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a.
\(\frac{1}{4}+\frac{1}{3}\div2x=-5\)
\(\frac{1}{3}\div2x=-5-\frac{1}{4}\)
\(\frac{1}{3}\div2x=-\frac{20}{4}-\frac{1}{4}\)
\(\frac{1}{3}\div2x=-\frac{21}{4}\)
\(2x=\frac{1}{3}\div\left(-\frac{21}{4}\right)\)
\(2x=\frac{1}{3}\times\left(-\frac{4}{21}\right)\)
\(2x=-\frac{4}{63}\)
\(x=-\frac{4}{63}\div2\)
\(x=-\frac{4}{63}\times\frac{1}{2}\)
\(x=-\frac{2}{63}\)
b.
\( \left(3x+\frac{1}{4}\right)\left(x+\frac{1}{2}\right)=0\)
TH1:
\(3x+\frac{1}{4}=0\)
\(3x=-\frac{1}{4}\)
\(x=-\frac{1}{4}\div3\)
\(x=-\frac{1}{4}\times\frac{1}{3}\)
\(x=-\frac{1}{12}\)
TH2:
\(x+\frac{1}{2}=0\)
\(x=-\frac{1}{2}\)
Vậy \(x=-\frac{1}{12}\) hoặc \(x=-\frac{1}{2}\)
c.
\(\left|2x-3,5\right|=28\)
\(2x-3,5=\pm28\)
TH1:
\(2x-3,5=28\)
\(2x=28+3,5\)
\(2x=31,5\)
\(x=31,5\div2\)
\(x=15,75\)
TH2:
\(2x-3,5=-28\)
\(2x=-28+3,5\)
\(2x=-24,5\)
\(x=-24,5\div2\)
\(x=-12,25\)
Vậy \(x=-12,25\) hoặc \(x=-15,75\)
Chúc bạn học tốt
a)
\(\begin{array}{l}\frac{2}{9}:x + \frac{5}{6} = 0,5\\\frac{2}{9}:x = \frac{1}{2} - \frac{5}{6}\\\frac{2}{9}:x = \frac{3}{6} - \frac{5}{6}\\\frac{2}{9}:x = \frac{{ - 2}}{6}\\x = \frac{2}{9}:\frac{{ - 2}}{6}\\x = \frac{2}{9}.\frac{{ - 6}}{2}\\x = \frac{{ - 2}}{3}\end{array}\)
Vậy \(x = \frac{{ - 2}}{3}\).
b)
\(\begin{array}{l}\frac{3}{4} - \left( {x - \frac{2}{3}} \right) = 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - \frac{4}{3}\\x - \frac{2}{3} = \frac{9}{{12}} - \frac{{16}}{{12}}\\x - \frac{2}{3} = \frac{{ - 7}}{{12}}\\x = \frac{{ - 7}}{{12}} + \frac{2}{3}\\x = \frac{{ - 7}}{{12}} + \frac{8}{{12}}\\x = \frac{1}{12}\end{array}\)
Vậy\(x = \frac{1}{12}\).
c)
\(\begin{array}{l}1\frac{1}{4}:\left( {x - \frac{2}{3}} \right) = 0,75\\\frac{5}{4}:\left( {x - \frac{2}{3}} \right) = \frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}:\frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}.\frac{4}{3}\\x - \frac{2}{3} = \frac{5}{3}\\x = \frac{5}{3} + \frac{2}{3}\\x = \frac{7}{3}\end{array}\)
Vậy \(x = \frac{7}{3}\).
d)
\(\begin{array}{l}\left( { - \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}\\ - \frac{5}{6}x + \frac{5}{4} = \frac{4}{3}.\frac{3}{2}\\ - \frac{5}{6}x + \frac{5}{4} = 2\\ - \frac{5}{6}x = 2 - \frac{5}{4}\\ - \frac{5}{6}x = \frac{8}{4} - \frac{5}{4}\\ - \frac{5}{6}x = \frac{3}{4}\\x = \frac{3}{4}:\left( { - \frac{5}{6}} \right)\\x = \frac{3}{4}.\frac{{ - 6}}{5}\\x = \frac{{ - 9}}{{10}}\end{array}\)
Vậy \(x = \frac{{ - 9}}{{10}}\).
a) Lm r nkoa!
