Rút gọn:
a,\(\left(m-2\right)\sqrt{\frac{5m}{4-m^2}}=-\sqrt{\frac{5m\left(2-m\right)}{m+2}}\)với 0<m<2
b,\( \left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)=\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
M=\(\left[\frac{x+2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}-1\right]:\frac{1-\sqrt{x}}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
a) Rút gọn M
b) Tính giá trị của A với x=1
c) Tìm x để M=0
M=(\(\frac{\sqrt{x}}{\sqrt{x}+1}\)-1): \(\frac{-1}{x+\sqrt{x}+1}\)
M=\(\frac{-1}{\sqrt{x}+1}\). -(x+\(\sqrt{x}\)+1)
M=\(\frac{x+\sqrt{x}+1}{\sqrt{x}+1}\)
b, x=1
M = \(\frac{3}{2}\)
c, M= 0
=> x +\(\sqrt{x}\)+1= 0
mặt khác x+\(\sqrt{x}\)+1 = (\(\sqrt{x}\)+0,5)2+0,75 >0
=> x vô nghiệm........
cho M =\(\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}-1}\right):\left(\frac{2}{x}-\frac{2-x}{x\left(\sqrt{x+1}\right)}\right)\)
a , rút gọn M
b , tính M với \(x=\frac{2}{2-\sqrt{3}}\)
Tìm điều kiện tham số m để phương trình: \(\frac{\left(4m+11\right)x-5m+7}{\sqrt{16-x^2}}=\frac{\left(4m+3\right)x+4m+5}{\sqrt{16-x^2}}\) có nghiệm.
1.Chứng minh
(m-2).\(\sqrt{\dfrac{5m}{4-m}}=-\sqrt{\dfrac{5m\left(2-m\right)}{2+m}}\)
B1 : Rút gọn :
\(6xy.\sqrt{\frac{9x^2}{16y^2}}\) \(\left(x< 0;y\ne0\right)\)
\(\sqrt{\frac{4+20a+25a^2}{b^4}}\)\(\left(b< 0;a\ge\frac{-2}{5}\right)\)
\(\left(m-n\right).\sqrt{\frac{m-n}{\left(m-n\right)^2}}\)\(\left(0< m< n\right)\)
B2 : Tính :
\(1.\left(2\sqrt{3}-\sqrt{12}\right):5\sqrt{3}\)
\(2.\sqrt{\frac{317^2-302^2}{1013^2-1012^2}}\)
\(3.\sqrt{27\left(1-\sqrt{3}\right)^2}:3\sqrt{75}\)
\(4.\left(5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{20}-\frac{5}{4}\sqrt{\frac{4}{5}}+\sqrt{5}\right):2\sqrt{5}\)
Bài 1 :
\(6xy\cdot\sqrt{\frac{9x^2}{16y^2}}=6xy\cdot\frac{3x}{4y}=\frac{18x^2y}{4y}=\frac{9}{2}x^2\)
\(\sqrt{\frac{4+20a+25a^2}{b^4}}=\sqrt{\frac{\left(2+5a\right)^2}{\left(b^2\right)^2}}=\frac{2+5a}{b^2}\)
\(\left(m-n\right).\sqrt{\frac{m-n}{\left(m-n\right)^2}}=\sqrt{\left(m-n\right)^2}\cdot\sqrt{\frac{1}{m-n}}=\sqrt{\frac{\left(m-n\right)^2}{m-n}}=\sqrt{m-n}\)
Bài 2 :
1. \(\left(2\sqrt{3}-\sqrt{12}\right):5\sqrt{3}=\left(2\sqrt{3}-2\sqrt{3}\right):5\sqrt{3}=0:5\sqrt{3}=0\)
2. \(\sqrt{\frac{317^2-302^2}{1013^2-1012^2}}=\frac{\sqrt{\left(317+302\right)\left(317-302\right)}}{\sqrt{\left(1013+1012\right)\left(1013-1012\right)}}=\frac{\sqrt{619}\cdot\sqrt{15}}{\sqrt{2025}}=\sqrt{\frac{619}{135}}\)(check lại)
3. \(\sqrt{27\left(1-\sqrt{3}\right)^2}:3\sqrt{75}\)
\(=\sqrt{27}\left(1-\sqrt{3}\right):15\sqrt{3}\)
\(=3\sqrt{3}\left(1-\sqrt{3}\right):15\sqrt{3}\)
