\(M=\left(\dfrac{x\sqrt{x}-1}{\sqrt{x}}-\dfrac{x-1}{\sqrt{x}+1}\right):\left(\sqrt{x}-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)
a) rút gọn và tìm x để M nguyên
cho biểu thức A=\(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\)với x≥0,x≠1
a)rút gọn A
b)tìm x nguyên để M =A.\(\dfrac{\sqrt{x}+2}{2\sqrt{x}+1}+\dfrac{x-\sqrt{x}-5}{\sqrt{x}+3}\)có giá trị nguyên
a: \(A=\dfrac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)
1) M=\(\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right):\left(\dfrac{2}{x}-\dfrac{2-x}{x\sqrt{x}+x}\right)\)
a) Rút gọn M ( đkxđ )
b) Tìm x để M= - 1/2
c) Tìm x để M >1 ; M<1
2. \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\)
a. Tim ĐKXĐ rồi rút gọn A
b. Tính giá trị của A với x =36
c. Tìm x để \(\left|A\right|>A\)
3. \(M=\left|\dfrac{1}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x}+3}\right|:\dfrac{3}{\sqrt{x}-3}\)
a. Tìm ĐKXĐ rồi rút gọn M
b. Tìm x để M > \(\dfrac{1}{3}\)
c. Tìm x để biểu thức M đạt được giá trị lớn nhất, tìm giá trị lớn nhất đó
help me
3:
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >9\end{matrix}\right.\)
\(M=\left(\dfrac{1}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x}+3}\right):\dfrac{3}{\sqrt{x}-3}\)
\(=\dfrac{\sqrt{x}+3-\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{3}\)
\(=\dfrac{6}{3\left(\sqrt{x}+3\right)}=\dfrac{2}{\sqrt{x}+3}\)
b: M>1/3
=>M-1/3>0
=>\(\dfrac{2}{\sqrt{x}+3}-\dfrac{1}{3}>0\)
=>\(\dfrac{6-\sqrt{x}-3}{3\left(\sqrt{x}+3\right)}>0\)
=>\(3-\sqrt{x}>0\)
=>\(\sqrt{x}< 3\)
=>0<=x<9
c: \(\sqrt{x}+3>=3\) với mọi x thỏa mãn ĐKXĐ
=>\(M=\dfrac{2}{\sqrt{x}+3}< =\dfrac{2}{3}\) với mọi x thỏa mãn ĐKXĐ
Dấu = xảy ra khi x=0
Cho biểu thức \(M=\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{6\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\) với \(x\ge0;x\ne1\)
a. Rút gọn M
b. Tìm số nguyên x để M có giá trị là số nguyên
a) \(M=\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{6\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\left(x\ge0,x\ne1\right)\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)-6\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{x-4\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\)
b) \(M=\dfrac{\sqrt{x}-3}{\sqrt{x}+2}=1-\dfrac{5}{\sqrt{x}+2}\in Z\)
\(\Rightarrow\sqrt{x}+2\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)
Do \(\sqrt{x}\ge0\forall x\)
\(\Rightarrow\sqrt{x}\in\left\{3\right\}\Rightarrow x=9\left(tm\right)\)
Cho biểu thức M = \(\left(\dfrac{2x+3\sqrt{x}}{x\sqrt{x}+1}+\dfrac{1}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right).\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)
a, Rút gọn biểu thức
b, So sánh M và 1
c, Tìm x ∈ R để M có giá trị là số nguyên
a) \(M=\left(\dfrac{2x+3\sqrt{x}}{x\sqrt{x}+1}+\dfrac{1}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right).\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\left(x>0\right)\)
\(=\left(\dfrac{2x+3\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}+1}\right).\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{2x+3\sqrt{x}+1-\left(x-\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}.\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{x+4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}.\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+4\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}.\dfrac{x-\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}+4}{\sqrt{x}+1}\)
b) Ta có: \(\sqrt{x}+4>\sqrt{x}+1\Rightarrow\dfrac{\sqrt{x}+4}{\sqrt{x}+1}>1\)
