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a) \(\dfrac{3}{4}-\dfrac{1}{2}x=-\dfrac{1}{4}\)
\(\Leftrightarrow3-2x=-1\)
\(\Leftrightarrow-2x=-1-3\)
\(\Leftrightarrow-2x=-4\)
\(\Leftrightarrow x=2\)
Vậy \(x=2\)
b) \(1\dfrac{2}{3}x+0,2=x-\dfrac{7}{8}\)
\(\Leftrightarrow\dfrac{5}{3}x+\dfrac{1}{5}=x-\dfrac{7}{8}\)
\(\Leftrightarrow200x+24=120x-105\)
\(\Leftrightarrow80x=-129\)
\(\Leftrightarrow x=-\dfrac{129}{80}\)
Vậy \(x=-\dfrac{129}{80}\)
c) \(\dfrac{3}{4}-\left|x+0,5\right|=\dfrac{1}{5}\)
\(\Leftrightarrow-\left|x+0,5\right|=\dfrac{1}{5}-\dfrac{3}{4}\)
\(\Leftrightarrow-\left|x+0,5\right|=-\dfrac{11}{20}\)
\(\Leftrightarrow\left|x+0,5\right|=\dfrac{11}{20}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+0,5=\dfrac{11}{20}\\x+0,5=-\dfrac{11}{20}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{20}\\x=-\dfrac{21}{20}\end{matrix}\right.\)
Vậy \(x_1=-\dfrac{21}{20};x_2=\dfrac{1}{20}\)
d) \(\left(x+0,2\right)^2+0,75=1\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2+\dfrac{3}{4}=1\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=1-\dfrac{3}{4}\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{1}{4}\)
\(\Leftrightarrow x+\dfrac{1}{5}=\pm\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{1}{2}\\x+\dfrac{1}{5}=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{10}\\x=-\dfrac{7}{10}\end{matrix}\right.\)
Vậy \(x_1=-\dfrac{7}{10};x_2=\dfrac{3}{10}\)
a, \(\dfrac{3}{4}-\dfrac{1}{2}x=-\dfrac{1}{4}\)
=>\(-\dfrac{1}{2}x=-\dfrac{1}{4}-\dfrac{3}{4}\)
=>\(-\dfrac{1}{2}x=-1\)
=>\(x=-1:(-\dfrac{1}{2})\)
=>\(x=2\)
vậy ...........
b,\(1\dfrac{2}{3}x+0,2=x-\dfrac{7}{8}\)
=>\(\dfrac{5}{3}x+0,2=x-\dfrac{7}{8}\)
=>\(\dfrac{5}{3}x-x=-\dfrac{7}{8}-0,2\)
=>\(\dfrac{2}{3}x=-\dfrac{43}{40}\)
=>\(x=-\dfrac{43}{40}:\dfrac{2}{3}\)
=>\(x=-\dfrac{192}{80}\)
vậy...................
c,\(\dfrac{3}{4}-\left|x+0,5\right|=\dfrac{1}{5}\)
=>\(-\left|x+0,5\right|=\dfrac{1}{5}-\dfrac{3}{4}\)
=>\(-\left|x+0,5\right|=-\dfrac{11}{20}\)
=>\(\left|x+0,5\right|=\dfrac{11}{20}\)
=>\(\left[{}\begin{matrix}x+0.5=\dfrac{11}{20}\\x+0,5=-\dfrac{11}{20}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{1}{20}\\x=-\dfrac{21}{20}\end{matrix}\right.\)
vậy ....... hoặc.....
d,\((x+0,2)^2+0,75=1\)
=>\(\left(x+0,2\right)^2=1-0,75\)
=>\(\left(x+0,2\right)^2=0,25\)
=>\(\left[{}\begin{matrix}x+0,2=0,5\\x+0,2=-0,5\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=0,3\\x=-0,7\end{matrix}\right.\)
vậy..........................
HỌC TỐT NHA !!!!!
