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6:ĐKXĐ: x>=0; x<>1/25

BPT=>\(\dfrac{3\sqrt{x}}{5\sqrt{x}-1}+3< =0\)

=>\(\dfrac{3\sqrt{x}+15\sqrt{x}-5}{5\sqrt{x}-1}< =0\)

=>\(\dfrac{18\sqrt{x}-5}{5\sqrt{x}-1}< =0\)

=>\(\dfrac{1}{5}< \sqrt{x}< =\dfrac{5}{18}\)

=>\(\dfrac{1}{25}< x< =\dfrac{25}{324}\)

7:

ĐKXĐ: x>=0

BPT \(\Leftrightarrow\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}>\dfrac{8}{3}:\dfrac{8}{3}=1\)

=>\(\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}-1>=0\)

=>\(\dfrac{\sqrt{x}+1-2\sqrt{x}-3}{2\sqrt{x}+3}>=0\)

=>\(-\sqrt{x}-2>=0\)(vô lý)

8:

ĐKXĐ: x>=0; x<>9/4

BPT \(\Leftrightarrow\dfrac{\sqrt{x}-2}{2\sqrt{x}-3}+4< 0\)

=>\(\dfrac{\sqrt{x}-2+8\sqrt{x}-12}{2\sqrt{x}-3}< 0\)

=>\(\dfrac{9\sqrt{x}-14}{2\sqrt{x}-3}< 0\)

TH1: 9căn x-14>0 và 2căn x-3<0

=>căn x>14/9 và căn x<3/2

=>14/9<căn x<3/2

=>196/81<x<9/4

TH2: 9căn x-14<0 và 2căn x-3>0

=>căn x>3/2 hoặc căn x<14/9

mà 3/2<14/9

nên trường hợp này Loại

9: 

ĐKXĐ: x>=0

\(BPT\Leftrightarrow\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}< =-\dfrac{1}{3}\)

=>\(\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}+\dfrac{1}{3}< =0\)

=>\(\dfrac{6\sqrt{x}+9+5\sqrt{x}+7}{3\left(5\sqrt{x}+7\right)}< =0\)

=>\(\dfrac{11\sqrt{x}+16}{3\left(5\sqrt{x}+7\right)}< =0\)(vô lý)

10: 

ĐKXĐ: x>=0; x<>1/49

\(BPT\Leftrightarrow\dfrac{6\sqrt{x}-2}{7\sqrt{x}-1}+6>0\)

=>\(\dfrac{6\sqrt{x}-2+42\sqrt{x}-6}{7\sqrt{x}-1}>0\)

=>\(\dfrac{48\sqrt{x}-8}{7\sqrt{x}-1}>0\)

=>\(\dfrac{6\sqrt{x}-1}{7\sqrt{x}-1}>0\)

TH1: 6căn x-1>0 và 7căn x-1>0

=>căn x>1/6 và căn x>1/7

=>căn x>1/6

=>x>1/36

TH2: 6căn x-1<0 và 7căn x-1<0

=>căn x<1/6 và căn x<1/7

=>căn x<1/7

=>0<=x<1/49

30 tháng 8 2023

câu 9 nhầm đề bài r bạn

 

27 tháng 6 2018

a) \(\sqrt{\sqrt{2\sqrt{6}+6+2\sqrt{2}+2\sqrt{3}-\sqrt{5+2\sqrt{6}}}}\)

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

b) \(A=\sqrt{x^2-6x+9}-\dfrac{x^2-9}{\sqrt{9-6x+x^2}}\)

\(=\left|x-3\right|-\dfrac{\left(x-3\right)\left(x+3\right)}{\left|x-3\right|}\)

Th1: x-3 < 0

\(A=\left(3-x\right)-\dfrac{\left(x-3\right)\left(x+3\right)}{3-x}=3-x+x-3=0\)

