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DT
Đỗ Thanh Hải
CTVVIP
21 tháng 8 2021
1 has changed
2 are driving
3 were made
4 went - saw
5 finish
6 have been
7 weren't
8 will write
9 doing
10 to work
11 will clean
12 tastes - are
13 has become
14 went
15 is learning
16 won't be
17 is cut
18 eats
19 stewing
20 to drink
21 is organised
22 built
23 were made
14 was set
25 haven't done
26 is speaking
27 are spoken
3 tháng 10 2021
a) \(\left(2x+1\right)^2-4\left(x+2\right)^2=9\\ \Rightarrow\left(2x+1\right)^2-\left[2\left(x+2\right)\right]^2=9\\ \Rightarrow\left(2x+1-2x-4\right)\left(2x+1+2x+4\right)=9\\ \Rightarrow-3\left(4x+5\right)=9\\ \Rightarrow-12x-15=9\\ \Rightarrow x=-2\)
b) \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\\ \Rightarrow x^2+6x+9-\left(x^2+4x-32\right)=1\\ \Rightarrow x^2+6x+9-x^2-4x+32=1\\ \Rightarrow2x=-40\\ \Rightarrow x=-20\)
3 tháng 10 2021
\(a,\Rightarrow4x^2+4x+1-4x^2-16x-16=9\\ \Rightarrow-12x=24\Rightarrow x=-2\\ b,\Rightarrow x^2+6x+9-x^2-4x+32=1\\ \Rightarrow2x=-40\Rightarrow x=-20\\ c,\Rightarrow3x^2+12x+12+4x^2-4x+1-7x^2+63=36\\ \Rightarrow8x=-40\Rightarrow x=-5\\ d,\Rightarrow x^3-27+4x-x^3=1\\ \Rightarrow4x=28\Rightarrow x=7\\ e,\Rightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6=-19\\ \Rightarrow12x=-15\Rightarrow x=-\dfrac{5}{4}\)
KT
6
NT
6 tháng 9 2021
c. \(\left|\dfrac{8}{4}-\left|x-\dfrac{1}{4}\right|\right|-\dfrac{1}{2}=\dfrac{3}{4}\)
\(\Rightarrow\left[{}\begin{matrix}\left|\dfrac{8}{4}-x+\dfrac{1}{4}\right|-\dfrac{1}{2}=\dfrac{3}{4}\\\left|\dfrac{8}{4}+x-\dfrac{1}{4}\right|-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left|\dfrac{9}{4}-x\right|-\dfrac{1}{2}=\dfrac{3}{4}\\\left|\dfrac{7}{4}+x\right|-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}\dfrac{9}{4}-x-\dfrac{1}{2}=\dfrac{3}{4}\\x=\dfrac{9}{4}-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\\\left[{}\begin{matrix}\dfrac{7}{4}+x-\dfrac{1}{2}=\dfrac{3}{4}\\-\dfrac{7}{4}-x-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=\dfrac{7}{2}\end{matrix}\right.\\\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-3\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{7}{2}\\x=-3\end{matrix}\right.\)
Ở nơi x=9/4-1/2 là x-9/4-1/2 nha
KK
6 tháng 9 2021
a. -1,5 + 2x = 2,5
<=> 2x = 2,5 + 1,5
<=> 2x = 4
<=> x = 2
b. \(\dfrac{3}{2}\left(x+5\right)-\dfrac{1}{2}=\dfrac{4}{3}\)
<=> \(\dfrac{3}{2}x+\dfrac{15}{2}-\dfrac{1}{2}=\dfrac{4}{3}\)
<=> \(\dfrac{9x}{6}+\dfrac{45}{6}-\dfrac{3}{6}=\dfrac{8}{6}\)
<=> 9x + 45 - 3 = 8
<=> 9x = 8 + 3 - 45
<=> 9x = -34
<=> x = \(\dfrac{-34}{9}\)
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.
VT
1
24 tháng 7 2021
2 dreamed/dreamt
3 were taking
4 was
5 were
6 were travelling
7 was
8 were sitting
9 were drinking
10 were dancing
11 was playing
12 had
13 arrived
14 was waiting
15 stayed
16 were beating
17 were singing
18 were dancing
19 was getting
20 heard
21 was
22 was
DT
2
21 tháng 10 2021
a: \(x^2-9y^2=\left(x-3y\right)\left(x+3y\right)\)
c: \(\left(x+5\right)^2-16=\left(x+1\right)\left(x+9\right)\)
e: \(\left(2x+3\right)^2-\left(x-7\right)^2\)
\(=\left(2x+3+x-7\right)\left(2x+3-x+7\right)\)
\(=\left(3x-4\right)\left(x+10\right)\)
13-C
14-b
15-d
16-a
13: C
14: B
15: D
16: A