tìm x, biết
a, x + \(\frac{5}{12}\)= -1\(\frac{2}{7}\)
b, 4\(\frac{1}{2}\)x : \(\frac{5}{12}\)= 0,5
c, 7,5 .1\(\frac{3}{4}\)x = 6\(\frac{2}{5}\)
1 tìm x biết ;
a, 0-|x + 1| = 5
b, 2 - | \(\frac{3}{4}\)- x | = \(\frac{7}{12}\)
c, 2 | \(\frac{1}{2}\)x - \(\frac{1}{3}\)| - \(\frac{3}{2}\)= \(\frac{1}{4}\)
d, | x - \(\frac{1}{3}\)| = \(\frac{5}{6}\)
e, \(\frac{3}{4}\)- 2 | 2x - \(\frac{2}{3}\)| = 2
f, \(\frac{2x-1}{2}\)= \(\frac{5+3x}{3}\)
d,
\(|x-\frac{1}{3}|=\frac{5}{6}\Rightarrow \left[\begin{matrix} x-\frac{1}{3}=\frac{5}{6}\\ x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{7}{6}\\ x=\frac{-1}{2}\end{matrix}\right.\)
e,
\(\frac{3}{4}-2|2x-\frac{2}{3}|=2\)
\(\Leftrightarrow 2|2x-\frac{2}{3}|=\frac{3}{4}-2=\frac{-5}{4}\)
\(\Leftrightarrow |2x-\frac{2}{3}|=-\frac{5}{8}<0\) (vô lý vì trị tuyệt đối của 1 số luôn không âm)
Vậy không tồn tại $x$ thỏa mãn đề bài.
f,
\(\frac{2x-1}{2}=\frac{5+3x}{3}\Leftrightarrow 3(2x-1)=2(5+3x)\)
\(\Leftrightarrow 6x-3=10+6x\)
\(\Leftrightarrow 13=0\) (vô lý)
Vậy không tồn tại $x$ thỏa mãn đề bài.
a,
$0-|x+1|=5$
$|x+1|=0-5=-5<0$ (vô lý do trị tuyệt đối của một số luôn không âm)
Do đó không tồn tại $x$ thỏa mãn điều kiện đề.
b,
\(2-|\frac{3}{4}-x|=\frac{7}{12}\)
\(|\frac{3}{4}-x|=2-\frac{7}{12}=\frac{17}{12}\)
\(\Rightarrow \left[\begin{matrix} \frac{3}{4}-x=\frac{17}{12}\\ \frac{3}{4}-x=\frac{-17}{12}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-2}{3}\\ x=\frac{13}{6}\end{matrix}\right.\)
c,
\(2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}\)
\(2|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{4}\)
\(|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{8}\)
\(\Rightarrow \left[\begin{matrix} \frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\ \frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{29}{12}\\ x=\frac{-13}{12}\end{matrix}\right.\)
1 tìm x biết ;
a, 0-|x + 1| = 5
b, 2 - | \(\frac{3}{4}\)- x | = \(\frac{7}{12}\)
c, 2 | \(\frac{1}{2}\)x - \(\frac{1}{3}\)| - \(\frac{3}{2}\)= \(\frac{1}{4}\)
d, | x - \(\frac{1}{3}\)| = \(\frac{5}{6}\)
e, \(\frac{3}{4}\)- 2 | 2x - \(\frac{2}{3}\)| = 2
f, \(\frac{2x-1}{2}\)= \(\frac{5+3x}{3}\)
1. Tính hợp lí :
a) \(\frac{3}{4}-0,25-\left[\frac{7}{3}+\left(\frac{-9}{2}\right)\right]-\frac{5}{6}\)
b) 7 + \(\left(\frac{7}{12}-\frac{1}{2}+3\right)-\left(\frac{1}{12}+5\right)\)
2. Tìm các sô nguyên x biết:
\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)
Câu 1:
a)\(\frac{3}{4}-0,25-\left[\frac{7}{3}+\left(-\frac{9}{2}\right)\right]-\frac{5}{6}\)
\(=\frac{3}{4}-\frac{1}{4}-\frac{14}{6}+\frac{27}{6}-\frac{5}{6}\)
\(=\frac{1}{2}-\frac{4}{3}\)
\(=-\frac{5}{6}\)
b)\(7+\left(\frac{7}{12}-\frac{1}{2}+3\right)-\left(\frac{1}{12}+5\right)\)
\(=7+\frac{1}{12}+3-\frac{1}{12}-5\)
\(=5\)
Câu 2:
\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)
\(-\frac{1}{12}\le\frac{x}{12}< 1-\frac{5}{12}\)
\(-\frac{1}{12}\le\frac{x}{12}< \frac{7}{12}\)
Vậy -1\(\le\)x<7
1) \(\frac{24}{-12}=\frac{x}{5}=\frac{-y}{3}\)Tìm x và y
2) \(\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-5}{25}\le\frac{x}{10}< \frac{-3}{4}+\frac{4}{14}+\frac{-2}{8}+\frac{-3}{5}+\frac{5}{7}\)Tìm x
3) \(\frac{8.x+18}{2.x+6}\)Tìm x
Tìm số nguyên x
a) \(\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{3}{5}+\frac{5}{7}\)
b)\(\frac{5}{17}+\frac{-9}{4}+\frac{-26}{31}+\frac{12}{17}+\frac{-11}{31}< \frac{x}{9}\le\frac{-3}{7}+\frac{7}{15}+\frac{4}{-7}+\frac{8}{15}\)
\(a)\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{3}{5}+\frac{5}{7}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{-2}{5}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{-1}{4}+\frac{2}{7}+\frac{5}{7}+\frac{3}{5}\)
\(\Rightarrow\frac{2}{6}+\frac{1}{6}+\frac{-3}{5}\le x< -1+1+\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}+\frac{-3}{5}\le x< \frac{3}{5}\)
\(\Rightarrow\frac{-1}{10}\le x< \frac{6}{10}\)
\(\Rightarrow-1\le x< 6\)
\(\Rightarrow x\in\left\{-1;0;1;2;3;4;5\right\}\)
Bài b tương tự
bạn ơi bạn giải câu b được ko. mk ko biết làm câu b
$\frac{4x+3}{5}$ -$\frac{6x-2}{7}$ =$\frac{5x+4}{3}$ +3
b.
