a/ \(\dfrac{-3}{5}\) - x = \(\dfrac{21}{10}\)
b/ x : \(\dfrac{2}{9}\) = \(\dfrac{9}{2}\)
c/ \(\dfrac{x}{9}\) = \(\dfrac{5}{3}\)
d/ x : \(\left(\dfrac{2}{5}\right)^3\)= \(\left(\dfrac{5}{2}\right)^3\)
\(\text{a)}\dfrac{-3}{5}-x=\dfrac{21}{10}\)
\(x=\dfrac{-3}{5}-\dfrac{21}{10}=\dfrac{-27}{10}\)
\(\text{b)}x:\dfrac{2}{9}=\dfrac{9}{2}\)
\(x\) \(=\dfrac{9}{2}.\dfrac{2}{9}=1\)
\(\text{c) }\dfrac{x}{9}=\dfrac{5}{3}\)
\(\Rightarrow x=\dfrac{9.5}{3}=15\)
\(\text{d)}x:\left(\dfrac{2}{5}\right)^3=\left(\dfrac{5}{2}\right)^3\)
\(x:\dfrac{8}{125}=\dfrac{125}{8}\)
\(x\) \(=\dfrac{125}{8}.\dfrac{8}{125}=1\)
a)\(\dfrac{2}{x+2}-\dfrac{1}{x+3}+\dfrac{2x+5}{\left(x+2\right)\left(x+3\right)}\)
b)\(\dfrac{2}{x+1}-\dfrac{1}{x+5}+\dfrac{2x+6}{\left(x+5\right)\left(x+1\right)}\)
c)\(\dfrac{-6}{x^2-9}-\dfrac{1}{x+3}+\dfrac{3}{x-3}\)
d)\(\dfrac{x}{x-2}-\dfrac{x}{x+2}+\dfrac{8}{x^2-4}\)
So sánh \(\dfrac{9}{170};\dfrac{9}{230};\dfrac{53}{144}\)
Số nguyên \(x\) thỏa mãn \(\left(\dfrac{3}{4}-\dfrac{2}{3}\right)+\dfrac{5}{6}\le x\le\dfrac{4}{5}-\left(\dfrac{3}{10}-\dfrac{5}{4}\right)\)
A. \(x=1\) B. \(x=0\) C. \(x=2\) D. \(x\in\left\{0;1\right\}\)
EM CẦN GẤP Ạ!
\(a,\left(\dfrac{37}{9}+\dfrac{13}{4}\right)x\dfrac{9}{4}+\dfrac{11}{4}\) b,\(1+\left(\dfrac{9}{10}-\dfrac{-4}{5}\right):\dfrac{19}{6}\)
c,\(\dfrac{1}{4}-\dfrac{3}{2}+\dfrac{1}{2}x\dfrac{12}{5}\)
Giúp mik nha:>
a: \(=\dfrac{37}{4}+\dfrac{117}{16}+\dfrac{1}{4}=\dfrac{19}{2}+\dfrac{117}{16}=\dfrac{269}{16}\)
b: \(=1+\left(\dfrac{9}{10}+\dfrac{8}{10}\right):\dfrac{19}{6}=1+\dfrac{17}{10}\cdot\dfrac{6}{19}=\dfrac{146}{95}\)
c: \(=\dfrac{1}{4}-\dfrac{6}{4}+\dfrac{6}{5}=\dfrac{-5}{4}+\dfrac{6}{5}=\dfrac{-1}{20}\)
Giải các phương trình sau:
\(a.\dfrac{5x-2}{3}=\dfrac{5-3x}{2}\)
\(b.\dfrac{10x+3}{12}=1+\dfrac{6+8x}{9}\)
\(c.2\left(x+\dfrac{3}{5}\right)=5-\left(\dfrac{13}{5}+x\right)\)
\(d.\dfrac{7}{8}x-5\left(x-9\right)=\dfrac{20x+1,5}{6}\)
\(e.\dfrac{7x-1}{6}+2x=\dfrac{16-x}{5}\)
\(f.\dfrac{x+4}{5}-x+4=\dfrac{x}{3}-\dfrac{x-2}{2}\)
a: =>10x-4=15-9x
=>19x=19
hay x=1
b: \(\Leftrightarrow3\left(10x+3\right)=36+4\left(8x+6\right)\)
=>30x+9=36+32x+24
=>30x-32x=60-9
=>-2x=51
hay x=-51/2
c: \(\Leftrightarrow2x+\dfrac{6}{5}=5-\dfrac{13}{5}-x\)
