a,\(\left(\frac{-1}{2}\right)^3-\left(\frac{2}{5}x+\frac{1}{3}\right)=3\)
b,\(-2\frac{1}{3}-\left(4\frac{1}{6}-\frac{4}{3}+1\frac{1}{2}\right)\)
AI GIAI DUOC MINH CHO 10 TICK.NHANH NHA MINH DANG CAN GAP
a,\(x^3-\frac{4}{25}x=0\)
b,\(\left(\frac{3}{8}+\frac{-3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
c\(4.\left(\frac{-1}{2}\right)^3-2.\left(\frac{-1}{2}\right)^2+3.\left(\frac{-1}{2}\right)-1.\left(\frac{-1}{2}\right)^0\)
AI GIAI DUOC MINH CHO 10 TICK
a) \(x^3-\frac{4}{25}x=0\)
\(\Leftrightarrow x\left(x+\frac{2}{5}\right)\left(x-\frac{2}{5}\right)=0\)
<=> x = 0
Xét 2 trường hợp:
\(\Leftrightarrow x+\frac{2}{5}=0\)
\(x=0-\frac{2}{5}\)
\(x=-\frac{2}{5}\)
\(\Leftrightarrow x-\frac{2}{5}=0\)
\(x=0+\frac{2}{5}\)
\(x=\frac{2}{5}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\pm\frac{2}{5}\end{cases}}\)
b) \(\left(\frac{3}{8}+\frac{-3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
\(=\left(\frac{3}{8}+\frac{-3}{4}+\frac{7}{12}\right):\frac{4}{3}\)
\(=\frac{13}{40}:\frac{4}{3}\)
\(=\frac{39}{120}=\frac{13}{40}\)
c) \(4\left(\frac{-1}{2}\right)^3-2\left(\frac{-1}{2}\right)^2+3\left(\frac{-1}{2}\right)-1\left(\frac{-1}{2}\right)^0\)
\(=4\left(\frac{-1}{2}\right)^3-2\left(\frac{-1}{2}\right)^3+3\left(\frac{-1}{2}\right)-1.1\)
\(=-\frac{1}{2}-\frac{1}{2}-\frac{3}{2}-1.1\)
\(=-\frac{5}{2}-1\)
\(=-\frac{7}{2}\)
a,\(\frac{2}{3}x-\frac{3}{2}\left(x-\frac{1}{2}\right)=\frac{5}{12}\)
b,\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.......+\frac{2}{x\left(x+1\right)}=\frac{2013}{2015}\)
AI GIAI DUOC MINH TICK CHO.NHO GIAI CHI TIET NHA
b,\(\Rightarrow\)\(\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}\right):2=\frac{2013}{2015}:2\)
\(\Rightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2013}{4030}\)
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{2013}{4030}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2013}{4030}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2013}{4030}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2015}\)
\(\Rightarrow\)\(x+1=2015\)
\(\Rightarrow x=2014\)
a, 2/3x -3/2.x-1/2x=5/12
x.(2/3-3/2-1/2)=5/12
x. -4/3=5/12
x=5/12:-4/3
x=-5/16
b,2/6+2/12+2/20+...+2/x.(x+1)=2013/2015
2/2.3+2/3.4+2/4.5+...+2/x.(x+1)=2013/2015
1/2(1-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1)=2013/2015
1/2(1-1/x+1)=2013/2015
1-1/x+1=2013/2015 : 1/2
1-1/x+1=4206/2015
suy ra đề sai
Tìm x biết :
\(\frac{x}{2}-\frac{2}{3}\left(3x-2\right)=\left(-\frac{1}{2}\right)^3\)
Ai biet giai thi giup minh voi minh like cho minh dang can gap .
