Mọi ngưòi giúp mik vs 

Bài 1:

Phần a bạn tự làm nha! (Đ/S: 0,5)

b, B = \(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\) với \(x\ge0;x\ne4;x\ne9\)

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

B = \(\dfrac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

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

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

Vậy ...

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

T = \(\dfrac{A}{B}\)= \(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)= 1 - \(\dfrac{3}{\sqrt{x}+1}\)

Ta có: x \(\ge\) 0 \(\Leftrightarrow\) \(\sqrt{x}\ge0\) \(\Leftrightarrow\) \(\sqrt{x}+1\ge1\) \(\Leftrightarrow\) \(\dfrac{3}{\sqrt{x}+1}\le3\) \(\Leftrightarrow\) \(-\dfrac{3}{\sqrt{x}+1}\ge-3\) \(\Leftrightarrow\) T \(\ge\) -2

Vậy ...

Bài 2: ĐK: x \(\ge\) 0

Giả sử: \(P\) < \(\sqrt{P}\)

\(\Leftrightarrow\) \(\dfrac{\sqrt{x}+2}{\sqrt{x}+5}< \dfrac{\sqrt{\sqrt{x}+2}}{\sqrt{\sqrt{x}+5}}\)

\(\Leftrightarrow\) \(\dfrac{\sqrt{\left(\sqrt{x}+2\right)\left(\sqrt{x}+5\right)}-\left(\sqrt{x}+2\right)}{\sqrt{x}+5}>0\)

\(\Leftrightarrow\) \(\sqrt{\left(\sqrt{x}+2\right)\left(\sqrt{x}+5\right)}-\left(\sqrt{x}+2\right)>0\) (\(\sqrt{x}+5>0\) với mọi x \(\ge\) 0)

\(\Leftrightarrow\) \(\sqrt{\left(\sqrt{x}+2\right)}\sqrt{\sqrt{x}+5-\sqrt{x}-2}>0\)

\(\Leftrightarrow\) \(\sqrt{\left(\sqrt{x}+2\right)}\sqrt{3}>0\)

\(\Leftrightarrow\) \(\sqrt{\sqrt{x}+2}>0\)

Vì x \(\ge\) 0 \(\Leftrightarrow\) \(\sqrt{x}+2\ge2\) \(\Leftrightarrow\) \(\sqrt{\sqrt{x}+2}\ge\sqrt{2}>0\) (Đpcm)

Vậy \(P\) < \(\sqrt{P}\)

Chúc bn học tốt!

g, PT \(\Leftrightarrow\dfrac{x+24}{1996}+1+\dfrac{x+25}{1995}+1+\dfrac{x+26}{1994}+1+\dfrac{x+27}{1993}+1+\dfrac{x+2036}{4}-4=0\)

\(\Leftrightarrow\left(x+2020\right)\left(\dfrac{1}{1996}+\dfrac{1}{1995}+\dfrac{1}{1994}+\dfrac{1}{1993}+\dfrac{1}{4}\right)=0\)

\(\Leftrightarrow x+2020=0\)

\(\Leftrightarrow x=-2020\)

Vậy ...

Ca + 2H₂O --> Ca(OH)₂ + H₂

NaoH + HCl --> NaCl + H₂O

Cu + 2AgNO₃ --> Cu(NO₃)₂ + 2Ag\(\downarrow\)

2KClO₃ --t^o--> 2KCl + 3O₂

Bài T5 nhé

ĐKXĐ : \(2-x^4\ge0\)

Áp dụng Cô si

\(\sqrt[4]{2-x^4}=\sqrt[4]{\left(2-x^4\right).1.1.1}\le\dfrac{2-x^4+1+1+1}{4}=\dfrac{5-x^4}{4}\)

\(VT\le\dfrac{x^2\left(5-x^4\right)}{4}-x^4+x^3-1=\dfrac{-\left(x-1\right)^2\left(\left(x^2+x\right)^2+6\left(x+\dfrac{2}{5}\right)^2\right)}{4}\le0=VP\)

Dấu "=" \(x=1\)

Vậy S = {1}

Khá căn bản ta dự đoán x=1.

`x^2\root{4}{2-x^4}-x^4+x^3-1=0`
`đk:-root{4}{2}<=x<root{4}{2}`
`<=>x^3-1+x^2\root{4}{2-x^4}-x^4=0`
`<=>(x-1)(x^2+x+1)+x^2(1-root{4}{2-x^4})=0`
`<=>(x-1)(x^2+x+1)+x^2((1-2+x^4)/(1+root{4}{2-x^4}))=0`
`<=>(x-1)(x^2+x+1)+(x^2(x-1)(x^3+x^2+x+1))/(1+root{4}{2-x^4})=0`
`<=>(x-1)(x^2+x+1+(x^2(x^3+x^2+x+1))/(1+root{4}{2-x^4}))=0`
`<=>x-1=0` do `x^2+x+1+(x^2(x^3+x^2+x+1))/(1+root{4}{2-x^4})>0`
`<=>x=1`

III

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to buy

to speak

making

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working

to call

to built

to do

IV

collection

bird-watching

photography

interesting

Carving

activity

usually

creative

V

was

got

went

buy

traveled

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bought

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VI

was

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