9/10 - ( 2/5 - 3/10 ) + 7/20.Mọi người giúp mình với ạ !
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GN
GV Nguyễn Trần Thành Đạt
Giáo viên
10 tháng 8 2023
\(\dfrac{21}{36}-\left(-\dfrac{11}{30}\right)=\dfrac{7}{12}+\dfrac{11}{30}=\dfrac{7.5+11.2}{60}=\dfrac{57}{60}=\dfrac{19}{20}\\ ----\\\dfrac{-4}{8}+\left(-\dfrac{3}{10}\right)=\dfrac{-1}{2}-\dfrac{3}{10}=\dfrac{-1.5-3}{10}=\dfrac{-8}{10}=-\dfrac{4}{5}\\ ----\\ \dfrac{7}{12}-\left(-\dfrac{9}{20}\right)=\dfrac{7}{12}+\dfrac{9}{20}=\dfrac{7.5+9.3}{60}=\dfrac{62}{60}=\dfrac{31}{30}\\ ---\\ \dfrac{-2}{5}+\left(-\dfrac{11}{30}\right)=-\dfrac{2}{5}-\dfrac{11}{30}=\dfrac{-2.6-11}{30}=-\dfrac{29}{30}\)
23 tháng 7 2023
`@` `\text {Ans}`
`\downarrow`
`a)`
\(\dfrac{7}{5}\cdot\dfrac{8}{19}+\dfrac{7}{5}\cdot\dfrac{12}{19}-\dfrac{7}{5}\cdot\dfrac{1}{19}\)
`=`\(\dfrac{7}{5}\cdot\left(\dfrac{8}{19}+\dfrac{12}{19}-\dfrac{1}{19}\right)\)
`=`\(\dfrac{7}{5}\cdot\dfrac{19}{19}=\dfrac{7}{5}\cdot1=\dfrac{7}{5}\)
`b)`
\(-\dfrac{3}{5}\cdot\dfrac{5}{7}+\left(-\dfrac{3}{5}\right)\cdot\dfrac{3}{7}+\left(-\dfrac{3}{5}\right)\cdot\dfrac{6}{7}\)
`=`\(-\dfrac{3}{5}\cdot\left(\dfrac{5}{7}+\dfrac{3}{7}+\dfrac{6}{7}\right)\)
`=`\(-\dfrac{3}{5}\cdot\dfrac{14}{7}\)
`=`\(-\dfrac{3}{5}\cdot2=-\dfrac{6}{5}\)
`c)`
\(10\dfrac{2}{9}+\left(2\dfrac{2}{5}-7\dfrac{2}{9}\right)\)
`=`\(10\dfrac{2}{9}+2\dfrac{2}{5}-7\dfrac{2}{9}\)
`=`\(\left(10\dfrac{2}{9}-7\dfrac{2}{9}\right)+2\dfrac{2}{5}\)
`=`\(3+2\dfrac{2}{5}=\dfrac{27}{5}\)
`d)`
\(6\dfrac{3}{10}-\left(3\dfrac{4}{7}+2\dfrac{3}{10}\right)\)
`=`\(6\dfrac{3}{10}-3\dfrac{4}{7}-2\dfrac{3}{10}\)
`=`\(\left(6\dfrac{3}{10}-2\dfrac{3}{10}\right)-3\dfrac{4}{7}\)
`=`\(4-3\dfrac{4}{7}=\dfrac{3}{7}\)
VN
Võ Ngọc Phương
VIP
23 tháng 7 2023
a) \(\dfrac{7}{5}.\dfrac{8}{19}+\dfrac{7}{5}.\dfrac{12}{19}-\dfrac{7}{5}.\dfrac{1}{19}\)
\(=\dfrac{7}{5}.\left(\dfrac{8}{19}+\dfrac{12}{19}-\dfrac{1}{19}\right)\)
\(=\dfrac{7}{5}.1\)
\(=\dfrac{7}{5}\)
b) \(\dfrac{-3}{5}.\dfrac{5}{7}+\dfrac{-3}{5}.\dfrac{3}{7}+\dfrac{-3}{5}.\dfrac{6}{7}\)
\(=\dfrac{-3}{5}.\left(\dfrac{5}{7}+\dfrac{3}{7}+\dfrac{6}{7}\right)\)
\(=\dfrac{-3}{5}.2\)
\(=\dfrac{-6}{5}\)
c) \(10\dfrac{2}{9}+\left(2\dfrac{2}{5}-7\dfrac{2}{9}\right)\)
\(=\dfrac{92}{9}+\dfrac{12}{5}-\dfrac{65}{9}\)
\(=\dfrac{92}{9}-\dfrac{65}{9}+\dfrac{12}{5}\)
\(=3+\dfrac{12}{5}\)
\(=\dfrac{15}{5}+\dfrac{12}{5}\)
\(=\dfrac{27}{5}\)
d) \(6\dfrac{3}{10}-\left(3\dfrac{4}{7}+2\dfrac{3}{10}\right)\)
\(=\dfrac{63}{10}-\dfrac{25}{7}-\dfrac{23}{10}\)
\(=\dfrac{63}{10}-\dfrac{23}{10}-\dfrac{25}{7}\)
\(=4-\dfrac{25}{7}\)
\(=\dfrac{28}{7}-\dfrac{25}{7}\)
\(=\dfrac{3}{7}\)
Chúc bạn học tốt
16 tháng 8 2023
a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)
\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)
\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)
\(\Rightarrow x=\dfrac{-5}{12}\)
b) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\dfrac{-1}{3}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)
\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{2}{5}\)
c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)
\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-77}{120}\)
d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)
\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{-7}{20}\)