c) \(\dfrac{\sqrt{x}+4}{\sqrt{x}+1}=1+\dfrac{3}{\sqrt{x}+1}\)
Ta có: \(\left\{{}\begin{matrix}3>0\\\sqrt{x}+1>0\end{matrix}\right.\Rightarrow1+\dfrac{3}{\sqrt{x}+1}>1\Rightarrow M>1\)
Lại có: \(\sqrt{x}+1>1\left(x>0\right)\Rightarrow\dfrac{3}{\sqrt{x}+1}< 3\Rightarrow1+\dfrac{3}{\sqrt{x}+1}< 4\Rightarrow M< 4\)
\(\Rightarrow1< M< 4\Rightarrow M\in\left\{2;3\right\}\)
\(M=2\Rightarrow1+\dfrac{3}{\sqrt{x}+1}=2\Rightarrow\dfrac{3}{\sqrt{x}+1}=1\Rightarrow\sqrt{x}+1=3\)
\(\Rightarrow\sqrt{x}=2\Rightarrow x=4\)
\(M=3\Rightarrow1+\dfrac{3}{\sqrt{x}+1}=3\Rightarrow\dfrac{3}{\sqrt{x}+1}=2\Rightarrow2\sqrt{x}+2=3\)
\(\Rightarrow2\sqrt{x}=1\Rightarrow\sqrt{x}=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{4}\)
\(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}+\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\times\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
rút gọn biểu thức và tìm a để N=7
\(N=\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}+\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1}{x-1}\)
\(=2+\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{2x+2}{x-1}=2+\dfrac{2x+2}{\sqrt{x}}=\dfrac{2x+2\sqrt{x}+2}{\sqrt{x}}\)
N=7
=>2x+2căn x+2=7căn x
=>2x-5căn x+2=0
=>(2căn x-1)(căn x-2)=0
=>x=4 hoặc x=1/4
4.A=\(\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right)\): \(\dfrac{2\left(x-2\sqrt{x}+1\right)}{x-1}\)
a) Rút gọn A
b)Tìm x nguyên để A có giá trị nguyên
a. \(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{2\left(x-2\sqrt{x}+1\right)}{x-1}\)
\(=\left(\dfrac{\left(x\sqrt{x}-1\right)\left(x+\sqrt{x}\right)-\left(x\sqrt{x}+1\right)\left(x-\sqrt{x}\right)}{\left(x-\sqrt{x}\right)\left(x+\sqrt{x}\right)}\right):\dfrac{2\left(\sqrt{x}-1\right)^2}{x-1}\)
\(=\left(\dfrac{\left(x\sqrt{x}-1\right)\left(x+\sqrt{x}\right)-\left(x\sqrt{x}+1\right)\left(x-\sqrt{x}\right)}{x^2-x}\right).\dfrac{x-1}{2\left(\sqrt{x}-1\right)^2}\)
\(=\left(\dfrac{x^2\sqrt{x}+x^2-x-\sqrt{x}-\left(x^2\sqrt{x}-x^2+x-\sqrt{x}\right)}{x^2-x}\right).\dfrac{x-1}{2\left(\sqrt{x}-1\right)^2}\)
\(=\left(\dfrac{x^2\sqrt{x}+x^2-x-\sqrt{x}-x^2\sqrt{x}+x^2-x+\sqrt{x}}{x^2-x}\right).\dfrac{x-1}{2\left(\sqrt{x}-1\right)^2}\)
\(=\dfrac{2x^2-2x}{x^2-x}.\dfrac{x-1}{2\left(\sqrt{x}-1\right)^2}\)
\(=\dfrac{2\left(x^2-x\right)}{x^2-x}.\dfrac{x-1}{2\left(\sqrt{x}-1\right)^2}\)
\(=2.\dfrac{x-1}{2\left(\sqrt{x}-1\right)^2}=\dfrac{x-1}{\left(\sqrt{x}-1\right)^2}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b. \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1+2}{\sqrt{x}-1}=1+\dfrac{2}{\sqrt{x}-1}\)
Để A có giá trị nguyên \(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}\in Z\) \(\Leftrightarrow2⋮\left(\sqrt{x}-1\right)\)\(\Leftrightarrow\left(\sqrt{x}-1\right)\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)\(\Leftrightarrow\sqrt{x}\in\left\{2;0;3;-1\right\}\)
Vì \(\sqrt{x}\ge0\Rightarrow\sqrt{x}\in\left\{2;0;3\right\}\Leftrightarrow x\in\left\{4;0;9\right\}\)
Vậy để A có giá trị nguyên thì \(x\in\left\{4;0;9\right\}\)
Cho B=\(\left(\dfrac{4\sqrt{x}}{2+\sqrt{x}}-\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)
a)Rút gọn B
b)Tìm m để với mọi giá trị x>9 ta có \(m\left(\sqrt{x}-3\right)B>x+1\)
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
\(\left(\dfrac{2\sqrt{x}}{\sqrt{x+3}}+\dfrac{\sqrt{x}}{\sqrt{x-3}}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{\sqrt{x-1}}{\sqrt{x-3}}-\dfrac{1}{2}\right)\)
rút gọn A
tìm các giá trị x nguyên để biểu thức trên nguyên