Tìm x :
a. \(15-2\left|x+\dfrac{1}{3}\right|=\dfrac{4}{7}\)
b. \(\dfrac{13}{11}.\dfrac{22}{26}-x^2=\dfrac{7}{16}\)
1: Tính hợp lí:
a, \(\dfrac{11}{125}\)- \(\dfrac{17}{18}-\dfrac{5}{7}+\dfrac{4}{9}\)+ \(\dfrac{17}{14}\)
b,\(-1\dfrac{5}{7}.15+\dfrac{2}{7}.\left(-15\right)+\left(-105\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\)
2: Tìm x:
a, \(\dfrac{11}{13}-\left(\dfrac{5}{42}-x\right)=-\left(\dfrac{15}{28}-\dfrac{11}{13}\right)\)
b,\(|x+\dfrac{4}{15}|-|-3,75|=-|-2,15|\)
c, \(^{7^{x+2}}+2.7^{x-1}=345\)
\(1,\)
\(a,\dfrac{11}{125}-\dfrac{17}{18}-\dfrac{5}{7}+\dfrac{4}{9}+\dfrac{17}{14}\)
\(=\dfrac{11}{125}+\left(\dfrac{4}{9}-\dfrac{17}{18}\right)+\left(\dfrac{17}{14}-\dfrac{5}{7}\right)\)
\(=\dfrac{11}{125}+\left(\dfrac{-1}{2}\right)+\dfrac{1}{2}\)
\(=\dfrac{11}{125}\)
\(b,-1\dfrac{5}{7}.15+\dfrac{2}{7}.\left(-15\right)+\left(-105\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\)
\(=\dfrac{-12}{7}.15+\dfrac{2}{7}.\left(-15\right)+\left(105\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\)
\(=-15.\left[\dfrac{12}{7}+\dfrac{2}{7}+\left(-5\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\right]\)
\(=-15.\left[2+\left(-5\right).\dfrac{1}{105}\right]\)
\(=-15.\left(2-\dfrac{1}{21}\right)\)
\(=-15.\dfrac{41}{21}=\dfrac{-615}{21}\)
\(2,\)
\(a,\dfrac{11}{13}-\left(\dfrac{5}{42}-x\right)=-\left(\dfrac{15}{28}-\dfrac{11}{13}\right)\)
\(\Leftrightarrow\dfrac{11}{13}-\dfrac{5}{42}+x=\dfrac{-15}{28}+\dfrac{11}{13}\)
\(\Leftrightarrow x=\dfrac{-15}{28}+\dfrac{11}{13}-\dfrac{11}{13}+\dfrac{5}{42}\)
\(\Leftrightarrow x=\left(\dfrac{11}{13}-\dfrac{11}{13}\right)+\left(\dfrac{5}{42}+\dfrac{-15}{28}\right)\)
\(\Leftrightarrow x=\dfrac{5}{12}\)
Vậy \(x=\dfrac{5}{12}\)
\(b,\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)
\(\Leftrightarrow\left|x+\dfrac{4}{15}\right|-3,75=-2,15\)
\(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2,15+3,75=1,6=\dfrac{16}{10}=\dfrac{8}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{4}{15}=\dfrac{8}{5}\\x+\dfrac{4}{15}=\dfrac{-8}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{5}-\dfrac{4}{15}=\dfrac{4}{3}\\x=\dfrac{-8}{5}-\dfrac{4}{15}=\dfrac{-28}{15}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{4}{3};\dfrac{-28}{15}\right\}\)
\(c,7^{x+2}+2.7^{x-1}=345\)
\(\Leftrightarrow7^{x-1}.\left(7^3+2\right)=345\)
\(\Leftrightarrow7^{x-1}.\left(343+2\right)=345\)
\(\Leftrightarrow7^{x-1}.345=345\)
\(\Leftrightarrow7^{x-1}=345:345=1\)
\(\Leftrightarrow x-1=0\)
\(x=0+1=1\)
Vậy \(x=1\)
a)\(\dfrac{-3}{7}+\dfrac{15}{26}-\left(\dfrac{2}{13}-\dfrac{3}{7}\right)\)
b)\(2.\dfrac{3}{7}+\left(\dfrac{2}{9}-1\dfrac{3}{7}\right)-\dfrac{5}{3}:\dfrac{1}{9}\)
c)\(\dfrac{-11}{23}.\dfrac{6}{7}+\dfrac{8}{7}.\dfrac{-11}{23}-\dfrac{1}{23}\)
d)\(\left(\dfrac{377}{-231}-\dfrac{123}{89}+\dfrac{34}{791}\right).\left(\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{24}\right)\)
