Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 2
a. \(-1\frac{2}{3}-|2x-1|:\frac{3}{5}=-2\)
\(|2x-1|:\frac{3}{5}=\frac{5}{3}-2\)
\(|2x-1|:\frac{3}{5}=-\frac{1}{3}\)
\(|2x-1|=-\frac{1}{5}\)
Vì giá trị tuyệt đối luôn \(\ge0\)với mọi x
mà \(-\frac{1}{5}< 0\)
=> \(x\in\varnothing\)
1,\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\left(7-\frac{1}{6}\right)+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\frac{41}{6}+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{41}{14}+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{137}{42}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{137}{42}-\frac{1}{2}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{58}{21}\)
\(\left(x-\frac{9}{4}\right)=\frac{5}{2}:\frac{2}{9}\)
\(\left(x-\frac{9}{4}\right)=\frac{45}{4}\)
\(x=\frac{45}{4}+\frac{9}{4}\)
\(x=\frac{27}{2}\)
\(1)\frac{1}{2}x-\frac{3}{5}=\frac{-4}{5}\)
\(\Rightarrow\frac{1}{2}x=\frac{-4}{5}+\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}x=\frac{-1}{5}\)
\(\Rightarrow x=\frac{-1}{5}:\frac{1}{2}=\frac{-1}{5}\cdot\frac{2}{1}=\frac{-2}{5}\)
\(\Leftrightarrow x=\frac{-2}{5}\)
\(2)3\frac{1}{5}-2\frac{1}{3}x=-1\frac{3}{5}+1\frac{7}{10}\)
\(\Rightarrow\frac{16}{5}-\frac{7}{3}x=-\frac{8}{5}+\frac{17}{10}\)
\(\Rightarrow\frac{7}{3}x=\frac{16}{5}-\frac{-8}{5}+\frac{17}{10}\)
\(\Rightarrow\frac{7}{3}x=\frac{16}{5}+\frac{8}{5}+\frac{17}{10}\)
\(\Rightarrow\frac{7}{3}x=\frac{24}{5}+\frac{17}{10}\)
\(\Rightarrow\frac{7}{3}x=\frac{48}{10}+\frac{17}{10}\)
Đến đây tìm được rồi nhé
3,4, áp dụng bài 1,2 rồi làm :v
a) \(\left(x-\frac{1}{2}\right).\frac{1}{3}+\frac{5}{7}=9\frac{5}{7}\)
\(\left(x-\frac{1}{2}\right).\frac{1}{3}=9\frac{5}{7}-\frac{5}{7}\)
\(x-\frac{1}{2}=9:\frac{1}{3}\)
\(x=27+\frac{1}{2}\)
\(x=\frac{55}{2}\)
b)\(\frac{1}{2}.x+150\%.x=2014\)
\(\left(\frac{1}{2}+150\%\right)x=2014\)
\(2x=2014\)
\(x=2014:2\)
\(x=1007\)
c) \(2-\left|\frac{3}{4}\right|=\frac{7}{2}\)???????
và bài này nữa
tính nhanh
a, \(5\frac{7}{5}\cdot4\frac{2}{7}+5\frac{5}{7}\cdot4\frac{2}{7}\)
b, \(b,\frac{-7}{11}\cdot\frac{4}{19}+\frac{-7}{19}\cdot\frac{7}{11}+2\frac{7}{19}\)
cảm ơn giúp nhiều
a
\(5\frac{4}{7}:x+=13\)
\(\frac{39}{7}:x=13\)
\(x=\frac{39}{7}:13\)
\(x=\frac{3}{7}\)
\(\frac{4}{7}x=\frac{9}{8}-0,125\)
\(\frac{4}{7}x=1\)
\(x=1:\frac{4}{7}\)
\(x=\frac{7}{4}=1\frac{3}{4}\)
1.
a) (—7/3)3:(—7/3)2=(—7/3)3–2=—7/3
b) (—4/9):(—4/9)3= (—4/9)1–3=(—4/9)—2=81/16
c) (1/5)10:(1/5)7=(1/5)10–7=(1/5)3=1/125
2.
a) —x/7 =1/—21
==> —x.(—21)=7.1
==> —x.(—21)=7
==> —x=7:(—21)
==> —x=—1/3
==> x=1/3
b) 4 2/5 . 0,5–1 3/7= 22/5 . 1/2 —10/7= 22.1/5.2–10/7= 11/5 —10/7= 77/35 — 50/35= 27/35
c) 3x2–2x=0
==> x3(3–2)=0
x3.1=0
x3=0:1
x3=0
==> x=0
c) 9x2–1=0
9x2=0+1
9x2=1
x2=1:9
x2=1/9
x2=12/32 hoặc x2=(—1/3)2
Vậy x=1/3 hoặc x=—1/3
B=\(6\frac{4}{9}-4\frac{4}{9}+3\frac{7}{11}\)
B=\(2+3\frac{7}{11}\)
B=\(5\frac{7}{11}\)
B = \(5\frac{7}{11}=\frac{62}{11}\)
C = 1
D = \(\frac{5}{2}=2\frac{1}{2}\)
\(\frac{2}{5}:x=-\frac{1}{4}\)
\(x=\frac{2}{5}:-\frac{1}{4}\)
\(x=\frac{2}{5}\cdot-\frac{4}{1}\)
\(x=-\frac{8}{5}\)
Vậy \(x=-\frac{8}{5}\)
\(\frac{4}{7}\cdot x-\frac{2}{3}=\frac{1}{5} \)
\(\frac{4}{7}\cdot x=\frac{1}{5}+\frac{2}{3}\)
\(\frac{4}{7}\cdot x=\frac{3}{15}+\frac{10}{15}\)
\(\frac{4}{7}\cdot x=\frac{13}{15}\)
\(x=\frac{13}{15}:\frac{4}{7}\)
\(x=\frac{13}{15}\cdot\frac{7}{4}\)
\(x=\frac{91}{60}\)
Vậy \(x=\frac{91}{60}\)
2/5 : x = -1/4
=> x = 2/5 : -1/4
=> x = -8/5
4/7.x - 2/3 = 1/5
=> 4/7x = 1/5-2/3
=> 4/7x = -7/15
=> x = -49/60