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
B = .................
Xét thừa số 63.1,2 - 21.3,6 = 0 nên B = 0
\(C=\left|\dfrac{4}{9}-\left(\dfrac{\sqrt{2}}{2}\right)^2\right|+\left|0,4+\dfrac{\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}}{\dfrac{2}{3}-\dfrac{4}{5}-\dfrac{6}{7}}\right|\)
\(C=\left|\dfrac{4}{9}-\dfrac{1}{2}\right|+\left|0,4+\dfrac{\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}}{2\left(\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}\right)}\right|\)
\(C=\left|\dfrac{4}{9}-\dfrac{1}{2}\right|+\left|0,4+\dfrac{1}{2}\right|=\dfrac{1}{18}+\dfrac{9}{10}=\dfrac{43}{45}\)
Mình làm câu 1,2 trước, câu 3 sau
Câu 1:
\(\sqrt{x^2}=0\)
=> \(\left(\sqrt{x^2}\right)^2=0^2\)
\(\Leftrightarrow x^2=0\Leftrightarrow x=0\)
Câu 2:
\(A=\left(0,75-0,6+\dfrac{3}{7}+\dfrac{3}{12}\right)\left(\dfrac{11}{7}+\dfrac{11}{3}+2,75-2,2\right)\)
\(A=\left(\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{13}\right)\left(\dfrac{11}{7}+\dfrac{11}{3}+\dfrac{11}{4}-\dfrac{11}{5}\right)\)
\(A=3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)\cdot11\left(\dfrac{1}{7}+\dfrac{1}{3}+\dfrac{11}{4}-\dfrac{11}{5}\right)\)
\(A=33\cdot\dfrac{491}{1820}\cdot\dfrac{221}{420}=\dfrac{3580863}{764400}\)
Câu 1: Thực hiện phép tính :
a) \(2.\left(\dfrac{-2}{3}\right)^2-\dfrac{7}{2}=2.\dfrac{4}{9}-\dfrac{7}{2}\)
\(=\dfrac{8}{9}-\dfrac{7}{2}\)
\(=\dfrac{16}{18}-\dfrac{63}{18}=\dfrac{-47}{18}\)
\(b,5\dfrac{4}{13}.\dfrac{-3}{4}+3\dfrac{9}{13}.\left(-0,75\right)=\dfrac{69}{13}.\dfrac{-3}{4}+\dfrac{48}{13}.\dfrac{-3}{4}\)
\(=\left(\dfrac{69}{13}+\dfrac{48}{13}\right).\dfrac{-3}{4}\)
\(=\dfrac{117}{13}.\dfrac{-3}{4}\)
\(=9.\dfrac{-3}{4}=\dfrac{-27}{4}\)
\(c,\left(-1\right)^{2017}+\left|\dfrac{-1}{13}\right|+\sqrt{\dfrac{144}{169}}=-1+\dfrac{1}{13}+\dfrac{12}{13}\)
\(=-1+\dfrac{13}{13}\)
\(=-1+1=0\)
Câu 3: Tìm x, biết:
a)\(\dfrac{3}{5}-x=25\)
\(x=\dfrac{3}{5}-\dfrac{125}{5}\)
\(x=\dfrac{-122}{5}\)
b)\(\dfrac{2}{3}\left|x-1\right|+\dfrac{1}{4}=\dfrac{5}{3}\)
\(\dfrac{2}{3}\left|x-1\right|=\dfrac{20}{12}-\dfrac{3}{12}\)
\(\dfrac{2}{3}\left|x-1\right|=\dfrac{17}{12}\)
\(\left|x-1\right|=\dfrac{17}{12}:\dfrac{2}{3}\)
\(\left|x-1\right|=\dfrac{17}{12}.\dfrac{3}{2}\)
