Question

[Lương Hà Linh]

Mọi người ơi giải giúp mk , mk cảm ơn ^_^

Answer 1

[Hoàng Như Quỳnh]

\(1:x< 0\left(B\right)\)

\(2:\left(D\right)\)

\(3:x< 2021\left(C\right)\)

\(4:x\ge15\left(D\right)\)

\(5:\)để pt có nghĩa thì 2x-5>0

\(2x>5< =>x>\frac{5}{2}\)

chọn (C)

\(6:\frac{1}{2}\sqrt{20}-\sqrt{\left(2-\sqrt{5}\right)^2}\)

\(\frac{1}{2}\sqrt{20}-\sqrt{5}+2\)

\(\sqrt{5}-\sqrt{5}+2=2\)

chọn (B)

\(7:\frac{6xy^2}{x^2-y^2}\sqrt{\frac{\left(x-y\right)^2}{\left(3xy^2\right)^2}}\)

\(\frac{6xy^2}{x^2-y^2}\frac{x-y}{3xy^2}\)

\(\frac{2}{x+y}\)

chọn (B)

\(8:\left(1+\frac{3-\sqrt{3}}{\sqrt{3}-1}\right)\left(\frac{3+\sqrt{3}}{\sqrt{3}+1}-1\right)\)

\(\left(1+\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right)\left(\frac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}-1\right)\)

\(\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\)

\(\sqrt{3}^2-1^2=3-1=2\)

chọn (D)

\(9:M=\left|1-\sqrt{3}\right|+\left|1-\sqrt{3}\right|\)

\(M=\sqrt{3}-1+\sqrt{3}-1\)

\(M=2\sqrt{3}-2\)

chọn (A)

\(10:\sqrt{4+\sqrt{x^2-1}}=2\)

\(4+\sqrt{x^2-1}=2^2=4\)

\(\sqrt{x^2-1}=0\)

\(x^2-1=0< =>x=1\)

chọn (A)

Answer 2

[Capheny Bản Quyền]

1 B 

2 D 

3 C 

4 D 

5 C 

6 B 

7 B 

8 D 

9 D 

10 B