=>3x=6/5
hay x=2/5
d: \(\Leftrightarrow\dfrac{7x}{8}-\dfrac{5\left(x-9\right)}{1}=\dfrac{20x+1.5}{6}\)
\(\Leftrightarrow21x-120\left(x-9\right)=4\left(20x+1.5\right)\)
=>21x-120x+1080=80x+60
=>-179x=-1020
hay x=1020/179
e: \(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)
=>35x-5+60x=96-6x
=>95x+6x=96+5
=>x=1
f: \(\Leftrightarrow6\left(x+4\right)+30\left(-x+4\right)=10x-15\left(x-2\right)\)
=>6x+24-30x+120=10x-15x+30
=>-24x+96=-5x+30
=>-19x=-66
hay x=66/19
Tìm x
\(a,3-x=x+1,8\)
\(b,2x-5=7x+35\)
\(c,2\left(x+10\right)=3\left(x-6\right)\)
\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)
\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)
\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)
\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)
\(k,7x^2-11=6x^2-2\)
\(m,5\left(x+3.2^3\right)=10^2\)
\(n,\dfrac{4}{9}-(\dfrac{1}{6^2})=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)
\(a,3-x=x+1,8\)
\(\Rightarrow-x-x=1,8-3\)
\(\Rightarrow-2x=-1,2\)
\(\Rightarrow x=0,6\)
\(b,2x-5=7x+35\)
\(\Rightarrow2x-7x=35+5\)
\(\Rightarrow-5x=40\)
\(\Rightarrow x=-8\)
\(c,2\left(x+10\right)=3\left(x-6\right)\)
\(\Rightarrow2x+20=3x-18\)
\(\Rightarrow2x-3x=-18-20\)
\(\Rightarrow-x=-38\)
\(\Rightarrow x=38\)
\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)
\(\Rightarrow8x-3+1=1+6x+x\)
\(\Rightarrow8x-3=7x\)
\(\Rightarrow8x-7x=3\)
\(\Rightarrow x=3\)
\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)
\(\Rightarrow-3x+x=\dfrac{4}{3}-\dfrac{2}{9}\)
\(\Rightarrow-2x=\dfrac{10}{9}\)
\(\Rightarrow x=-\dfrac{5}{9}\)
\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2}x-\dfrac{3}{4}x=-\dfrac{1}{2}-\dfrac{5}{6}\)
\(\Rightarrow-\dfrac{1}{4}x=-\dfrac{4}{3}\)
\(\Rightarrow x=\dfrac{16}{3}\)
\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)
\(\Rightarrow x-4=5-x\)
\(\Rightarrow x+x=5+4\)
\(\Rightarrow2x=9\)
\(\Rightarrow x=\dfrac{9}{2}\)
\(k,7x^2-11=6x^2-2\)
\(\Rightarrow7x^2-6x^2=-2+11\)
\(\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
\(m,5\left(x+3\cdot2^3\right)=10^2\)
\(\Rightarrow5\left(x+24\right)=100\)
\(\Rightarrow x+24=20\)
\(\Rightarrow x=-4\)