Th2: x-3 > 0

\(A=x-3-\dfrac{\left(x-3\right)\left(x+3\right)}{x-3}=x-3-\left(x+3\right)=-6\)

c)

Đk: x >/ 1 \(B=\dfrac{\sqrt{x+\sqrt{4\left(x-1\right)}}-\sqrt{x-\sqrt{4\left(x-1\right)}}}{\sqrt{x^2-4\left(x-1\right)}}\cdot\left(\sqrt{x-1}-\dfrac{1}{\sqrt{x-1}}\right)\)

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

\(=\dfrac{\sqrt{x-1}+1-\left|\sqrt{x-1}-1\right|}{\left|x-2\right|}\cdot\dfrac{x-2}{\sqrt{x-1}}\)

Th1: \(x-2\ge0\Leftrightarrow x\ge2\)

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

Th2: \(x-2\le0\Leftrightarrow x\le2\)

kết hợp với đk, ta được: 1 \< x \< 2

\(=\dfrac{\sqrt{x-1}+1-\sqrt{x-1}-1}{2-x}\cdot\dfrac{x-2}{\sqrt{x-1}}=0\)

d) \(A=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}=\sqrt{x-2}+\sqrt{2}+\left|\sqrt{x-2}-\sqrt{2}\right|=\sqrt{x-2}+\sqrt{2}-\sqrt{x-2}+\sqrt{2}=2\sqrt{2}\)

chẳng biết có sai sót gì 0 nữa, xin lỗi tớ 0 xem lại đâu vì chán quá!

6: \(\Leftrightarrow2x^2+3x+9+\sqrt{2x^2+3x+9}-42=0\)

Đặt \(\sqrt{2x^2+3x+9}=a\left(a>=0\right)\)

Phương trình sẽ trở thành là: a^2+a-42=0

=>(a+7)(a-6)=0

=>a=-7(loại) hoặc a=6(nhận)

=>2x^2+3x+9=36

=>2x^2+3x-27=0

=>2x^2+9x-6x-27=0

=>(2x+9)(x-3)=0

=>x=3 hoặc x=-9/2

8: \(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)
=>\(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)

=>\(\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=4\\z-3=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\\z=12\end{matrix}\right.\)

Bài 2: 

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

Ta có: \(P=x^2-2x+2020\)

\(=4+2\sqrt{3}-2\left(\sqrt{3}-1\right)+2020\)

\(=4+2\sqrt{3}-2\sqrt{3}+2+2020\)

=2026

Bài 1: 

\(A=-\dfrac{3}{4}\cdot\sqrt{9-4\sqrt{5}}\cdot\sqrt{\left(-8\right)^2\cdot\left(2+\sqrt{5}\right)^2}\)

\(=\dfrac{-3}{4}\cdot8\cdot\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)\)

=-6

Ta có: \(P=\left(\dfrac{4\sqrt{x}}{2+\sqrt{x}}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)

\(=\left(\dfrac{4\sqrt{x}\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}+\dfrac{8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)

\(=\dfrac{8\sqrt{x}-8x+8x}{\left(\sqrt{x}+2\right)\left(2-\sqrt{x}\right)}:\dfrac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\dfrac{-8\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)

\(=\dfrac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)

17 tháng 4 2021

ta có : \(P=\left(\dfrac{4\sqrt{x}}{2+\sqrt{x}}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)

=\(\left(\dfrac{4\sqrt{x}\left(2-\sqrt{x}\right)}{4-x}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2\left(\sqrt{x}-x\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)

=\(\dfrac{8\sqrt{x}-4x+8x}{4-x}:\dfrac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

=\(\dfrac{8\sqrt{x}+4x}{4-x}:\dfrac{3-\sqrt{x}}{\sqrt{x}\left(\sqrt{x-2}\right)}\) =\(\dfrac{4\sqrt{x}\left(2+\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}:\dfrac{3-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

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

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

này mới đúng !!

