giúp mik vs ạ , đang cần gấp ạ
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23 tháng 12 2021
e: \(E=\dfrac{x^2-9-x^2+4-x^2+9}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x+2}{x+3}\)
a: \(A=\dfrac{4x^2+x^2-2x+1+x^2+2x+1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{6x^2+2}{\left(x-1\right)\left(x+1\right)}\)
R
1
RY
24 tháng 4 2022
a. f(\(\dfrac{-1}{2}\)) = \(4.\left(\dfrac{-1}{2}\right)^2+3.\left(\dfrac{-1}{2}\right)-2\)
= \(4.\dfrac{1}{4}-\left(\dfrac{-3}{2}\right)-\dfrac{4}{2}\)
= \(\dfrac{2}{2}+\dfrac{3}{2}-\dfrac{4}{2}\)
= \(\dfrac{1}{2}\)
DT
Đỗ Thanh Hải
CTVVIP
5 tháng 7 2021
1 are
2 am
3 is
4 are
5 are
6 are
7 is
8 is
9 is
10 are
IV
1 is writing
2 are losing
3 is having
4 is staying
5 am not lying
6 is always using
7 are having
8 Are you playing
9 are not touching
10 Is - listening
11 Is- winning
12 am not staying
13 is not working
14 is not reading
15 isn't raining
16 am not listening
17 Are they making
18 Are you doing
19 Is - sitting
20 is - doing
21 are-putting
22 are-wearing
23 is-studying
5 tháng 7 2021
2, am
3, is
4,are
5,are
6,are
7,is
8,is
9,is
10,are
IV
1,2,7 OK
3,is having
4,has stayed
5,am not lying
6,always uses
8,Are-playing
9,not to touch
10,Is-listening
11,Are-winning
12,am not staying
13,isn't working
14,isn't reading
15,isn't raining
16,am not listening
17,Are-making
18,Are-doing
19,Is-sitting
20,is-doing
21,do-putting
22,do-wear
23,is-studying
29 tháng 12 2021
5: \(=\dfrac{1}{2}\cdot10-\dfrac{1}{2}=\dfrac{1}{2}\cdot9=\dfrac{9}{2}\)
TN
1
10 tháng 1 2022
a/ Tam giác AMN cân tại A (gt). \(\Rightarrow\) \(\widehat{AMN}=\widehat{ANM};AM=AN.\)
Xét tam giác AMB và tam giác ANC có:
+ AM = AN (cmt).
+ \(\widehat{AMB}=\widehat{ANC}\left(\widehat{AMN}=\widehat{ANM}\right).\)
+ MB = NC (gt).
\(\Rightarrow\) Tam giác AMB = Tam giác ANC (c - g - c).
\(\Rightarrow\) AB = AC (cặp cạnh tương ứng).
Xét tam giác ABC có: AB = AC (cmt).
\(\Rightarrow\) Tam giác ABC cân tại A.
b/ Tam giác ABC cân tại A (cmt) \(\Rightarrow\) \(\widehat{ABC}=\widehat{ACB}.\)
Mà \(\widehat{ABC}=\widehat{MBH;}\widehat{ACB}=\widehat{NCK}\text{}\) (đối đỉnh).
\(\Rightarrow\) \(\widehat{MBH}=\widehat{NCK}.\)
Xét tam giác MBH và tam giác NCK \(\left(\widehat{BHM}=\widehat{CKN}=90^o\right)\)có:
+ MB = NC (gt).
+ \(\widehat{MBH}=\widehat{NCK}\left(cmt\right).\)
\(\Rightarrow\) Tam giác MBH = Tam giác NCK (cạnh huyền - góc nhọn).
c/ Tam giác MBH = Tam giác NCK (cmt).
\(\Rightarrow\) \(\widehat{BMH}=\widehat{CNK}\) (cặp góc tương ứng).
Xét tam giác OMN có: \(\widehat{NMO}=\widehat{MNO}\) (do \(\widehat{BMH}=\widehat{CNK}\)).
\(\Rightarrow\) Tam giác OMN tại O.
PL
2
30 tháng 8 2021
Câu 3:
a: Ta có: \(2x\left(3x-1\right)-\left(x-3\right)\left(6x+2\right)\)
\(=6x^2-2x-6x^2-2x+18x+6\)
=14x+6
b: Ta có: \(2x\left(x+7\right)-3x\left(x+1\right)\)
\(=2x^2+14x-3x^2-3x\)
\(=-x^2+11x\)
30 tháng 8 2021
Câu 2:
a: Ta có: \(\left(-8x^5+12x^3-16x^2\right):4x^2\)
\(=-8x^5:4x^2+12x^3:4x^2-16x^2:4x^2\)
\(=-2x^3+3x-4\)
b: Ta có: \(\left(12x^3y^3-18x^2y+9xy^2\right):6xy\)
\(=12x^3y^3:6xy-18x^2y:6xy+9xy^2:6xy\)
\(=2x^2y^2-3x+\dfrac{3}{2}y\)
c: Ta có: \(\dfrac{x^3-11x^2+27x-9}{x-3}\)
\(=\dfrac{x^3-3x^2-8x^2+24x+3x-9}{x-3}\)
\(=x^2-8x+3\)
d: Ta có: \(\dfrac{6x^4-13x^3+7x^2-x-5}{3x+1}\)
\(=\dfrac{6x^4+2x^3-15x^3-5x^2+12x^2+4x-5x-\dfrac{5}{3}-\dfrac{10}{3}}{3x+1}\)
\(=2x^3-5x^2+4x-\dfrac{5}{3}-\dfrac{\dfrac{10}{3}}{3x+1}\)
TC
0
20 tháng 9 2021
4) \(\left|\dfrac{5}{18}-x\right|-\dfrac{7}{24}=0\)
\(\Leftrightarrow\left|\dfrac{5}{18}-x\right|=\dfrac{7}{24}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{18}-x=\dfrac{7}{24}\\\dfrac{5}{18}-x=-\dfrac{7}{24}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{72}\\x=\dfrac{41}{72}\end{matrix}\right.\)
b) \(\dfrac{2}{5}-\left|\dfrac{1}{2}-x\right|=6\)
\(\Leftrightarrow\left|\dfrac{1}{2}-x\right|=-\dfrac{28}{5}\)( vô lý do \(\left|\dfrac{1}{2}-x\right|\ge0\forall x\))
Vậy \(S=\varnothing\)