\(n,\dfrac{4}{9}-\left(\dfrac{1}{6^2}\right)=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{4}{9}-\dfrac{1}{36}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2=0\)
\(\Rightarrow x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)
#\(Urushi\text{☕}\)
a: 3-x=x+1,8
=>-2x=-1,2
=>x=0,6
b: 2x-5=7x+35
=>-5x=40
=>x=-8
c: 2(x+10)=3(x-6)
=>3x-18=2x+20
=>x=38
d; 8(x-3/8)+1=6(1/6+x)+x
=>8x-3+1=1+6x+x
=>8x-2=7x+1
=>x=3
e: =>-3x+x=4/3-2/9
=>-2x=12/9-2/9=10/9
=>x=-5/9
g: =>3/4x-1/2x=5/6+1/2
=>1/4x=5/6+3/6=8/6=4/3
=>x=4/3*4=16/3
h: =>x-4=-x+5
=>2x=9
=>x=9/2
Số nguyên \(x\) thỏa mãn \(\left(\dfrac{3}{4}-\dfrac{2}{3}\right)+\dfrac{5}{6}< x\le\dfrac{4}{5}-\left(\dfrac{3}{10}-\dfrac{5}{4}\right)\) là:
A. \(x=1\) B. \(x=0\) C. \(x=2\) D. \(x\in\left\{0;1\right\}\)
So sánh 3 phân số: \(\dfrac{9}{170};\dfrac{9}{230};\dfrac{53}{144}\)
Câu 1: D
Câu 3: 53/144>9/170>9/230
Tìm x:
a) \(\dfrac{1}{3}.x+\dfrac{2}{5}\left(x-1\right)=0\)
b)\(-5.\left(x+\dfrac{1}{5}\right)-\dfrac{1}{2}.\left(x-\dfrac{2}{3}\right)=x\)
c)\(\left(x+\dfrac{1}{2}\right).\left(\dfrac{2}{3}-2x\right)=0\)
d)\(9.\left(3x+1\right)^2=16\)
a: =>1/3x+2/5x-2/5=0
=>11/15x-2/5=0
=>11/15x=2/5
=>x=2/5:11/15=2/5*15/11=30/55=6/11
b: =>-5x-1-1/2x+1/3=x
=>-11/2x-2/3-x=0
=>-13/2x=2/3
=>x=-2/3:13/2=-2/3*2/13=-4/39
c: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=1/3 hoặc x=-1/2
d: 9(3x+1)^2=16
=>(3x+1)^2=16/9
=>3x+1=4/3 hoặc 3x+1=-4/3
=>3x=1/3 hoặc 3x=-7/3
=>x=1/9 hoặc x=-7/9
Tìm số x , biết
a) \(\dfrac{2}{3}x-\dfrac{3}{2}x=\dfrac{5}{12}\)
b) \(\dfrac{2}{5}+\dfrac{3}{5}.\left(3x-3,7\right)=-\dfrac{53}{10}\)
c) \(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)+\dfrac{5}{9}=\dfrac{23}{27}\)
d) \(\dfrac{-2}{3}.x+\dfrac{1}{5}=\dfrac{3}{10}\)
e) \(\left|x\right|-\dfrac{3}{4}=\dfrac{5}{3}\)
f) \(\left|2x-\dfrac{1}{3}\right|+\dfrac{5}{6}=1\)
a) \(\dfrac{2}{3}x-\dfrac{3}{2}x=\dfrac{5}{12}\)
\(-\dfrac{5}{6}x=\dfrac{5}{12}\)
\(x=-\dfrac{1}{2}\)
b) \(\dfrac{2}{5}+\dfrac{3}{5}\cdot\left(3x-3.7\right)=-\dfrac{53}{10}\)
\(\dfrac{3}{5}\left(3x-3.7\right)=-\dfrac{57}{10}\)
\(3x-3.7=-\dfrac{19}{2}\)
\(3x=-5.8\)
\(x=-\dfrac{29}{15}\)
c) \(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)+\dfrac{5}{9}=\dfrac{23}{27}\)
