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22 tháng 9 2017
Bài 1: a) 4 . ( \(\dfrac{1}{3}\) - x ) + \(\dfrac{1}{2}\) = \(\dfrac{5}{6}\) . x
=> \(\dfrac{4}{3}\) - 4.x + \(\dfrac{1}{2}\) = \(\dfrac{5}{6}\) . x
=> \(\dfrac{11}{6}\) = \(\dfrac{29}{6}\) . x
=> x = \(\dfrac{11}{29}\) .
b) \(\dfrac{5}{2}\) - 3 . ( \(\dfrac{1}{3}\) - x ) = \(\dfrac{1}{4}\) - 7.x
=> \(\dfrac{1}{4}\) - \(\dfrac{3}{2}\) = 7.x + 3.x
=> \(\dfrac{-5}{4}\) = 10.x
=> x = \(\dfrac{-1}{8}\).
25 tháng 7 2017
a, \(\left(x-\dfrac{1}{3}\right)^2=0\)
=> \(x-\dfrac{1}{3}=0\)
=>\(x=\dfrac{1}{3}\)
Vậy \(x=\dfrac{1}{3}\)
b, \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
=>\(\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{8}\right)^2\)
=> \(x+\dfrac{1}{2}=\dfrac{1}{8}\)
=> \(x=-\dfrac{3}{8}\)
c, (2x - 1)^3 = 8
=> (2x - 1)^3 = 2^3
=> 2x - 1 = 2
=> 2x = 3
=> x = 3/2
25 tháng 7 2017
a) (x - \(\dfrac{1}{3}\))2=0
=> x- \(\dfrac{1}{3}\)=0
x=\(\dfrac{1}{3}\)
b) (x + \(\dfrac{1}{2}\))2=\(\dfrac{1}{16}\)
=> (x+\(\dfrac{1}{2}\)) 2= (\(\dfrac{1}{4}\))2=(\(\dfrac{-1}{4}\))2
TH1: x+ \(\dfrac{1}{2}\)=\(\dfrac{1}{4}\)
x= \(\dfrac{-1}{4}\)
TH2 : x + \(\dfrac{1}{2}\)= \(\dfrac{-1}{4}\)
x = \(\dfrac{-3}{4}\)
Vậy x = \(\dfrac{-1}{4}\); \(\dfrac{-3}{4}\)
c) (2x-1)3 =8
=> 2x - 1 = 2
2x = 3
x = \(\dfrac{3}{2}\)
24 tháng 9 2022
1: \(\Leftrightarrow3x+4=2\)
=>3x=-2
=>x=-2/3
2: \(\Leftrightarrow7x-7=6x-30\)
=>x=-23
3: =>\(5x-5=3x+9\)
=>2x=14
=>x=7
4: =>9x+15=14x+7
=>-5x=-8
=>x=8/5
C
24 tháng 8 2018
\(a,\dfrac{2}{3}-\dfrac{1}{3}\left(x-\dfrac{3}{2}\right)-\dfrac{1}{2}\left(2x+1\right)=5\)
\(\dfrac{2}{3}-\dfrac{1}{3}x-\dfrac{1}{2}-x+\dfrac{1}{2}=5\)
\(\dfrac{2}{3}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}x-x=5\)
\(\dfrac{2}{3}-\dfrac{1}{3}x-x=5\)
\(\dfrac{2}{3}-\dfrac{4}{3}x=5\)
\(\dfrac{4}{3}x=\dfrac{2}{3}-5\)
\(\dfrac{4}{3}x=-\dfrac{13}{3}\)
\(x=-\dfrac{13}{3}:\dfrac{4}{3}\)
\(x=-\dfrac{13}{4}\)
Vậy...............
\(b,\left(x+\dfrac{1}{2}\right)\left(\dfrac{3}{4}-x\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{3}{4}-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy................
\(c,\dfrac{2x-1}{-3+2}=0\)
\(\Rightarrow2x-1=0\)
\(\Rightarrow x=\dfrac{1}{2}\)
Vậy.............
