đề là rút gọn hả bạn:

\(\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)

\(\frac{x+2+\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(\frac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

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

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

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

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

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

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

\(5:\)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)

1 B 

2 D 

3 C 

4 D 

5 C 

6 B 

7 B 

8 D 

9 D 

10 B 

\(\sqrt{x^2-4x+4}-\sqrt{x^2+2x+1}=-3\)

\(\Leftrightarrow\sqrt{\left(x-2\right)^2}-\sqrt{\left(x+1\right)^2}=-3\)

\(\Leftrightarrow\left|x-2\right|-\left|x+1\right|=-3\)(1)

Có: \(\left|x-2\right|-\left|x+1\right|=\left|x-2\right|-\left|x-2+3\right|\ge\left|x-2\right|-\left(\left|x-2\right|+3\right)=-3\)

Dấu \(=\)khi \(3\left(x-2\right)\ge0\Leftrightarrow x\ge2\).

Do đó nghiệm của (1) là \(x\ge2\).

Vậy nghiệm phương trình đã cho là \(x\ge2\).

<=>\(\sqrt{\left(x-2\right)^2}\)-\(\sqrt{\left(x+1\right)^2}=-3\)

<=>\(|x-2|-|x+1|=-3\)(1)

nếu \(\hept{\begin{cases}x-2\ge0\\x+1\ge0\end{cases}=>\hept{\begin{cases}x\ge2\\x\ge-1\end{cases}=>x\ge}2}\)

(1)<=> x-2-x-1+3=0

<=>0x=0(đúng với mọi x)

=>x\(\in\left\{x|x\ge2\right\}\)

nếu \(\hept{\begin{cases}x-2< 0\\x+1< 0\end{cases}< =>\hept{\begin{cases}x< 2\\x< -1\end{cases}< =>x< -1}}\)

(1)<=>2-x+x+1+3=0

<=>0x=-3(vô lí)

vậyphương trình đã cho có tập nghiêm là \(x\in\left\{x|x\ge2\right\}\)

\(P=\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\)

áp dụng bunhia - cốpxki

\(P^2=\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\le\left(1+1+1\right)\left(a+b+b+c+c+a\right)\)

\(=6\left(a+b+c\right)\)

\(=6.2021=12126< =>P=\sqrt{12126}\)

vậy MAX P=\(\sqrt{12126}\)

\(P=\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\)

\(\Rightarrow P^2=\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\)

Áp dụng BĐT Bunyakovsky ta có:

\(P^2\le\left(1^2+1^2+1^2\right)\left(a+b+b+c+c+a\right)=6\left(a+b+c\right)=6\cdot2021\)

\(\Rightarrow P\le\sqrt{6\cdot2021}=\sqrt{12126}\)

Dấu "=" xảy ra khi: \(a=b=c=\frac{2021}{3}\)

Vậy \(Max\left(P\right)=\sqrt{12126}\Leftrightarrow a=b=c=\frac{2021}{3}\)

a 

\(\sqrt{3}\cdot\sqrt{75}=\sqrt{3\cdot75}=\sqrt{225}=15\)   

b 

\(\sqrt{72}\cdot\sqrt{18}=6\sqrt{2}\cdot3\sqrt{2}=18\cdot2=36\)   

c 

\(\sqrt{2,5}\cdot\sqrt{30}\cdot\sqrt{48}=\sqrt{2,5\cdot30}\cdot\sqrt{48}=\sqrt{75}\cdot\sqrt{48}=5\sqrt{3}\cdot4\sqrt{3}=20\cdot3=60\)   

d 

\(\sqrt{\frac{5}{49}}\cdot\sqrt{\frac{16}{125}}=\sqrt{\frac{5}{49}\cdot\frac{16}{125}}=\sqrt{\frac{16}{49\cdot25}}=\frac{4}{7\cdot5}=\frac{4}{35}\)

a) Xét tam giác \(BDC\): 

\(\widehat{DBC}=180^o-\widehat{BDC}-\widehat{DCB}=180^o-30^o-60^o=90^o\)

Do đó tam giác \(BDC\)vuông tại \(B\).

Có \(\widehat{BDC}=30^o\)nên \(BC=\frac{1}{2}DC\Rightarrow AB=AC=\frac{1}{2}DC\Rightarrow DC=12\left(cm\right)\).