nhanh lên giúp mình vs
a) Ta có: \(\dfrac{-3}{7}+\dfrac{15}{26}-\left(\dfrac{2}{13}-\dfrac{3}{7}\right)\)
\(=\dfrac{-3}{7}+\dfrac{15}{26}-\dfrac{2}{13}+\dfrac{3}{7}\)
\(=\dfrac{15}{26}-\dfrac{4}{26}\)
\(=\dfrac{11}{26}\)
b) Ta có: \(2\cdot\dfrac{3}{7}+\left(\dfrac{2}{9}-1\dfrac{3}{7}\right)-\dfrac{5}{3}:\dfrac{1}{9}\)
\(=\dfrac{6}{7}+\dfrac{2}{9}-\dfrac{10}{7}-\dfrac{5}{3}\cdot9\)
\(=\dfrac{-4}{7}+\dfrac{2}{9}-15\)
\(=\dfrac{-36}{63}+\dfrac{14}{63}-\dfrac{945}{63}\)
\(=\dfrac{-967}{63}\)
c) Ta có: \(\dfrac{-11}{23}\cdot\dfrac{6}{7}+\dfrac{8}{7}\cdot\dfrac{-11}{23}-\dfrac{1}{23}\)
\(=\dfrac{-11}{23}\cdot\left(\dfrac{6}{7}+\dfrac{8}{7}\right)-\dfrac{1}{23}\)
\(=\dfrac{-11}{23}\cdot2-\dfrac{1}{23}\)
\(=-1\)
d) Ta có: \(\left(\dfrac{377}{-231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot\left(\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{24}\right)\)
\(=\left(\dfrac{-377}{231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot\left(\dfrac{4}{24}-\dfrac{3}{24}-\dfrac{1}{24}\right)\)
\(=\left(\dfrac{-377}{231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot0\)
=0
\(A.\dfrac{-15}{28}x\dfrac{7}{25}\\ B.\dfrac{-5}{14}x\dfrac{7}{-3}\\ C.\dfrac{-1}{5}-\dfrac{7}{15}x\dfrac{9}{35}\\ D.\dfrac{-3}{4}-(\dfrac{-1}{2})^2\\ E.\dfrac{-4}{5}-\dfrac{-4}{5}x\dfrac{15}{16}\\F.(\dfrac{3}{4}+\dfrac{-7}{2})x(\dfrac{2}{11}+\dfrac{12}{22})\)
a: \(A=\dfrac{-7}{28}\cdot\dfrac{15}{25}=\dfrac{-1}{4}\cdot\dfrac{3}{5}=\dfrac{-3}{20}\)
b: \(B=\dfrac{-5\cdot7}{14\cdot\left(-3\right)}=\dfrac{35}{42}=\dfrac{5}{6}\)
c: \(C=\dfrac{-1}{5}-\dfrac{1}{5}\cdot\dfrac{3}{5}=\dfrac{-1}{5}-\dfrac{3}{25}=\dfrac{-8}{25}\)
d: \(D=\dfrac{-3}{4}-\dfrac{1}{4}=-1\)
e: \(E=\dfrac{-4}{5}\left(1-\dfrac{15}{16}\right)=\dfrac{-4}{5}\cdot\dfrac{1}{16}=\dfrac{-1}{20}\)
f: \(F=\dfrac{6-7}{4}\cdot\dfrac{4+12}{22}=\dfrac{-1}{4}\cdot\dfrac{8}{11}=\dfrac{-2}{11}\)
a.,\(\dfrac{4}{5}+5\dfrac{1}{2}\text{x }\left(4,5-2\right)=\dfrac{7}{10}\) b,125%x\(\dfrac{17}{4}:\left(1\dfrac{5}{16}-0,5\right)+2008\)
c,\(\dfrac{5}{11}+\left(\dfrac{16}{11}+1\right)\) d, \(\dfrac{3}{17}+\dfrac{11}{4}+\dfrac{5}{8}+\dfrac{14}{17}+\dfrac{3}{8}\)
`a)4/5+5 1/2 xx (4,5-2)+7/10`
`=4/5+11/2*2,5+7/10`
`=0,8+2,2+0,7`
`=3+0,7=3,7`
`b)125%xx 17/4:(1 5/16-0,5)+2008`
`=1,25xx4,25:13/16+2008`
`=85/13+2008`
`=2014 7/13`
`c)5/11+(16/11+1)`
`=5/11+1+5/11+1`
`=2+10/11=32/11`
`d)3/17+11/4+5/8+14/17+3/8`
`=3/17+14/17+5/8+3/8+11/4`
`=1+1+11/4`
`=19/4`
a)
\(\dfrac{4}{5}+5\dfrac{1}{2}x\left(4,5-2\right)=\dfrac{7}{10}\)
<=> \(\dfrac{11}{2}x\times2,5=\dfrac{7}{10}-\dfrac{4}{5}=\dfrac{-1}{10}\)
<=> \(\dfrac{55}{4}x=\dfrac{-1}{10}< =>x=\dfrac{-2}{275}\)
b) \(125\%\times\dfrac{17}{4}:\left(1\dfrac{5}{16}-0,5\right)+2008\)
= \(\dfrac{85}{16}:\left(\dfrac{21}{16}-\dfrac{1}{2}\right)+2008=\dfrac{85}{16}:\dfrac{13}{16}+2008=\dfrac{26189}{13}\)
c) \(\dfrac{5}{11}+\left(\dfrac{16}{11}+1\right)\)
