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a,=11/12 x 8/15 - 1/6:2
= 22/45-1/2
= 73/180
b, =9/4:0.01
= 225
K nha
\(\frac{2}{3}\cdot x-\frac{1}{4}\left(1\frac{1}{3}-\frac{3}{4}\right)=0,25\cdot\frac{2}{3}\)
\(\Rightarrow\frac{2}{3}\cdot x-\frac{1}{4}\left(\frac{4}{3}-\frac{3}{4}\right)=\frac{1}{4}\cdot\frac{2}{3}\)
\(\Rightarrow\frac{2}{3}\cdot x-\frac{1}{4}\cdot\frac{7}{12}=\frac{1}{6}\)
\(\Rightarrow\frac{2}{3}.x-\frac{7}{48}=\frac{1}{6}\)
\(\Rightarrow\frac{2}{3}x=\frac{5}{16}\)
\(\Rightarrow x=\frac{15}{32}\)
\(\frac{2}{3}\times x-\frac{1}{4}:\left(1\frac{1}{3}-\frac{3}{4}\right)=0,25\times\frac{2}{3}\)
\(\frac{2}{3}\times x-\frac{1}{4}:\left(\frac{4}{3}-\frac{3}{4}\right)=\frac{1}{4}\times\frac{2}{3}\)
\(\frac{2}{3}\times x-\frac{1}{4}:\frac{7}{12}=\frac{1}{6}\)
\(\frac{2}{3}\times x-\frac{3}{7}=\frac{1}{6}\)
\(\frac{2}{3}\times x=\frac{1}{6}+\frac{3}{7}\)
\(\frac{2}{3}\times x=\frac{25}{42}\)
\(x=\frac{25}{42}:\frac{2}{3}\)
\(x=\frac{25}{28}\)
\(x-0.25=-\frac{6}{11}\cdot\left(\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right)\)
\(x-\frac{1}{4}=-\frac{6}{11}\cdot\left(\frac{6}{12}+\frac{9}{12}-\frac{4}{12}\right)\)
\(x-\frac{1}{4}=-\frac{6}{11}\cdot\frac{11}{12}\)
\(x-\frac{1}{4}=-\frac{1}{2}\)
\(x=-\frac{1}{2}+\frac{1}{4}\)
\(x=-\frac{2}{4}+\frac{1}{4}\)
\(x=-\frac{1}{4}\)
Vậy .......
\(x-0,25=-\frac{6}{11}.\left(\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right)\)
\(x-0,25=-\frac{6}{11}.\frac{11}{12}\)
\(x-0,25=-\frac{1}{2}\)
\(x=-\frac{1}{2}+0,25=-\frac{1}{4}\)
học tốt ~~~
\(\frac{3}{4}+\frac{1}{4}\cdot x+x-\frac{7}{6}\cdot x=\frac{5}{12}\)
\(\frac{3}{4}+\frac{1}{4}\cdot x-\frac{7}{6}\cdot x+x\cdot1=\frac{5}{12}\)
\(\frac{3}{4}+x\left(\frac{1}{4}-\frac{7}{6}+1\right)=\frac{5}{12}\)
\(\frac{3}{4}+x\cdot\frac{1}{12}=\frac{5}{12}\)
\(x\cdot\frac{1}{12}=\frac{5}{12}-\frac{3}{4}\)
\(x\cdot\frac{1}{12}=\frac{5}{12}-\frac{9}{12}\)
\(x\cdot\frac{1}{12}=\frac{-1}{3}\)
\(x=\frac{-1}{3}\text{ : }\frac{1}{12}\)
\(x=\frac{-1}{3}\cdot12\)
\(x=\frac{-12}{3}\)
\(x=-4\)
\(\text{b, }0,25\cdot x-\frac{2}{3}\cdot x=-1\frac{1}{6}\)
\(\frac{1}{4}\cdot x-\frac{2}{3}\cdot x=\frac{-7}{6}\)
\(x\cdot\left(\frac{1}{4}-\frac{2}{3}\right)=\frac{-7}{6}\)
\(x\cdot\frac{-5}{12}=\frac{-7}{6}\)
\(x=\frac{-7}{6}\text{ : }\frac{-5}{12}\)
\(x=\frac{-7}{6}\cdot\frac{12}{-5}\)
\(x=\frac{-14}{-5}\)
1) \(\left|4-2x\right|.\dfrac{1}{3}=\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}:\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}.3\)
\(\left|4-2x\right|=1\)
=>\(4-2x=\pm1\)
+)\(TH1:4-2x=1\) +)\(TH2:4-2x=-1\)
\(2x=4-1\) \(2x=4-\left(-1\right)\)
\(2x=3\) \(2x=4+1\)
\(x=3:2\) \(2x=5\)
\(x=1,5\) \(x=5:2\)
Vậy x=1,5 \(x=2,5\)
Vậy x=2,5
2) \(\left(-3\right)^2:\left|x+\left(-1\right)\right|=-3\)
\(9:\left|x+\left(-1\right)\right|=-3\)
\(\left|x+\left(-1\right)\right|=9:\left(-3\right)\)
\(\left|x+\left(-1\right)\right|=-3\)
=> \(x+\left(-1\right)\) sẽ không có giá trị nào ( Vì giá trị tuyệt đối luôn luôn lớn hơn hoặc bằng 0 )
Vậy x = \(\varnothing\)
Bài làm
|x-2/3|+0,25=3/4
|x-2/3|+1/4=3/4
|x-2/3| =3/4-1/4
|x-2/3| =2/4=1/2
|x| =1/2+2/3
|x| =7/6
x =-7/6 hoặc 7/6 !
|x-2/3| + 0,25 = 3/4
|x- 2/3| = 3/4 - 0,25
|x - 2/3| = 1/2
=> \(\orbr{\begin{cases}x-\frac{2}{3}=-\frac{1}{2}\\x-\frac{2}{3}=\frac{1}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{1}{2}+\frac{2}{3}\\x=\frac{1}{2}+\frac{2}{3}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{6}\\x=\frac{7}{6}\end{cases}}\)