac{5}{9}\end{matrix}\right.\)
Vậy.....................
Bài 1:
a, \(2y.\left(y-\dfrac{1}{7}\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}2y=0\\y-\dfrac{1}{7}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=0\\y=\dfrac{1}{7}\end{matrix}\right.\)
Vậy \(y\in\left\{0;\dfrac{1}{7}\right\}\)
b, \(\dfrac{-2}{5}+\dfrac{2}{3}y+\dfrac{1}{6}y=\dfrac{-4}{15}\)
\(\Rightarrow\dfrac{5}{6}y=\dfrac{-4}{15}+\dfrac{2}{5}\)
\(\Rightarrow\dfrac{5}{6}y=\dfrac{2}{15}\)
\(\Rightarrow y=\dfrac{4}{25}\)
Vậy \(y=\dfrac{4}{25}\)
Chúc bạn học tốt!!!
Bài 1:
a, \(2y\left(y-\dfrac{1}{7}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2y=0\\y-\dfrac{1}{7}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=0\\y=\dfrac{1}{7}\end{matrix}\right.\)
Vậy...
b, \(\dfrac{-2}{5}+\dfrac{2}{3}y+\dfrac{1}{6}y=\dfrac{-4}{15}\)
\(\Rightarrow\dfrac{5}{6}y=\dfrac{2}{15}\)
\(\Rightarrow y=\dfrac{4}{25}\)
Vậy...
Bài 2:
a, \(x\left(x-\dfrac{4}{7}\right)>0\)
\(\Rightarrow\left\{{}\begin{matrix}x>0\\x-\dfrac{4}{7}>0\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x< 0\\x-\dfrac{4}{7}< 0\end{matrix}\right.\)
\(\Rightarrow x>\dfrac{4}{7}\left(x\ne0\right)\) hoặc \(x< \dfrac{4}{7}\left(x\ne0\right)\)
Vậy...
Các phần còn lại tương tự nhé
a: \(\left|x\right|=3+\dfrac{1}{5}=\dfrac{16}{5}\)
mà x<0
nên x=-16/5
b: \(\left|x\right|=-2.1\)
nên \(x\in\varnothing\)
c: \(\left|x-3.5\right|=5\)
=>x-3,5=5 hoặc x-3,5=-5
=>x=8,5 hoặc x=-1,5
d: \(\left|x+\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
=>|x+3/4|=1/2
=>x+3/4=1/2 hoặc x+3/4=-1/2
=>x=-1/4 hoặc x=-5/4
Câu 1:
b: \(\Leftrightarrow\left|x-1\right|=-3x+1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{1}{3}\\\left(-3x+1-x+1\right)\left(-3x+1+x-1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{1}{3}\\\left(-4x+2\right)\cdot\left(-2x\right)=0\end{matrix}\right.\Leftrightarrow x=0\)
c: \(\Leftrightarrow\left[{}\begin{matrix}2x-1=2x+3\\2x+3=1-2x\end{matrix}\right.\Leftrightarrow4x=2\Leftrightarrow x=\dfrac{1}{2}\)
d: \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
e: \(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\\left[x\left(x^2-\dfrac{5}{4}\right)-x\right]\left[x\left(x^2-\dfrac{5}{4}\right)+x\right]=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\x\left(x^2-\dfrac{9}{4}\right)\cdot x\cdot\left(x^2-\dfrac{1}{4}\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{0;\dfrac{1}{2};\dfrac{3}{2}\right\}\)
a.
