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Answer :
1. writes
2. did
3. did (you) go
4. are singing
5. did (you) do
~Study well~
#KSJ
#Sai thì thôi nhé bn
Chia động từ:
1 sofar she ( write ) has written 50 pages of this book
2 she ( do) did her homework at 7.00 last night
3 where did you ( go ) go yesterday ?
4 they are singing ( sing ) karaoke at the moment
5 what did you do( do ) at this time yesterday ?
P/s: Sofar đi với thì HTHT chứ không phải HTĐ nhé!
1. This book can't be taken away. (mà sao this lại thêm s vào book thế kia?)
2. Your brother is not strong enough to lift that box.
3. Neither of the chairs is comfortable.
4. We haven't seen him since 1980
đây là nơi hok toán, lần sau đừng hỏi nhé ^^
1. Lam has a lot af stamps in her collection
2. You must be careful when playing electronic game because the can be addictive
Lần sau viết rõ đề hộ
Ta có : \(\frac{6}{11}x=\frac{9}{2}y\)=> \(\frac{12x}{22}=\frac{99y}{22}\)=> 12x = 99y => 4x = 33y => \(\frac{x}{33}=\frac{y}{4}\)
\(\frac{9}{2}y=\frac{15}{5}z\)=> \(\frac{45y}{10}=\frac{30z}{10}\)=> 45y = 30z => 3y = 2z => \(\frac{y}{2}=\frac{z}{3}\)
=> \(\frac{x}{33}=\frac{y}{4};\frac{y}{2}=\frac{z}{3}\)
=> \(\frac{x}{66}=\frac{y}{4};\frac{y}{4}=\frac{z}{12}\)
=> \(\frac{x}{66}=\frac{y}{4}=\frac{z}{12}\)và y - x + z = -120
Áp dụng t/c dãy tỉ số bằng nhau ta có :
\(\frac{x}{66}=\frac{y}{4}=\frac{z}{12}=\frac{y-x+z}{4-66+12}=\frac{-120}{-50}=\frac{12}{5}\)
=> \(\hept{\begin{cases}\frac{x}{66}=\frac{12}{5}\\\frac{y}{4}=\frac{12}{5}\\\frac{z}{12}=\frac{12}{5}\end{cases}}\)=> \(\hept{\begin{cases}x=\frac{792}{5}\\y=\frac{48}{5}\\z=\frac{144}{5}\end{cases}}\)
\(\frac{5^3\cdot3^7}{25\cdot9^4}=\frac{5^3\cdot3^7}{5^2\cdot3^8}=\frac{5}{3}\)
\(\left|a+2\right|=a\)
\(\Rightarrow a+2=\hept{\begin{cases}a\\-a\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}a-a=2\\-a-a=2\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}0=2\left(loai\right)\\-2a=2\end{cases}}\)
\(\Rightarrow a=-1\)
*\(M+\left(5x^2-2xy\right)=6x^2+9xy-y^2\)
\(M=6x^2+9xy-y^2-\left(5x^2-2xy\right)\)
\(M=6x^2+9xy-y^2-5x^2+2xy\)
\(M=\left(6-5\right)x^2+\left(9+2\right)xy-y^2\)
\(M=x^2+11xy-y^2\)
* \(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\)
Ta có : \(\hept{\begin{cases}\left(2x-5\right)^{2018}\ge0\forall x\\\left(3y+4\right)^{2020}\ge0\forall y\end{cases}\Rightarrow}\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\ge0\forall x,y\)
Mà đề cho \(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\)
=> \(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}=0\)
=> \(\hept{\begin{cases}2x-5=0\\3y+4=0\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{5}{2}\\y=-\frac{4}{3}\end{cases}}\)
Thay x = 5/2 ; y = -4/3 vào M ta được :
\(M=\left(\frac{5}{2}\right)^2+11\cdot\frac{5}{2}\cdot\left(-\frac{4}{3}\right)-\left(-\frac{4}{3}\right)^2\)
\(M=\frac{25}{4}+\frac{-110}{3}-\frac{16}{9}\)
\(M=\frac{-1159}{36}\)
Vậy giá trị của M = -1159/36 khi x = 5/2 ; y = -4/3
Không chắc nha