\(\left|x-1\right|=\dfrac{17}{8}\)
Ta có 2 TH: TH1:\(x-1=\dfrac{17}{8}\) TH2:\(x-1=\dfrac{-17}{8}\) \(x=\dfrac{17}{8}+1\) \(x=\dfrac{-17}{8}+1\) \(x=\dfrac{17}{8}+\dfrac{8}{8}=\dfrac{25}{8}\) \(x=\dfrac{-17}{8}+\dfrac{8}{8}=\dfrac{-9}{8}\) Vậy x∈\(\left\{\dfrac{25}{5};\dfrac{-9}{8}\right\}\)
MÌNH GHI KẾT QUẢ THÔI NHÉ
A,2
b,\(\dfrac{-2}{3}\)hoặc2
c,\(\dfrac{13}{15}\)hoặc\(\dfrac{17}{-15}\)
a: =>|x-1/4|=3/4
=>x-1/4=3/4 hoặc x-1/4=-3/4
=>x=1 hoặc x=-1/2
b: \(\left|x+\dfrac{1}{2}\right|=\dfrac{1}{2}-\dfrac{9}{4}=\dfrac{2-9}{4}=-\dfrac{7}{4}\)(vô lý)
c: \(\Leftrightarrow\left[{}\begin{matrix}2x+5=1-x\\2x+5=x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-4\\x=-6\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{4}{3};-6\right\}\)
e: =>|3/2-x|=0
=>3/2-x=0
hay x=3/2
a: TH1: x>=0
=>x+x=1/3
=>x=1/6(nhận)
TH2: x<0
Pt sẽ là -x+x=1/3
=>0=1/3(loại)
b: \(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\x^2-x-2=0\end{matrix}\right.\Leftrightarrow x=2\)
c: \(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-8}+\dfrac{1}{x-8}-\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{2}{x-20}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{x-20-2x+2}{\left(x-1\right)\left(x-20\right)}=\dfrac{-3}{4}\)
\(\Leftrightarrow-3\left(x^2-21x+20\right)=4\left(-x-18\right)\)
\(\Leftrightarrow3x^2-63x+60=4x+72\)
=>3x^2-67x-12=0
hay \(x\in\left\{22.51;-0.18\right\}\)
a.x-\(\dfrac{3}{5}\)=\(\left(\dfrac{-1}{2}\right)^2\)
=>x-\(\dfrac{3}{5}\)=\(\dfrac{1}{4}\)
=>x=\(\dfrac{1}{4}\)+\(\dfrac{3}{5}\)=\(\dfrac{17}{20}\)
vậy x=\(\dfrac{17}{20}\)
b.0.75-/\(\dfrac{1}{3}\)x+1/=/-0.5/
=>\(\dfrac{3}{4}\)-/\(\dfrac{1}{3}\)x+1/=\(\dfrac{1}{2}\)
=>/\(\dfrac{1}{3}\)x+1/=\(\dfrac{3}{4}\)-\(\dfrac{1}{2}\)=\(\dfrac{1}{4}\)
=>\(\left\{{}\begin{matrix}\dfrac{1}{3}x+1=\dfrac{-1}{4}=>\dfrac{1}{3}x=\dfrac{-1}{4}-1=\dfrac{-5}{4}=>x=\dfrac{-5}{4}:\dfrac{1}{3}=\dfrac{-15}{4}\\\dfrac{1}{3}x+1=\dfrac{1}{4}=>\dfrac{1}{3}x=\dfrac{1}{4}-1=\dfrac{-3}{4}=>x=\dfrac{-3}{4}:\dfrac{1}{3}=\dfrac{-9}{4}\end{matrix}\right.\)
a) x- \(\dfrac{3}{5}\) = \(\left(-\dfrac{1}{2}\right)^2\)
x- \(\dfrac{3}{5}\) = \(\dfrac{1}{4}\)
x = \(\dfrac{1}{4}+\dfrac{3}{5}\)
x= \(\dfrac{5}{20}\)+\(\dfrac{12}{20}\)
x=\(\dfrac{17}{20}\)