| x | = 5,6
=>\(\left[{}\begin{matrix}x=5,6\\x=-5,6\end{matrix}\right.\)
Vậy \(x\in\left\{-5,6;5,6\right\}\)
b, \(\left|x-3,5\right|=5\)
=>\(\left[{}\begin{matrix}x-3,5=5\\x-3,5=-5\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=8,5\\x=-1,5\end{matrix}\right.\)
Vậy \(x\in\left\{-1,5;8,5\right\}\)
c,\(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
=> \(\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
=>\(\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{1}{4};\dfrac{5}{4}\right\}\)
d,\(\left|4x\right|-\left(\left|-13,5\right|\right)=\left|\dfrac{1}{4}\right|\)
=> \(\left|4x\right|-13,5=\dfrac{1}{4}\)
=> \(\left|4x\right|=13,75\)
=>\(\left[{}\begin{matrix}4x=13,75\\4x=-13,75\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=3,4375\\x=-3,4375\end{matrix}\right.\)
Vậy \(x\in\left\{-3,4375;3,4375\right\}\)
e, ( x - 1 ) 3 = 27
=> x - 1 = 3
=> x = 4
Vậy x = 4
f, ( 2x - 3)2 = 36
=> \(\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=4,5\\x=-1,5\end{matrix}\right.\)
Vậy x\(\in\left\{-1,5;4,5\right\}\)
g, \(5^{x+2}=625\)
=> \(5^{x+2}=5^4\)
=> x + 2 = 4
=> x = 2
Vậy x = 2
h, ( 2x - 1)3 = -8
=> 2x - 1 = -2
=> x = \(\dfrac{-1}{2}\)
Vậy x = \(\dfrac{-1}{2}\)
i, \(\dfrac{1}{4}.\dfrac{2}{6}.\dfrac{3}{8}.\dfrac{4}{10}.\dfrac{5}{12}...\dfrac{30}{62}.\dfrac{31}{64}=2^x\)
=> \(\dfrac{1.2.3.4.5...30.31}{4.6.8.10.12...62.64}=2^x\)
=>\(\dfrac{1.2.3.4.5...30.31}{\left(2.3.4.5...30.31.32\right)\left(2.2.2.2...2.2_{ }\right)}=2^x\)(có 31 số 2)
=> \(\dfrac{1}{32.2^{31}}=2^x\)
=> \(\dfrac{1}{2^{36}}=2^x\)
=> x = -36
Vậy x = -36
1.
a) \(-\dfrac{4}{9}+\left(-\dfrac{5}{6}\right)-\dfrac{17}{4}=-\dfrac{16}{36}-\dfrac{30}{36}-\dfrac{153}{36}\)
\(=-\dfrac{199}{36}\)
b) \(5\dfrac{1}{2}+\left(-3\right)=5\dfrac{1}{2}-3=\dfrac{11}{2}-\dfrac{6}{2}=\dfrac{5}{2}\)
c) \(4\dfrac{9}{11}+\left(-2\dfrac{1}{11}\right)=\dfrac{53}{11}-\dfrac{23}{11}=\dfrac{30}{11}\)
2.
a) \(4,3-\left(1,2\right)=3,1\)
b) \(0-\left(-0,4\right)=0+0,4=0,4\)
c) \(-\dfrac{2}{3}-\dfrac{1}{3}=-\dfrac{3}{3}=-1\)
d) \(-\dfrac{1}{2}-\dfrac{-1}{6}=-\dfrac{1}{2}+\dfrac{1}{6}=-\dfrac{3}{6}+\dfrac{1}{6}=-\dfrac{2}{6}=-\dfrac{1}{3}\)
3a) A=\(\dfrac{5}{x+xy+xyz}+\dfrac{5}{y+yz+1}+\dfrac{5xyz}{z+xz+xyz}\)
=\(\dfrac{5}{x\left(1+y+yz\right)}+\dfrac{5}{y+yz+1}+\dfrac{5xy}{1+x+xy}\)
=\(\dfrac{5}{x\left(1+y+zy\right)}+\dfrac{5x}{x\left(1+zy+y\right)}+\dfrac{5xy}{x\left(1+y+zy\right)}\)
=\(\dfrac{5+5x+5xy}{x\left(1+yz+y\right)}\)
=\(\dfrac{5x\left(yz+1+y\right)}{x\left(1+yz+y\right)}=5\)
a) |x-3,5| = 7,5
TH1: => x - 3,5 = 7,5
=> x = 7,5 + 3,5 = 11
TH2 : x - 3,5 = -7,5
=> x = -7,5 + 3.5 = -4
b) 3,6 - | x - 0,4| = 0
=> | x - 0,4| = 3,6 - 0 = 3,6
Th1: x - 0,4 = 3,6
=> x = 0,4 + 3,6 = 4
th2: x - 0,4 = -3,6
=> x = 0,4 + (-3,6) = -3,2
c) |x - 3,5| + |4,5 - x | = 0
= a + a = 0 ( loại bỏ vì nếu vậy thì phép tính trên sẽ ko hợp lí)
= -a + a = 0
Ta có: x - 3,5 = -a
4,5 - x = a
=> 3,5 + -a = 4,5 - a = 4,5 + (-a)
Vậy , không có số x nào thỏa mãn đk trên (theo mk là thế!)
