rut gon bieu thuc:
a,\(\dfrac{\sqrt{3^2}+\sqrt{39^2}}{\sqrt{7^2}+\sqrt{91^2}}1\)
b, \(\sqrt{\left(2,5-0,7\right)^2}\)
cho 2 bieu thuc:
A=(\(\sqrt{20}\) -\(\sqrt{45}\) +3\(\sqrt{5}\) ).\(\sqrt{5}\) va B=\(\dfrac{x+1-2\sqrt{x}}{\sqrt{x}-1}\) +\(\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\) (Dieu kien: x>0, x khac 1
a) Rut gon bieu thuc A va B
b)Tim cac gia tri cua x de gia tri cua bieu thuc A bang 2lan gia tri B
a: \(A=\left(2\sqrt{5}-3\sqrt{5}+3\sqrt{5}\right)\cdot\sqrt{5}=2\sqrt{5}\cdot\sqrt{5}=10\)
\(B=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\sqrt{x}-1+\sqrt{x}=2\sqrt{x}-1\)
b: A=2B
=>\(10=4\sqrt{x}-2\)
=>\(4\sqrt{x}=12\)
=>x=9(nhận)
rút gọn biểu thức:
a, \(\sqrt{\left(2,5-0,7\right)^2}\)
b, \(\dfrac{\sqrt{3^2}+\sqrt{39^2}}{\sqrt{7^2}+\sqrt{91^2}}1\)
Giải:
a) \(\sqrt{\left(2,5-0,7\right)^2}\)
\(=\left|2,5-0,7\right|\)
\(=\left|1,8\right|=1,8\)
Vậy ...
b) \(\dfrac{\sqrt{3^2}+\sqrt{39^2}}{\sqrt{7^2}+\sqrt{91^2}}\)
\(=\dfrac{3+39}{7+91}\)
\(=\dfrac{42}{98}=\dfrac{3}{7}\)
Vậy ...
(\(\dfrac{1}{\sqrt{x}-\sqrt{x-1}}-\dfrac{x-3}{\sqrt{x-1}-\sqrt{2}})\left(\dfrac{2}{2-\sqrt{x}}-\dfrac{\sqrt{x}+\sqrt{2}}{2\sqrt{x}-x}\right)\)
Rut gon bieu thuc
\(=\left(\sqrt{x}+\sqrt{x-1}-\sqrt{x-1}+\sqrt{2}\right)\cdot\left(\dfrac{2\sqrt{x}-\sqrt{x}-\sqrt{2}}{\sqrt{x}\left(2-\sqrt{x}\right)}\right)\)
\(=\dfrac{\left(\sqrt{x}+\sqrt{2}\right)}{-\sqrt{x}}\)
1) Rut gon bieu thuc A = \(\left(4+\sqrt{7}\right).\left(\sqrt{2}-\sqrt{14}\right).\sqrt{4-\sqrt{7}}\)
rut gon bieu thuc
A=\(\left(\dfrac{\sqrt{X}+2}{X-9}-\dfrac{\sqrt{X}-2}{X+6\sqrt{X}+9}\right)\left(\sqrt{X}\dfrac{9}{\sqrt{X}}\right)\)
XEM CÓ SAI ĐỀ BÀI KHÔNG, MK RÚT GỌN RA TO LẮM
\(=\dfrac{x+5\sqrt{x}+6-x+5\sqrt{x}-6}{\left(\sqrt{x}+3\right)^2\cdot\left(\sqrt{x}-3\right)}\cdot\dfrac{x-9}{\sqrt{x}}\)
\(=\dfrac{10\sqrt{x}}{\sqrt{x}}\cdot\dfrac{1}{\sqrt{x}+3}=\dfrac{10}{\sqrt{x}+3}\)
Cho bieu thuc: \(p=\left(\frac{1}{\sqrt{x}-\sqrt{x-1}}-\frac{x-3}{\sqrt{x-1}-\sqrt{2}}\right)\left(\frac{2}{\sqrt{2}-\sqrt{x}}-\frac{\sqrt{x}+\sqrt{2}}{\sqrt{2x}-x}\right)\)
a) Tim DKXD cua bieu thuc p
b) Rut gon bieu thuc p
rut gon bieu thuc: \(\frac{\sqrt{\sqrt{\frac{x-1}{x+1}}+\sqrt{\frac{x+1}{x-1}}-2}\left(2x+\sqrt{x^2+1}\right)}{\sqrt{\left(x+1\right)^3}-\sqrt{\left(x-1\right)^2}}\)
\(\left(3+\frac{a-2\sqrt{a}}{\sqrt{a}-2}\right)\left(3-\frac{3a+\sqrt{a}}{3\sqrt{a}+1}\right).Rut\:gon\:bieu\:thuc\:nay\)
IQ vô cực mà , bn tự làm đc cái biểu thức dễ ợt này mà
a,\(\dfrac{5^4.20^4}{25^5.4^5}\) b,\(\dfrac{\left(5^4-5^3\right)^3}{125^4}\) c,\(\sqrt{\left(2,5-0,7\right)^2}\) d,\(\dfrac{\sqrt{3^2+\sqrt{39^2}}}{\sqrt{7^2+\sqrt{91^2}}}1\)
Ai biết rút gọn biểu thức này k ạ??
Giups mình với!!! Cảm ơn nhiều ạ!!
a: \(=\dfrac{5^4\cdot5^4\cdot4^4}{5^{10}\cdot4^5}=\dfrac{1}{5^2}\cdot\dfrac{1}{4}=\dfrac{1}{100}\)
b: \(=\dfrac{\left[5^3\left(5-1\right)\right]^3}{5^{12}}=\dfrac{5^9}{5^{12}}\cdot\dfrac{4^3}{1}=\dfrac{4^3}{5^3}\)
c: \(=\sqrt{1.8^2}=1.8\)