\(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)=\dfrac{8}{27}\)
\(2+\dfrac{3}{4}x=\dfrac{21}{8}\)
\(\dfrac{3}{4}x=\dfrac{5}{8}\)
\(x=\dfrac{5}{6}\)
d) \(-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\)
\(-\dfrac{2}{3}x=\dfrac{1}{10}\)
\(x=-\dfrac{3}{20}\)
\(\dfrac{2}{3}x-\dfrac{3}{2}x=\dfrac{5}{12}\)
\(\left(\dfrac{2}{3}-\dfrac{3}{2}\right)x=\dfrac{5}{12}\)
\(\dfrac{-5}{6}.x=\dfrac{5}{12}\)
-> x = \(\dfrac{-1}{2}\)
a) \(\dfrac{2}{3}x-\dfrac{3}{2}x=\dfrac{5}{12}\\ < =>x\left(\dfrac{2}{3}-\dfrac{3}{2}\right)=\dfrac{5}{12}\\ < =>-\dfrac{5}{6}x=\dfrac{5}{12}\\ =>x=\dfrac{\dfrac{5}{12}}{-\dfrac{5}{6}}=-\dfrac{1}{2}\)
b) \(\dfrac{2}{5}+\dfrac{3}{5}.\left(3x-3,7\right)=\dfrac{-53}{10}\\ < =>\dfrac{2}{5}+\dfrac{9}{5}x-\dfrac{111}{50}=-\dfrac{53}{10}\\ < =>\dfrac{9}{5}x=-\dfrac{2}{5}+\dfrac{111}{50}-\dfrac{53}{10}\\ < =>\dfrac{9}{5}x=-\dfrac{87}{25}\\ =>x=\dfrac{\dfrac{-87}{25}}{\dfrac{9}{5}}=-\dfrac{29}{15}=-1\dfrac{14}{15}\)
c) \(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)+\dfrac{5}{9}=\dfrac{23}{27}\\ < =>\dfrac{\dfrac{7}{9}}{2}+\dfrac{\dfrac{7}{9}}{\dfrac{3}{4}x}+\dfrac{5}{9}=\dfrac{23}{27}\\ < =>\dfrac{7}{18}+\dfrac{28}{27}x+\dfrac{5}{9}=\dfrac{23}{27}\\ < =>\dfrac{28}{27}x=-\dfrac{7}{18}-\dfrac{5}{9}+\dfrac{23}{27}\\ < =>\dfrac{28}{27}x=\dfrac{1}{54}\\ =>x=\dfrac{\dfrac{1}{54}}{\dfrac{28}{27}}=\dfrac{1}{56}\)
d) \(-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\\ < =>-\dfrac{2}{3}x=\dfrac{3}{10}-\dfrac{1}{5}\\ < =>-\dfrac{2}{3}x=\dfrac{1}{10}\\ =>x=\dfrac{\dfrac{1}{10}}{-\dfrac{2}{3}}=-\dfrac{3}{20}\)
e) \(\left|x\right|-\dfrac{3}{4}=\dfrac{5}{3}\\ =>\left|x\right|=\dfrac{5}{3}+\dfrac{3}{4}\\ =>\left|x\right|=\dfrac{29}{12}\)
Vậy: x= 29/12 hoặc x= -29/12
f) \(\left|2x-\dfrac{1}{3}\right|+\dfrac{5}{6}=1\\ =>\left|2x-\dfrac{1}{3}\right|=1-\dfrac{5}{6}=\dfrac{1}{6}\\ =>\left[{}\begin{matrix}\left|2x-\dfrac{1}{3}\right|=\dfrac{1}{6}\\\left|2x-\dfrac{1}{3}\right|=-\dfrac{1}{6}\end{matrix}\right.\\ =>x=\left[{}\begin{matrix}x=\dfrac{\dfrac{1}{6}+\dfrac{1}{3}}{2}\\x=\dfrac{\dfrac{-1}{6}+\dfrac{1}{3}}{2}\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{1}{12}\end{matrix}\right.\)
Vậy: x= 1/4 hoặc x= 1/12
