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tìm x a)
\(\dfrac{7}{2}\)-\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-5}{4}\)
\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-5}{4}\) + \(\dfrac{7}{2}\)
\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-5}{12}+\dfrac{7}{12}\)
\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-12}{12}=1\)
\(x+\dfrac{7}{10}\)= 1 . \(\dfrac{6}{5}\)
*Rồi tự làm phần tt đi
Các bạn ơi,mình ghi thiếu,còn 3 câu nữa nha!!!~~nya
e)| \(\dfrac{5}{2}\)x-\(\dfrac{1}{2}\) |-(-22).\(\dfrac{1}{3}\)(0,75-\(\dfrac{1}{7}\))=\(\dfrac{-5}{13}\):2\(\dfrac{9}{13}\)-0,5.(\(\dfrac{-2}{3}\))
f)| 5x+21 | = | 2x -63 |
g) -45 - |-3x-96 | - 54=-207
Làm ơn giúp mình với ạ!Mình đang cần gấp lắm trong ngày hôm nay ạ!!!Mình xin cảm ơn các bạn nhiều nhiều lắm luôn đó!!!Thank you very much!!!(^-^)
a, (\(\dfrac{2}{9}\)(6x - \(\dfrac{3}{4}\)) - 3(\(\dfrac{1}{4}x-\dfrac{1}{5}\)) = \(\dfrac{-8}{15}\)
<=> (\(\dfrac{4}{3}x-\dfrac{1}{6}\)) - (\(\dfrac{3}{4}x-\dfrac{3}{5}\)) = \(\dfrac{-8}{15}\)
<=> \(\dfrac{4}{3}x-\dfrac{1}{6}-\dfrac{3}{4}x+\dfrac{3}{5}=\dfrac{-8}{15}\)
<=> \(\dfrac{7}{12}x+\dfrac{13}{30}=\dfrac{-8}{15}\)
<=> \(\dfrac{7}{12}x=\dfrac{-8}{15}-\dfrac{13}{30}\)
<=> \(\dfrac{7}{12}x=-\dfrac{29}{30}\)
<=> x = \(-\dfrac{58}{35}\)
@Nguyễn Gia Hân
bài 1:
a) \(4\dfrac{1}{2}x:\dfrac{5}{12}=0,5\) ; b)\(1,5+1\dfrac{1}{4}x=\dfrac{2}{3}\)
\(\dfrac{9}{2}x:\dfrac{5}{12}=\dfrac{1}{2}\) \(\dfrac{3}{2}+\dfrac{5}{4}x=\dfrac{2}{3}\)
\(\dfrac{9}{2}x\) \(=\dfrac{1}{2}.\dfrac{5}{12}\) \(\dfrac{5}{4}x=\dfrac{2}{3}-\dfrac{3}{2}\)
\(\dfrac{9}{2}x\) \(=\dfrac{5}{24}\) \(\dfrac{5}{4}x=\dfrac{-5}{6}\)
\(x\) \(=\dfrac{5}{24}:\dfrac{9}{2}\) \(x=\dfrac{-5}{6}:\dfrac{5}{4}\)
\(x\) \(=\dfrac{5}{108}\) \(x=\dfrac{-2}{3}\)
c) Cho mình hỏi x ở đâu vậy ???
