a) ĐKXĐ: \(x\ge\dfrac{3}{2}\)
\(\sqrt{2x-3}=5\)
\(\Leftrightarrow2x-3=25\)
\(\Leftrightarrow2x=25+3\)
\(\Leftrightarrow2x=28\)
\(\Leftrightarrow x=\dfrac{28}{2}\)
\(\Leftrightarrow x=14\) (nhận)
Vậy \(S=\left\{14\right\}\)
b) \(\sqrt{x^2-6x+9}=5\)
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=5\) (1)
ĐKXĐ: \(x\in R\)
(1) \(\Leftrightarrow\left(x-3\right)^2=25\)
\(\Leftrightarrow x-3=5\) hoặc \(x-3=-5\)
*) \(x-3=5\)
\(x=5+3\)
\(x=8\)
*) \(x-3=-5\)
\(x=-5+3\)
\(x=-2\)
Vậy \(S=\left\{-2;8\right\}\)
c) \(\sqrt{x-1}+\sqrt{9x-9}-\sqrt{4x-4}=4\)
\(\Leftrightarrow\sqrt{x-1}+3\sqrt{x-1}-2\sqrt{x-1}=4\)
\(\Leftrightarrow2\sqrt{x-1}=4\) (2)
ĐKXĐ:
(2) \(\Leftrightarrow\sqrt{x-1}=2\)
\(\Leftrightarrow x-1=4\)
\(\Leftrightarrow x=5\) (nhận)
Vậy \(S=\left\{5\right\}\)
`a, sqrt(2x-3) = 5`
`<=> 2x - 3 = 25`
`<=> 2x = 28`
`<=> x = 14`
Vậy `x = 14`
`b, sqrt(x^2-6x+9) = 5`
`<=> x - 3 = 5` hoặc `3 - x = 5`
`<=> x = 8`hoặc `x = -2`
Vậy `x = 8` hoặc `x = -2`
`c, sqrt(x-1) + sqrt(9x-9) - sqrt(4x-4) = 4`
`<=> sqrt(x-1) + 3sqrt(x-1) - 2sqrt(x-1) = 4`
`<=> 2 sqrt(x-1) = 4`
`<=> sqrt(x-1) = 2`
`<=> x - 1 = 4`
`<=> x = 5`
Vậy `x = 5`
`d, 2x^2+8x+14 = 2(x+4)sqrt(x^2+7)`
`x^2 + 8x + 16 - 2(x+4)sqrt(x^2+7) + x^2 + 7 = 9`
`<=> ((x+4) - sqrt(x^2+7))^2 = 9`
Đến đây bạn tự làm tiếp nhé.
