Căn bậc hai - Hỏi đáp

1) ta có : \(x\sqrt{x}+\sqrt{x}-x-1=\sqrt{x}\left(x+1\right)-\left(x+1\right)\)

\(=\left(\sqrt{x}-1\right)\left(x+1\right)\)

2) ta có : \(\sqrt{ab}-\sqrt{a}-\sqrt{b}+1=\sqrt{a}\left(\sqrt{b}-1\right)-\left(\sqrt{b}-1\right)\)

\(=\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)\)

3) ta có : \(x-\sqrt{x}-2=x+\sqrt{x}-2\sqrt{x}-2\)

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

4) ta có : \(x-3\sqrt{x}+2=x-\sqrt{x}-2\sqrt{x}+2\)

\(=\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)=\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)\)

5) ta có : \(-6x+5\sqrt{x}+1=-6x+6\sqrt{x}-\sqrt{x}+1\)

\(=6\sqrt{x}\left(1-\sqrt{x}\right)+\left(1-\sqrt{x}\right)=\left(6\sqrt{x}+1\right)\left(1-\sqrt{x}\right)\)

6) ta có : \(x+4\sqrt{x}+3=x+\sqrt{x}+3\sqrt{x}+3\)

\(=\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}+1\right)=\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)\)

7) ta có : \(3\sqrt{a}-2a-1=-2a+2\sqrt{a}+\sqrt{a}-1\)

\(=-2\sqrt{a}\left(\sqrt{a}-1\right)+\left(\sqrt{a}-1\right)=\left(1-2\sqrt{a}\right)\left(\sqrt{a}-1\right)\)

8) ta có : \(x+2\sqrt{x-1}=x-1+2\sqrt{x-1}+1\)

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

9) ta có : \(7\sqrt{x}-6x-2=-6x+3\sqrt{x}+4\sqrt{x}-2\)

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

10) ta có : \(x-5\sqrt{x}+6=x-2\sqrt{x}-3\sqrt{x}+6\)

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

11) ta có : \(x-2+\sqrt{x^2-4}=\sqrt{\left(x-2\right)^2}+\sqrt{\left(x-2\right)\left(x+2\right)}\)

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

bài 1) ta có : \(\dfrac{\left(5-x\right)\sqrt{5-x}+\left(3-x\right)\sqrt{3-x}}{\sqrt{5-x}+\sqrt{3-x}}=2\) đk : \(x\le3\)

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

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

\(\Leftrightarrow5-x+\sqrt{\left(5-x\right)\left(3-x\right)}+3-x=2\)

\(\Leftrightarrow\sqrt{\left(5-x\right)\left(3-x\right)}=2x-6\) đk : \(x\ge3\) \(\Rightarrow x=3\)

thử lại ta thấy \(x=3\) là nghiệm

vậy \(x=3\)

bài 2) ta có : \(\sqrt{x-5}-\dfrac{x-14}{3+\sqrt{x-5}}=3\) đk : \(x\ge5\)

\(\Leftrightarrow\dfrac{3\sqrt{x-5}+x-5-x+14}{3+\sqrt{x-5}}=3\)

\(\Leftrightarrow3\sqrt{x-5}+9=9+3\sqrt{x-5}\) (luôn đúng)

vậy \(x\ge5\)

\(\sqrt{4+\sqrt{8}}\). \(\sqrt{\left(2+\sqrt{2+\sqrt{2}}\right)\left(2-\sqrt{2+\sqrt{2}}\right)}\)

= \(\sqrt{4+\sqrt{8}}\).\(\sqrt{4-2-\sqrt{2}}\)

=\(\sqrt{2\left(2+\sqrt{2}\right)}\).\(\sqrt{2-\sqrt{2}}\)

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

=\(\sqrt{2\left(4-2\right)}\)

=\(\sqrt{4}\)=2

\(....=\dfrac{\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}}{\sqrt{x^2-4x+4}}\left(\dfrac{x-2}{x-1}\right)\)

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

\(=\dfrac{\left|\sqrt{x-1}-1\right|+\sqrt{x-1}+1}{\left|x-2\right|}.\dfrac{x-2}{x-1}\)

Kết hợp điều kiện, ta xét các khoảng sau

\(x>2\) thì

\(..=\dfrac{\sqrt{x-1}-1+\sqrt{x-1}+1}{x-2}.\dfrac{x-2}{x-1}\)

\(=\dfrac{2\sqrt{x-1}}{x-1}\)

\(1< x< 2\) thì

\(..=\dfrac{1-\sqrt{x-1}+\sqrt{x-1}+1}{2-x}.\dfrac{x-2}{x-1}\)

\(=\dfrac{-2}{x-1}\)

Vậy.....

