Question

Giải giúp giùm mình nha

1/Tính:

a) \(\sqrt{48}-6\sqrt{\dfrac{1}{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}\)

b) \(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}-\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}\)

2/Tìm x:

a) \(\sqrt{4\left(1-x\right)^2}-12=0\)

b)\(\sqrt{4x^2-12x+9}=5\)

Answer 1

2.

a) \(\sqrt{4\left(1-x\right)^2}-12=0\)

\(\Leftrightarrow\sqrt{4}.\sqrt{\left(1-x\right)^2}-12=0\)

\(\Leftrightarrow2.\left|1-x\right|-12=0\) (1)

TH1: \(1-x\ge0\Leftrightarrow x\le1\)

(1) \(\Leftrightarrow2\left(1-x\right)-12=0\)

\(\Leftrightarrow2-2x-12=0\)

\(\Leftrightarrow-10-2x=0\)

\(\Leftrightarrow-2x=10\)

\(\Leftrightarrow x=-5\left(TMĐK\right)\)

TH2: \(1-x< 0\Leftrightarrow x>0\)

(1) \(\Leftrightarrow2\left(x-1\right)-12=0\)

\(\Leftrightarrow2x-2-12=0\)

\(\Leftrightarrow2x-14=0\)

\(\Leftrightarrow2x=14\)

\(\Leftrightarrow x=7\left(TMĐK\right)\)

Vậy phương trình có tập nghiệm: \(S=\left\{7;-5\right\}\)

Answer 2

2.

b) \(\sqrt{4x^2-12x+9}=5\)

\(\Leftrightarrow\sqrt{\left(2x-3\right)^2}=5\)

\(\Leftrightarrow\left|2x-3\right|=5\)

TH1: \(2x-3\ge0\Leftrightarrow x\ge\dfrac{3}{2}\)

\(\Leftrightarrow2x-3=5\Leftrightarrow x=4\left(TMĐK\right)\)

TH2: \(2x-3< 0\Leftrightarrow x< \dfrac{3}{2}\)

\(\Leftrightarrow3-2x=5\Leftrightarrow x=-1\left(TMĐK\right)\)

Vậy phương trình có tập nghiệm là: \(S=\left\{-1;4\right\}\)