5 tháng 6 2022
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
30 tháng 6 2017
bạn sử dụng 7 hằng đẳnng thức đó
a. A – B)3 = A3 – 3A2B + 3AB2 – B3
b. A2 – B2 = (A – B)(A + B)
c. (A + B)3 = A3 + 3A2B + 3AB2 + B3
d. A3 – B3 = (A – B)(A2 + AB + B2)
27 tháng 11 2022
b: =>(3x-1)(3x+1)(2x+3)=0
hay \(x\in\left\{\dfrac{1}{3};-\dfrac{1}{3};-\dfrac{3}{2}\right\}\)
c: \(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{5}{6}+\dfrac{3}{4}=\dfrac{19}{12}\)
=>2x-1/3=19/12 hoặc 2x-1/3=-19/12
=>2x=23/12 hoặc 2x=-15/12=-5/4
=>x=23/24 hoặc x=-5/8
d: \(\Leftrightarrow-\dfrac{5}{6}\cdot x+\dfrac{3}{4}=-\dfrac{3}{4}\)
=>-5/6x=-3/2
=>x=3/2:5/6=3/2*6/5=18/10=9/5
e: =>2/5x-1/2=3/4 hoặc 2/5x-1/2=-3/4
=>2/5x=5/4 hoặc 2/5x=-1/4
=>x=5/4:2/5=25/8 hoặc x=-1/4:2/5=-1/4*5/2=-5/8
f: =>14x-21=9x+6
=>5x=27
=>x=27/5
h: =>(2/3)^2x+1=(2/3)^27
=>2x+1=27
=>x=13
i: =>5^3x*(2+5^2)=3375
=>5^3x=125
=>3x=3
=>x=1
19 tháng 5 2022
a: \(\left(2x-\dfrac{1}{2}\right)^3\)
\(=\left(2x\right)^3-3\cdot4x^2\cdot\dfrac{1}{2}+3\cdot2x\cdot\dfrac{1}{4}-\dfrac{1}{8}\)
\(=8x^3-6x^2+\dfrac{3}{2}x-\dfrac{1}{8}\)
b: \(\left(\dfrac{1}{2}x-y\right)\left(\dfrac{1}{2}x+y\right)=\dfrac{1}{4}x^2-y^2\)
c: \(\left(x+\dfrac{1}{3}\right)^3\)
\(=x^3+3\cdot x^2\cdot\dfrac{1}{3}+3\cdot x\cdot\dfrac{1}{9}+\dfrac{1}{27}\)
\(=x^3+x^2+\dfrac{1}{3}x+\dfrac{1}{27}\)
d: \(\left(x-2\right)\left(x^2+2x+4\right)=x^3-8\)
TD
19 tháng 3 2017
a) Ta có: |2x-5| \(\ge\)0 với mọi x
mà |2x-5|=-4
=> x\(\in\varnothing\)
b)\(\dfrac{1}{3}-\left|\dfrac{5}{4}-2x\right|=\dfrac{1}{4}\)
=>\(\left|\dfrac{5}{4}-2x\right|=\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{1}{12}\)
=>\(\left[{}\begin{matrix}\dfrac{5}{4}-2x=\dfrac{1}{12}\\\dfrac{5}{4}-2x=-\dfrac{1}{12}\end{matrix}\right.=>\left[{}\begin{matrix}2x=\dfrac{5}{4}-\dfrac{1}{12}=\dfrac{7}{6}\\2x=\dfrac{5}{4}+\dfrac{1}{12}=\dfrac{4}{3}\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=\dfrac{7}{12}\\x=\dfrac{2}{3}\end{matrix}\right.\)
phần c và d cũng tương tự bạn tự làm nha
Ta có
TH1 : \(\dfrac{1}{2}\)x = 3-2x TH2 : \(\dfrac{1}{2}\)x = -3+2x
\(\dfrac{5}{2}\)x = 3 \(\dfrac{-3}{2}\) x = -3
x = \(\dfrac{6}{5}\) < Thỏa mãn > x = 2 < Loại >
Vậy x = \(\dfrac{6}{5}\)
Chúc bn làm bài tốt ❤❤❤❤❤❤
|\(\dfrac{1}{2}\)\(x\)| = 3 - 2\(x\) (3 - 2\(x\) > 0; \(x< \dfrac{3}{2}\))
\(\left[{}\begin{matrix}\dfrac{1}{2}x=3-2x\\\dfrac{1}{2}x=-3+2x\end{matrix}\right.\)
\(\left[{}\begin{matrix}\dfrac{1}{2}x+2x=3\\\dfrac{1}{2}x-2x=-3\end{matrix}\right.\)
\(\left[{}\begin{matrix}\dfrac{5}{2}x=3\\-\dfrac{3}{2}x=-3\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=3:\dfrac{5}{2}\\x=-3:-\dfrac{3}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{6}{5}\\x=2\end{matrix}\right.\)
vì \(x< \dfrac{3}{2}\) nên \(x=2\) loại
Vậy \(x\) = \(\dfrac{6}{5}\)