Question

Tìm x biết

a. x - \(\sqrt{x}\) - 6 = 0

b. 2x - 5\(\sqrt{x}\) + 3 = 0

c. x - 5\(\sqrt{x}\) +4 = 0

d. 3x - 10\(\sqrt{x}\) + 3 = 0

Answer 1

a) x≥0;∆(√x)=1+24=25

√x=(1-5)/2(l)

√x=(1+5)/2=3;x=9

b)

x≥0

( a+b+c=0)

√x=1; x=1

√x=3/2=>x=9/4

Answer 2

c/ ĐK: x ≥ 0

pt <=> x - √x - 4√x + 4 = 0

<=> √x(√x -1) - 4(√x - 1) = 0

<=> (√x - 1)(√x - 4) = 0

<=> \(\left[{}\begin{matrix}\sqrt{x}-1=0\\\sqrt{x}-4=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=16\end{matrix}\right.\) (tm)

d/ 3x - 10√x + 3= 0

ĐK: x >= 0

pt <=> 3x - √x - 9√x +3 =0

<=> √x (3√x - 1) - 3(3√x -1) = 0

<=> (3√x - 1)(√x - 3) = 0

<=> x = 9; x = 1/9 (tm)

Vậy......................................