 

30 tháng 7 2021

a) \(B=\left(\dfrac{x-3\sqrt{x}}{x-9}-1\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)

\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-1\right):\left(\dfrac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)

\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-1\right):\dfrac{9-x+\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

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

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

b) \(\sqrt{x}=\sqrt{7-4\sqrt{3}}=\sqrt{2^2-2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(2-\sqrt{3}\right)^2}\)

\(=\left|2-\sqrt{3}\right|=2-\sqrt{3}\)

Thế vào B \(\Rightarrow B=\dfrac{3}{2-\sqrt{3}-2}=\dfrac{3}{-\sqrt{3}}=-\sqrt{3}\)

a) Ta có: \(B=\left(\dfrac{x-3\sqrt{x}}{x-9}-1\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)

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

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

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

3 tháng 9 2021

\(C=\left(\dfrac{\sqrt{x}+1}{x-4}-\dfrac{\sqrt{x}-1}{x+4\sqrt{x}+4}\right).\dfrac{x\sqrt{x}+2x-4\sqrt{x}-8}{\sqrt{x}-2}\)

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

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

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

\(=\dfrac{6\sqrt{x}}{\sqrt{x}-2}\)

3 tháng 9 2021

\(C=\left[\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+2\right)^2}\right].\dfrac{\sqrt{x}\left(x-4\right)+2\left(x-4\right)}{\sqrt{x}-2}\) (\(x\ge0,x\ne4,x\ne9\))
\(C=\left[\dfrac{\sqrt{x}+1-\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)^2}\right].\dfrac{\left(\sqrt{x}+2\right)\left(x-4\right)}{\sqrt{x}-2}\)
\(C=\dfrac{2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)^2}.\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+2\right)\left(\sqrt{x-2}\right)}{\sqrt{x}-2}\)
\(C=\dfrac{2}{\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)^2}.\left(\sqrt{x}+2\right)^2\)
\(C=\dfrac{2}{\sqrt{x}-2}\)
 

24 tháng 9 2021

1)\(=\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{26^2}=\sqrt{5}-2+26=24-\sqrt{5}\)

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

3) \(=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{\left|x-1\right|}{x-1}\)\(=\left[{}\begin{matrix}1\left(x>1\right)\\-1\left(x< 1\right)\end{matrix}\right.\)

4) \(=\sqrt{\left(\sqrt{\dfrac{7}{2}}+\sqrt{\dfrac{1}{2}}\right)^2}-\sqrt{\left(\sqrt{\dfrac{7}{2}}-\sqrt{\dfrac{1}{2}}\right)^2}=\sqrt{\dfrac{7}{2}}+\sqrt{\dfrac{1}{2}}-\sqrt{\dfrac{7}{2}}+\sqrt{\dfrac{1}{2}}=2\sqrt{\dfrac{1}{2}}=\sqrt{2}\)

24 tháng 9 2021

2. \(\dfrac{x^2-5}{x+\sqrt{5}}=\dfrac{x^2-\left(\sqrt{5}\right)^2}{x+\sqrt{5}}=\dfrac{\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)}{x+\sqrt{5}}=x-\sqrt{5}\)

3. \(\dfrac{\sqrt{x^2-2x+1}}{x-1}=\dfrac{\sqrt{x^2-2.x.1+1^2}}{x-1}=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{|x-1|}{x-1}=\left[{}\begin{matrix}x-1>0\left(x>1\right)\\x-1< 0\left(x< 1\right)\end{matrix}\right.=\left[{}\begin{matrix}=1\\=\dfrac{x+1}{x-1}\end{matrix}\right.\)

1) Ta có: \(\left\{{}\begin{matrix}2x+y=5\\3x-2y=11\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x+3y=15\\6x-4y=22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7y=-7\\2x+y=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=-1\\2x=5-y=5-\left(-1\right)=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)

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

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

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

\(=\dfrac{3x-6\sqrt{x}}{\sqrt{x}-2}\)

\(=3\sqrt{x}\)