b) \(\left(0,25x\right):3=\frac{5}{6}:0,125\)
\(=\left(0,25x\right):3=\frac{20}{3}\)
\(\Rightarrow0,25x=\frac{20}{3}\cdot3=20\)
\(\Rightarrow x=20:0,25=80\)
\(\Rightarrow x=80\)
c) \(0,01:2,5=\left(0,75x\right):0,75\)
\(=\frac{1}{250}=\left(0,75x\right):0,75\)
\(\Rightarrow0,75x=\frac{1}{250}\cdot0,75=\frac{3}{1000}\)
\(\Rightarrow x=\frac{3}{1000}:0,75=\frac{1}{250}\)
\(\Rightarrow x=\frac{1}{250}\)
d) \(1\frac{1}{3}:0,8=\frac{2}{3}:\left(0,1x\right)\)
\(=\frac{5}{3}=\frac{2}{3}:\left(0,1x\right)\)
\(\Rightarrow0,1x=\frac{2}{3}:\frac{5}{3}=\frac{2}{5}\)
\(\Rightarrow x=\frac{2}{5}:0,1=4\)
\(\Rightarrow4\)
Chưa có ai trả lời câu hỏi này, hãy gửi một câu trả lời để giúp Nguyễn Hải Đăng giải bài toán này.
\(A=\frac{\frac{1}{2}:\left(\frac{1}{3}\right)^2\cdot\left(\frac{3}{2}\right)^2}{-0,75:\left(\frac{1}{4}\right)^2\cdot\left(\frac{4}{3}\right)^3}\)
\(=\frac{\frac{81}{8}}{-\frac{256}{9}}=-\frac{729}{2048}\)
Bài 2:
\(\left(\frac{-2}{3}\right)^3:\frac{3}{4}+\left(\frac{-2}{3}\right)^4:\left(\frac{3}{2}\right)^2\)
\(=\left(\frac{-2}{3}\right)^3\cdot\frac{4}{3}+\left[\left(\frac{-2}{3}\right)^3\cdot\frac{4}{3}\right]\cdot\frac{-2}{3}\cdot\frac{1}{3}\)
\(=\frac{-32}{81}+\frac{-32}{81}\cdot\frac{-2}{9}\)
\(=\frac{-32}{81}\left(1+\frac{-2}{9}\right)=\frac{-32}{81}\cdot\frac{7}{9}=-\frac{224}{729}\)
Bài 3:
Xét 2 trường hợp:
TH1: \(\text{3-2x=0}\Rightarrow x=\frac{3}{2}\)(thỏa mãn)
TH2: \(x=\frac{1}{2}\)(thỏa mãn)
Bài 4:
Điều kiện: \(y\ge\frac{1}{3}:2=\frac{1}{6}\)
Xét \(\frac{1}{6}\le y\le\frac{1}{2}\) ta có:
\(\frac{1}{2}-y=2y-\frac{1}{3}\Rightarrow3y=\frac{5}{6}\Rightarrow y=\frac{5}{18}\)(chọn)
\(\Rightarrow y^3=\frac{125}{5832}\)
Xét \(y>\frac{1}{2}\)ta có:
\(y-\frac{1}{2}=2y-\frac{1}{3}\Rightarrow y=\frac{-1}{6}\) (loại)
\(\Rightarrow y^3=-\frac{1}{216}\)
1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)
=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)
b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c) TT
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)
=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)
=> \(\left|50x-140\right|=\left|25x+24\right|\)
=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)
Bài 2 : a. |2x - 5| = x + 1
TH1 : 2x - 5 = x + 1
=> 2x - 5 - x = 1
=> 2x - x - 5 = 1
=> 2x - x = 6
=> x = 6
TH2 : -2x + 5 = x + 1
=> -2x + 5 - x = 1
=> -2x - x + 5 = 1
=> -3x = -4
=> x = 4/3
Ba bài còn lại tương tự
a. 3,5- / 2x-1/2 /=-0,75
=>/ 2x-1/2 /=4,25
=> 2x- 1/2= 4,25 hoặc -4,25
+) nếu 2x-1/2 = 4,25=> x=2,375
+) nếu 2x- 1/2=-4,25=> x=-1,875
vậy x= 2,375 hoặc x=-1,875
b. /x-2/ -4=1
=> /x-2/=5
=> x-2= 5 hoặc-5
+) nếu x-2=5=>x=7
+) nếu x-2=-5=>x=-3
vậy x=7 hoặc x=-3