Question

2) giải pt

a) \(\sqrt{4-2x}=5\)

b) \(\sqrt{25\left(x+1\right)}+\sqrt{9x+9}=16\)

c) \(\sqrt{4x^2+12x+9}=4\)

giúp mk vs ạ mk cần gấp

Answer 1

a) ĐKXĐ: x <= 2

pt --> 4 - 2x = 25 <=> x = -21/2 (thỏa)

Answer 2

??

Đề kiểu gì vậy ?

Answer 3

b) ĐKXĐ: x >= -1

pt <=> 8sqrt(x + 1)=16 <=> sqrt(x+1)=2 --> x + 1 = 4 <=> x = 3

Answer 4

c) pt <=> |2x+3|=4

TH1: 2x+3=4 <=> x=1/2

TH2: 2x+3=-4 <=> x=-7/2

Answer 5

\(a.\sqrt{4-2x}=5\)  Đk x<=2

\(4-2x=25\)

\(2x=-21\)

\(x=-\dfrac{21}{2}\)

b) \(\sqrt{25\left(x+1\right)}+\sqrt{9x+9}=16\) ĐK x>=-1

\(5\sqrt{x+1}+3\sqrt{x+1}=16\)

\(\sqrt{x+1}=2\)

\(x+1=4\Leftrightarrow x=3\)

c)\(4x^2+12x+9=16\)

\(4x^2+12x-7=0\)

\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)