Tíc nhá!
1)
a) \(|x-3,5|=7,5\)
\(\Rightarrow x-3,5=7,5\)
hay \(x-3,5=-7,5\)
TH1 : \(x-3,5=7,5\Rightarrow x=7,5+3,5=11\)
TH2 : \(x-3,5=-7,5\Rightarrow x=-7,5+3,5=-4\)
b) \(|x+\dfrac{4}{5}|-\dfrac{1}{2}=0\)
\(\Rightarrow\left(x+\dfrac{4}{5}\right)-\dfrac{1}{2}=0\) (chỉ có 1 TH vì số 0 ko phải dương or âm)
\(\left(x+\dfrac{4}{5}\right)=0+\dfrac{1}{2}=\dfrac{1}{2}\)
\(x=\dfrac{1}{2}-\dfrac{4}{5}=\dfrac{5-8}{10}=\dfrac{-3}{10}\)
c) \(3,6-|x-0,4|=0\)
\(\Rightarrow3,6-\left(x-0,4\right)=0\) ( giải thích giống câu b )
\(\Rightarrow-\left(x-0,4\right)=0-3,6\)
\(\Rightarrow-\left(x-0,4\right)=-3,6\)
\(\Rightarrow-x+0,4=-3,6\) ( Phá dấu )
\(\Rightarrow-x=-3,6-0,4=-3,6+\left(-0,4\right)=-4\)
\(\Rightarrow x=4\)
d) \(-\dfrac{5}{12}:|\dfrac{-5}{6}:x|=\dfrac{-5}{9}\)
\(\Rightarrow-\dfrac{5}{12}:|\dfrac{-5}{6}:x|=\dfrac{-5}{9}\)
hay \(\Rightarrow-\dfrac{5}{12}:|\dfrac{-5}{6}:x|=\dfrac{5}{9}\)
TH1 : \(-\dfrac{5}{12}:\left(-\dfrac{5}{6}:x\right)=\dfrac{-5}{9}\Rightarrow\left(-\dfrac{5}{6}:x\right)=-\dfrac{5}{12}:\left(-\dfrac{5}{9}\right)\)
\(\Rightarrow\left(-\dfrac{5}{6}:x\right)=\dfrac{5}{12}.\dfrac{9}{5}=\dfrac{9}{12}=\dfrac{3}{4}\)
\(\Rightarrow x=-\dfrac{5}{6}:\dfrac{3}{4}=-\dfrac{5.4}{6.3}=-\dfrac{5.2}{3.3}=-\dfrac{10}{9}\)
TH2 : \(\Rightarrow-\dfrac{5}{12}:\left(-\dfrac{5}{6}:x\right)=\dfrac{5}{9}\)
\(\Rightarrow\)\(\left(-\dfrac{5}{6}:x\right)=-\dfrac{5}{12}:\dfrac{5}{9}=-\dfrac{5.9}{12.5}=-\dfrac{9}{12}=-\dfrac{3}{4}\)
\(\Rightarrow x=-\dfrac{5}{6}:\left(-\dfrac{3}{4}\right)=\dfrac{5}{6}.\dfrac{4}{3}=\dfrac{10}{9}\)
Vậy x = ....
e)
Vì \(|x-3,5|\ge0;|4,5-x|\ge0\) với mọi x
Do đó : \(|x-3,5|+|4,5-x|=0\)
\(\Rightarrow|x-3,5|=0;|4,5-x|=0\)
\(\Rightarrow x-3,5=0\) và \(4,5-x=0\)
\(\Rightarrow x=0+3,5=3,5\) và \(-x=0+4,5=4,5\Rightarrow x=-4,5\)
( không đồng thời xảy ra)
\(\Rightarrow\) Không tồn tại x thuộc Q để \(|x-3,5|+|4,5-x|=0\)
2)
a) Đề sai
b) (45,3 + 7,3) + (-22)
= 52,6 + (-22) = 30,6
c) [(-11.7) + (11.7)] + [5.5+10]
= 0 + 15.5 = 15.5
( Câu c bạn cho rối quá )
d) [(-6.8) + 2.8] + [(-56.9) + 5.9 ]
= (-4) + (-51) = 55