= \(\dfrac{21}{11}+1=\dfrac{32}{11}\)
d) \(\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)+\dfrac{11}{4}\)
= 1 + 1 + \(\dfrac{11}{4}\) = \(\dfrac{19}{4}\)
1, A= \(\dfrac{-3}{4}.\left(0,125-1\dfrac{1}{2}\right):\dfrac{33}{16}-25\%\)
2, B= \(1\dfrac{13}{15}.0,75-\left(\dfrac{11}{20}+25\%\right):\dfrac{7}{3}\)
3, C= \(\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)
4, D= \(6\dfrac{5}{12}:2\dfrac{5}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\)
\(1,A=-\dfrac{3}{4}.\left(0,125-1\dfrac{1}{2}\right):\dfrac{33}{16}-25\%\)
\(A=-\dfrac{3}{4}.\left(0,125-\dfrac{3}{2}\right):\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=-\dfrac{3}{4}.\left(-\dfrac{11}{8}\right):\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=\dfrac{33}{32}:\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=\dfrac{33}{32}.\dfrac{16}{33}-\dfrac{1}{4}\)
\(A=\dfrac{1}{2}-\dfrac{1}{4}\)
\(A=\dfrac{2}{4}-\dfrac{1}{4}\)
\(A=\dfrac{1}{4}\)
\(D=6\dfrac{5}{12}:2\dfrac{5}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\)
\(D=\dfrac{77}{12}:\dfrac{13}{4}+\dfrac{45}{4}.\dfrac{2}{15}\)
\(D=\dfrac{77}{39}+\dfrac{3}{2}\)
\(D=\dfrac{271}{78}\)
\(C=\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)
\(C=\dfrac{5}{16}:0,125-\left(\dfrac{9}{4}-0,6\right).\dfrac{10}{11}\)
\(C=\dfrac{5}{16}:0,125-\dfrac{33}{20}.\dfrac{10}{11}\)
\(C=\dfrac{5}{2}-\dfrac{3}{2}\)
\(C=1\)
a) \(\dfrac{11}{13}-\left(\dfrac{5}{42}-x\right)=\left(\dfrac{15}{28}-\dfrac{11}{13}\right)\)
b) \(\left(\dfrac{7}{2}-2x\right)x3\dfrac{2}{5}+1\dfrac{4}{5}=7\dfrac{6}{5}\)
c) \(\left|2x-\dfrac{1}{3}\right|=\dfrac{5}{3}-2\)
d) (2x-1)3= -27
e) \(\dfrac{16}{2x}=1\)
a: =>5/42-x=11/13-15/28+11/13=421/364
=>x=-1193/1092
b: =>\(\dfrac{7}{2}-2x=7+\dfrac{6}{5}-3-\dfrac{2}{5}-1-\dfrac{4}{5}=3\)
=>2x=1/2
=>x=1/4
c: =>|2x-1/3|=-1/3(vô lý)
d: =>2x-1=-3
=>2x=-2
hay x=-1
e: =>2x=16
hay x=8
Tìm x, biết:
a) \(\dfrac{11}{13}-\left(\dfrac{5}{42}-x\right)=-\left(\dfrac{15}{28}-\dfrac{11}{13}\right)\)
b) \(2x.\left(x-\dfrac{1}{7}\right)=0\)
c) \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
a. \(\dfrac{11}{13}-\left(\dfrac{5}{42}-x\right)=-\left(\dfrac{15}{28}-\dfrac{11}{13}\right)\)
\(\Rightarrow\dfrac{11}{13}-\left(\dfrac{5}{42}-x\right)=-\left(\dfrac{-113}{364}\right)=\dfrac{113}{364}\)
\(\Rightarrow\left(\dfrac{5}{42}-x\right)=\dfrac{11}{13}-\dfrac{113}{364}\)
\(\Rightarrow\left(\dfrac{5}{42}-x\right)=\dfrac{15}{28}\)
\(\Rightarrow x=\dfrac{5}{42}-\dfrac{15}{28}=\dfrac{-5}{12}\)
Vậy..............
b. \(2x.\left(x-\dfrac{1}{7}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-1}{7}\end{matrix}\right.\)
Vậy............
c. \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Rightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{4}:x=\dfrac{-7}{20}\)
\(\Rightarrow x=\dfrac{1}{4}:\dfrac{-7}{20}=\dfrac{-5}{7}\)
Vậy...........