\(=\frac{1-\sqrt{3}}{5}\)
4.\(\left(5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{20}-\frac{5}{4}\sqrt{\frac{4}{5}}+\sqrt{5}\right):2\sqrt{5}\)
\(=\left(\frac{5}{\sqrt{5}}+\frac{\sqrt{20}}{2}-\frac{\frac{5}{4}\cdot2}{\sqrt{5}}+\sqrt{5}\right):2\sqrt{5}\)
\(=\left(\sqrt{5}+\frac{2\sqrt{5}}{2}-\frac{\frac{5}{2}}{\sqrt{5}}+\sqrt{5}\right):2\sqrt{5}\)
\(=\left(\sqrt{5}+\sqrt{5}+\frac{\sqrt{5}}{2}+\sqrt{5}\right):2\sqrt{5}\)
\(=\frac{7}{2}\sqrt{5}:2\sqrt{5}\)
\(=\frac{7}{4}\)
Rút gon
A = \(\left(\sqrt{6x^2-12xy^2+6y^3}+\sqrt{24x^2y}\right):\sqrt{6y}\)
B = \(\frac{\sqrt{343xy^3\left(x-y\right)^2}}{\sqrt{28xy}}\) với x, y>0 , x<y
C= \(\sqrt{\frac{m}{1-2x+x^2}}:\frac{\sqrt{81}}{4m^3\left(x^2-2x+1\right)}\) với m>0 , m khác 1
\(A=\left(\sqrt{6\left(x^2-2xy^2+y^3\right)}+\sqrt{6.4x^2y}\right).\frac{1}{\sqrt{6y}}\)
\(=\left(\sqrt{6\left(x^2-xy^2+y^3\right)}+2x\sqrt{6y}\right).\frac{1}{\sqrt{6y}}\)
\(=\left[\sqrt{6}\left(\sqrt{x^2-xy^2+y^3}+2x\sqrt{y}\right)\right].\frac{1}{\sqrt{6y}}=\sqrt{6}\left(\sqrt{x^2-xy^2+y^3}-2x\sqrt{y}\right).\frac{1}{\sqrt{6}\sqrt{y}}\)
\(=\frac{x^2-xy^2+y^3}{\sqrt{y}}-\frac{2x\sqrt{y}}{\sqrt{y}}=\frac{x^2-xy^2+y^3}{\sqrt{y}}-2x\)
mik chỉ lm đến đây đc thui
\(B=\frac{7y\left(y-x\right)\sqrt{7xy}}{2\sqrt{7xy}}=7y^2-7x\)
\(C=\frac{\sqrt{m}}{\sqrt{\left(x-1\right)^2}}.\frac{4m^3\left(x-1\right)^2}{9}=\frac{\sqrt{m}}{\left(x-1\right)}.\frac{4m^3\left(x-1\right)^2}{9}=\frac{4m^3\sqrt{m}\left(x-1\right)}{9}\)
Cho: \(P=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8-x}{4-x}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)
a. Rút gọn
b. Với x > 9. Tìm m để \(m\left(\sqrt{x}-3\right).P>x+1\)
viết lại pt dưới dạng thần thánh
\(x^2-\frac{2mx}{\left(m-1\right)}+\frac{\left(c+1\right)}{4\left(m-1\right)}=0.\)
\(\left(x^2-\frac{2mx}{\left(m-1\right)}+\frac{m^2}{\left(m-1\right)^2}\right)+\frac{\left(c+1\right)}{4\left(m-1\right)}-\frac{m^2}{\left(m-1\right)^2}=0\)
\(\left(x-\frac{m}{\left(m-1\right)}\right)^2=\frac{4m^2-\left(c+1\right)\left(m-1\right)}{4\left(m-1\right)^2}\)
vậy pt có 2 nghiệm phân biệt :
\(\Leftrightarrow\hept{\begin{cases}\left(x-\frac{m}{m-1}\right)=\sqrt{\frac{4m^2-\left(c+1\right)\left(m-1\right)}{4\left(m-1\right)^2}}\\\left(x-\frac{m}{m-1}\right)=-\sqrt{\frac{4m^2-\left(c+1\right)\left(m-1\right)}{4\left(m-1\right)^2}}\end{cases}}\) " sủa lên nào em
a) \(Q=\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+2x\sqrt{x}+y\sqrt{y}}{x\sqrt{x}+y\sqrt{y}}\left(x>0,y>0\right)\)
Rút Gọn
b) \(M=\frac{x^2-\sqrt{2}}{x^4+\left(\sqrt{3}-\sqrt{2}\right)x^2-\sqrt{6}}\)
Rút Gọn