Bài 70 trang 40 SGK Toán 9 tập 1
Bài 70. Tìm giá trị các biểu thức sau bằng cách biến đổi, rút gọn thích hợp
(a)sqrt {{{25} over {81}}.{{16} over {49}}.{{196} over 9}})
(b)sqrt {3{1 over {16}}.2{{14} over {25}}2{{34} over {81}}})
(c){{sqrt {640} .sqrt {34,3} } over {sqrt {567} }})
(d)sqrt {21,6} .sqrt {810.} sqrt {{{11}^2} – {5^2}})
Giải
a)
(eqalign{ & sqrt {{{25} over {81}}.{{16} over {49}}.{{196} over 9}} cr & = sqrt {{{25} over {81}}} .sqrt {{{16} over {49}}} .sqrt {{{196} over 9}} cr & = {5 over 9}.{4 over 7}.{{14} over 3} = {{40} over {27}} cr} )
b)
(eqalign{ & sqrt {3{1 over {16}}.2{{14} over {25}}2{{34} over {81}}} cr & = sqrt {{{49} over {16}}.{{64} over {25}}.{{196} over {81}}} cr & = sqrt {{{49} over {16}}} .sqrt {{{64} over {25}}} .sqrt {{{196} over {81}}} cr & = {7 over 4}.{8 over 5}.{{14} over 9} = {{196} over {45}} cr} )
c)
(eqalign{ & {{sqrt {640} .sqrt {34,3} } over {sqrt {567} }} cr & = sqrt {{{640.34,3} over {567}}} cr & = sqrt {{{64.49} over {81}}} cr & = {{sqrt {64} .sqrt {49} } over {sqrt {81} }} = {{8.7} over 9} = {{56} over 9} cr} )
d)
(eqalign{ & sqrt {21,6} .sqrt {810.} sqrt {{{11}^2} – {5^2}} cr & = sqrt {21,6.810.left( {{{11}^2} – {5^2}} right)} cr & = sqrt {216.81.left( {11 + 5} right)left( {11 – 5} right)} cr & = sqrt {{{36}^2}{{.9}^2}{{.4}^2}} = 36.9.4 = 1296 cr} )