Bài 1: Căn bậc hai

1: \(\dfrac{12}{5\sqrt{6}}=\dfrac{12\sqrt{6}}{30}=\dfrac{2\sqrt{6}}{5}\)

2: \(\dfrac{3}{2+\sqrt{6}}=\dfrac{-6+3\sqrt{6}}{2}\)

ĐK:\(x\ge0;y\ge1\)

\(\Leftrightarrow\left(x-4\sqrt{x}+4\right)+\left(y-1-6\sqrt{y-1}+9\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+\left(\sqrt{y-1}-3\right)^2=0\)

Nhận thấy VT\(\ge0\)\(\forall x,y\) thỏa mãn đk

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\sqrt{x}-2=0\\\sqrt{y-1}=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=10\end{matrix}\right.\)(tm)

Vậy...

\(2\sqrt{3x}=12\Leftrightarrow\sqrt{3x}=12:2=6\Leftrightarrow3x=\sqrt{6}\Leftrightarrow x=3\sqrt{6}\)

= \(\sqrt{3x}\)=6

= (\(\sqrt{3x}\))²=6²

= 3x=36

= x=12

Ta có: \(2\sqrt{3x}=12\)

\(\Leftrightarrow\sqrt{3x}=6\)

\(\Leftrightarrow3x=36\)

hay x=12

 x=-0,5 hoạc x=0,5

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\(3x^2=0,75\Leftrightarrow x^2=0,75:3=0,25\Rightarrow x=\sqrt{0,25}=0,5\)

Chúc bạn học tốt

Ta có: \(3x^2=0.75\)

\(\Leftrightarrow x^2=\dfrac{1}{4}\)

hay \(x\in\left\{\dfrac{1}{2};-\dfrac{1}{2}\right\}\)

\(\sqrt{15}-1< \sqrt{10}\)

\(\sqrt{15}\)−1<\(\sqrt{10}\)

\(\sqrt{15}-1< \sqrt{16}-1=3\)

\(3< \sqrt{10}\)

Do đó: \(\sqrt{15}-1< \sqrt{10}\)

Ta có: \(\sqrt{x-2}+\sqrt{4x-8}+\sqrt{9x-18}=\sqrt{25x-50}+9\)

\(\Leftrightarrow\sqrt{x-2}+2\sqrt{x-2}+3\sqrt{x-2}-5\sqrt{x-2}=9\)

\(\Leftrightarrow x-2=81\)

hay x=83

Ta có: 

hay x=83

1: Ta có: \(\sqrt{x}=3\)

nên x=9

2: Ta có: \(\sqrt{x}=\sqrt{7}\)

nên x=7

3: Ta có: \(\sqrt{x}=-5\)

nên \(x\in\varnothing\)

4: Ta có: \(\sqrt{x}=0\)

nên x=0

6: Ta có: \(\sqrt{2x+1}+3=0\)

\(\Leftrightarrow\sqrt{2x+1}=-3\)

hay \(x\in\varnothing\)

4x+1−|8−x|=x−9

⇔ −|8−x|=x−9-4x-1

⇔ −|8−x|=-3x−10

⇔ |8−x|=3x+10

TH1: 8−x≥0 ⇔ x≤8

8−x=3x+10

⇔ -x-3x=10-8

⇔ -4x=2

⇔ x=\(\dfrac{-1}{2}\) (TMĐK)

TH2: 8−x<0 ⇔ x>8

x-8=3x+10

⇔ x-3x=10+8

⇔ -2x=18

⇔ x=-9 (KTMĐK)

Vậy x=\(\dfrac{-1}{2}\)