rút gọn biểu thức sau:
\(\dfrac{x^2+\sqrt{x}}{x\sqrt{x}+1}+1-\dfrac{2x+\sqrt{x}}{\sqrt{x}}\left(x>0\right)\)
Câu 1: Rút gọn biểu thức: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{2}}+\dfrac{6}{x+3\sqrt{x}}\right)\) với x > 0
Câu 2: Rút gọn biểu thức:
\(P=\dfrac{x\sqrt{2}}{2\sqrt{x}+x\sqrt{2}}+\dfrac{\sqrt{2x}-2}{x-2}\) với x > 0; x \(\ne\) 2
Câu 3: Rút gọn biểu thức:
\(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\) với a > 0; a \(\ne\) 4
Câu 1:
Sửa đề: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
Ta có: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}\right):\left(\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\sqrt{x}-6+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x+\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=1\)
Câu 3:
Ta có: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)
\(=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\)
\(=\dfrac{a+\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-2}{1}\)
\(=\sqrt{a}\left(\sqrt{a}-2\right)\)
\(=a-2\sqrt{a}\)
Rút gọn biểu thức:
1, \(B=\left(\dfrac{x.\sqrt{x}+x+\sqrt{x}}{x.\sqrt{x}-1}-\dfrac{\sqrt{x}+3}{1-\sqrt{x}}\right).\dfrac{x-1}{2x+\sqrt{x}-1}\)với x>-0, x khác 1, x khác \(\dfrac{1}{4}\)
2, \(A=\dfrac{\left(\sqrt{x}-1\right)^2.\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{3\sqrt{x}+1}{x-1}\) với x\(\ge\)0:x\(\ne\)0
Rút gọn Biểu thức sau:
\(P=\left(\dfrac{1}{2\sqrt{x}}-\dfrac{\sqrt{x}}{2}\right)^2.\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right)\)với x lớn hơn 0 và x khác 1
\(P=\left(\dfrac{1}{2\sqrt{x}}-\dfrac{x}{2\sqrt{x}}\right)^2.\left(\dfrac{\left(\sqrt{x}-1\right)^2}{x-1}-\dfrac{\left(\sqrt{x}+1\right)^2}{x-1}\right)\)
\(=\left(\dfrac{1-x}{2\sqrt{x}}\right)^2.\left(\dfrac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{x-1}\right)\)
\(=\dfrac{\left(1-x\right)^2}{2\sqrt{x}}.\dfrac{-4\sqrt{x}}{-\left(1-x\right)}\)
\(=\left(1-x\right).2\sqrt{x}\)
\(=2\sqrt{x}-2x\sqrt{x}\)
34. Cho biểu thức A= \(\left(\dfrac{x+4\sqrt{x}+4}{x+\sqrt{x}-2}+\dfrac{x+\sqrt{x}}{1-x}\right):\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{1-\sqrt{x}}\right)\)
a. Rút gọn A
c. Tính A với x thỏa mãn \(2x-3\sqrt{x}-5=0\)
a: \(A=\left(\dfrac{x+4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}-1+\sqrt{x}+1}{x-1}\)
\(=\dfrac{x+4\sqrt{x}+4-x-2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{x-1}{2\sqrt{x}}\)
\(=\dfrac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{2\sqrt{x}+2}{\sqrt{x}}\)
c: 2x-3căn x-5=0
=>2x-5căn x+2căn x-5=0
=>2căn x-5=0
=>x=25/4
Khi x=25/4 thì \(A=\dfrac{2\cdot\dfrac{5}{4}+2}{\dfrac{5}{4}}=\dfrac{18}{5}\)
Rút gọn biểu thức:
\(B=\left(\dfrac{2x+1}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{x}{x+\sqrt{x}+1}\right)\) với \(0\le x\ne1\)
Ta có: \(B=\left(\dfrac{2x+1}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{x}{x+\sqrt{x}+1}\right)\)
\(=\dfrac{2x\sqrt{x}-2x+\sqrt{x}-1-x\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}:\dfrac{x+\sqrt{x}+1-x}{x+\sqrt{x}+1}\)
\(=\dfrac{x\sqrt{x}-2x+\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{\sqrt{x}+1}\)
\(=\dfrac{\left(x+1\right)\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2\cdot\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x+\sqrt{x}+1\right)}{\left(x-1\right)\left(x-\sqrt{x}+1\right)}\)
1) cho biểu thức A= \(\dfrac{x^2-\sqrt{x}}{x-\sqrt{x}+1}\) - \(\dfrac{2x+\sqrt{x}}{\sqrt{x}}\) + \(\dfrac{2.\left(x-1\right)}{\sqrt{x}-1}\) ( x>0; x ≠1)
a) Rút gọn biểu thức A
b) Tìm giá trị nhỏ nhất của 4
Lời giải:
a.
\(A=\frac{\sqrt{x}(\sqrt{x^3}-1)}{x+\sqrt{x}+1}-\frac{\sqrt{x}(2\sqrt{x}+1)}{\sqrt{x}}+\frac{2(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}(\sqrt{x}-1)(x+\sqrt{x}+1)}{x+\sqrt{x}+1}-(2\sqrt{x}+1)+2(\sqrt{x}+1)\)
\(=\sqrt{x}(\sqrt{x}-1)-2\sqrt{x}-1+2\sqrt{x}+2\\ =x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\\ =x-\sqrt{x}+1\)
b.
