check ib mình vs ạ

is your father's age ?

I do not play the guitar as well as him

a, \(V_{CO}=\left(\dfrac{2,8}{28}\right).22,4=2,24l\\ V_{NO}=0,1.22,4=2,24l\\ V_{hh}=2,24+2,24=4,48l\)

b, \(V_{hh}=1,12+8,96=10,08l\)

When was the black car sold 

The reported was finished by Nam this morning

The eletricty system is being checked by us now

1 D

2 B

3 D

4 B

5 A

1 D

2 B

3 D

4 B

5 A

representatives

natural - include

recycling

mixture

environmentally

packaging

Dạng 2:

k: =(x-4)(x-2)

i: =(x-4)(x+3)

Bài 5: 

a) Ta có: \(A=\left(\dfrac{4x+5\sqrt{x}-1}{x\sqrt{x}+2x-\sqrt{x}-2}-\dfrac{3\sqrt{x}+1}{x+\sqrt{x}-2}\right):\dfrac{x+4\sqrt{x}+4}{x-1}\)

\(=\left(\dfrac{4x+5\sqrt{x}-1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\left(3\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{\left(\sqrt{x}+2\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{4x+5\sqrt{x}-1-3x-3\sqrt{x}-\sqrt{x}-1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+2\right)^2}\)

\(=\dfrac{x+\sqrt{x}-2}{\left(\sqrt{x}+2\right)^3}\)

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

b) Ta có: \(A-1=\dfrac{\sqrt{x}-1-x-4\sqrt{x}-4}{x+4\sqrt{x}+4}\)

\(=\dfrac{-\left(x+3\sqrt{x}+5\right)}{x+4\sqrt{x}+4}\)

\(=\dfrac{-\left(x+2\cdot\sqrt{x}\cdot\dfrac{3}{2}+\dfrac{9}{4}\right)-\dfrac{11}{4}}{x+4\sqrt{x}+4}< 0\forall x\) thỏa mãn ĐKXĐ

nên A<1

1: Ta có: \(\sqrt{x^2+3x+2}=1\)

\(\Leftrightarrow x^2+3x+2=1\)

\(\Leftrightarrow x^2+3x+1=0\)

\(\text{Δ}=3^2-4\cdot1\cdot1=5\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-3-\sqrt{5}}{2}\\x_2=\dfrac{-3+\sqrt{5}}{2}\end{matrix}\right.\)