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
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: \(\left(x-3\right)^2-\left(x-3\right)\left(3-x^2\right)=0\)
=>\(\left(x-3\right)\left(x-3-3+x^2\right)=0\)
=>\(\left(x-3\right)\left(x^2+x-6\right)=0\)
=>(x-3)(x+3)(x-2)=0
=>\(\left[\begin{array}{l}x-3=0\\ x+3=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=-3\\ x=2\end{array}\right.\)
b: \(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)=11\)
=>\(4\left(x^2+2x+1\right)+4x^2-4x+1-8\left(x^2-1\right)=11\)
=>\(4x^2+8x+4+4x^2-4x+1-8x^2+8=11\)
=>4x+13=11
=>4x=-2
=>\(x=-\frac24=-\frac12\)
c: \(\left(x-3\right)^2-\left(x+3\right)\left(x-5\right)=3x+4\)
=>\(x^2-6x+9-\left(x^2-5x+3x-15\right)=3x+4\)
=>\(x^2-6x+9-\left(x^2-2x-15\right)=3x+4\)
=>\(x^2-6x+9-x^2+2x+15=3x+4\)
=>-4x+24=3x+4
=>-7x=-20
=>\(x=\frac{20}{7}\)
d: \(3\left(x-5\right)^2+2x\left(x-5\right)=0\)
=>\(\left(x-5\right)\left\lbrack3\left(x-5\right)+2x\right\rbrack=0\)
=>(x-5)(3x-15+2x)=0
=>(x-5)(5x-15)=0
=>5(x-5)(x-3)=0
=>(x-5)(x-3)=0
=>\(\left[\begin{array}{l}x-5=0\\ x-3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5\\ x=3\end{array}\right.\)
e: \(\left(2x-1\right)^2-\left(x+3\right)^2=0\)
=>(2x-1-x-3)(2x-1+x+3)=0
=>(x-4)(3x+2)=0
=>\(\left[\begin{array}{l}x-4=0\\ 3x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4\\ x=-\frac23\end{array}\right.\)
f: \(4\left(3x+2\right)\left(3x-2\right)-\left(6x+1\right)^2=7\)
=>\(4\left(9x^2-4\right)-36x^2-12x-1=7\)
=>\(36x^2-16-36x^2-12x-1=7\)
=>-12x-17=7
=>-12x=24
=>x=-2
g: \(3\left(2x-1\right)^2-6x\left(2x-3\right)=6\)
=>\(3\left(4x^2-4x+1\right)-6x\left(2x-3\right)=6\)
=>\(12x^2-12x+3-12x^2+18x=6\)
=>6x+3=6
=>6x=3
=>\(x=\frac36=\frac12\)
h: \(\left(2x+3\right)^2-4\left(x-1\right)^2=16\)
=>\(4x^2+12x+9-4\left(x^2-2x+1\right)=16\)
=>\(4x^2+12x+9-4x^2+8x-4=16\)
=>20x+5=16
=>20x=11
=>\(x=\frac{11}{20}\)