Question

Answer 1

a: Ta có: \(\sqrt{3x-2}=3\)

\(\Leftrightarrow3x-2=9\)

hay \(x=\dfrac{11}{3}\)

b: Ta có: \(\sqrt{7-5x}=2\)

\(\Leftrightarrow7-5x=4\)

hay \(x=\dfrac{3}{5}\)

Answer 2

help

 

Answer 3

a. \(\sqrt{3x-2}=3\)

<=> 3x - 2 = 3²

<=> 3x = 9 + 2

<=> 3x = 11

<=> x = \(\dfrac{11}{3}\)

b. \(\sqrt{7-5x}=2\)

<=> 7 - 5x = 2²

<=> -5x = 4 - 7

<=> -5x = -3

<=> x = \(\dfrac{3}{5}\)

c. \(\sqrt{\left(4x+5\right)^2}=9\)

<=> 4x + 5 = 9

<=> 4x = 4

<=> x = 1

d. \(\sqrt{x^2-6x+9}=4\)

<=> x² - 6x + 9 = 16

<=> (x - 3)² = 16

<=> x - 3 = 4

<=> x = 7

Answer 4

\(g,\sqrt{x+2}-\sqrt{x-6}=2\left(x\ge6\right)\\ \Leftrightarrow x+2+x-6-2\sqrt{\left(x+2\right)\left(x-6\right)}=4\\ \Leftrightarrow2x-8-2\sqrt{x^2-4x-12}=0\\ \Leftrightarrow x-4-\sqrt{x^2-4x-12}=0\\ \Leftrightarrow\sqrt{x^2-4x-12}=x-4\\ \Leftrightarrow x^2-4x-12=x^2-8x+16\\ \Leftrightarrow4x=28\\ \Leftrightarrow x=7\left(TM\right)\)

Answer 5

g. \(\sqrt{x+2}-\sqrt{x-6}=2\)

<=> \(\sqrt{x+2}=2+\sqrt{x-6}\)

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

<=> x + 2 = 4 + \(4\sqrt{x-6}\) + (x - 6)

<=> 4\(\sqrt{x-6}\) = 4 + x - 6 - x - 2

<=> \(4\sqrt{x-6}\) = -4 

<=> \(\sqrt{x-6}\) = -1

<=> PT vô nghiệm

Answer 6

f. \(\sqrt{2x+38}=x-5\)

Ta có \(\sqrt{2x+38}\) phải ≥ 0 ( vì biểu thức trong căn ko bao giờ được mang giá trị âm ) 

==> x - 5 phải ≥ 0 ==> x ≥ 5

f.\(\sqrt{2x+38}=x-5\)

<=> 2x + 38 = (x  -  5)²

<=> 2x  +  38 = x² - 10x + 25
<=> 2x + 38 - x² + 10x - 25 = 0
<=> 12x + 13 - x² = 0
<=> 12x + 1 + 12 - x² = 0
<=> 12.( x + 1 ) + ( 1 - x² ) = 0
<=> 12. ( x + 1 ) + ( 1 - x ) . ( 1 + x ) = 0
<=> ( 12 + 1 - x ) . ( x + 1) = 0
<=> ( 13 - x) . ( x + 1 ) = 0
<=> \(\left\{{}\begin{matrix}x=13\\x=-1\left(KTM\right)\end{matrix}\right.\)

==> x = 13
Em hơi ngáo tí =)), có j em sai chỗ nào mong mn thông cảm =))