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GN
GV Nguyễn Trần Thành Đạt
Giáo viên
6 tháng 1
Bài 2:
\(a,\dfrac{-1}{3}.\dfrac{5}{7}=\dfrac{-5}{21}\\ b,\dfrac{1}{2}.\dfrac{-3}{4}=\dfrac{-3}{8}\\ c,\dfrac{19}{7}.\dfrac{7}{15}=\dfrac{19.1}{1.15}=\dfrac{19}{15}\\ d,\dfrac{5}{11}:\left(-9\right)=\dfrac{5}{11}.\dfrac{-1}{9}=\dfrac{-5}{99}\\ e,-1:\dfrac{3}{5}=-1.\dfrac{5}{3}=-\dfrac{5}{3}\\ f,\dfrac{-2}{9}:\dfrac{2}{9}:\dfrac{-9}{14}=-\left(\dfrac{2}{9}:\dfrac{2}{9}\right).\dfrac{-14}{9}=-1.\dfrac{-14}{9}=\dfrac{14}{9}\\ k,\left(-\dfrac{2}{7}\right)^2=\left(-1\right)^2.\left(\dfrac{2}{7}\right)^2=1.\dfrac{2^2}{7^2}=\dfrac{4}{49}\\ l,\dfrac{1}{3}.\left(\dfrac{1}{3}\right)^3=\left(\dfrac{1}{3}\right)^4=\dfrac{1^4}{3^4}=\dfrac{1}{243}\\ m,\left(-\dfrac{1}{2}\right)^2.\left(\dfrac{1}{2}\right)^3=\left(-1\right)^2.\left(\dfrac{1}{2}\right)^{2+3}=1.\left(\dfrac{1}{2}\right)^5=1.\dfrac{1^5}{2^5}=\dfrac{1}{32}\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
6 tháng 1
Bài 1:
\(a,\dfrac{1}{-8}+\dfrac{-5}{8}=\dfrac{-1}{8}+\dfrac{-5}{8}=\dfrac{-\left(1+5\right)}{8}=-\dfrac{6}{8}=\dfrac{-6:2}{8:2}=-\dfrac{3}{4}\\ b,\dfrac{1}{7}+\dfrac{-3}{7}=\dfrac{1-3}{7}=-\dfrac{2}{7}\\ c,\dfrac{-12}{35}+\dfrac{-7}{35}=\dfrac{-\left(12+7\right)}{35}=\dfrac{-19}{35}\\ d,\dfrac{1}{6}+\dfrac{2}{5}=\dfrac{1.5+2.6}{6.5}=\dfrac{5+12}{30}=\dfrac{17}{30}\\ e,\dfrac{3}{5}+\dfrac{-7}{4}=\dfrac{3.4-7.5}{5.4}=\dfrac{12-35}{20}=\dfrac{-23}{20}\\ g,-2-\left(-\dfrac{1}{5}\right)=-2+\dfrac{1}{5}=\dfrac{-2.5+1}{5}=\dfrac{-9}{5}\\ h,\dfrac{2}{3}-\left(-1\right)=\dfrac{2}{3}+1=\dfrac{5}{3}\\ i,4-\dfrac{2}{3}=\dfrac{4.3-2}{3}=\dfrac{12-2}{3}=\dfrac{10}{3}\\ j,\dfrac{3}{4}-2=\dfrac{3-2.4}{4}=\dfrac{-5}{4}\\ k,-1-\left(-\dfrac{2}{3}\right)=-1+\dfrac{2}{3}=\dfrac{-1.3+2}{3}=\dfrac{-3+2}{3}=-\dfrac{1}{3}\)
DT
Đỗ Thanh Hải
CTVVIP
22 tháng 11 2021
16 C
17 A
18 B
19 A
20 C
21 B
22 B
23 C
24 B
25 D
26 D
27 B
28 C
29 A
30 D
31 C
9 tháng 2 2022
Ta có: \(7x+4=x-2m\left(1\right)\)
Thay \(x=6\) vào \(\left(1\right)\) ta có:
\(7.6+4=6-2m\)
\(\Rightarrow46=6-2m\)
\(\Rightarrow-2m=40\)
\(\Rightarrow m=-20\)
23 tháng 1
a: Xét ΔEBF và ΔECD có
\(\widehat{EBF}=\widehat{ECD}\)(hai góc so le trong, BF//CD)
