Bài 1: Căn bậc hai

a) 4x - 6 = x + 3

<=> 4x - x = 3 + 6

<=> 3x = 9

<=> x = 3

Vậy phương trình có tập nghiệm là S = {3}.

b) 2x(x + 2) - 8(x + 2) = 0

<=> (2x - 8)(x + 2) = 0

<=> 2x - 8 = 0 hoặc x + 2 = 0

<=> x = 4 hoặc x = -2

Vậy phương trình có tập nghiệm là S = {4;-2}.

c) |2x| = x + 5

\(\Leftrightarrow\left\{{}\begin{matrix}x+5\ge0\\\left[{}\begin{matrix}2x=x+5\\2x=-x-5\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-5\\\left[{}\begin{matrix}x=5\\3x=-5\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-5\\\left[{}\begin{matrix}x=5\\x=-\dfrac{5}{3}\end{matrix}\right.\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=5\\x=-\dfrac{5}{3}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là \(S=\left\{5;-\dfrac{5}{3}\right\}.\)

d) 4x + 2(x - 1) > 6x - (x - 1)

<=> 4x + 2x - 2 > 6x - x + 1

<=> 6x - 2 > 5x + 1

<=> 6x - 5x > 1 + 2

<=> x > 3

Vậy bất phương trình có tập nghiệm là: \(\left\{x|x>3\right\}\)

Điều kiện: \(x\ge0\)

\(x+\sqrt{x}-6=0\)

\(\Leftrightarrow\left(x-2\sqrt{x}\right)+\left(3\sqrt{x}-6\right)=0\)

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

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

\(\Leftrightarrow\sqrt{x}-2=0\) (vì \(\sqrt{x}+3\ge3\))

\(\Leftrightarrow\sqrt{x}=2\)

\(\Leftrightarrow x=4\) (nhận)

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

ĐKXĐ: x + 3 khác 0

x khác -3

\(=\left(2\sqrt{7}+\sqrt{7}-2\sqrt{14}\right)\cdot\sqrt{7}+14\sqrt{2}\)

\(=21-2\sqrt{98}+2\sqrt{98}=21\)

Để A=1/2 thì \(\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{1}{2}\)

=>4 căn x-2=căn x+1

=>x=1

\(B=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{x-1}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}\cdot\dfrac{x-1}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)

`c, -x + x - 4 + 6 = 0`

`=> 2 = 0`

`->` Vô lý.

c) \(-x+\sqrt{x^2-8x+16}+6=0\)

\(\Leftrightarrow\sqrt{\left(x-4\right)^2}=x-6\)

\(\Leftrightarrow\left|x-4\right|=x-6\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-6\ge0\\\left[{}\begin{matrix}x-4=x-6\\x-4=6-x\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge6\\\left[{}\begin{matrix}0x=-2\left(PTVN\right)\\x=5\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow x\in\varnothing\)

- Vậy \(S=\varnothing\)

d) \(\sqrt{x^2+9}-\sqrt{x^2-7}=2\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2-7\ge0\\\sqrt{x^2+9}=\sqrt{x^2-7}+2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-\sqrt{7}\right)\left(x+\sqrt{7}\right)\ge0\\x^2+9=x^2-7+4\sqrt{x^2-7}+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge\sqrt{7}\\x\le-\sqrt{7}\end{matrix}\right.\\4\sqrt{x^2-7}=12\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge\sqrt{7}\\x\le-\sqrt{7}\end{matrix}\right.\\\sqrt{x^2-7}=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge\sqrt{7}\\x\le-\sqrt{7}\end{matrix}\right.\\x^2-7=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge\sqrt{7}\\x\le-\sqrt{7}\end{matrix}\right.\\\left(x-4\right)\left(x+4\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge\sqrt{7}\\x\le-\sqrt{7}\end{matrix}\right.\\\left[{}\begin{matrix}x=4\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

- Vậy \(S=\left\{4;-4\right\}\)

e) \(x-5\sqrt{x-2}=-2\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2\ge0\\\left(x-2\right)-5\sqrt{x-2}+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\\left(x-2\right)-\sqrt{x-2}-4\sqrt{x-2}+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2\ge0\\\sqrt{x-2}\left(\sqrt{x-2}-1\right)-4\left(\sqrt{x-2}-1\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\\left(\sqrt{x-2}-1\right)\left(\sqrt{x-2}-4\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2\ge0\\\left[{}\begin{matrix}\sqrt{x-2}-1=0\\\sqrt{x-2}-4=0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\\left[{}\begin{matrix}x=3\left(nhận\right)\\x=18\left(nhận\right)\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=18\end{matrix}\right.\)

- Vậy \(S=\left\{3;18\right\}\)

\(c,=\dfrac{6\left(\sqrt{3}+3\right)}{3-9}-\dfrac{5\left(\sqrt{2}+\sqrt{3}\right)}{2-3}+\dfrac{4\left(\sqrt{3}-1\right)}{3-1}\\ =-\sqrt{3}-3+5\sqrt{2}+5\sqrt{3}+2\sqrt{3}-2\\ =6\sqrt{3}+5\sqrt{2}-5\)

a: =>(x²-3)(2x²-5x-1)=0

=>\(x\in\left\{\sqrt{3};-\sqrt{3};\dfrac{5+\sqrt{33}}{4};\dfrac{5-\sqrt{33}}{4}\right\}\)

b: \(\Leftrightarrow\left(x^2+x-3\right)\left(3x^2-x-1\right)=0\)

hay \(x\in\left\{\dfrac{-1+\sqrt{13}}{2};\dfrac{-1-\sqrt{13}}{2};\dfrac{1+\sqrt{13}}{6};\dfrac{1-\sqrt{13}}{6}\right\}\)

b: \(3\sqrt{12}< 3\cdot5=15\)

c: \(\sqrt{\dfrac{5}{7}}>\sqrt{\dfrac{3}{5}}\)

nên \(-\sqrt{\dfrac{5}{7}}< -\sqrt{\dfrac{3}{5}}\)