d)\(\left(x-5\right):\dfrac{1}{3}=\dfrac{2}{5}\) e)\(\left(4,5-2x\right):\dfrac{3}{4}=1\dfrac{1}{3}\)
\(\left(x-5\right)\) \(=\dfrac{2}{5}.\dfrac{1}{3}\) \(\left(\dfrac{9}{2}-2x\right):\dfrac{3}{4}=\dfrac{4}{3}\)
\(x-5\) \(=\dfrac{2}{15}\) \(\dfrac{9}{2}-2x\) =\(\dfrac{4}{3}.\dfrac{3}{4}\)
\(x\) \(=\dfrac{2}{15}+5\) \(\dfrac{9}{2}-2x=1\)
\(x\) \(=\dfrac{77}{15}\) \(2x=\dfrac{9}{2}-1\)
f) \(\left(2,7x-1\dfrac{1}{2}x\right):\dfrac{2}{7}=\dfrac{-21}{7}\) \(2x=\dfrac{7}{2}\)
\(\left(\dfrac{27}{10}x-\dfrac{3}{2}x\right):\dfrac{2}{7}=-3\) \(x=\dfrac{7}{2}:2\)
\(\left[x\left(\dfrac{27}{10}-\dfrac{3}{2}\right)\right]=-3.\dfrac{2}{7}\) \(x=\dfrac{7}{4}\)
\(x.\dfrac{6}{5}=\dfrac{-6}{7}\)
\(x=\dfrac{-6}{7}:\dfrac{6}{5}\)
\(x=\dfrac{-5}{7}\)
bài 2:
Theo bài ra ta có :\(\dfrac{a}{27}=\dfrac{-5}{9}=\dfrac{-45}{b}\)
\(\Rightarrow9a=27.\left(-5\right)\Rightarrow a=\dfrac{27.\left(-5\right)}{9}=-15\)
\(\Rightarrow\left(-5\right)b=\left(-45\right).9\Rightarrow b=\dfrac{\left(-45\right).9}{-5}=81\)
Vậy \(a=-15;b=81\)
9) \(\dfrac{x}{4}=\dfrac{9}{x}\)
Theo định nghĩa về hai phân số bằng nhau, ta có:
\(4\cdot9=x^2\\ 36=x^2\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
8)
\(x:\dfrac{5}{3}+\dfrac{1}{3}=-\dfrac{2}{5}\\ x:\dfrac{5}{3}=-\dfrac{2}{5}+\dfrac{1}{3}\\ x:\dfrac{5}{3}=-\dfrac{1}{15}\\ x=\dfrac{1}{15}\cdot\dfrac{5}{3}\\ x=\dfrac{1}{9}\)
7)
\(2x-16=40+x\\ 2x-x=40+16\\ x\left(2-1\right)=56\\ x=56\)
6)
\(1\dfrac{1}{2}+x=\dfrac{3}{2}-7\\ \dfrac{3}{2}+x=\dfrac{3}{2}-7\\ \dfrac{3}{2}-\dfrac{3}{2}=-7-x\\ -7-x=0\\ x=-7-0\\ x=-7\)
5)
\(3\dfrac{1}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\\ \dfrac{7}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\\ \dfrac{1}{2}x=\dfrac{7}{2}-\dfrac{2}{3}\\ \dfrac{1}{2}x=\dfrac{17}{6}\\ x=\dfrac{17}{6}:\dfrac{1}{2}\\ x=\dfrac{17}{3}\)
4)
\(x\cdot\left(x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
3)
\(\left(\dfrac{2x}{5}+2\right):\left(-4\right)=-1\dfrac{1}{2}\\ \left(\dfrac{2x}{5}+2\right):\left(-4\right)=-\dfrac{3}{2}\\ \dfrac{2x}{5}+2=-\dfrac{3}{2}\cdot\left(-4\right)\\ \dfrac{2x}{5}+2=6\\ \dfrac{2x}{5}=6-2\\ \dfrac{2x}{5}=4\\ 2x=4\cdot5\\ 2x=20\\ x=20:2\\ x=10\)
2)
\(\dfrac{1}{3}+\dfrac{1}{2}:x=-0,25\\ \dfrac{1}{3}+\dfrac{1}{2}:x=-\dfrac{1}{4}\\ \dfrac{1}{2}:x=-\dfrac{1}{4}-\dfrac{1}{3}\\ \dfrac{1}{2}:x=-\dfrac{7}{12}\\ x=\dfrac{1}{2}:-\dfrac{7}{12}\\ x=-\dfrac{6}{7}\)
1)
\(\dfrac{4}{3}+x=\dfrac{2}{15}\\ x=\dfrac{2}{15}-\dfrac{4}{3}x=-\dfrac{6}{5}\)
