Question

Tìm x biết:

a, \(\sqrt{16x}\) = 8

b, \(\sqrt{x^2}\) = 2x -1

c, \(\sqrt{9\left(x-1\right)}\) = 21

d, \(\sqrt{4\left(1-x\right)^2}\) - 6 = 0

e, \(\sqrt{4\left(x^2-1\right)}\) - \(2\sqrt{15}\) = 0

f, \(\sqrt{x^2-25}-\sqrt{x-5}=0\)

g, \(\sqrt{9\left(2-3x\right)^2}=6\)

h, \(\sqrt{x+5}+\sqrt{5-x}=4\)

Answer 1

\(f,\sqrt{x^2-25}-\sqrt{x-5}=0\)

=> \(\sqrt{x^2-25}=\sqrt{x-5}\)

=>\(x^2-25=x-5\)

=>\(x^2-x=25-5=20\)

=>( đến đoạn này mình xin chịu )

Answer 2

\(a,\sqrt{16x}=8\)

=>\(16x=8^2\)

=>\(16x=64\)

=>\(x=64:16=4\)

Vậy \(x\in\left\{4\right\}\)

\(b,\sqrt{x^2}=2x-1\)

=>\(x=2x-1\)

=>\(2x-x=1\)

=>\(x=1\)

Vậy \(x\in\left\{1\right\}\)

\(c,\sqrt{9.\left(x-1\right)}=21\)

=>\(9.\left(x-1\right)=21^2=441\)

=> \(x-1=441:9=49\)

=>\(x=49+1=50\)

Vậy \(x\in\left\{50\right\}\)

\(d,\sqrt{4\left(1-x\right)^2}-6=0\)

=>\(\sqrt{4\left(1-x\right)^2}=0+6=6\)

=> \(4\left(1-x\right)^2=6^2=36\)

=>\(\left(1-x\right)^2=36:4=9\)

=>\(1-x=\sqrt{9}=3\)

=>\(x=1-3=-2\)

Vậy \(x\in\left\{-2\right\}\)

\(g,\sqrt{9\left(2-3x\right)^2}=6\)

=> \(9.\left(2-3x\right)^2=6^2=36\)

=> \(\left(2-3x\right)^2=36:9=4\)

=> \(2-3x=\sqrt{4}=2\)

=>\(3x=2-2=0\)

=>\(x=0:3=0\)

Vậy \(x\in\left\{0\right\}\)

( còn các bài còn lại mình sẽ nghĩ tiếp , HS6-7 làm bài )