\(\left(\frac{-5}{12}+\frac{7}{4}-\frac{3}{8}\right)-\left[4\frac{1}{2}-7\frac{1}{3}\right]-\left(\frac{1}{4}-\frac{5}{2}\right)\)
\(\left[2\frac{1}{4}-5\frac{3}{2}\right]-\left(\frac{3}{10}-1\right)-5\frac{1}{2}+\left(\frac{1}{3}-\frac{5}{6}\right)\)
\(\frac{4}{7}-\left(3\frac{2}{5}-1\frac{1}{2}\right)-\frac{5}{21}+\left[3\frac{1}{2}-4\frac{2}{3}\right]\)
\(\frac{1}{8}-1\frac{3}{4}+\left(\frac{7}{8}-3\frac{7}{2}+\frac{3}{4}\right)-\left[\frac{7}{4}-\frac{5}{8}\right]\)
\(\left(\frac{3}{5}-2\frac{1}{10}+\frac{11}{20}\right)-\left[\frac{-3}{4}+1\frac{7}{2}\right]\)
\(\left[-2\frac{1}{5}-2\frac{2}{3}\right]-\left(\frac{1}{15}-5\frac{1}{2}\right)+\left[\frac{-1}{6}+\frac{1}{3}\right]\)
\(1\frac{1}{8}-\left(\frac{1}{15}-\frac{1}{2}+\frac{-1}{6}\right)+\left[\frac{5}{4}+\frac{3}{2}\right]\)
\(\frac{5}{6}-\left(1\frac{1}{3}-1\frac{1}{2}\right)+\left[\frac{5}{12}-\frac{3}{4}-\frac{1}{6}\right]\)
\(1\frac{1}{4}-\left(\frac{7}{12}-\frac{2}{3}-1\frac{3}{8}\right)+\left[\frac{5}{24}-2\frac{1}{2}\right]-\frac{1}{6}-\left[\frac{-3}{4}\right]\)
\(-2\frac{1}{5}+2\frac{3}{10}-\left(\frac{6}{20}-\left[\frac{2}{8}-1\frac{1}{2}\right]\right)+\left[\frac{7}{20}-1\frac{1}{4}\right]\)
\(-\left[1\frac{2}{3}-3\frac{1}{2}+\frac{1}{4}\right]+\left(\frac{2}{6}-\frac{5}{12}\right)-\left(\frac{1}{3}-\left[\frac{1}{4}-\frac{1}{3}\right]\right)\)
\(-\frac{4}{5}-\left(1\frac{1}{10}-\frac{7}{10}\right)+\left[\frac{3}{4}-1\frac{1}{5}\right]+1\frac{1}{2}\)
\(\frac{3}{21}-\frac{5}{14}+\left[1\frac{1}{3}-5\frac{1}{2}+\frac{5}{14}\right]-\left(\frac{1}{6}-\frac{3}{7}+\frac{1}{3}\right)\)
\(-1\frac{2}{5}+\left[1\frac{3}{10}-\frac{7}{20}-1\frac{1}{4}\right]-\left(\frac{1}{5}-\left[\frac{3}{4}-1\frac{1}{2}\right]\right)\)
\(2\frac{1}{3}-\left(\frac{1}{2}-2\frac{1}{6}+\frac{3}{4}\right)+\left[\frac{5}{12}-1\frac{1}{3}\right]-\frac{7}{8}+3\frac{1}{2}\)
\(2\frac{1}{4}-1\frac{3}{5}-\left(\frac{9}{20}-\frac{7}{10}\right)+\left[1\frac{3}{5}-2\frac{1}{2}\right]+\frac{3}{4}\)
\(\left[\frac{8}{3}-5\frac{1}{4}+\frac{1}{6}\right]-\frac{7}{4}+\frac{-5}{12}-\left(1-1\frac{1}{2}+\frac{1}{3}\right)\)
\(\left(\frac{1}{4}-\left[1\frac{1}{4}-\frac{7}{10}\right]+\frac{1}{2}\right)-2\frac{1}{5}-1\frac{3}{10}+\left[1-\frac{1}{2}\right]\)
TRÌNH BÀY GIÚP MÌNH NHA
\(A=\left(6:\frac{3}{5}-1\frac{1}{6}x\frac{6}{7}\right):\left(4\frac{1}{5}x\frac{10}{11}+5\frac{2}{11}\right)\)\(B=\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{4}\right)x.......x\left(1-\frac{1}{2015}\right)x\left(1-\frac{1}{2016}\right)\)
