giúp em giải bài này với ạ
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TT
2
DT
Đỗ Thanh Hải
CTVVIP
24 tháng 5 2021
1 The meeting was canceled 3 days ago
2 She told me she was watching a film with her sister then
3 I admire the guitarist who is perfroming on the stage
4 Had it not been for Pauline's interest, the project would have been abandoned
KN
2
HN
0
PA
mọi người giải giúp em bài này với ạ em đang cần gấp ạ
1
26 tháng 9 2021
a) \(\dfrac{A}{x-2}=\dfrac{x^2+3x+2}{x^2-4}\)
\(\Leftrightarrow\dfrac{A}{x-2}=\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\dfrac{A}{x-2}=\dfrac{x+1}{x-2}\Leftrightarrow A=x+1\)
b) \(\dfrac{M}{x-1}=\dfrac{x^2+3x+2}{x+1}\)
\(\Leftrightarrow\dfrac{M}{x-1}=\dfrac{\left(x+1\right)\left(x+2\right)}{x+1}\)
\(\Leftrightarrow\dfrac{M}{x-1}=x+2\Leftrightarrow M=\left(x-1\right)\left(x+2\right)=x^2+x-2\)
KY
4 tháng 8 2021
asked how much that computer was
wanted to know if I had the keys
asked Sam why he hadn't come to the party
asked Jane where she was going for her holidays
asked me if I spoke English
asked how old her mother was
asked me whether I was British or American
DT
Đỗ Thanh Hải
CTVVIP
4 tháng 8 2021
1 Tom asked how much that computer was
2 The officer wanted to know if I had wanted the key
3 Ann asked Sam why he hadn't come to her party
4 He asked Jane where she was going for her holiday
5 He asked me if I spoke E
6 He asked how old her mother was
7 He asked me whether I was British or American
NL
Nguyễn Lê Phước Thịnh
CTVHS
16 tháng 3
Xét ΔBAC có \(cosB=\frac{BA^2+BC^2-AC^2}{2\cdot BA\cdot BC}\)
=>\(BA^2+BC^2-AC^2=2\cdot BA\cdot BC\cdot cosB\)
=>\(27,11^2+BC^2-20,16^2=2\cdot27,11\cdot BC\cdot cos38\)
=>\(BC^2+328,5265-54,22\cdot BC\cdot cos38=0\)
=>BC≃32,67005129(cm)
Xét ΔABC có \(\frac{BC}{\sin BAC}=\frac{AC}{\sin B}\)
=>\(\sin BAC=BC\cdot\sin B:AC\) ≃0,998
=>\(\hat{BAC}\) ≃86 độ 7p
RH
2 tháng 1 2022
a) x(x + 4) - 3x - 12 = 0
x(x + 4) - 3(x + 4) = 0
(x + 4)(x - 3) = 0
x = -4 hoặc x = 3
b) 2x(x - 4) - x + 4 = 0
2x(x - 4) - (x - 4) = 0
(x - 4)(2x - 1) = 0
x = 4 hoặc x = 1/2
c) 5x(x - 3) - x + 3 = 0
5x(x - 3) - (x - 3) = 0
(x - 3)(5x - 1) = 0
x = 3 hoặc x = 1/5.
2 tháng 1 2022
a) x(x + 4) - 3x - 12 = 0
x(x + 4) - 3(x + 4) = 0
(x + 4)(x - 3) = 0
x = -4 hoặc x = 3
b) 2x(x - 4) - x + 4 = 0
2x(x - 4) - (x - 4) = 0
(x - 4)(2x - 1) = 0
x = 4 hoặc x = 1/2
c) 5x(x - 3) - x + 3 = 0
5x(x - 3) - (x - 3) = 0
(x - 3)(5x - 1) = 0
x = 3 hoặc x = 1/5.
