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Câu 4: ĐKXĐ: x>=1/2
Ta có: \(2\left(x-\sqrt{2x^2+5x-3}\right)=1+x\left(\sqrt{2x-1}-2\sqrt{x+3}\right)\)
=>\(2x-2\sqrt{2x^2+5x-3}=1+x\cdot\sqrt{2x-1}-2x\cdot\sqrt{x+3}\)
=>\(2x-1-2\cdot\sqrt{\left(2x-1\right)\left(x+3\right)}-x\cdot\sqrt{2x-1}+2x\cdot\sqrt{x+3}=0\)
=>\(2x-1-x\cdot\sqrt{2x-1}-2\cdot\sqrt{\left(2x-1\right)\left(x+3\right)}+2x\cdot\sqrt{x+3}=0\)
=>\(\sqrt{2x-1}\left(\sqrt{2x-1}-x\right)-2\cdot\sqrt{x+3}\left(\sqrt{2x-1}-x\right)=0\)
=>\(\left(\sqrt{2x-1}-\sqrt{4x+12}\right)\left(\sqrt{2x-1}-x\right)=0\)
TH1: \(\sqrt{2x-1}-\sqrt{4x+12}=0\)
=>\(\sqrt{2x-1}=\sqrt{4x+12}\)
=>4x+12=2x-1
=>2x=-13
=>\(x=-\frac{13}{2}\) (loại)
TH2: \(\sqrt{2x-1}-x=0\)
=>\(\sqrt{2x-1}=x\)
=>\(\begin{cases}2x-1=x^2\\ x\ge0\end{cases}\Rightarrow\begin{cases}x^2-2x+1=0\\ x\ge0\end{cases}\Rightarrow\begin{cases}\left(x-1\right)^2=0\\ x\ge0\end{cases}\)
=>x-1=0
=>x=1(nhận)
VD6: \(\overrightarrow{AM}=3\cdot\overrightarrow{AB}-2\cdot\overrightarrow{AC}\)
=>\(\overrightarrow{AM}=3\left(\overrightarrow{AM}+\overrightarrow{MB}\right)-2\left(\overrightarrow{AM}+\overrightarrow{MC}\right)\)
=>\(\overrightarrow{AM}=3\cdot\overrightarrow{AM}+3\cdot\overrightarrow{MB}-2\cdot\overrightarrow{AM}-2\cdot\overrightarrow{MC}\)
=>\(3\cdot\overrightarrow{MB}-2\cdot\overrightarrow{MC}=\overrightarrow{0}\)
=>\(3\cdot\overrightarrow{MB}=2\cdot\overrightarrow{MC}\)
=>\(\overrightarrow{MB}=\frac23\cdot\overrightarrow{MC}\)
=>\(\overrightarrow{MB};\overrightarrow{MC}\) là hai vecto cùng phương
VD7: \(\frac{\overrightarrow{BC}}{\overrightarrow{AB}}=\frac{3\left(2\cdot\overrightarrow{a}+3\cdot\overrightarrow{b}\right)}{2\cdot\overrightarrow{a}+3\cdot\overrightarrow{b}}=3\)
=>\(\overrightarrow{BC}=3\cdot\overrightarrow{AB}=-3\cdot\overrightarrow{BA}\)
=>B,A,C thẳng hàng
4. \(\dfrac{-3}{2}+x-\dfrac{5}{4}=\dfrac{-1}{3}-2x\)
<=> \(\dfrac{-18}{12}+\dfrac{12x}{12}-\dfrac{15}{12}=\dfrac{-4}{12}-\dfrac{24x}{12}\)
<=> -18 + 12x - 15 = -4 - 24x
<=> 12x + 24x = 18 + 15 - 4
<=> 36x = 29
<=> x = \(\dfrac{29}{36}\)
6. \(\dfrac{3}{4}x-\dfrac{3}{2}=\dfrac{5}{6}+\dfrac{3}{8}x\)
<=> \(\dfrac{18x}{24}-\dfrac{36}{24}=\dfrac{20}{24}+\dfrac{9x}{24}\)
<=> 18x - 36 = 20 + 9x
<=> 18x - 9x = 20 + 36
<=> 9x = 56
<=> x = \(\dfrac{56}{9}\)
7. \(3-\left(\dfrac{1}{2}+2x\right)=\dfrac{2}{3}-x\)
<=> \(3-\dfrac{1}{2}-2x=\dfrac{2}{3}-x\)
<=> \(\dfrac{18}{6}-\dfrac{3}{6}-\dfrac{12x}{6}=\dfrac{4}{6}-\dfrac{6x}{6}\)
<=> 18 - 3 - 12x = 4 - 6x
<=> 15 - 4 = 12x - 6x
<=> 11 = 6x
<=> x = \(\dfrac{11}{6}\)
1. John was sad yesterday.
2. The birds were in the cage.
3. The children were in the class.
4. My cat was on the chair.
5. I was at the theater yesterday.
6. It was foggy this morning.
7. They were at home last Sunday.
8. I was very happy yesterday.
9. There were a lot of balloons.
10. You were at the park last night.
a) Hiệu suất thụ tinh của trứng và tinh trùng lak 100%
-> Số trứng tham gia thụ tinh : 16 trứng
Số tinh trùng tham gia thụ tinh : 16 tinh trùng
b) Số tb sinh tinh : 16 : 4 = 4 (tb) (mak mik hỏi tí lak chữ "tg" sau chỗ tb sinh tinh lak viết tắt j v mik ko đọc đc)
c) Kỳ sau giảm phân II :
Số NST : 2n đơn = 16 NST
Số tâm động: 2n = 16 tâm động
Số cromatit : 0 cromatit
I like playing soccer, but I must do my homework before.
We are saving money because we want to buy a new house.
I like English and French books.
Lan saves energy, so she turns off the lights.
1 I like playing soccer but I must do my homework before
2 We are saving money because we want to buy a new house
3 I like English books and French books
4 Lan saves money so she turns off the lights
Ta có: \(AB^2+HC^2=\left(AA'^2+A'B^2\right)+\left(A'H^2+A'C^2\right)\)
\(=\left(AA'^2+A'C^2\right)+\left(A'B^2+A'H^2\right)=AC^2+HB^2\)
Lại có: \(BC^2+HA^2=\left(BB'^2+B'C^2\right)+\left(B'H^2+B'A^2\right)\)
\(=\left(BB'^2+B'A^2\right)+\left(B'C^2+B'H^2\right)=AB^2+HC^2\)
\(\Rightarrow AB^2+HC^2=AC^2+HB^2=BC^2+HA^2\)











Bài 2:
b: 2x-4y=5
=>4y=2x+5
=>y=1/2x+5/4
Vậy: Hệ số góc là 1/2