Hộ em vs ạ :(
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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) \(\dfrac{8}{3}-2x=\dfrac{8}{5}-1\)
\(\dfrac{8}{3}-2x=\dfrac{3}{5}\)
\(2x=\dfrac{8}{3}-\dfrac{3}{5}\)
\(2x=\dfrac{31}{15}\)
\(x=\dfrac{31}{15}\div2\)
\(x=\dfrac{31}{30}\)
1. How much is this watch?
2. How much are these beautiful scarves?
3. Can you tell me how to get to Dong Nai post office?
4. Can you tell me how to get to the station?
5. Peter works hard
6. There aren't any bottles on the shelf
7. We don't have time to prepare the speech
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 \(x=\dfrac{1}{2}\)
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
\(v_o=36km/h=10m/s\\ v=54km/h=15m/s\\ s=625m\)
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:
\(a=\dfrac{v-v_o}{t}\rightarrow t=\dfrac{v-v_o}{a}=\dfrac{15-10}{0,1}=50\left(s\right)\)
\(-x^2+4=0\\ \Leftrightarrow-x^2=-4\\ \Leftrightarrow x^2=4\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
\(a,A\left(x\right)=x^2+3x^4-4x+7+x^4\)
\(=x^2+4x^4-4x+7\)
Sắp xếp : \(4x^4+x^2-4x+7\)
\(B\left(x\right)=x^4-2x^2+\left(1-5x^4+4x-4\right)\)
\(=x^4-2x^2+1-5x^4+4x-4\)
\(=-4x^4-2x^2+4x-3\) ( Đã sắp xếp )
\(b,\) \(C\left(x\right)-A\left(x\right)=B\left(x\right)\)
\(\Rightarrow C\left(x\right)=A\left(x\right)+B\left(x\right)\)
\(=4x^4+x^2-4x+7-4x^4-2x^2+4x-3\)
\(=-x^2+4\)
Đặt \(C\left(x\right)=0\Rightarrow-x^2+4=0\Rightarrow x^2=-4\left(ktm\right)\)
Vậy C(x) vô nghiệm
42
bố 42 con 12