\(BC^2+BD^2=CD^2\)(định lí Pythagore) 

\(\Leftrightarrow BD^2=CD^2-BC^2=12^2-6^2=108\)

\(\Leftrightarrow BD=6\sqrt{3}\left(cm\right)\)

b) \(S_{ABD}=S_{DBC}-S_{ABC}=\frac{1}{2}.6.6\sqrt{3}-\frac{6^2\sqrt{3}}{4}=9\sqrt{3}\left(cm^2\right)\)

\(a,ĐKXĐ:x\ge0;x\ne1\)

\(P=\left(\frac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+x\right)}{1-\sqrt{x}}+\sqrt{x}\right)\left(\frac{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}+x\right)}{1+\sqrt{x}}-\sqrt{x}\right)\)

\(P=\left(1+\sqrt{x}+x+\sqrt{x}\right)\left(1-\sqrt{x}+x-\sqrt{x}\right)\)

\(P=\left(x+2\sqrt{x}+1\right)\left(x-2\sqrt{x}+1\right)\)

\(P=\left(x+1\right)^2\left(x-1\right)^2\)

\(P=\left[\left(x+1\right)\left(x-1\right)\right]^2\)

\(P=\left(x^2+x-x-1\right)^2\)

\(P=\left(x^2-1\right)^2\)

b, \(7-4\sqrt{3}=2^2-4\sqrt{3}+\sqrt{3}\)

\(\left(2-\sqrt{3}\right)^2\)

\(P=\left(x^2-1\right)^2< \left(2-\sqrt{3}\right)^2\)

\(x^2-1< 2-\sqrt{3}\)

\(x^2< 3-\sqrt{3}\)

\(x< \sqrt{3-\sqrt{3}}\)

a) ĐKXĐ: \(\hept{\begin{cases}x\ge0\\1-\sqrt{x}\ne0\\1+\sqrt{x}\ne0\end{cases}}\) <=> \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

Ta có: \(P=\left(\frac{1-x\sqrt{x}}{1-\sqrt{x}}+\sqrt{x}\right)\left(\frac{1+x\sqrt{x}}{1+\sqrt{x}}-\sqrt{x}\right)\)

\(P=\left(\frac{\left(1-\sqrt{x}\right)\left(x+\sqrt{x}+1\right)}{1-\sqrt{x}}+\sqrt{x}\right)\left(\frac{\left(1+\sqrt{x}\right)\left(x-\sqrt{x}+1\right)}{\left(1+\sqrt{x}\right)}-\sqrt{x}\right)\)

\(P=\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)^2=\left(x-1\right)^2\)

b) Với x > = 0 và x khác 1

Ta có: \(P< 7-4\sqrt{3}\)

<=> \(\left(x-1\right)^2< \left(2-\sqrt{3}\right)^2\)

<=> \(\left(x-1-2+\sqrt{3}\right)\left(x-1+2-\sqrt{3}\right)< 0\)

<=> \(\left(x-3+\sqrt{3}\right)\left(x+1-\sqrt{3}\right)< 0\)

<=> \(\hept{\begin{cases}x-3+\sqrt{3}< 0\\x+1-\sqrt{3}>0\end{cases}}\) hoặc \(\hept{\begin{cases}x-3+\sqrt{3}>0\\x+1-\sqrt{3}< 0\end{cases}}\)

<=> \(\hept{\begin{cases}x< 3-\sqrt{3}\\x>\sqrt{3}-1\end{cases}}\) hoặc \(\hept{\begin{cases}x>3-\sqrt{3}\\x< \sqrt{3}-1\end{cases}}\)

<=> \(\sqrt{3}-1< x< 3-\sqrt{3}\)

\(ĐKXĐ:x\ge0;x\ne1;0\)

\(A=\frac{2x+2}{\sqrt{x}}+\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(A=\frac{2x+2}{\sqrt{x}}+\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}\)

\(A=\frac{2x+2+x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\)

\(A=\frac{2x+2+2\sqrt{x}}{\sqrt{x}}\)

\(A=2\sqrt{x}+\frac{2}{\sqrt{x}}+2\)

a/d bđt cauchy 

\(2\sqrt{x}+\frac{2}{\sqrt{x}}\ge2\sqrt{2.2}=2.2=4\)

\(A\ge4+2=6\)

\(< =>A>5\)

dấu "=" xảy ra khi x=1

a, \(B=\left(\frac{\sqrt{a}+2}{a+2\sqrt{a}+1}-\frac{\sqrt{a}-2}{a-1}\right)\frac{\sqrt{a}+1}{\sqrt{a}}\)ĐKXĐ : \(a>0;a\ne1\)

\(=\left(\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}\right)\frac{\sqrt{a}+1}{\sqrt{a}}\)

\(=\left(\frac{a+\sqrt{a}-2-a+\sqrt{a}+2}{a-1}\right)\frac{1}{\sqrt{a}}\)

\(=\frac{2\sqrt{a}}{\left(a-1\right)\sqrt{a}}=\frac{2}{a-1}\)

b, quá rõ ràng rồi nhé 

a) $ĐKXĐ:x>0$

$A=\frac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+1$

$\iff A=\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+1$

$\iff A=x+\sqrt{x}-2\sqrt{x}-1+1$

$\iff A=x-\sqrt{x}$

b) Để A = 0

$\iff x-\sqrt{x}=0$

$\iff \sqrt{x}\left(\sqrt{x}-1\right)=0$

$\iff [\begin{array}{c}\sqrt{x}=0\\ \sqrt{x}=1\end{array}$

$\iff [\begin{array}{c}x=0\left(ktm\right)\\ x=1\left(tm\right)\end{array}$

vậy ...