\(C=5\frac{9}{10}:\frac{3}{2}-\left(2\frac{1}{3}x4\frac{1}{2}-2x2\frac{1}{3}\right):\frac{7}{4}\)
\(3\frac{1}{2}-4\frac{2}{3}+\left[\frac{3}{4}-2\frac{1}{3}\right]-\left(\frac{5}{6}-\frac{7}{4}\right)+5\frac{1}{2}-3\)
\(2\frac{2}{3}-1\frac{2}{5}+1\frac{3}{10}-\left(\frac{2}{5}-\frac{5}{6}\right)+\frac{4}{15}-1\frac{1}{3}\)
\(\left[2\frac{1}{3}-1\frac{4}{3}\right]-\left(\frac{5}{4}-\frac{7}{12}+\frac{-11}{6}\right)+\frac{4}{3}-\frac{3}{4}\)
\(-3\frac{3}{2}+5\frac{4}{3}-\left(\frac{7}{6}-1\frac{3}{4}\right)+\left[\frac{2}{3}-2\frac{1}{4}\right]\)
\(2\frac{2}{3}-\frac{5}{12}-\left(1\frac{3}{4}-2\frac{1}{4}\right)-\left[1-1\frac{1}{6}\right]+\left[\frac{-5}{3}\right]\)
\(1\frac{1}{3}-5\frac{1}{2}-\left[\frac{5}{6}-2\frac{2}{3}\right]+\left[\frac{7}{12}-\frac{5}{6}\right]\)
\(\frac{8}{15}-\left(\frac{2}{5}-3\frac{1}{3}+\left[\frac{-5}{6}\right]\right)+\left[\frac{1}{2}-\frac{4}{5}\right]-\left(\frac{1}{6}-1\frac{1}{3}\right)\)
\(-2\frac{3}{2}+\left[\frac{5}{6}-1\frac{1}{3}\right]-\left(\frac{5}{12}-\frac{7}{6}\right)+\left[\frac{4}{3}-3\frac{1}{4}\right]\)
\(\frac{9}{10}-1\frac{2}{5}-\left(\frac{5}{6}-3\frac{1}{2}\right)-\left[2\frac{1}{4}-5\frac{2}{36}\right]-\left[1-2\frac{1}{15}\right]\)
\(\frac{5}{7}-\frac{5}{21}+1\frac{2}{3}-\left(1\frac{1}{2}-\frac{5}{14}-\frac{1}{3}\right)+\left[\frac{1}{6}-\frac{4}{3}\right]\)
\(\frac{5}{7}-\frac{5}{21}+1\frac{2}{3}-\left(1\frac{1}{2}-\frac{5}{14}-\frac{1}{3}\right)+\left[\frac{1}{6}-\frac{4}{3}\right]\)
\(1\frac{1}{5}-\left(\frac{-9}{10}-2\frac{1}{2}+\frac{3}{4}\right)+\left[\frac{1}{5}-2\frac{1}{2}\right]+\frac{7}{10}-\left(\frac{1}{2}-\frac{1}{4}\right)\)
\(2\frac{1}{3}-\left(5\frac{1}{2}-2\frac{2}{3}\right)+\left[1\frac{1}{6}-2\frac{1}{2}\right]-\frac{5}{12}+\left(\frac{1}{4}-\frac{1}{8}\right)\)
tim x :
\(a.\left|x+5\right|=\frac{1}{7}-\left|\frac{4}{3}-\frac{1}{6}\right|\)|
\(b.\left|x+\frac{2}{3}\right|=\frac{1}{2}-\left[\frac{1}{4}-\frac{2}{3}\right]\)đau [ ] là dấu ngoặc đơn nha!
\(c.\left|x-\frac{1}{5}\right|=\left|\frac{-5}{2}\right|-\left[\frac{1}{4}-\frac{2}{3}\right]\)đau [ ] là dấu ngoặc đơn nha!
\(d.\left|x-\frac{5}{2}\right|=\frac{4}{3}-\left[\frac{2}{3}-\frac{1}{6}\right]\)đau [ ] là dấu ngoặc đơn nha!
ai nhanh nhất mk tk !
cau a dau nhi cuoi cung k phai j dau nha ! mk an lom !