a: \(\frac{x}{x-y}+\frac{y}{x+y}+\frac{2y^2}{x^2-y^2}\)
\(=\frac{x}{x-y}+\frac{y}{x+y}+\frac{2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\frac{x\left(x+y\right)+y\left(x-y\right)+2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\frac{x^2+xy+xy-y^2+2y^2}{\left(x-y\right)\left(x+y\right)}=\frac{x^2+2xy+y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\frac{\left(x+y\right)^2}{\left(x-y\right)\left(x+y\right)}=\frac{x+y}{x-y}\)
b: \(\frac{x^3+2x}{x^3+1}+\frac{2x}{x^2-x+1}+\frac{1}{x+1}\)
\(=\frac{x^3+2x}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{2x}{x^2-x+1}+\frac{1}{x+1}\)
\(=\frac{x^3+2x+2x\left(x+1\right)+x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{x^3+x^2+x+1+2x^2+2x}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\frac{x^3+3x^2+3x+1}{\left(x+1\right)\cdot\left(x^2-x+1\right)}=\frac{\left(x+1\right)^3}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{\left(x+1\right)^2}{x_{}^2-x+1}\)
c: \(\frac{x^2+2}{x^3-1}+\frac{x}{x^2+x+1}+\frac{1}{1-x}\)
\(=\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x}{x^2+x+1}-\frac{1}{x-1}\)
\(=\frac{x^2+2+x\left(x-1\right)-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{-x+1+x^2-x}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{x^2-2x+1}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{x-1}{x^2+x+1}\)
d: \(\frac{x}{x+3}+\frac{2x}{x-3}+\frac{9-3x^2}{x^2-9}\)
\(=\frac{x}{x+3}+\frac{2x}{x-3}+\frac{9-3x^2}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x\left(x-3\right)+2x\left(x+3\right)+9-3x^2}{\left(x-3\right)\left(x+3\right)}=\frac{x^2-3x+2x^2+6x+9-3x^2}{\left(x-3\right)\left(x+3\right)}=\frac{3x+9}{\left(x-3\right)\left(x+3\right)}=\frac{3}{x-3}\)
e: \(\frac{2}{x-3}+\frac{2x}{x^2-4x+3}+\frac{x}{x-1}\)
\(=\frac{2}{x-3}+\frac{2x}{\left(x-3\right)\left(x-1\right)}+\frac{x}{x-1}\)
\(=\frac{2\left(x-1\right)+2x+x-3}{\left(x-1\right)\left(x-3\right)}=\frac{2x-2+3x-3}{\left(x-1\right)\left(x-3\right)}=\frac{5x-5}{\left(x-1\right)\left(x-3\right)}=\frac{5}{x-3}\)
f: \(\frac{2}{x-1}+\frac{2x-1}{x^2+x+1}+\frac{x^2+6x+2}{x^3-1}\)
\(=\frac{2}{x-1}+\frac{2x-1}{x^2+x+1}+\frac{x^2+6x+2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{2\left(x^2+x+1\right)+\left(2x-1\right)\left(x-1\right)+x^2+6x+2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{2x^2+2x+2+2x^2-3x+1+x^2+6x+2}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{5x^2+5x+5}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{5}{x-1}\)
g: \(\frac{x}{x^2-4}+\frac{2\left(x^2+2x+4\right)}{8-x^3}+\frac{1}{x+2}\)
\(=\frac{x}{\left(x-2\right)\left(x+2\right)}+\frac{2\left(x^2+2x+4\right)}{\left(2-x\right)\left(x^2+2x+4\right)}+\frac{1}{x+2}\)
\(=\frac{x}{\left(x-2\right)\left(x+2\right)}+\frac{2}{2-x}+\frac{1}{x+2}\)
\(=\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2}{x-2}+\frac{1}{x+2}\)
\(=\frac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}=\frac{2x-2-2x-4}{\left(x-2\right)\left(x+2\right)}=-\frac{6}{x^2-4}\)
h: \(\frac{x+2}{x-2}+\frac{1}{x}+\frac{-8}{x^2-2x}\)
\(=\frac{x\left(x+2\right)+x-2-8}{x\left(x-2\right)}=\frac{x^2+2x+x-10}{x\left(x-2\right)}\)
\(=\frac{x^2+3x-10}{x\left(x-2\right)}=\frac{\left(x+5\right)\left(x-2\right)}{x\left(x-2\right)}=\frac{x+5}{x}\)