$A=x-\sqrt{x}+1=(x-\sqrt{x}+\frac{1}{4})+\frac{3}{4}$
$=(\sqrt{x}-\frac{1}{2})^2+\frac{3}{4}\geq 0+\frac{3}{4}=\frac{3}{4}$
$\Rightarrow A_{\min}=\frac{3}{4}$
Giá trị này đạt tại $\sqrt{x}-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{4}$
cho biểu thức P=\(\left(\dfrac{1}{\sqrt{x}-\sqrt{x-1}}-\dfrac{x-3}{\sqrt{x-1}-\sqrt{2}}\right)\left(\dfrac{2}{\sqrt{2}-\sqrt{x}}-\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{2x}-x}\right)\)
a,rút gọn biểu thức
b,tính giá trị của biểu thức với x=3 - \(2\sqrt{2}\)
a) Ta có: \(P=\left(\dfrac{1}{\sqrt{x}-\sqrt{x-1}}-\dfrac{x-3}{\sqrt{x-1}-\sqrt{2}}\right)\left(\dfrac{2}{\sqrt{2}-\sqrt{x}}-\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{2x}-x}\right)\)
\(=\left(\dfrac{\sqrt{x}+\sqrt{x-1}}{x-\left(x-1\right)}-\dfrac{\left(\sqrt{x-1}-\sqrt{2}\right)\left(\sqrt{x-1}+\sqrt{2}\right)}{\sqrt{x-1}-\sqrt{2}}\right)\cdot\left(\dfrac{2}{\sqrt{2}-\sqrt{x}}-\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\right)\)
\(=\left(\sqrt{x}+\sqrt{x-1}-\sqrt{x-1}-\sqrt{2}\right)\cdot\left(\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}-\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\right)\)
\(=\left(\sqrt{x}-\sqrt{2}\right)\cdot\dfrac{2\sqrt{x}-\sqrt{x}-\sqrt{2}}{-\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\left(\sqrt{x}-\sqrt{2}\right)\cdot\dfrac{\sqrt{x}-\sqrt{2}}{-\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{2}-\sqrt{x}}{\sqrt{x}}\)
b) Ta có: \(x=3-2\sqrt{2}\)
\(=2-2\cdot\sqrt{2}\cdot1+1\)
\(=\left(\sqrt{2}-1\right)^2\)
Thay \(x=\left(\sqrt{2}-1\right)^2\) vào biểu thức \(P=\dfrac{\sqrt{2}-\sqrt{x}}{\sqrt{x}}\), ta được:
\(P=\dfrac{\sqrt{2}-\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(\sqrt{2}-1\right)^2}}\)
\(=\dfrac{\sqrt{2}-\left(\sqrt{2}-1\right)}{\sqrt{2}-1}\)
\(=\dfrac{\sqrt{2}-\sqrt{2}+1}{\sqrt{2}-1}\)
\(=\dfrac{1}{\sqrt{2}-1}\)
\(=\sqrt{2}+1\)
Vậy: Khi \(x=3-2\sqrt{2}\) thì \(P=\sqrt{2}+1\)
rút gọn biểu thức sau :
a. \(A=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
b. \(B=7:\left(a+b\right)+8:\left(a-b\right)-16b:\left(a^2-b^2\right)\)
\(A=\dfrac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\\ A=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\\ A=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2=x-\sqrt{x}+1\)
\(B=\dfrac{7a-7b+8a+8b-16b}{\left(a+b\right)\left(a-b\right)}=\dfrac{15a-15b}{\left(a-b\right)\left(a+b\right)}\\ B=\dfrac{15\left(a-b\right)}{\left(a-b\right)\left(a+b\right)}=\dfrac{15}{a+b}\)
Rút gọn biểu thức
\(\left(\dfrac{1}{\sqrt{x}+2}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{2}{\sqrt{x}}+1\right)\) (x>0)
Với `x>0` có:
\(\left(\dfrac{1}{\sqrt{x}+2}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{2}{\sqrt{x}}+1\right)\\ =(\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}+2\right)})\left(\dfrac{2}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}}\right)\)
\(=\left(\dfrac{\sqrt{x}-\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\right)\left(\dfrac{2+\sqrt{x}}{\sqrt{x}}\right)\\ =\dfrac{-2\left(2+\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}+2\right).\sqrt{x}}=-\dfrac{2}{x}\)
\(=\dfrac{\sqrt{x}-\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{2+\sqrt{x}}{\sqrt{x}}=\dfrac{-2}{x}\)
\(\left(\dfrac{1}{\sqrt{x}+2}-\dfrac{1}{\sqrt{x}}\right)\cdot\left(\dfrac{2}{\sqrt{x}}+1\right)\\ =\left(\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}+2\right)}\right).\left(\dfrac{2}{\sqrt{x}}+1\right)\\=\dfrac{\sqrt{x}-\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}.\left(\dfrac{2}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}}\right)\\ =\dfrac{-2}{\sqrt{x}\left(\sqrt{x}+2\right)}.\dfrac{2+\sqrt{x}}{\sqrt{x}}\\ =\dfrac{-2.\left(2+\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\sqrt{x}}\\=\dfrac{-2}{\sqrt{x}.\sqrt{x}}\\ =-\dfrac{2}{x}\)
\(\left(\dfrac{2x}{\sqrt{x^3}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+\sqrt{x^3}}{1+\sqrt{x}}-\sqrt{x}\right)\)
rút gọn biểu thức trên
\(=\dfrac{2x-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\left(\sqrt{x}-1\right)^2\)
\(=\dfrac{x\left(x-1\right)}{x+\sqrt{x}+1}\)