a) 11/12-(2/5+x)=2/3
b) 2x.(x-1/7)=0
c) 3/4+1/4:x=2/5
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PT
23 tháng 6 2021
1/
a, \(A=\dfrac{2}{3}+\dfrac{3}{4}.\left(-\dfrac{4}{9}\right)=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)
b, \(B=2\dfrac{3}{11}.\dfrac{11}{12}.\left(-2,2\right)=\dfrac{25}{11}.\dfrac{11}{12}.\left(-\dfrac{11}{5}\right)=-\dfrac{55}{12}\)
c, \(C=\left(\dfrac{3}{4}-0,2\right):\left(0,4-\dfrac{4}{5}\right)=\left(\dfrac{3}{4}-\dfrac{1}{5}\right):\left(\dfrac{2}{5}-\dfrac{4}{5}\right)=\dfrac{11}{20}:\left(-\dfrac{2}{5}\right)=-\dfrac{11}{8}\)
2/
a, \(\dfrac{11}{12}-x=\dfrac{2}{3}+\dfrac{1}{4}\\ \Rightarrow\dfrac{11}{12}-x=\dfrac{11}{12}\\ \Rightarrow x=0\)
b, \(2x\left(x-\dfrac{1}{7}\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)
c, \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\\ \Rightarrow\dfrac{1}{4}:x=-\dfrac{7}{20}\\ \Rightarrow x=-\dfrac{5}{7}\)
NA
1
SN
20 tháng 7 2021
a. 2x+\(\dfrac{4}{5}\)=0 hoặc 3x-\(\dfrac{1}{2}\)=0
2x=- 4/5 hoặc 3x=1/2
x=-2/5 hoặc x=\(\dfrac{1}{6}\)
b. x-\(\dfrac{2}{5}\)=0 hoặc x+\(\dfrac{4}{7}\)=0
x=2/5 hoặc x=-\(\dfrac{4}{7}\)
d. x(1+5/8-12/16)=1
\(\dfrac{7}{8}\)x=1=> x=8/7
10 tháng 12 2021
\(a,\Leftrightarrow2x^2+10x-2x^2=12\Leftrightarrow x=\dfrac{12}{10}=\dfrac{6}{5}\\ b,\Leftrightarrow\left(5-2x-4\right)\left(5-2x+4\right)=0\\ \Leftrightarrow\left(1-2x\right)\left(9-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\\ d,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ e,\Leftrightarrow4x^2-4x+1-4x^2+196=0\\ \Leftrightarrow-4x=-197\Leftrightarrow x=\dfrac{197}{4}\)
\(f,\Leftrightarrow x^2+8x+16-x^2+1=16\Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\\ g,Sửa:\left(3x+1\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(3x+1-x-1\right)\left(3x+1+x+1\right)=0\\ \Leftrightarrow2x\left(4x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\\ h,\Leftrightarrow x^2+8x-x-8=0\\ \Leftrightarrow\left(x+8\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-8\end{matrix}\right.\\ i,\Leftrightarrow2x^2-13x+15=0\\ \Leftrightarrow2x^2+2x-15x-15=0\\ \Leftrightarrow\left(x+1\right)\left(2x-15\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{15}{2}\end{matrix}\right.\)
16 tháng 2 2022
\(a)\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1.\left(x\ne2;4\right).\\ \Leftrightarrow\dfrac{\left(x-3\right)\left(x-4\right)+\left(x-2\right)^2}{\left(x-2\right)\left(x-4\right)}=-1.\\ \Rightarrow x^2-4x-3x+12+x^2-4x+4+x^2-4x-2x+8=0.\\ \Leftrightarrow3x^2-17x+24=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}.\\x=3.\end{matrix}\right.\) (TM).
\(b)3x+12=0.\\ \Leftrightarrow3x=-12.\\ \Leftrightarrow x=-4.\)
\(c)5+2x=x-5.\\ \Leftrightarrow2x-x=-5-5.\\ \Leftrightarrow x=-10.\)
\(d)2x\left(x-2\right)+5\left(x-2\right)=0.\\ \Leftrightarrow\left(2x+5\right)\left(x-2\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{2}.\\x=2.\end{matrix}\right.\)
\(e)\dfrac{3x-4}{2}=\dfrac{4x+1}{3}.\\ \Rightarrow3\left(3x-4\right)-2\left(4x+1\right)=0.\\ \Leftrightarrow9x-12-8x-2=0.\\ \Leftrightarrow x=14.\)
\(f)\dfrac{2x}{x-1}-\dfrac{x}{x+1}=1.\left(x\ne\pm1\right).\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x}{x^2-1}=1.\\ \Leftrightarrow x^2+3x-x^2+1=0.\\ \Leftrightarrow3x+1=0.\\ \Leftrightarrow x=\dfrac{-1}{3}.\)
\(g)\dfrac{2x}{x-1}+\dfrac{3-2x}{x+2}=\dfrac{6}{\left(x-1\right)\left(x+2\right)}.\left(x\ne1;-2\right).\\ \Leftrightarrow\dfrac{2x^2+4x+\left(3-2x\right)\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}=\dfrac{6}{\left(x-1\right)\left(x+2\right)}.\\ \Rightarrow2x^2+4x+3x-3-2x^2+2x-6=0.\\ \Leftrightarrow9x=9.\)
