Giúp m với ạ.Mình cảm ơn
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GA
Giúp mình với ạ.Mình cảm ơn.
0
DT
Đỗ Thanh Hải
CTVVIP
30 tháng 9 2021
1 is watered by my father everymorning
2 is prepared by her mother in the kitchen
3 should be cleaned twice a day
4 repainting before Tet
5 was invited to his birthday party by John last night
6 of the car has to be repaired
7 won't be hung in the living room
8 the most valuable painting in the national museum been stolen
9 were told this story by their grandmother when they visited her last week
10 taken by a man when we were......
11 given up smoking since last year
12 read War and Peace since 2005
31 tháng 5 2023
C nguyên
=>2căn x+2+3 chia hết cho căn x+1
=>căn x+1 thuộc Ư(3)
=>căn x+1=1 hoặc căn x+1=3
=>x=4 hoặc x=0
NH
1
6 tháng 12 2021
b: \(\Leftrightarrow\dfrac{-m^2+3}{m}=\dfrac{13}{4}\)
\(\Leftrightarrow-4m^2+12=13m\)
\(\Leftrightarrow4m^2-13m-12=0\)
\(\Leftrightarrow\left(m-4\right)\left(4m+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}m=4\\m=-\dfrac{3}{4}\end{matrix}\right.\)
25 tháng 2 2022
a: \(E=x^4z^3y\left(\dfrac{2}{5}-0+1\right)=\dfrac{7}{5}x^4yz^3\)
b: \(E=\dfrac{7}{5}x^4yz^3=\dfrac{7}{5}\cdot2^4\cdot\dfrac{1}{2}\cdot\left(-1\right)^3=\dfrac{7}{5}\cdot8\cdot\left(-1\right)=-\dfrac{56}{5}\)
25 tháng 2 2022
a, \(=\left(\dfrac{2}{5}+1\right)x^4z^3y=\dfrac{7}{5}x^4z^3y\)
b, Thay x = 2 ; y = 1/2 ; z = -1 vào ta được
\(\dfrac{7}{5}.16.\dfrac{-1}{2}.1=-\dfrac{56}{5}\)
1.\(x=4-2\sqrt{3}-\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1^2=\left(\sqrt{3}-1\right)^2\)
\(A=\dfrac{2x-3\sqrt{x}-2}{\sqrt{x}-2}=\dfrac{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}{\sqrt{x}-2}=2\sqrt{x}+1=2\sqrt{\left(\sqrt{3}-1\right)^2}+1=2\left(\sqrt{3}-1\right)+1=2\sqrt{3}-2+1=2\sqrt{3}-1\)
2.\(B=\dfrac{\sqrt{x^3}-\sqrt{x}+2x-2}{\sqrt{x}+2}=\dfrac{\sqrt{x}^2\left(\sqrt{x}+2\right)-\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)}=\dfrac{\left(\sqrt{x}+2\right)\left(x-1\right)}{\sqrt{x}+2}=x-1\)
\(B=A+1\)
\(\Leftrightarrow x-1=2\sqrt{x}+1+1\)
\(\Leftrightarrow x-1=2\sqrt{x}+2\)
\(\Leftrightarrow x-2x\sqrt{x}-3=0\)
\(\Leftrightarrow\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=-1\left(vô.lý\right)\end{matrix}\right.\) \(\rightarrow x=\sqrt{3}\)