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1a) 25 .8 - 12.5 + 272 :17 - 8
= 200 - 60 + 16 - 8
= 140+16-8
= 156 - 8
= 148
b) 125: 25 +14 - 142 : 71
= 5 + 14 - 2
= 19 - 2 = 17
c) 13. 17 - 256: 16 + 14 : 7 - 1
= 221 - 16 + 2 - 1
= 205 + 2 - 1
= 207 - 1 = 206
d) 15 . 24 - 14 . 5 (145 : 5 - 27)
= 360 - 70 (29- 27)
= 360 - 70 . 2
= 360 - 140
= 220
2a) 37 + x = 55
x= 55 - 37
x = 18
b) x- 31 = 24
x = 24 + 31
x = 55
c) 21 (x-11) = 21
x - 11 = 21 : 21
x = 1 +11
x = 12
d ) 9. (x - 29)= 0
x - 29 = 0 :9
x = 29
e) (x - 1954) . 5 = 50
x - 1954 = 50 : 5
x = 1964
f ) 30 . 60 - x = 30
1800 - x = 30
x = 1800 - 30
x = 1770
Viết mỏi tay quá huhuh
d) Ta có: \(32\%-0.25:x=-\dfrac{17}{5}\)
\(\Leftrightarrow0.25:x=\dfrac{8}{25}+\dfrac{17}{5}=\dfrac{93}{25}\)
hay \(x=\dfrac{25}{372}\)
Vậy: \(x=\dfrac{25}{372}\)
e) Ta có: \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{2}{5};-\dfrac{4}{5}\right\}\)
f) Ta có: \(-\dfrac{32}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)
\(\Leftrightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-8}{27}\)
\(\Leftrightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)
\(\Leftrightarrow3x=\dfrac{1}{9}\)
hay \(x=\dfrac{1}{27}\)
g) Ta có: \(60\%\cdot x+0.4x+x:3=2\)
\(\Leftrightarrow\dfrac{4}{3}x=2\)
hay \(x=\dfrac{3}{2}\)
Vậy: \(x=\dfrac{3}{2}\)
h) PT \(\Leftrightarrow\left|\dfrac{20}{9}-x\right|=\dfrac{2}{9}\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{20}{9}-x=\dfrac{2}{9}\\x-\dfrac{20}{9}=\dfrac{2}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{22}{9}\end{matrix}\right.\)
Vậy ...
i) PT \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}x=\dfrac{16}{5}\) \(\Leftrightarrow\dfrac{2}{5}x=\dfrac{8}{5}\) \(\Leftrightarrow x=4\)
Vậy ...
a) x=50-37
x= 13
b) 2.x=11+3
2.x=14
x=14:2=7
c) 2+x=6.5
2.x=30
x=30:2=15
e) \(^{^{^{ }}x^9}\)=25
\(\Rightarrow\)x=5
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!!
\(a)9.4.25.8.125\)
\(=9.\left(4.25\right).\left(8.125\right)\)
\(=9.100.1000\)
\(=900.1000\)
\(=900000\)
\(a,9.4.25.8.125=9.\left(4.25\right).\left(8.125\right)=9.100.1000=900000\)
\(b,25.5.27.2\)(câu này hình như thiếu đề, ko tính nhanh đc)
\(c,37.7+80.3+43.7\)\(=7.\left(37+43\right)+80.3\)\(=7.80+3.80=80\left(7+3\right)=80.10=800\)
\(d,113.38+113.62+87.62+87.38\)\(=113\left(38+62\right)+87\left(62+38\right)\)
\(=113.100+87.100=100.\left(113+87\right)=100.200=20000\)
Viết gọn lũy thừa:
\(a,x.x.y.y.x.y.x=\left(x.x.x.x\right).\left(y.y.y\right)=x^4.y^4\)
\(b,7^5:343=7^5:7^3=7^{5-3}=7^2\)
\(c,A^{12}:A^{18}=A^{12-18}=A^{-6}\)
\(d,x^7.x^4.x=x^{7+4+1}=x^{12}\)
2 bài này dễ mà bn . bn áp dụng vào công thức này mà làm nhé !
- Khi bỏ dấu ngoặc có dấu "-" đằng trước, ta phải đổi dấu tất cả các số hạng trong dấu ngoặc: dấu "+" thành dấu "-" và dấu "-" thành dấu "+".
- Khi bỏ dấu ngoặc có dấu "+" đằng trước thì dấu các số hạng trong ngoặc vẫn giữ nguyên.