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26 tháng 6 2018

b) \(\dfrac{16}{\sqrt{x-3}}+\dfrac{4}{\sqrt{y-1}}+\dfrac{1225}{\sqrt{z-665}}=82-\sqrt{x-3}-\sqrt{y-1}-\sqrt{z-665}\) (*)

Đk: \(\left\{{}\begin{matrix}x>3\\y>1\\z>665\end{matrix}\right.\)

(*) \(\Leftrightarrow\dfrac{16}{\sqrt{x-3}}+\dfrac{4}{\sqrt{y-1}}+\dfrac{1225}{\sqrt{z-665}}=82-\dfrac{x-3}{\sqrt{x-3}}-\dfrac{y-1}{\sqrt{y-1}}-\dfrac{z-665}{\sqrt{z-665}}\)

\(\Leftrightarrow\dfrac{16}{\sqrt{x-3}}+\dfrac{4}{\sqrt{y-1}}+\dfrac{1225}{\sqrt{z-665}}-82+\dfrac{x-3}{\sqrt{x-3}}+\dfrac{y-1}{\sqrt{y-1}}+\dfrac{z-665}{\sqrt{z-665}}=0\)

\(\Leftrightarrow\left(\dfrac{x-3}{\sqrt{x-3}}-\dfrac{8\sqrt{x-3}}{\sqrt{x-3}}+\dfrac{16}{\sqrt{x-3}}\right)+\left(\dfrac{y-1}{\sqrt{y-1}}-\dfrac{4\sqrt{y-1}}{\sqrt{y-1}}+\dfrac{4}{\sqrt{y-1}}\right)+\left(\dfrac{z-665}{\sqrt{z-665}}-\dfrac{70\sqrt{z-665}}{\sqrt{z-665}}+\dfrac{1225}{\sqrt{z-665}}\right)=0\)

\(\Leftrightarrow\dfrac{\left(\sqrt{x-3}-4\right)^2}{\sqrt{x-3}}+\dfrac{\left(\sqrt{y-1}-2\right)^2}{\sqrt{y-1}}+\dfrac{\left(\sqrt{z-665}-35\right)^2}{\sqrt{z-665}}=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-3}-4=0\\\sqrt{y-1}-2=0\\\sqrt{z-665}-35=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=19\\y=5\\z=1890\end{matrix}\right.\)

Kl: x=19, y= 5, z=1890

26 tháng 6 2018

c) \(\sqrt{x-5}-\dfrac{x-14}{3+\sqrt{x-5}}=3\) (*)

Đk: \(x\ge5\)

(*) \(\Leftrightarrow3\sqrt{x-5}+x-5-x+14=9+3\sqrt{x-5}\)

\(\Leftrightarrow0x=0\) (luôn đúng)

Vậy nghiệm của phương trình (*) là \(x\ge5\)

25 tháng 8 2018

ĐK : \(x\ge3;y\ge1;z\ge665\)

\(\dfrac{16}{\sqrt{x-3}}+\dfrac{4}{\sqrt{y-1}}+\dfrac{1225}{\sqrt{z-665}}=82-\sqrt{x-3}-\sqrt{y-1}-\sqrt{z-665}\)

\(\Leftrightarrow\left(\dfrac{16}{\sqrt{x-3}}+\sqrt{x-3}\right)+\left(\dfrac{4}{\sqrt{y-1}}+\sqrt{y-1}\right)+\left(\dfrac{1225}{\sqrt{z-665}}+\sqrt{z-665}\right)=82\)

Theo BĐT Cô Si cho các số dương ta có :

\(\left\{{}\begin{matrix}\dfrac{16}{\sqrt{x-3}}+\sqrt{x-3}\ge2\sqrt{\dfrac{16\sqrt{x-3}}{\sqrt{x-3}}}=2\sqrt{16}=8\\\dfrac{4}{\sqrt{y-1}}+\sqrt{y-1}\ge2\sqrt{\dfrac{4\sqrt{y-1}}{\sqrt{y-1}}}=2\sqrt{4}=4\\\dfrac{1225}{\sqrt{z-665}}+\sqrt{z-665}\ge2\sqrt{\dfrac{1225\sqrt{z-665}}{\sqrt{z-665}}}=2\sqrt{1225}=70\end{matrix}\right.\)

\(\Rightarrow\left(\dfrac{16}{\sqrt{x-3}}+\sqrt{x-3}\right)+\left(\dfrac{4}{\sqrt{y-1}}+\sqrt{y-1}\right)+\left(\dfrac{1225}{\sqrt{z-665}}+\sqrt{z-665}\right)\ge82\)

Dấu \("="\) hiển nhiên xảy ra khi :

\(\left\{{}\begin{matrix}\dfrac{16}{\sqrt{x-3}}=\sqrt{x-3}\\\dfrac{4}{\sqrt{y-1}}=\sqrt{y-1}\\\dfrac{1225}{\sqrt{z-665}}=\sqrt{z-665}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-3=16\\y-1=4\\z-665=1225\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=19\\y=5\\z=1890\end{matrix}\right.\)

8 tháng 7 2023

\(a,\dfrac{3}{5}-\dfrac{1}{2}\sqrt{1\dfrac{11}{25}}=\dfrac{3}{5}-\dfrac{1}{2}\sqrt{\dfrac{36}{25}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{\sqrt{6^2}}{\sqrt{5^2}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{6}{5}=\dfrac{3}{5}-\dfrac{6}{10}=\dfrac{3}{5}-\dfrac{3}{5}=0\)