a: \(\Leftrightarrow x\cdot\dfrac{62}{7}=\dfrac{29}{9}\cdot\dfrac{56}{3}=\dfrac{1624}{27}\)
hay \(x=\dfrac{1624}{27}:\dfrac{62}{7}=\dfrac{5684}{837}\)
b: \(\Leftrightarrow\dfrac{1}{5}:x=\dfrac{12}{35}\)
nên \(x=\dfrac{1}{5}:\dfrac{12}{35}=\dfrac{1}{5}\cdot\dfrac{35}{12}=\dfrac{7}{12}\)
c: \(\Leftrightarrow\left|2x+\dfrac{1}{3}\right|=\dfrac{30-7}{42}=\dfrac{23}{42}\)
=>2x+1/3=23/42 hoặc 2x+1/3=-23/42
=>2x=3/14 hoặc 2x=-37/42
=>x=3/28 hoặc x=-37/84
\(4)\)
\(\dfrac{-\left(-x\right)}{5}-\dfrac{2}{10}=\dfrac{1}{-5}-\dfrac{7}{50}\)
\(\Leftrightarrow\dfrac{x}{5}-\dfrac{2}{10}=\dfrac{1}{-5}-\dfrac{7}{50}\)
\(\dfrac{2x}{10}-\dfrac{2}{10}=\dfrac{-10}{50}-\dfrac{7}{50}\)
\(\Leftrightarrow\dfrac{2x-2}{10}=\dfrac{-10-7}{50}\)
\(\dfrac{2x-2}{10}=\dfrac{-17}{50}\)
\(\Leftrightarrow50\left(2x-2\right)=-17.10\)
\(100x-100=-170\)
\(100x=-170+100=-70\)
\(x=-70:100=\dfrac{-7}{10}\)
\(\dfrac{x+1}{5}=\dfrac{7}{x-1}\)
\(\left(x+1\right)\left(x-1\right)5.7\)
\(x\left(x-1\right)+1\left(x-1\right)=35\)
\(x^2-x+x-1=35\)
\(x^2-1=35\)
\(x^2=36\)
\(\Leftrightarrow x=\left\{\pm6\right\}\)
bạn có thể giải đc các bài còn lại k ? K phải mk ép bạn đâu nhưng nếu bạn lm đc thì giúp mk nha
a) Để phân số \(\dfrac{3}{x-1}\) có giá trị là số nguyên thì:
3 \(⋮\) x - 1
Vì x \(\in\) Z => x - 1 \(\in\) Z
=> x \(\in\) Ư(3) = {1;3;-1;-3}
Ta lập bảng sau
| x | x - 1 | Điều kiện x \(\in\) Z |
| 2 | 1 | TM |
| 4 | 3 | TM |
| 0 | -1 | TM |
| -2 | -3 |
TM |
Vậy x \(\in\) {2;4;0;-2} là giá trị cần tìm
b)Để phân số \(\dfrac{x-2}{x+3}\) có giá trị là số nguyên thì:
x - 2 \(⋮\) x + 3
=> \(\left\{{}\begin{matrix}x-2⋮x+3\\x+3⋮x+3\end{matrix}\right.\)
=> 5\(⋮\)x+3
Vì x \(\in\) Z => x + 3 \(\in\) Z
=> x + 3 \(\in\) Ư(5) = {1;5;-1;-5}
Ta lập bảng sau
| x | x + 3 | ĐK x\(\in\) Z |
| -2 | 1 | TM |
| 2 | 5 | TM |
| -4 | -1 | TM |
| -8 | -5 | TM |
Vậy x \(\in\) {-2;2;-4;-8} là giá trị cần tìm
a) Để phân số \(A=\dfrac{3}{x-1}\) có giá trị là 1 số nguyên thì :
\(3⋮x-1\)
\(\Rightarrow x-1\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)
Ta có bảng :
| \(x-1\) | \(1\) | \(-1\) | \(3\) | \(-3\) |
| \(x\) | \(2\) | \(0\) | \(4\) | \(-2\) |
| \(Đk\) \(x\in Z\) | \(TM\) | \(TM\) | \(TM\) | \(TM\) |
Vậy \(x\in\left\{2;0;4;-2\right\}\) là giá trị cần tìm
b) Để phân số \(A=\dfrac{x-2}{x+3}\) có giá trị là 1 số nguyên thì :
\(x-2⋮x+3\)
\(\Rightarrow x+3-5⋮x+3\)
\(\Rightarrow-5⋮x+3\)
\(\Rightarrow x+3\inƯ\left(-5\right)=\left\{1;-1;5;-5\right\}\)
