Câu 6: Để hàm số y=(1-m)x+3 nghịch biến trên R thì 1-m<0

=>m>1

=>Chọn B

Câu 7: D

Câu 10: (D)//(D')

=>\(\left\{{}\begin{matrix}3m+1=2\left(m+1\right)\\-2\ne-2\left(loại\right)\end{matrix}\right.\Leftrightarrow m\in\varnothing\)

=>Chọn D

Câu 11: \(x^2+2x+2=\left(x+1\right)^2+1>=1>0\forall x\)

=>\(\sqrt{x^2+2x+2}\) luôn xác định với mọi số thực x

=>Chọn A

Câu 12: Để hai đường thẳng y=x+3m+2 và y=3x+2m+3 cắt nhau tại một điểm trên trục tung thì \(\left\{{}\begin{matrix}1\ne3\left(đúng\right)\\3m+2=2m+3\end{matrix}\right.\)

=>3m+2=2m+3

=>m=1

=>Chọn C

a) \(\dfrac{\left(a+b\right)^2-\left(a-b\right)^2}{4}=ab\)

\(\Leftrightarrow\dfrac{a^2+2ab+b^2-a^2+2ab-b^2}{4}=ab\)

\(\Leftrightarrow\dfrac{4ab}{4}=ab\left(đúng\right)\)

b) \(2\left(x^2+y^2\right)=\left(x+y\right)^2+\left(x-y\right)^2\)

\(\Leftrightarrow2x^2+2y^2=x^2+2xy+y^2+x^2-2xy+y^2\)

\(\Leftrightarrow2x^2+2y^2=2x^2+2y^2\left(đúng\right)\)

c) \(\left(x+y\right)^2-\left(x-y\right)\left(x+y\right)=2y\left(x+y\right)\)

\(\Leftrightarrow\left(x+y\right)\left(x+y-x+y\right)=2y\left(x+y\right)\)

\(\Leftrightarrow\left(x+y\right).2y=2y\left(x+y\right)\left(đúng\right)\)

X= 16

\(A^2=8+2\sqrt{16-10-2\sqrt{5}}\\ A^2=8+2\sqrt{6-2\sqrt{5}}\\ A^2=8+2\left(\sqrt{5}-1\right)=6+2\sqrt{5}\\ A=\sqrt{6+2\sqrt{5}}=\sqrt{5}+1\)

Đặt \(\sqrt{10+2\sqrt5}\)= a. Ta có A = \(\sqrt{4+a}+\sqrt{4-a}\)

=> A² = 4 + a  + 4 - a + 2\(\sqrt{(4+a)(4-a)}\)

=> A² = 8 + 2\(\sqrt{16-a^2}\)

=> A² = 8 + 2\(\sqrt{16 - 10 + 2\sqrt5}\)

=> A² = 8 + 2\(\sqrt{5+2\sqrt{5}+1}\)

=> A² = 8 + 2\(\sqrt{(\sqrt{5}+1)^2}\)

=> A² = 8 + 2\(\sqrt5\) + 2

=> A = \(\sqrt{2\sqrt{5}+10}\)

Ta có: \(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\ge0\)

\(A^2=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}\)

\(=8+2\sqrt{16-10-2\sqrt{5}}=8+2\sqrt{6-2\sqrt{5}}\)

\(=8+2\sqrt{\left(\sqrt{5}-1\right)^2}=8+2\left(\sqrt{5}-1\right)=6+2\sqrt{5}\)

\(\Leftrightarrow A=\sqrt{6+2\sqrt{5}}=\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}+1\)(do \(A\ge0\))

\(a,A=\left(5x-1+1-x\right)^2=16x^2\\ B=x^3-x^3+4x=4x\\ c,A=B\Leftrightarrow16x^2-4x=0\\ \Leftrightarrow4x\left(4x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{4}\end{matrix}\right.\)

a) \(\Rightarrow A=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)

\(\Rightarrow A=\sqrt{x}-1+\sqrt{x}\)

\(\Rightarrow A=2\sqrt{x}-1\)

b) \(\Rightarrow A=2.4-1=7\)

a) \(A=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}=\sqrt{x}-1+\sqrt{x}=2\sqrt{x}-1\)

b) \(A=2\sqrt{x}-1=2\sqrt{16}-1=2.4-1=7\)

B ơi có thể chia làm 2 hong chứ tui nhìn tui nản quá tui k làm nổi :<

1. Tom wishes he hadn't missed Mr. Baker's class yesterday afternoon.

2. I wish there were some red pepper in my food.

3. She wishes she sent me her recent photos.

4. I wish I had had time to see all the things we wanted to see.

5. Mary wishes she could sing well.

6. He wishes he could pick his children up.

7. He wishes he would attend the conference.

8. My sister wishes she spent a lot of time chatting with her friends.