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: \(\frac{3x}{x+1}-\frac{1}{1+x}=\frac{3x}{x+1}-\frac{1}{x+1}=\frac{3x-1}{x+1}\)
b: \(\frac{3y-2x}{x-2y}-\frac{x-y}{2y-x}\)
\(=\frac{-2x+3y}{x-2y}+\frac{x-y}{x-2y}=\frac{-2x+3y+x-y}{x-2y}=\frac{-x+2y}{x-2y}\)
=-1
c: \(\frac{1}{x-2}-\frac{1}{x+1}\)
\(=\frac{x+1-\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}\)
\(=\frac{x+1-x+2}{\left(x+1\right)\left(x-2\right)}=\frac{3}{\left(x+1\right)\left(x-2\right)}\)
d: \(\frac{12}{x^2-9}-\frac{2}{x-3}\)
\(=\frac{12}{\left(x-3\right)\left(x+3\right)}-\frac{2}{x-3}\)
\(=\frac{12-2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{12-2x-6}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{-2x+6}{\left(x-3\right)\left(x+3\right)}=\frac{-2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=-\frac{2}{x+3}\)
e: \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}\)
\(=\frac{1}{x\left(y-x\right)}-\frac{1}{y\left(y-x\right)}\)
\(=\frac{y-x}{xy\left(y-x\right)}=\frac{1}{xy}\)
f: \(\frac{x}{x^2-25}-\frac{x-5}{x^2+5x}\)
\(=\frac{x}{\left(x-5\right)\left(x+5\right)}-\frac{x-5}{x\left(x+5\right)}\)
\(=\frac{x^2-\left(x-5\right)}{x\left(x+5\right)\left(x-5\right)}=\frac{\left(x-x+5\right)\left(x+x-5\right)}{x\left(x+5\right)\left(x-5\right)}=\frac{5\left(2x-5\right)}{x\left(x+5\right)\left(x-5\right)}\)
g: \(\frac{2x^3+x^2-x}{x^3-1}-\frac{2x-1}{x-1}\)
\(=\frac{2x^3+x^2-x}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x-1}{x-1}\)
\(=\frac{2x^3+x^2-x}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{\left(2x-1\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{2x^3+x^2-x-\left(2x-1\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{2x^3+x^2-x-2x^3-2x^2-2x+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{-2x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)
h: \(\frac{x}{x-1}-\frac{2}{x+1}-\frac{2}{x^2-1}\)
\(=\frac{x}{x-1}-\frac{2}{x+1}-\frac{2}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x\left(x+1\right)-2\left(x-1\right)-2}{\left(x-1\right)\left(x+1\right)}=\frac{x^2+x-2x+2-2}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2-x}{\left(x-1\right)\left(x+1\right)}=\frac{x}{x+1}\)
i: \(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\)
\(=\frac{8x^2}{4x^2\left(x-2\right)}-\frac{3x}{3\left(x^2-4\right)}-\frac{1}{x+2}\)
\(=\frac{2}{x-2}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{1}{x+2}\)
\(=\frac{2\left(x+2\right)-x-x+2}{\left(x-2\right)\left(x+2\right)}=\frac{2x+4-2x+2}{\left(x-2\right)\left(x+2\right)}=\frac{6}{x^2-4}\)
j: \(\frac{x^3+1}{x^2-9}-\frac{3}{x+3}-x\)
\(=\frac{x^3+1}{\left(x-3\right)\left(x+3\right)}-\frac{3}{x+3}-x\)
\(=\frac{x^3+1-3\left(x-3\right)-x\left(x^2-9\right)}{\left(x-3\right)\left(x+3\right)}=\frac{x^3+1-3x+9-x^3+9x}{\left(x-3\right)\left(x+3\right)}=\frac{6x+10}{\left(x-3\right)\cdot\left(x+3\right)}\)
k: \(\frac{2x}{x^2-1}-\frac{3}{2x+2}+\frac{1}{2-2x}\)
\(=\frac{2x}{\left(x-1\right)\left(x+1\right)}-\frac{3}{2\left(x+1\right)}-\frac{1}{2\left(x-1\right)}\)
\(=\frac{4x-3\left(x-1\right)-x-1}{2\left(x+1\right)\left(x-1\right)}=\frac{3x-1-3x+3}{2\left(x+1\right)\left(x-1\right)}=\frac{2}{2\left(x^2-1\right)}=\frac{1}{x^2-1}\)
l: \(x+\frac{1}{x+1}-1=x-1+\frac{1}{x+1}=\frac{\left(x-1\right)\left(x+1\right)+1}{x+1}\)
\(=\frac{x^2-1+1}{x+1}=\frac{x^2}{x+1}\)