Ta có bảng :
| \(x+3\) | \(1\) | \(-1\) | \(5\) | \(-5\) |
| \(x\) | \(-2\) | \(-4\) | \(2\) | \(-8\) |
| \(Đk\) \(x\in Z\) | \(TM\) | \(TM\) | \(TM\) | \(TM\) |
Vậy \(x\in\left\{-2;-4;2;-8\right\}\) là giá trị cần tìm
Bài 1:
\(\left(-\dfrac{72}{40}-\dfrac{144}{60}-2\dfrac{1}{3}\right):\left(\dfrac{45}{100}-\dfrac{25}{60}+-\dfrac{75}{25}\right)\)
\(=\left(-\dfrac{9}{5}-\dfrac{12}{5}-\dfrac{7}{3}\right):\left(\dfrac{9}{20}-\dfrac{5}{12}+-3\right)\)
\(=\left(-\dfrac{27}{15}-\dfrac{36}{15}-\dfrac{21}{15}\right):\left(\dfrac{27}{60}-\dfrac{25}{60}+-3\right)\)
\(=\left(-\dfrac{28}{5}\right):\left(-\dfrac{89}{30}\right)\)
\(=\left(-\dfrac{28}{5}\right).\left(-\dfrac{30}{89}\right)\)
\(=\dfrac{168}{89}\)
Khó thế
làm thôi chứ mình nghĩ quá với lớp 6
\(\left\{{}\begin{matrix}\dfrac{4}{7}< \dfrac{a}{b}< \dfrac{2}{3}\\7a+4b=1994\end{matrix}\right.\)
\(7a+4b=1994\Rightarrow a=\dfrac{1994-4b}{7}=\dfrac{1998-4-4b}{7}\)
\(\Rightarrow\left\{{}\begin{matrix}b=7n-2\\a=286-4n\end{matrix}\right.\)(*)
\(\left\{{}\begin{matrix}\dfrac{a}{b}< \dfrac{2}{3}\left(1\right)\\\dfrac{a}{b}>\dfrac{4}{7}\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow3a< 2b\Rightarrow3\left(286-4n\right)< 2\left(7n-2\right)\)
\(858-12n< 14n-4\Rightarrow n>\dfrac{862}{26}=33,15..\)
\(\left(2\right)\Leftrightarrow7a>4b\Rightarrow7\left(286-4n\right)>2\left(7n-2\right)\)
\(\Leftrightarrow1932-28n>14n-4\Rightarrow N< \dfrac{1936}{42}=46,09..\)
\(\)Tập hợp giá trị phân số a/b thỏa mãn là:
\(\left\{{}\begin{matrix}n\in N\\33< n\le46\\\dfrac{a}{b}=\dfrac{286-4n}{7n-2}\end{matrix}\right.\)
thích bao nhiêu thay vào
ví dụ
\(\left\{{}\begin{matrix}n=34\\\dfrac{a}{b}=\dfrac{286-4.34}{7.34-2}=\dfrac{150}{236}\\7.150+4.236=1050+944=1994\end{matrix}\right.\)
cảm ơn bn nhìu nhé!!!!!!!!ko có bn mk ko bít làm thế nào lun á!!!!!!!!!
Thái Nhữ đừng tức giận , ♂ ♀Thanh ღ Lê ｡◕‿◕｡ ( ♥ ILTKM ♥) ♫ ♪ k sao đâu , nhất thì làm lại là xong mà , nhỉ !!!
a,(6x-72):2-84=5628
=>(6x-72):2=5628+84
=>(6x-72):2=5712
=>6x-72=5712.2
=>6x-72=11424
=>6x=11424+72
=>6x=11496
=>x=11496:6
=>x=1916
b,\(\left(\dfrac{9}{2}-2x\right):75\%=1\dfrac{1}{3}\)
=>\(\left(\dfrac{9}{2}-2x\right):\dfrac{3}{4}=\dfrac{4}{3}\)
=>\(\dfrac{9}{2}-2x=\dfrac{4}{3}.\dfrac{3}{4}\)
=>\(\dfrac{9}{2}-2x=1\)
=>\(2x=\dfrac{9}{2}-1\)
=>\(2x=\dfrac{9}{2}-\dfrac{2}{2}\)
=>\(2x=\dfrac{7}{2}\)
=>\(x=\dfrac{7}{2}:2\)
=>\(x=\dfrac{7}{2}.\dfrac{1}{2}\)
=>x\(x=\dfrac{7}{4}\)