\(\sqrt{x}=\sqrt{5^2}+25=>\sqrt{x}=5+25=30=>x=30^2=>x=900\)
vậy x=900
=>\(\sqrt{x}=\sqrt{25}+25\)
=>\(\sqrt{x}=5+25\)
=>\(\sqrt{x}=30\)
=>x=302
=>x=900
ta co :\(\sqrt{25=5};-\sqrt{25=-5;\sqrt{\left(-5\right)^2}=\sqrt{25=5}}\)
a) \(\sqrt{36}\)
b) -\(\sqrt{16}\)
c) \(\sqrt{\frac{9}{25}}\)
d)\(\sqrt{3^2}\)
e)\(\sqrt{\left(-3\right)^2}\)
\(\sqrt{x}=\sqrt{5^2+25}\)
Ta co \(\sqrt{25}\)=5 ; -\(\sqrt{25}\)= -5 ; \(\sqrt{\left(-5\right)^2}\)= \(\sqrt{25}\)=5
Theo mau tren , hay tinh :
a) \(\sqrt{36}\)
b) - \(\sqrt{16}\)
c) \(\sqrt{\frac{9}{25}}\)
d) \(\sqrt{3^2}\)
e) \(\sqrt{\left(-3\right)^2}\)
Tìm GTLN
A=\(\sqrt{x+1}\)+ 5
B=3-\(\sqrt{x^2-25}\)
Tính
1. \(\sqrt{\left(\frac{2}{3}\right)^2}-\sqrt{0,09}+\sqrt{\frac{9}{25}}\)
2. \(\sqrt{\left(\frac{-2}{5}\right)^2}+\sqrt{1,44}-\sqrt{\frac{25}{4}}\)
Tính nhanh:
\(\frac{\left(\frac{1}{14}-\frac{\sqrt{2}}{7}+\frac{3\sqrt{2}}{25}\right).\left(-\frac{4}{15}\right)}{\:\left(\frac{1}{10}+\frac{3\sqrt{2}}{25}-\frac{\sqrt{2}}{5}\right).\left(\frac{5}{7}\right)}\)
a)22+2x+3=144
b)(\(\sqrt{9}+\sqrt{4}\)).\(\sqrt{x}\)=10
c)(x+\(\dfrac{1}{2}\))2=\(\dfrac{4}{25}\)
1) Rút gọn biểu thức theo là cách hợp lý:
A = \(\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}\)
2) Tính hợp lý:
M = \(1-\frac{5}{\sqrt{196}}-\frac{5}{\left(2\sqrt{21}\right)^2}-\frac{\sqrt{25}}{204}-\frac{\left(\sqrt{5}\right)^2}{374}\)
3) Có hay không giá trị của x thỏa mãn điều kiện sau:
\(2002.\sqrt{\left(1+x\right)^2}+2003.\sqrt{\left(1-x\right)^2}=0\)
4) Tìm các số x, y, z thỏa mãn đẳng thức:
\(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+\left|x+y+z\right|=0\)
Rút gọc A = \(1-\frac{5}{\sqrt{196}}-\frac{5}{\left(2x.\sqrt{21}\right)^2}-\frac{\sqrt{25}}{204}-\frac{\left(\sqrt{5}\right)^2}{374}\)
