Giải hộ em vs ạ!
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O
16 tháng 3 2022
Bài 1
a) \(\dfrac{3}{8}+\dfrac{5}{8}=\dfrac{8}{8}=1\)
b) \(\dfrac{3}{4}+\dfrac{-7}{16}=\dfrac{12}{16}+\dfrac{-7}{16}=\dfrac{5}{16}\)
c) \(2\dfrac{17}{20}-\dfrac{1}{2}+3\dfrac{3}{20}=\dfrac{57}{20}-\dfrac{1}{2}+\dfrac{63}{20}\)\(=\dfrac{47}{20}+\dfrac{63}{20}=\dfrac{110}{20}=\dfrac{11}{2}\)
d) \(\dfrac{2}{3}-2\dfrac{1}{8}+\dfrac{7}{24}=\dfrac{2}{3}-\dfrac{17}{8}+\dfrac{7}{24}=\dfrac{16}{24}-\dfrac{51}{24}+\dfrac{7}{24}=\dfrac{16-51+7}{24}=\dfrac{-28}{24}=\dfrac{-7}{6}\)
Bài 2 :
a) \(x-\dfrac{7}{4}=3\)
\(x=3+\dfrac{7}{4}\)
\(x=\dfrac{19}{4}\)
b) \(x-\dfrac{1}{2}=\dfrac{4}{16}\cdot\dfrac{8}{3}\)
\(x-\dfrac{1}{2}=\dfrac{2}{3}\)
\(x=\dfrac{2}{3}+\dfrac{1}{2}\)
\(x=\dfrac{5}{6}\)
c) \(\dfrac{15}{11}\div x=\dfrac{45}{22}\)
\(x=\dfrac{15}{11}\div\dfrac{45}{22}\)
\(x=\dfrac{2}{3}\)
d) \(\dfrac{8}{3}-2x=\dfrac{8}{5}-1\)
\(\dfrac{8}{3}-2x=\dfrac{3}{5}\)
\(2x=\dfrac{8}{3}-\dfrac{3}{5}\)
\(2x=\dfrac{31}{15}\)
\(x=\dfrac{31}{15}\div2\)
\(x=\dfrac{31}{30}\)
DT
Đỗ Thanh Hải
CTVVIP
16 tháng 4 2021
1 that => which
2 he => who
3 hardly => hard
4 quick => quickly
5 turning off => turning on
6 to => and
7 For => therefore
8 therefore => so
9 to have => having
10 to swim => swimming
11 will go => go
12 don't => won't
13 was => is
14 for => since
15 didn't => haven't
25 tháng 9 2021
\(v_o=36km/h=10m/s\\ v=54km/h=15m/s\\ s=625m\)
a) Gia tốc của xe là:
\(v^2-v_o^2=2as\rightarrow a=\dfrac{v^2-v_o^2}{2s}=\dfrac{15^2-10^2}{2.625}=0,1\left(m/s\right)\)
b) Thời gian tăng tốc:
\(a=\dfrac{v-v_o}{t}\rightarrow t=\dfrac{v-v_o}{a}=\dfrac{15-10}{0,1}=50\left(s\right)\)
14 tháng 2 2022
1A 2D 3A 4B 5A 6A 7D 8C 9D 10B 11A 12D 13A 14B 15C 16A 17A 18D 19A
Bài 1:
a) Ta có: \(A=x^2-2x+7\)
\(=x^2-2x+1+6\)
\(=\left(x-1\right)^2+6\ge6\forall x\)
Dấu '=' xảy ra khi x=1
b) Ta có: \(B=5x^2-20x\)
\(=5\left(x^2-4x+4-4\right)\)
\(=5\left(x-2\right)^2-20\ge-20\forall x\)
Dấu '=' xảy ra khi x=2
d) Ta có: \(D=4x^2+4x+11\)
\(=4x^2+4x+1+10\)
\(=\left(2x+1\right)^2+10\ge10\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{1}{2}\)
Bài 2:
a) Ta có: \(A=-x^2+10x-2\)
\(=-\left(x^2-10x+2\right)\)
\(=-\left(x^2-10x+25-23\right)\)
\(=-\left(x-5\right)^2+23\le23\forall x\)
Dấu '=' xảy ra khi x=5
b) Ta có: \(B=-2x^2+2x+3\)
\(=-2\left(x^2-x-\dfrac{3}{2}\right)\)
\(=-2\left(x^2-x+\dfrac{1}{4}-\dfrac{7}{4}\right)\)
\(=-2\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{2}\le\dfrac{7}{2}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)