\(a,\)\(\left|x+5\right|=\frac{1}{7}-\left|\frac{4}{3}-\frac{1}{6}\right|\)
\(\Leftrightarrow\left|x+5\right|=\frac{1}{7}-\frac{7}{6}\)
\(\Leftrightarrow\left|x+5\right|=\frac{-43}{42}\)
ta có |x+5| \(\ge\)0 \(\forall x\)
Mà \(-\frac{43}{42}< 0\)nên ko có giá trị x thoả mãn
b,
\(\left|x+\frac{2}{3}\right|=\frac{1}{2}-\left(\frac{1}{4}+\frac{2}{3}\right)\)
\(\Leftrightarrow\left|x+\frac{2}{3}\right|=\frac{11}{12}\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{2}{3}=\frac{11}{12}\forall x\ge-\frac{2}{3}\\-x-\frac{2}{3}=\frac{11}{12}\forall< -\frac{2}{3}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=-\frac{19}{12}\end{cases}}\)(thoả mãn đk)
Bài 1: Cho A=\(\left(\frac{1}{3}+\frac{3}{x^2-3x}\right):\left(\frac{x^2}{27-3x^2}+\frac{1}{x+3}\right)\)
a, Rút gọn A
b, Tìm x để A<-1
cac ban oi giup minh di minh dang can gap lam. ai giup minh hên nhat nam 2016
a. \(A=\left[\frac{1}{3}+\frac{3}{x.\left(x-3\right)}\right]:\left[\frac{x^2}{3.\left(9-x^2\right)}+\frac{1}{x+3}\right]\)
\(=\left[\frac{x.\left(x-3\right)}{3.x.\left(x-3\right)}+\frac{3.3}{x\left(x-3\right).3}\right]:\left[\frac{x^2}{3.\left(3-x\right)\left(3+x\right)}+\frac{1}{x+3}\right]\)
\(=\left[\frac{x^2-3x+9}{3x.\left(x-3\right)}\right]:\left[\frac{x^2}{3.\left(3-x\right)\left(3+x\right)}+\frac{\left(3-x\right).3}{\left(x+3\right).\left(3-x\right).3}\right]\)
\(=\frac{x^2-3x+9}{3x.\left(x-3\right)}:\left[\frac{x^2+9-3x}{3.\left(3-x\right)\left(3+x\right)}\right]\)
\(=\frac{x^2-3x+9}{3x.\left(x-3\right)}.\frac{3.\left(3-x\right)\left(3+x\right)}{x^2-3x+9}\)
\(=\frac{-\left(x-3\right)\left(3+x\right)}{x-3}=-\left(3+x\right)\)
b. Để A < -1 thì:
-(3+x) < -1
=> -3 - x < -1
=> x < -3 - (-1) = -2
Vậy x < -2 thì A < -1.
1, \(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\left(7-\frac{1}{6}\right)+\frac{1}{3}\)
2, \(2.\left(\frac{3}{2}-x\right)-\frac{1}{3}=7x-\frac{1}{4}\)
3,\(-\frac{3}{2}.\left(5-\frac{1}{6}\right)+4.\left(x-\frac{1}{2}\right)=\frac{1}{2}+x\)
4,\(-\frac{5}{7}.\left(\frac{2}{5}-x\right)-\frac{1}{3}=\frac{1}{5}-\frac{3}{10}\)
5,\(4-\frac{2}{3}.\left(x-3\right)=2-\frac{1}{2}+\frac{2}{3}\)
6,\(\frac{2}{3}-\frac{5}{3}.x=\frac{7}{10}.x+\frac{5}{6}\)
7,\(3.\left(x-\frac{5}{3}\right)+\frac{1}{2}=2\left(x-\frac{1}{4}\right)+\frac{5}{2}\)
Phần nào có bn giải rầu các men đừng giải lại nha mk sẽ ko tk đâu chỉ tik những phần chưa lm
1,\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\left(7-\frac{1}{6}\right)+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\frac{41}{6}+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{41}{14}+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{137}{42}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{137}{42}-\frac{1}{2}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{58}{21}\)
\(\left(x-\frac{9}{4}\right)=\frac{5}{2}:\frac{2}{9}\)
\(\left(x-\frac{9}{4}\right)=\frac{45}{4}\)
\(x=\frac{45}{4}+\frac{9}{4}\)
\(x=\frac{27}{2}\)
Bước cưối 58/21 minh man viết nhầm nên sai
\(\left(x-\frac{9}{4}\right)=\frac{58}{21}:\frac{2}{9}\)
\(\left(x-\frac{9}{4}\right)=\frac{87}{7}\)
\(x=\frac{87}{7}+\frac{9}{4}\)
\(x=\frac{411}{28}\)