4:

a: Vì ON<OM

nên N nằm giữa O và M

b: Vì N nằm giữa O và M

nên ON+NM=OM

=>NM=3,5cm=ON

=>N là trung điểm của OM

tự chép đi. thế thì tự đi mà trả lời

\(a.PTHH:Ba+2H_2O\rightarrow Ba\left(OH\right)_2+H_2\)

\(b.n_{Ba}=\dfrac{m}{M}=\dfrac{13,7}{137}=0,1\left(mol\right)\)

\(\Rightarrow n_{H_2O}=\dfrac{1}{2}.n_{Ba}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\\ \Rightarrow m_{H_2O}=n.M=0,9\left(g\right)\)

\(c.n_{Ba}=0,1\left(mol\right)\Rightarrow n_{H_2}=n_{Ba}=0,1\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=n.22,4=0,1.22,4=2,24\left(l\right)\)

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Cau 23: B

Câu 24: A

23 là b

còn 24 là a

Câu 8: A

hong hiểu

8 là a

Câu 13: D

Câu 14: C

15A 

mí câu nì k bt lm thì thế số vô tính cx đựt :>

I: Trắc Nghiệm

Câu 1: D

Câu 2: C

Câu 3: D

Câu 4: D

Bài 1:
a.

\(=(6\sqrt{5}-3\sqrt{5}+3\sqrt{5}-6\sqrt{5}):\sqrt{5}=0:\sqrt{5}=0\)

b.

\(=3\sqrt{a}-\frac{1}{2a}\sqrt{(3a)^2.a}+\sqrt{a^2}.\sqrt{4^2}.\sqrt{\frac{1}{a}}-\frac{2}{a^2}.\sqrt{(6a^2)^2.a}\)

\(=3\sqrt{a}-\frac{1}{2a}.3a\sqrt{a}+4\sqrt{a^2.\frac{1}{a}}-\frac{2}{a^2}.6a^2\sqrt{a}\)

\(=3\sqrt{a}-1,5\sqrt{a}+4\sqrt{a}-12\sqrt{a}=-6,5\sqrt{a}\)

Bài 2:
a. ĐKXĐ: $x\geq \frac{3}{2}$

PT $\Leftrightarrow \sqrt{2x-3}+\sqrt{4}.\sqrt{2x-3}=15$

$\Leftrightarrow \sqrt{2x-3}+2\sqrt{2x-3}=15$

$\Leftrightarrow 3\sqrt{2x-3}=15$

$\Leftrightarrow \sqrt{2x-3}=5$

$\Leftrightarrow 2x-3=25$

$\Leftrightarrow x=14$ (tm)

b. ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow \sqrt{25}.\sqrt{x-4\sqrt{x}+4}=30$

$\Leftrightarrow 5\sqrt{(\sqrt{x}-2)^2}=30$

$\Leftrightarrow 5|\sqrt{x}-2|=30$

$\Leftrightarrow |\sqrt{x}-2|=6$

$\Leftrightarrow \sqrt{x}-2=\pm 6$

$\Leftrightarrow \sqrt{x}=8$ (chọn) hoặc $\sqrt{x}=-4$ (loại)

$\Leftrightarrow x=64$ (chọn)

\(a,\) Hàm số bậc nhất \(\Leftrightarrow2m-3\ne0\Leftrightarrow m\ne\dfrac{3}{2}\)

\(b,\) Để \(\left(d\right)\) tạo với Ox một góc nhọn thì:

\(2m-3>0\Leftrightarrow m>\dfrac{3}{2}\)

\(c,m=3\Leftrightarrow y=3x+2\)

\(x=0\Leftrightarrow y=2\Leftrightarrow A\left(0;2\right)\\ x=1\Leftrightarrow y=5\Leftrightarrow B\left(1;5\right)\)