\(\Leftrightarrow x=1\left(koTM\right).\)
VC
1
14 tháng 2 2022
a, \(\dfrac{x}{2}+\dfrac{3x}{5}=-\dfrac{3}{2}\Rightarrow5x+6x=-15\Leftrightarrow x=-\dfrac{15}{11}\)
b, TH1 : \(\dfrac{2}{3}x-\dfrac{4}{7}=0\Leftrightarrow x=\dfrac{6}{7}\);TH2 : \(\dfrac{1}{2}-\dfrac{3}{7x}=0\Rightarrow7x-6=0\Leftrightarrow x=\dfrac{6}{7}\)
c, TH1 : \(\dfrac{4}{5}-2x=0\Leftrightarrow x=\dfrac{4}{5}:2=\dfrac{2}{5}\)
TH2 : \(\dfrac{1}{3}+\dfrac{3}{5x}=0\Rightarrow5x+9=0\Leftrightarrow x=-\dfrac{9}{5}\)
9 tháng 9 2021
a: Ta có: \(5\left(4x-1\right)+2\left(1-3x\right)-6\left(x+5\right)=10\)
\(\Leftrightarrow20x-5+2-6x-6x-30=10\)
\(\Leftrightarrow8x=43\)
hay \(x=\dfrac{43}{8}\)
b: ta có: \(2x\left(x+1\right)+3\left(x-1\right)\left(x+1\right)-5x\left(x+1\right)+6x^2=0\)
\(\Leftrightarrow2x^2+2x+3x^2-3-5x^2-5x+6x^2=0\)
\(\Leftrightarrow6x^2-3x-3=0\)
\(\Leftrightarrow2x^2-x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)
LQ
2
11 tháng 10 2021
\(a,\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{5}=0\\\dfrac{8}{5}+2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{4}{5}\end{matrix}\right.\)
\(b,\dfrac{x-\dfrac{4}{7}}{x+\dfrac{1}{2}}>0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-\dfrac{4}{7}>0\\x+\dfrac{1}{2}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-\dfrac{4}{7}< 0\\x+\dfrac{1}{2}< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{4}{7}\\x< -\dfrac{1}{2}\end{matrix}\right.\)
\(c,\dfrac{2x-3}{x+\dfrac{7}{4}}< 0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-3< 0\\x+\dfrac{7}{4}>0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-3>0\\x+\dfrac{7}{4}< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{3}{2}\\x >-\dfrac{7}{4}\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{3}{2}\\x< -\dfrac{7}{4}\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-\dfrac{7}{4}< x< \dfrac{3}{2}\\x\in\varnothing\end{matrix}\right.\Leftrightarrow-\dfrac{7}{4}< x< \dfrac{3}{2}\)
LT
2
15 tháng 2 2022
a, đk : x khác -1 ; x khác 2
\(\Rightarrow4x-8=3x+3\Leftrightarrow x=11\left(tm\right)\)
b, đk : x khác 2 ; -2
\(\Rightarrow\left(x+2\right)^2-8x=0\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x=2\left(ktm\right)\)
-> vậy pt vô nghiệm
c, đk : x khác 3 ; 0
\(\Rightarrow x-5\left(x-3\right)=3x+7\Leftrightarrow-4x+15=3x+7\Leftrightarrow-7x=-8\Leftrightarrow x=\dfrac{8}{7}\left(tm\right)\)
TD
15 tháng 2 2022
a)Với \(x\ne-1;x\ne2\)
\(\dfrac{4}{x+1}=\dfrac{3}{x-2}\)
<=>4(x-2)=3(x+1)
<=>4x-8=3x+3
<=>x=11(TM)
b)Với\(x\ne\pm2\)
\(\dfrac{x+2}{2x-4}-\dfrac{4x}{x^2-4}=0\)
<=>\(\dfrac{x+2}{2\left(x-2\right)}-\dfrac{4x}{\left(x+2\right)\left(x-2\right)}=0\)
<=>\(\dfrac{\left(x+2\right)^2-8x}{2\left(x+2\right)\left(x-2\right)}=0\)
<=>\(x^2+4x+4-8x=0\left(Vĩx\ne\pm2\right)\)
<=>\(x^2-4x+4=0\)
<=>\(\left(x-2\right)^2=0\)
<=>x-2=0
<=>x=2(Không thỏa mãn)
c)Với \(x\ne3,x\ne0\)
\(\dfrac{1}{x-3}-\dfrac{5}{x}=\dfrac{3x+7}{x\left(x-3\right)}\)
<=>\(\dfrac{x-5\left(x-3\right)}{x\left(x-3\right)}=\dfrac{3x+7}{x\left(x-3\right)}\)
<=>x-5x+15=3x+7(Vì \(x\ne0,x\ne3\))
<=>7x=8
<=>x=\(\dfrac{8}{7}\left(TM\right)\)
TN
1
19 tháng 9 2021
\(a,\Rightarrow\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}=\dfrac{1}{4}\Rightarrow x=\dfrac{1}{4}-\dfrac{2}{5}=-\dfrac{3}{20}\\ b,\Rightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\\ c,\Rightarrow\dfrac{1}{4}:x=-\dfrac{2}{5}-\dfrac{3}{4}=-\dfrac{23}{20}\\ \Rightarrow x=\dfrac{1}{4}:\left(-\dfrac{23}{20}\right)=-\dfrac{5}{23}\)
a)-3/20
b)1/7 hoặc 0
c) -29/140
a) -3/20
b) 1/7 hoặc 0
c) -29/140