Ta có: x(x+1)(x+2)(x+3)+5

= x(x+3)(x+1)(x+2)+5 = (x² +3x)(x² +3x +2)+5 = (x² +3x +1 -1)(x² +3x +1 +1) +5

= (x² +3x +1)² -1 +5 = (x² +3x +1)² +4

Vì (x² +3x +1)² ≥0 (∀x) nên (x² +3x +1) +4 ≥4

⇒ C ≥ √4 ⇒ C ≥ 2

⇒ C đạt GTNN bằng 2 ⇔ (x² +3x +1)² =0

Bạn làm tiếp nha......

Phan Văn KhởiPhùng Khánh LinNguyễn Thị Ngọc ThơhHoàng Ngọc Anh giup mk voi

d)

ĐKXĐ: \(x^2+5x+2\ge 0\)

\((x+1)(x+4)-3\sqrt{x^2+5x+2}=6\)

\(\Leftrightarrow (x^2+5x+4)-3\sqrt{x^2+5x+2}=6\)

Đặt \(\sqrt{x^2+5x+2}=a(a\geq 0)\Rightarrow x^2+5x+2=a^2\)

PT trở thành:

\(a^2+2-3a=6\)

\(\Leftrightarrow a^2-3a-4=0\Leftrightarrow (a-4)(a+1)=0\)

\(\Rightarrow a=4\) vì \(a\geq 0\)

\(\Rightarrow x^2+5x+2=a^2=16\)

\(\Rightarrow x^2+5x-14=0\Leftrightarrow (x-2)(x+7)=0\)

\(\Rightarrow \left[\begin{matrix} x=2\\ x=-7\end{matrix}\right.\) (đều thỏa mãn)

Vậy................

c) ĐK: \(x\in \mathbb{R}\)

Đặt \(\sqrt{6x^2-12x+7}=a(a\geq 0)\Rightarrow 6x^2-12x+7=a^2\)

\(\Rightarrow 6(x^2-2x)=a^2-7\Rightarrow x^2-2x=\frac{a^2-7}{6}\)

Khi đó:

\(2x-x^2+\sqrt{6x^2-12x+7}=0\)

\(\Leftrightarrow \frac{7-a^2}{6}+a=0\)

\(\Leftrightarrow 7-a^2+6a=0\)

\(\Leftrightarrow -a(a+1)+7(a+1)=0\Leftrightarrow (a+1)(7-a)=0\)

\(\Rightarrow \left[\begin{matrix} a=-1\\ a=7\end{matrix}\right.\) \(\Rightarrow a=7\) vì \(a\geq 0\)

\(\Rightarrow 6x^2-12x+7=a^2=49\)

\(\Rightarrow 6x^2-12x-42=0\Leftrightarrow x^2-2x-7=0\)

\(\Leftrightarrow (x-1)^2=8\Rightarrow x=1\pm 2\sqrt{2}\)

(đều thỏa mãn)

Vậy..........

\(B^2=8+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}\\ B^2=8+2\sqrt{\left(16-\left(10+2\sqrt{5}\right)\right)}\\ B^2=8+2\sqrt{6-2\sqrt{5}}\\ B^2=8+2\sqrt{\left(\sqrt{5}-1\right)^2}\\ B^2=8+2\left(\sqrt{5}-1\right)\\ \Rightarrow B=\sqrt{\left|8+2\left(\sqrt{5}-1\right)\right|}=\sqrt{6+2\sqrt{5}}\\ B=\sqrt{\left(1+\sqrt{5}\right)^2}=1+\sqrt{5}\)

ĐKXĐ: x ≥ 0; x ≠ 4

Khi đó ta có:

\(\dfrac{2+\sqrt{x}}{2-\sqrt{ }x}\) \(-\dfrac{2-\sqrt{x}}{2+\sqrt{x}}\)\(-\dfrac{4}{x-4}\)

= \(\dfrac{\left(2+\sqrt{x}\right)\left(2+\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)\(-\dfrac{\left(2-\sqrt{x}\right)\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)\(+\dfrac{4}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)

= \(\dfrac{\left(4+4\sqrt{x}+x\right)-\left(4-4\sqrt{x}+x\right)+4}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)

= \(\dfrac{8\sqrt{x}+4}{x-4}\)

mk nghỉ ở giữa 2 ngoặc là dấu chia mới đúng chứ :

đk : \(x\ge0;x\ne9\)

\(D=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

\(D=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\dfrac{2\sqrt{x}-2-\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\right)\)

\(D=\left(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-\left(3x+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)

\(D=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

\(D=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

\(D=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\)

Với câu c, Thiên Anh nên thêm điều kiện để phần kết luận là: \(0\le x< 2.\)

a. 225

b. 49

c. 1<x<2

d. 1<x<7