Question

[Ng Ngọc Gia Hân]

 giúp tớ vs mn ơi gấp lắm roiii

Answer 1

[Nguyễn Lê Phước Thịnh]

Câu 3:

1: \(\sqrt{\dfrac{1}{4}}=\dfrac{\sqrt{1}}{\sqrt{4}}=\dfrac{1}{2}\)

2: \(\sqrt{\dfrac{25}{49}}=\dfrac{\sqrt{25}}{\sqrt{49}}=\dfrac{5}{7}\)

3: \(\sqrt{\dfrac{64}{81}}=\dfrac{\sqrt{64}}{\sqrt{81}}=\dfrac{8}{9}\)

4: \(\sqrt{\dfrac{100}{9}}=\dfrac{\sqrt{100}}{\sqrt{9}}=\dfrac{10}{3}\)

5: \(\sqrt{\dfrac{17+8}{16}}=\sqrt{\dfrac{25}{16}}=\dfrac{\sqrt{25}}{\sqrt{16}}=\dfrac{5}{4}\)

6: \(\sqrt{\dfrac{36}{100-36}}=\sqrt{\dfrac{36}{64}}=\sqrt{\dfrac{9}{16}}=\dfrac{\sqrt{9}}{\sqrt{16}}=\dfrac{3}{4}\)

7: \(\sqrt{1-\dfrac{11}{36}}=\sqrt{\dfrac{36}{36}-\dfrac{11}{36}}=\sqrt{\dfrac{25}{36}}=\dfrac{\sqrt{25}}{\sqrt{36}}=\dfrac{5}{6}\)

8: \(\sqrt{2+\dfrac{1}{4}}=\sqrt{\dfrac{9}{4}}=\dfrac{\sqrt{9}}{\sqrt{4}}=\dfrac{3}{2}\)

Câu 5:

1: ĐKXĐ: x>=0

\(\sqrt{x}+\dfrac{1}{3}=\dfrac{1}{4}\)

=>\(\sqrt{x}=\dfrac{1}{4}-\dfrac{1}{3}=-\dfrac{1}{12}\)<0(vô lý)

=>Phương trình vô nghiệm

2: ĐKXĐ: x>=0

\(2-3\sqrt{x}=-7\)

=>\(3\sqrt{x}=2+7=9\)

=>\(\sqrt{x}=3\)

=>\(x=3^2=9\)(nhận)

3: ĐKXĐ: x+1>=0

=>x>=-1

\(\sqrt{x+1}=1\)

=>\(x+1=1^2=1\)

=>x=1-1=0(nhận)

4: ĐKXĐ: x>=0

\(\dfrac{3}{5}\sqrt{x}-\dfrac{2}{3}=\dfrac{4}{5}\)

=>\(\dfrac{3}{5}\sqrt{x}=\dfrac{2}{3}+\dfrac{4}{5}=\dfrac{10}{15}+\dfrac{12}{15}=\dfrac{22}{15}\)

=>\(\sqrt{x}=\dfrac{22}{15}:\dfrac{3}{5}=\dfrac{22}{15}\cdot\dfrac{5}{3}=\dfrac{110}{45}=\dfrac{22}{9}\)

=>\(x=\left(\dfrac{22}{9}\right)^2=\dfrac{264}{81}\)

5: ĐKXĐ: 2x-7>=0

=>x>=7/2

\(\sqrt{2x-7}=5\)

=>\(2x-7=5^2=25\)

=>2x=7+25=32

=>x=32/2=16(nhận)

6: ĐKXĐ: 2-3x>=0

=>3x<=2

=>\(x< =\dfrac{2}{3}\)

\(\sqrt{2-3x}=4\)

=>\(2-3x=4^2=16\)

=>3x=2-16=-14

=>\(x=-\dfrac{14}{3}\left(nhận\right)\)