\(b,\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)=5^2-\left(2\sqrt{6}\right)^2=25-2^2.\sqrt{6^2}=25-4.6=25-24=1\)

\(c,\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\\ =\left|2-\sqrt{3}\right|+\sqrt{\sqrt{3^2}-2\sqrt{3}+1}\\ =2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\\ =2-\sqrt{3}+\left|\sqrt{3}-1\right|\\ =2-\sqrt{3}+\sqrt{3}-1\\ =1\)

\(d,\dfrac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\left(dk:x,y>0\right)\\ =\dfrac{\left(\sqrt{x^2}.\sqrt{y}+\sqrt{y^2}.\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\sqrt{x^2}-\sqrt{y^2}\\ =\left|x\right|-\left|y\right|\\ =x-y\)

12 tháng 10 2022

a: \(=4+\sqrt{11}+\dfrac{3}{2}-\dfrac{1}{2}\sqrt{7}-4-2\sqrt{7}-\dfrac{1}{2}\sqrt{7}+\dfrac{5}{2}\)

\(=4+\sqrt{11}-3\sqrt{7}\)

b: \(VT=\dfrac{x+2\sqrt{xy}+y-x+2\sqrt{xy}-y+2x+2y}{2\left(x-y\right)}\)

\(=\dfrac{2x+4\sqrt{xy}+2y}{2\left(x-y\right)}=\dfrac{x+2\sqrt{xy}+y}{x-y}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)

20 tháng 8 2021

giups mình với

 

Bài 1:

\(\sqrt{\left(4-\sqrt{5}\right)^2}+\sqrt{5+2\sqrt{5}+1}\)

\(=\left|4-\sqrt{5}\right|+\sqrt{\left(\sqrt{5}+1\right)^2}\)

\(=4-\sqrt{5}+\sqrt{5}+1=5\)

Bài 2:

a: ĐKXĐ: x>=3

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

=>x-3=36

=>x=36+3=39(nhận)

b: ĐKXĐ: \(x\in R\)

\(\sqrt{\left(x-3\right)^2}=12\)

=>\(\left|x-3\right|=12\)

=>\(\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)

Bài 3:

a: \(P=\left(\dfrac{3-x\sqrt{x}}{3-\sqrt{x}}+\sqrt{x}\right)\cdot\left(\dfrac{3-\sqrt{x}}{3-x}\right)\)

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

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

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

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

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

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

c: \(A=\sqrt{3x-1}+3\cdot\sqrt{12x-4}-\sqrt{6^2\left(3x-1\right)}+\sqrt{5}\)

\(=\sqrt{3x-1}+6\sqrt{3x-1}-6\sqrt{3x-1}+\sqrt{5}\)

\(=\sqrt{3x-1}+\sqrt{5}\)

d: \(A=\left(\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\dfrac{a+2}{a-2}\)

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

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

\(=\dfrac{2\left(a-2\right)}{a+2}\)

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

 

a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)

\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)

\(\Leftrightarrow3\sqrt{x+5}=6\)

\(\Leftrightarrow x+5=4\)

hay x=-1

b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)

\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

\(\Leftrightarrow\sqrt{x-1}=17\)

\(\Leftrightarrow x-1=289\)

hay x=290

28 tháng 8 2021

\(1,ĐKx\ge5\)

\(\sqrt{\left(x-5\right)\left(x+5\right)}+2\sqrt{x-5}=3\sqrt{x+5}+6\)

\(\Rightarrow\sqrt{x-5}\left(\sqrt{x+5}+2\right)-3\left(\sqrt{x+5}+2\right)=0\)

\(\Rightarrow\left(\sqrt{x+5}+2\right)\left(\sqrt{x-5}-3\right)=0\)

\(\left[{}\begin{matrix}\sqrt{x+5}=-2loại\\\sqrt{x-5}=3\end{matrix}\right.\)\(\Rightarrow x-5=9\Rightarrow x=14\)(TMĐK)

2a,ĐK \(x\ge0;x\ne9\)

,\(B=\dfrac{7\left(3-\sqrt{x}\right)-12}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}=\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}\)

\(M=\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{x-6\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)

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

 

 

 

a) Ta có: \(\dfrac{6}{\sqrt{5}+1}+\sqrt{\dfrac{2}{3-\sqrt{5}}}-\dfrac{10}{\sqrt{5}}\)

\(=\dfrac{6\left(\sqrt{5}-1\right)}{4}+\sqrt{\dfrac{2\left(3+\sqrt{5}\right)}{4}}-2\sqrt{5}\)

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

\(=\dfrac{3}{2}\sqrt{5}-\dfrac{3}{2}-2\sqrt{5}+\dfrac{\sqrt{5}+1}{2}\)

\(=-\dfrac{1}{2}\sqrt{5}-\dfrac{3}{2}+\dfrac{1}{2}\sqrt{5}+\dfrac{1}{2}\)

=-1

 

Bài 1: 

a) Thay \(x=\dfrac{1}{4}\)vào B, ta được:

\(B=1:\left(\dfrac{1}{4}\cdot\dfrac{1}{2}+27\right)=1:\left(27+\dfrac{1}{8}\right)=\dfrac{8}{217}\)

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

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

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

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

\(=\dfrac{3}{\sqrt{x}+3}\)

c) Để \(A>\dfrac{1}{2}\) thì \(A-\dfrac{1}{2}>0\)

\(\Leftrightarrow\dfrac{6-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>0\)

\(\Leftrightarrow3-\sqrt{x}>0\)

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

hay x<9

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne4\end{matrix}\right.\)