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MC
15 tháng 7 2021
a) \(\left(\frac{4}{13}.\frac{6}{5}+\frac{4}{13}.\frac{2}{5}\right).\left(2x+1\right)^2=\frac{10}{13}\)
\(\left(\frac{4}{13}.\frac{8}{5}\right).\left(2x+1\right)^2=\frac{10}{13}\)
\(\frac{32}{65}.\left(2x+1\right)^2=\frac{10}{13}\)
\(\left(2x+1\right)^2=\frac{10}{13}\div\frac{32}{65}\)
\(\left(2x+1\right)^2=\frac{25}{16}\)
\(\Rightarrow2x+1\in\left\{\frac{5}{4};-\frac{5}{4}\right\}\)
\(\hept{\begin{cases}2x+1=\frac{5}{4}\\2x+1=-\frac{5}{4}\end{cases}\Rightarrow\hept{\begin{cases}2x=\frac{1}{4}\\2x=-\frac{9}{4}\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1}{8}\\x=-\frac{9}{8}\end{cases}}}\)
Vậy \(x\in\left\{\frac{1}{8};-\frac{9}{8}\right\}\)
MC
15 tháng 7 2021
\(x^3-\frac{9}{16}.x=0\)
\(x\left(x^2-\frac{9}{16}\right)=0\)
\(\hept{\begin{cases}x=0\\x^2-\frac{9}{16}=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x^2=\frac{9}{16}\end{cases}\Rightarrow}\hept{\begin{cases}x=0\\x=\pm\frac{3}{4}\end{cases}}}\)
Vậy \(x\in\left\{0;\frac{3}{4};-\frac{3}{4}\right\}\)
BT
2 tháng 6 2017
a) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)
\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
b) \(\dfrac{39}{7}:x=13\)
\(x=\dfrac{\dfrac{39}{7}}{13}=\dfrac{3}{7}\)
c) \(\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)
\(\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)
\(\dfrac{14}{5}x=34+50=84\)
\(x=\dfrac{84}{\dfrac{14}{5}}=30\)
d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
e) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)
\(\dfrac{1}{6}x=\dfrac{5}{12}\)
\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)
g) \(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\dfrac{11}{5}-\dfrac{3}{7}=-2\)
\(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)
\(x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{11}{7}:\dfrac{11}{5}=-\dfrac{5}{7}\)
\(\dfrac{44}{7}x=-\dfrac{5}{7}-\dfrac{3}{7}=-\dfrac{8}{7}\)
\(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)
h) \(\dfrac{13}{4}x+\left(-\dfrac{7}{6}\right)x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{25}{12}\)
\(x=1\)
Mỏi tay woa bn làm nốt nha!!
9 tháng 2 2021
l) (x + 9) . (x2 – 25) = 0
<=> (x + 9) . (x – 5) . (x + 5) = 0
<=> \(\left[{}\begin{matrix}\text{x + 9 = 0}\\x-5=0\\x+5=0\end{matrix}\right.\left[{}\begin{matrix}x=-9\\x=5\\x=-5\end{matrix}\right.\)
Vậy S = \(\left\{-9,5,-5\right\}\)
e) |x - 4 |< 7
<=> \(\left[{}\begin{matrix}x-4=7\\x-4=-7\end{matrix}\right.< =>\left[{}\begin{matrix}x=11\\x=-3\end{matrix}\right.\)
Vậy S = \(\left\{11;-3\right\}\)
9 tháng 2 2021
I,(x+9).(x^2-25)=0
tương đương:x+9=0
x^2-25=0
tương đương : x=-9
x=5
e,\(\left|x-4\right|\)=7
tương đương x-4=4
x-4=-4
tương đương :x=0
x=-8
4 tháng 1 2019
a) 89-(73-x) = 20
=> 73+x =89-20
=> 73+x=69
=> x=69-73
=> x=-4
b) (x+7)-25=13
=> x+7=13+25
=> x+7=38
=> x=38-7
=> x=31
c) 98-(x+4)=20
=> x+4=98-20
=> x+4=78
=> x=78-4
=> x=73
4 tháng 1 2019
d) 140:(x-8)=7
=> x-8=140:7
=> x-8=20
=> x=20+8
=> x=28
e) 4(x+41)=400
=> x+41=400:4
=> x+41=100
=> x=100-41
=> x=59
f) x-[42+(-28)]=-8
=> x-14=-8
=> x=-8+14
=> x=6
30 tháng 5 2021
\(1)x+\frac{5}{6}\times2\frac{2}{5}-1\frac{1}{4}=35\%\)
\(x+\frac{5}{6}\times\frac{12}{5}-\frac{5}{4}=\frac{7}{12}\)
\(x+\frac{5}{6}\times\frac{12}{5}=\frac{7}{12}+\frac{5}{4}\)
\(x+\frac{5}{6}.\frac{12}{5}=\frac{8}{5}\)
\(x+\frac{5}{6}=\frac{8}{5}:\frac{12}{5}\)
\(x+\frac{5}{6}=\frac{2}{3}\)
\(x=\frac{2}{3}-\frac{5}{6}\)
\(x=-\frac{1}{6}\)
HỌC TỐT !
O
30 tháng 5 2021
\(2\)) \(\left|x-\frac{1}{2}\right|-\frac{3}{4}=0\)
\(\left|x-\frac{1}{2}\right|\) \(=0+\frac{3}{4}\)
\(\left|x-\frac{1}{2}\right|\) \(=\frac{3}{4}\)
\(x-\frac{1}{2}\) \(=\frac{3}{4}\)hoặc \(-\frac{3}{4}\)
Ta xét 2 trường hợp :
Trường hợp 1 : \(x-\frac{1}{2}=\frac{3}{4}\)
\(x\) \(=\frac{3}{4}+\frac{1}{2}\)
\(x\) \(=\frac{5}{4}\)
Trường hợp 2 : \(x-\frac{1}{2}=-\frac{3}{4}\)
\(x\) \(=-\frac{3}{4}+\frac{1}{2}\)
\(x\) \(=-\frac{1}{4}\)
Vậy \(x\in\text{{}\frac{5}{4};-\frac{1}{4}\)}
5 tháng 4 2021
Bài 1:
Ta có: \(x-35\%\cdot x=\dfrac{1}{25}\)
\(\Leftrightarrow65\%\cdot x=\dfrac{1}{25}\)
\(\Leftrightarrow x=\dfrac{1}{25}:\dfrac{13}{20}=\dfrac{1}{25}\cdot\dfrac{20}{13}=\dfrac{4}{65}\)
Vậy: \(x=\dfrac{4}{65}\)
5 tháng 4 2021
Bài 2:
a) Ta có: \(17\dfrac{2}{31}-\left(\dfrac{15}{17}+6\dfrac{2}{31}\right)\)
\(=17\dfrac{2}{31}-\dfrac{15}{17}-6\dfrac{2}{31}\)
\(=11+\dfrac{2}{31}-\dfrac{15}{17}\)
\(=\dfrac{5366}{527}\)
\(a,-\dfrac{4}{7}-x=\dfrac{3}{5}-2x\\ \Leftrightarrow x=\dfrac{41}{35}\)
\(b,\dfrac{5}{7}-\dfrac{1}{13}+\dfrac{1}{4}=\dfrac{31}{2}-x\\ \Leftrightarrow x=\dfrac{5319}{364}\)