\(\widehat{BEF}=\widehat{CED}\)(hai góc đối đỉnh)
Do đó: ΔEBF~ΔECD(2)
Xét ΔEBF và ΔDAF có
\(\widehat{F}\) chung
\(\widehat{EBF}=\widehat{DAF}\)(hai góc đồng vị, BE//AD)
Do đó: ΔEBF~ΔDAF(1)
Từ (1) và (2) suy ra ΔECD~ΔDAF
b: BE+CE=BC
=>BE+4=6
=>BE=2(cm)
Xét ΔFAD có BE//AD
nên \(\dfrac{FB}{FA}=\dfrac{EB}{AD}\)
=>\(\dfrac{FB}{BF+15}=\dfrac{2}{6}=\dfrac{1}{3}\)
=>\(3BF=BF+15\)
=>2BF=15
=>BF=7,5(cm)
AF=AB+BF=15+7,5=22,5(cm)
c: Ta có: ΔECD~ΔDAF
=>\(\dfrac{EC}{DA}=\dfrac{DE}{DF}\)
=>\(EC\cdot DF=DE\cdot DA\)
Ta có: ΔECD~ΔDAF
=>\(\dfrac{CD}{AF}=\dfrac{EC}{DA}\)
=>\(EC\cdot AF=CD\cdot DA\)
giúp mình với, cảm ơn ạaa
DT
Đỗ Thanh Hải
CTVVIP
7 tháng 5 2021
1 I'd rather you didn't go out this Christmas
2 I'd rather the children went to bed
3 I'd rather you helped your mother with housework
4 I'd rather you didn't come to class late
5 I'd rather you kept silent in the classroom
It's time
1 It's time for you to dress yourself
2 It's about time the children went to bed
3 It's time we hurried up or we will be late for the train
4 It's high time children went to bed
5 It's about time for us to go home
16 tháng 1 2022
C
\(SO_2+H_2O⇌H_2SO_3\)
\(K_2O+H_2O\rightarrow2KOH\)
\(BaO+H_2O\rightarrow Ba\left(OH\right)_2\)
\(SO_3+H_2O\rightarrow H_2SO_4\)
KY
23 tháng 5 2022
where my father worked then
the good performance, she lost the match
, who sings Western folk songs very well, can compost songs
the first time I've ever read such an interesting book
Câu 22:
\(n_{CH_3COOH}=\dfrac{45}{60}=0,75\left(mol\right)\)
PT: \(CH_3COOH+C_2H_5OH⇌CH_3COOC_2H_5+H_2O\) (xt: H2SO4, to)
Theo PT: \(n_{CH_3COOC_2H_5\left(LT\right)}=n_{CH_3COOH}=0,75\left(mol\right)\)
Mà: H = 80%
\(\Rightarrow n_{CH_3COOC_2H_5\left(TT\right)}=0,75.80\%=0,6\left(mol\right)\)
\(\Rightarrow m_{CH_3COOC_2H_5\left(TT\right)}=0,6.88=52,8\left(g\right)\)
Câu 23:
mCaCO3 = 20.80% = 16 (g) \(\Rightarrow n_{CaCO_3}=\dfrac{16}{100}=0,16\left(mol\right)\)
PT: \(CaCO_3+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Ca+CO_2+H_2O\)
a, Theo PT: \(n_{CH_3COOH}=2n_{CaCO_3}=0,32\left(mol\right)\Rightarrow m_{CH_3COOH}=0,32.60=19,2\left(g\right)\)
b, \(n_{CO_2}=n_{CaCO_3}=0,16\left(mol\right)\Rightarrow V_{CO_2}=0,16.22,4=3,584\left(l\right)\)