Bài 1:
a,\(|x-3|+|2-x|=0\)
b,\(\left(2-\dfrac{3}{4}x\right).\left(x+1\right)=0\)
bài 2:
a,A=\(\dfrac{\dfrac{-6}{7}+\dfrac{6}{13}-\dfrac{6}{29}}{\dfrac{9}{7}-\dfrac{9}{13}+\dfrac{9}{29}}\)
b,B=\(\dfrac{\dfrac{2}{15}-\dfrac{2}{21}+\dfrac{2}{39}}{0,25-\dfrac{5}{28}+\dfrac{5}{52}}\)
c,C=\(\dfrac{50-\dfrac{4}{15}+\dfrac{2}{15}-\dfrac{2}{17}}{100-\dfrac{8}{13}+\dfrac{4}{15}-\dfrac{4}{17}}:\dfrac{1+\dfrac{2}{21}-\dfrac{5}{121}}{\dfrac{65}{121}-\dfrac{26}{71}-13}\)
1.a) Dễ nhận thấy đề toán chỉ giải được khi đề là tìm x,y. Còn nếu là tìm x ta nhận thấy ngay vô nghiệm. Do đó: Sửa đề: \(\left|x-3\right|+\left|2-y\right|=0\)
\(\Leftrightarrow\left|x-3\right|=\left|2-y\right|=0\)
\(\left|x-3\right|=0\Rightarrow\left\{{}\begin{matrix}x-3=0\\-\left(x-3\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\) (1)
\(\left|2-y\right|=0\Rightarrow\left\{{}\begin{matrix}2-y=0\\-\left(2-y\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\) (2)
Từ (1) và (2) có: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x_1=3\\x_2=-3\end{matrix}\right.\\\left\{{}\begin{matrix}y_1=2\\y_2=-2\end{matrix}\right.\end{matrix}\right.\)
Bài 2:
a: \(=\dfrac{-6\left(\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{29}\right)}{9\left(\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{29}\right)}=\dfrac{-6}{9}=\dfrac{-2}{3}\)
b: \(=\dfrac{\dfrac{2}{15}-\dfrac{2}{21}+\dfrac{2}{39}}{\dfrac{10}{40}-\dfrac{10}{56}+\dfrac{10}{104}}\)
\(=\dfrac{\dfrac{2}{3}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{13}\right)}{\dfrac{10}{8}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{13}\right)}=\dfrac{2}{3}:\dfrac{5}{4}=\dfrac{2}{3}\cdot\dfrac{4}{5}=\dfrac{8}{15}\)
c: \(=\dfrac{2\left(25-\dfrac{2}{13}+\dfrac{1}{15}-\dfrac{1}{17}\right)}{4\left(25-\dfrac{2}{13}+\dfrac{1}{15}-\dfrac{1}{17}\right)}:\dfrac{1+\dfrac{2}{21}-\dfrac{5}{121}}{13\left(\dfrac{5}{121}-\dfrac{2}{21}-1\right)}\)
=2/4:(-1)/13=2/4x(-13)=-13/2
Tìm x, biết :
a) \(\left(\dfrac{31}{20}-\dfrac{26}{45}\right).\dfrac{-36}{35}< x< \left(\dfrac{51}{56}+\dfrac{8}{21}+\dfrac{1}{3}\right).\dfrac{8}{13}\)
b) \(\dfrac{-3}{5}.x=\dfrac{1}{4}+0,75\)
c) \(\left(\dfrac{1}{7}-\dfrac{1}{3}\right).x=\dfrac{28}{3}.\left(\dfrac{1}{4}-\dfrac{1}{7}\right)\)
d) \(\dfrac{5}{7}.x=\dfrac{9}{8}-0,125\)
e)\(\left(\dfrac{2}{11}+\dfrac{1}{3}\right).x=\left(\dfrac{1}{7}-\dfrac{1}{8}\right).56\)
Mấy bài này bạn tự làm đi, chuyển vế tìm x gần giống cấp I mà.
b)\(\dfrac{-3}{5}.x=\dfrac{1}{4}+0,75\)
=>\(\dfrac{-3}{5}.x=1\)
=>\(x=1:\dfrac{-3}{5}\)
=>\(x=\dfrac{-5}{3}\)
Vậy \(x=\dfrac{-5}{3}\)