$\frac{x+4}{5}$ -x+4=$\frac{x}{3}$ -$\frac{x-2}{2}$
c.$\frac{5x+2}{6}$ -$\frac{8x-1}{3}$ =$\frac{4x+2}{5}$ -5
d.$\frac{2x+3}{3}$ =$\frac{5-4}{2}$
e. $\frac{5x+3}{12}$ =$\frac{1+2x}{9}$
f.$\frac{7x-1}{6}$ =$\frac{16-x}{5}$
g. $\frac{x-3}{5}$ =6-$\frac{1-2x}{3}$
h. $\frac{3x-2}{6}$ -5=$\frac{3-2(x+7)}{4}$
giúp vs ạ, cần gấp
d: =>4x+6=15x-12
=>4x-15x=-12-6=-18
=>-11x=-18
hay x=18/11
e: =>\(45x+27=12+24x\)
=>21x=-15
hay x=-5/7
f: =>35x-5=96-6x
=>41x=101
hay x=101/41
g: =>3(x-3)=90-5(1-2x)
=>3x-9=90-5+10x
=>3x-9=10x+85
=>-7x=94
hay x=-94/7
Tìm x, biết:
a) 60%x + 0,4x + x :3 =2
b)1-\(\left(5\frac{3}{8}+x-7\frac{5}{24}\right):\left(-16\frac{2}{3}\right)\)
c)\(3\frac{1}{4}x-\frac{7}{6}x=\frac{-5}{12}+1\frac{2}{3}\)
Bài 2: Tính:
a) A= \(\frac{-45.58-45.42}{2+4+6+...+16+18}\)
b)1-2-3+4+5-6-7+...+601-602-603+604
b) \(\frac{\left(140\frac{3}{7}-138\frac{5}{12}\right):18\frac{1}{6}}{0,002}\)
Bài 3: Cho A và B, biết:
A=\(\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\) và B= \(\frac{4}{35}+\frac{4}{63}+\frac{4}{99}+\frac{4}{143}+\frac{4}{195}\)
Hãy so sánh A & B
Tìm x:
a) \(\frac{-15}{12}x+\frac{3}{7}=\frac{6}{4}x-\frac{1}{2}\)
b) \(\frac{2}{5}\left(x+1\right)-\frac{4}{5}x=0\)
a)\(-\frac{15}{12}x+\frac{3}{7}=\frac{6}{4}x-\frac{1}{2}\)
\(\Leftrightarrow\frac{15}{12}x+\frac{6}{4}x=\frac{1}{2}+\frac{3}{7}\)
\(\Leftrightarrow\frac{11}{4}x=\frac{9}{14}\Leftrightarrow x=\frac{36}{154}\)
b) \(\frac{2}{5}\left(x+1\right)-\frac{4}{5}x=0\)
\(\Leftrightarrow\frac{2}{5}x+\frac{2}{5}-\frac{4}{5}x=0\)
\(\Leftrightarrow-\frac{2}{5}x=-\frac{2}{5}\Leftrightarrow x=1\)
câu 1:thực hiện phép tính (tính nhanh nếu có thể):
a, \(\frac{-7}{25}.\frac{11}{13}+\frac{-7}{25}.\frac{2}{13}-\frac{18}{25}\)
b,\(\frac{5}{7}.\frac{1}{3}-\frac{5}{7}.\frac{1}{4}-\frac{5}{7}.\frac{1}{12}\)
c,\(5\frac{2}{5}.4\frac{2}{7}+5\frac{5}{7}.5\frac{2}{5}\)
d,\(75\%-1\frac{1}{2}+0,5:\frac{5}{12}-\left(\frac{-1}{2}\right)^2\)
câu 2:tìm x biết:
a, \(x+\frac{1}{6}=\frac{-3}{8}\)
b, \(\frac{1}{2}x+\frac{1}{8}x=\frac{3}{4}\)
c,\(2-\left|\frac{3}{4}-x\right|=\frac{7}{12}\)
d,\(\left(2x-4,5\right):\frac{3}{4}-\frac{1}{3}=1\)
GIÚP MIK VS MAI MIK PHẢI NỘP BÀI RỒI!
CACS BN GIÚP MIK TRẢ LỜI TRONG TỐI NAY ĐC K