e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)
\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-59}{105}\)
16 tháng 8 2023
g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-13}{12}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
21 tháng 2 2024
-x + 20 = - (-15) - (8) + 13
-x + 20 = 15 - 8 + 13
-x + 20 = 7 + 13
- x + 20 = 20
x = 20 - 20
x = 0
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
21 tháng 2 2024
-(-10) + x = -13 + (-9) + (-6)
10 + x = -13 - 9 - 6
10 + x = -28
x = -28 - 10
x = -38
BA
5 tháng 10 2019
\(7^{13}:49^2=7^{13}:7^4=7^9\)
\(27^{16}:9^{10}=3^{48}:3^{20}=3^{28}\)
\(5^{20}\cdot9^{10}=5^{20}\cdot3^{20}=15^{20}\)
\(7^7\cdot13+7^7\cdot36=7^7\cdot\left(13+36\right)=7^7\cdot49=7^7\cdot7^2=7^9\)
\(5^{12}\cdot37-5^{12}\cdot12=5^{12}\cdot\left(37-12\right)=5^{12}\cdot25=5^{12}\cdot5^2=5^{14}\)
MH
29 tháng 12 2015
1. -x+20 = -(-15)-8+13
=> -x=15-8+13-20
=> -x=0
=> x=0
2. -(-10)+x=-13+(-9)+(-6)
=> 10+x=-13-9-6
=> x = -13-9-6-10
=> x = -38
3. 8-(-12)+10=-(-14)-x
=> 8+12+10=14-x
=> x = 14-8-12-10
=> x = -16
4. -(+12)+(-x)-(-3)=5-(-7)
=> -12-x+3=5+7
=> -x=5+7+12-3
=> -x=21
=> x=-21
5. 14-x+(-10)=-(-9)+(+15)
=> 14-x-10=9+15
=> -x=9+15-14+10
=> -x=20
=> x=-20
6. 12-(-17)+(-3)=-5+x
=> 12+17-3+5=x
=> x=31
7. x-(-19)-(+32)=14-(+16)
=> x+19-32=14-16
=> x=14-16+32-19
=> x=11
8. x-|-15|-|7|=-(-9)+|-5|
=> x-15-7=9+5
=> x=9+5+7+15
=> x=36
9. 15-x+17=13-(-21)
=> 15-x+17=13+21
=> -x=13+21-15-17
=> -x=2
=> x=-2
10. -|-5|-(-x)+4=3-(-25)
=> -5+x+4=3+25
=> x=3+25-4+5
=> x=29
3 tháng 9 2018
\(Q_{\left(x\right)}=x^{14}-10x^{13}+10x^{12}-10x^{11}+...+10x^2-10x+10\)
\(=x^{14}-\left(x+1\right)x^{13}+\left(x+1\right)x^{12}-\left(x+1\right)x^{11}+..+\left(x+1\right)x^2-\left(x+1\right)x+x+1\)
\(=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-x^{12}-x^{11}+...+x^3+x^2-x^2-x+x+1\)
\(=1\)
T
7 tháng 7 2020
\(a.P(x)=x^7-80x^6+80x^5-80x^4+....+80x+15\)
\(=x^7-79x^6-x^6+79x^5+x^5-79x^4-....-x^2+79x+x+15\)
\(=x^6(x-79)-x^5(x-79)+x^4(x-79)-....-x(x-79)+x+15\)
\(=(x-79)(x^6-x^5+x^4-....-x)+x+15\)
Thay x = 79 vào biểu thức trên , ta có
\(P(79)=(79-79)(79^6-79^5+79^4-...-79)+79+15\)
\(=0+79+15\)
\(=94\)
Vậy \(P(x)=94\)khi x = 79
\(b.Q(x)=x^{14}-10x^{13}+10x^{12}-.....+10x^2-10x+10\)
\(=x^{14}-9x^{13}-x^{13}+9x^{12}+.....-x^3+9x^2+x^2-9x-x+10\)
\(=x^{13}(x-9)-x^{12}(x-9)+.....-x^2(x-9)+x(x-9)-x+10\)
\(=(x-9)(x^{13}-x^{12}+.....-x^2+x)-x+10\)
Thay x = 9 vào biểu thức trên , ta có
\(Q(9)=(9-9)(9^{13}-9^{12}+.....-9^2+9)-9+10\)
\(=0-9+10\)
\(=1\)
Vậy \(Q(x)=1\)khi x = 9
\(c.R(x)=x^4-17x^3+17x^2-17x+20\)
\(=x^4-16x^3-x^3+16x^2+x^2-16x-x+20\)
\(=x^3(x-16)-x^2(x-16)+x(x-16)-x+20\)
\(=(x-16)(x^3-x^2+x)-x+20\)
Thay x = 16 vào biểu thức trên , ta có
\(R(16)=(16-16)(16^3-16^2+16)-16+20\)
\(=0-16+20\)
\(=4\)
Vậy \(R(x)=4\)khi x = 16
\(d.S(x)=x^{10}-13x^9+13x^8-13x^7+.....+13x^2-13x+10\)
\(=x^{10}-12x^9-x^9+12x^8+.....+x^2-12x-x+10\)
\(=x^9(x-12)-x^8(x-12)+....+x(x-12)-x+10\)
\(=(x-12)(x^9-x^8+....+x)-x+10\)
Thay x = 12 vào biểu thức trên , ta có
\(S(12)=(12-12)(12^9-12^8+....+12)-12+10\)
\(=0-12+10\)
\(=-2\)
Vậy \(S(x)=-2\)khi x = 12
Hình như đây là toán lớp 7 có trong phần trắc nghiệm của thi HSG huyện
Chúc bạn học tốt , nhớ kết bạn với mình
E
15
(9 / 10) - ((2 / 5) - (3 / 10)) + (7 / 20) =
1,15