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

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

b: \(m\left(\sqrt{x}-3\right)\cdot B>x+1\)

=>\(-4xm\left(3\sqrt{x}-2\right)>\left(\sqrt{x}+2\right)\cdot\left(x+1\right)\)

=>\(-12m\cdot x\sqrt{x}+8xm>x\sqrt{x}+2x+\sqrt{x}+2\)

=>\(x\sqrt{x}\left(-12m-1\right)+x\left(8m-2\right)-\sqrt{x}-2>0\)

Để BPT luôn đúng thì m<-0,3

23 tháng 6 2021

a) \(P=\dfrac{4\sqrt{x}\left(2-\sqrt{x}\right)+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\dfrac{\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ =\dfrac{8\sqrt{x}+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\dfrac{3-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ =\dfrac{8\sqrt{x}+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}=\dfrac{4x}{\sqrt{x}-3}\)

\(\left(x\ge0;x\ne4;9\right)\)

b)\(P=-1\Leftrightarrow4x+\sqrt{x}-3=0\Leftrightarrow\sqrt{x}=\dfrac{3}{4}\Leftrightarrow x=\dfrac{9}{16}\)

c) bpt đưa về dạng \(4mx>x+1\Leftrightarrow\left(4x-1\right)x>1\)

Nếu \(4m-1\le0\) thì tập nghiệm không thể chứa mọi giá trị \(x>9\); Nếu \(4m-1>0\) thì tập nghiệm bpt là \(x>\dfrac{1}{4m-1}\). Do đó bpt tm mọi \(x>9\Leftrightarrow9\ge\dfrac{1}{4m-1}\) và \(4m-1>0\). ta có \(m\ge\dfrac{5}{18}\)

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

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

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

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

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

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

7 tháng 7 2021

a) \(A=\sqrt{28}-\sqrt{63}+\dfrac{7+\sqrt{7}}{\sqrt{7}}-\sqrt{\left(\sqrt{7}+1\right)^2}\)

\(=2\sqrt{7}-3\sqrt{7}+\dfrac{\sqrt{7}\left(\sqrt{7}+1\right)}{\sqrt{7}}-\left|\sqrt{7}+1\right|\)

\(=-\sqrt{7}+\sqrt{7}+1-\sqrt{7}-1=-\sqrt{7}\)

\(B=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}-3}\right)\dfrac{4\sqrt{x}+12}{\sqrt{x}}\)

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

\(=\dfrac{8}{\sqrt{x}-3}\)

b) \(A>B\Rightarrow-\sqrt{7}>\dfrac{8}{\sqrt{x}-3}\Rightarrow\dfrac{8}{\sqrt{x}-3}+\sqrt{7}< 0\)

\(\Rightarrow\dfrac{\sqrt{7x}+8-3\sqrt{7}}{\sqrt{x}-3}< 0\)

Ta có: \(\left\{{}\begin{matrix}8=\sqrt{64}\\3\sqrt{7}=\sqrt{63}\end{matrix}\right.\Rightarrow8-3\sqrt{7}>0\Rightarrow8-3\sqrt{7}+\sqrt{7x}>0\)

\(\Rightarrow\sqrt{x}-3< 0\Rightarrow\sqrt{x}< 3\Rightarrow x< 9\Rightarrow0< x< 9\)

 

6 tháng 9 2021

rút gọn R ?

6 tháng 9 2021

P

 

25 tháng 5 2023

\(a,\) Rút gọn 

\(A=\dfrac{3}{\sqrt{7}-2}+\sqrt{\left(\sqrt{7}-3\right)^2}\)

\(=\dfrac{3}{\sqrt{7}-2}+\left|\sqrt{7}-3\right|\)

\(=\dfrac{3}{\sqrt{7}-2}+3-\sqrt{7}\)

\(=\dfrac{3+\left(3-\sqrt{7}\right)\left(\sqrt{7}-2\right)}{\sqrt{7}-2}\)

\(=\dfrac{3+3\sqrt{7}-6-7+2\sqrt{7}}{\sqrt{7}-2}\)

\(=\dfrac{5\sqrt{7}-10}{\sqrt{7}-2}\)

\(=\dfrac{5\left(\sqrt{7}-2\right)}{\sqrt{7}-2}\)

\(=5\)

Vậy \(A=5\)

\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}}{x-\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{x-1}\left(dkxd:x\ge0,x\ne1\right)\)

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

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

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

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

Vậy \(B=\sqrt{x}-1\)

\(b,\) Để \(B< A\) thì \(\sqrt{x}-1< 5\)

\(\Leftrightarrow\sqrt{x}< 6\)

\(\Leftrightarrow x< 36\)

10 tháng 7 2021

a) \(B=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\left(x\ge0,x\ne1\right)\)

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

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

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

 

10 tháng 7 2021

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AH
Akai Haruma
Giáo viên
13 tháng 7 2018

Lời giải:

ĐK: \(x>0; x\neq 4\)

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

\(=\frac{8\sqrt{x}-4x+8x}{(2+\sqrt{x})(2-\sqrt{x})}: \frac{\sqrt{x}-1-2(\sqrt{x}-2)}{\sqrt{x}(\sqrt{x}-2)}\)

\(=\frac{8\sqrt{x}+4x}{(2+\sqrt{x})(2-\sqrt{x})}.\frac{\sqrt{x}(\sqrt{x}-2)}{-\sqrt{x}+3}\)

\(=\frac{4\sqrt{x}(2+\sqrt{x})}{2+\sqrt{x}}. \frac{-\sqrt{x}}{3-\sqrt{x}}=\frac{-4\sqrt{x}.\sqrt{x}}{3-\sqrt{x}}=\frac{4x}{\sqrt{x}-3}\)

b)

\(K=-1\Leftrightarrow \frac{4x}{\sqrt{x}-3}=-1\Rightarrow 4x=-(\sqrt{x}-3)\)

\(\Leftrightarrow 4x+\sqrt{x}-3=0\)

\(\Leftrightarrow (4\sqrt{x}-3)(\sqrt{x}+1)=0\)

\(\sqrt{x}+1>0\Rightarrow 4\sqrt{x}-3=0\Rightarrow x=\frac{9}{16}\)

c) \(m(\sqrt{x}-3)K>x+1\)

\(\Leftrightarrow m. (\sqrt{x}-3).\frac{4x}{\sqrt{x}-3}>x+1\)

\(\Leftrightarrow m> \frac{x+1}{4x}\)

\(\Leftrightarrow m> max(\frac{4x}{x+1}), \forall x< 9\)

Với đk đã cho thì ta thấy \(\frac{4x}{x+1}\) có min thôi.

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

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

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

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

ĐKXĐ: x>=0; x<>1

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

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

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

b: Khi x=4-2căn 3=(căn 3-1)^2 thì \(B=\dfrac{\sqrt{3}-1}{\sqrt{3}-1+1}=\dfrac{\sqrt{3}-1}{\sqrt{3}}=\dfrac{3-\sqrt{3}}{3}\)

c: B=2/3

=>căn x/căn x+1=2/3

=>căn x=2

=>x=4

d: \(B-1=\dfrac{\sqrt{x}-\sqrt{x}-1}{\sqrt{x}+1}=-\dfrac{1}{\sqrt{x}+1}< 0\)

=>B<1

e: B>1

=>-1/căn x+1>0

=>căn x+1<0(vô lý)

=>KO có x thỏa mãn

f: B nguyên khi căn x chia hết cho căn x+1

=>căn x+1-1 chia hết cho căn x+1

=>căn x+1=1 hoặc căn x+1=-1(loại)

=>căn x=0

=>x=0