\(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}+4}{4-x}\)
rút gọn biểu thức. Mình cần gấp lắm rồi
giúp mik với mik đag cần gấp !
Câu 1:Cho biểu thức P=\(\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2}{x-\sqrt{x}}\right)\):\(\left(\dfrac{1}{\sqrt{x+1}}-\dfrac{2}{1-x}\right)\)
a) Rút gọn biểu thức P
b) Tính giá trị của biểu thức P khi x=4+2\(\sqrt{3}\)
\(a,P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2}{1-x}\right)\left(dkxd:x\ge0,x\ne1\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
\(=\dfrac{\sqrt{x}.\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\dfrac{x-2}{\sqrt{x}}\)
\(b,x=4+2\sqrt{3}\Rightarrow P=\dfrac{\left(4+2\sqrt{3}\right)-2}{\sqrt{4+2\sqrt{3}}}\)
\(=\dfrac{2\sqrt{3}+4-2}{\sqrt{\sqrt{3}^2+2\sqrt{3}+1}}\)
\(=\dfrac{2\sqrt{3}+2}{\sqrt{\left(\sqrt{3}+1\right)^2}}\)
\(=\dfrac{2\left(\sqrt{3}+1\right)}{\left|\sqrt{3}+1\right|}\)
\(=\dfrac{2\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=2\)
a: \(P=\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{x-1}\)
\(=\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{x-1}{\sqrt{x}+1}=\dfrac{x-2}{\sqrt{x}}\)
b: Khi x=4+2căn 3 thì \(P=\dfrac{2+2\sqrt{3}}{\sqrt{3}+1}=2\)
Rút gọn biểu thức sau
B=\(\dfrac{\sqrt{x}+2}{\sqrt{x}}:\)\(\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)
giải chi tiết hộ mình ak
\(ĐK:x>0;x\ne4\\ B=\dfrac{\sqrt{x}+2}{\sqrt{x}}\cdot\dfrac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\\ B=\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\\ B=\dfrac{x+2\sqrt{x}+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ B=\dfrac{x+4\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
Ta có: \(B=\dfrac{\sqrt{x}+2}{\sqrt{x}}\cdot\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)
\(=\dfrac{x+3\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+4\sqrt{x}}{x-4}\)
Rút gọn biểu thức sau
P=\(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+1:\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{2\sqrt{x+7}}{4-x}\right)\)
giải chi tiết hộ mình vs ạ !!!
\(P=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+1:\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{2\sqrt{x}+7}{x-4}\right)\)
\(=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+1:\left(\dfrac{x+2\sqrt{x}-x+\sqrt{x}+2-2\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\)
\(=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-5}\)
\(=\dfrac{-x+8\sqrt{x}-15+\left(x-4\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{-x+8\sqrt{x}-15+x\sqrt{x}-2x-4\sqrt{x}+8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{x\sqrt{x}-3x+4\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)
\(ĐK:x\ge0;x\ne4\\ P=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+1:\dfrac{x+2\sqrt{x}-x+\sqrt{x}+2-2\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ P=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-5}\\ P=\dfrac{\left(3-\sqrt{x}\right)\left(\sqrt{x}-5\right)+\left(x-4\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\\ P=\dfrac{8\sqrt{x}-15-x+x\sqrt{x}-2x-4\sqrt{x}+8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\\ P=\dfrac{x\sqrt{x}-3x+4\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)
1) cho biểu thức A= \(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{3-11\sqrt{x}}{9-x}\)
B= \(\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\) với \(0\le x\ne9\)
b) rút gọn A
c) tìm số nguyên x để P= A. B là số nguyên
giúp mk vs ah mk cần gấp lắm
\(A=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{3-11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)\(A=\dfrac{2x-6\sqrt{x}+x+\sqrt{x+}3\sqrt{x}+3+3-11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)\(A=\dfrac{3x-13\sqrt{x}+6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
1. Cho biểu thức: A=\(\left(\sqrt{x}+\dfrac{4\sqrt{x}}{\sqrt{x}-2}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{4}{2\sqrt{x}-x}\right)\)
Rút gọn biểu thức trên
\(A=\left(\sqrt{x}+\dfrac{4\sqrt{x}}{\sqrt{x}-2}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{4}{2\sqrt{x}-x}\right)\)ĐK : x > 0 ; x \(\ne\)4
\(=\left(\dfrac{x+2\sqrt{x}}{\sqrt{x}-2}\right):\left(\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)=\dfrac{x\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(x-4\right)}\)
\(=\dfrac{x}{\sqrt{x}-2}\)
Ta có: \(A=\left(\sqrt{x}+\dfrac{4\sqrt{x}}{\sqrt{x}-2}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{4}{2\sqrt{x}-x}\right)\)
\(=\dfrac{x-2\sqrt{x}+4\sqrt{x}}{\sqrt{x}-2}:\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x}{\sqrt{x}-2}\)
1.cho biểu thức A=\(\dfrac{\sqrt{x}+2}{\sqrt{x}+3}-\dfrac{5}{x+\sqrt{x}-6}-\dfrac{1}{\sqrt{x}-2}\)với(x≥0;x≠4)
a)rút gọn A
b)tính A khi x=6+4\(\sqrt{2}\)
2.cho biểu thức P=\(\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}-\dfrac{8x}{x-4}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-2}+3\right)\)với x≥0;x≠1;x≠4
a)rút gọn P
b)tìm x để P=-4
Rút gọn biểu thức:
\(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)
\(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+2\sqrt{x}\left(\sqrt{x}-2\right)-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
Ta có: \(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}+2}{4-x}\)
\(=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
rút gọn biểu thức sau
B=\(\left(\dfrac{2}{\sqrt{x}+2}-\dfrac{\sqrt{x}-5}{x-4}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)
giải chi tiết hộ mình vs ạ !!!
Ta có: \(B=\left(\dfrac{2}{\sqrt{x}+2}-\dfrac{\sqrt{x}-5}{x-4}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)
\(=\dfrac{2\sqrt{x}-4-\sqrt{x}+5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
\(=\dfrac{1}{\sqrt{x}+2}\)
\(B=\left(\dfrac{2}{\sqrt{x}+2}-\dfrac{\sqrt{x}-5}{x-4}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\left(x\ge0;x\ne4\right)\\ B=\dfrac{2\sqrt{x}-4-\sqrt{x}+5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\\ B=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\cdot\dfrac{1}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}+2}\)
1) Rút gọn biểu thức : A=\(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\) + \(\dfrac{2\sqrt{x}}{\sqrt{x}+2}\) + \(\dfrac{2+5\sqrt{x}}{4-x}\) với x≥0 ; x≠4