Bài 7: Biến đối đơn giản biểu thức chứa căn bậc hai (Tiếp theo)

a) \(\sqrt{x^2+2x+1}=9\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x}+1\right)^2}=9\)

\(\Leftrightarrow\left|\sqrt{x}+1\right|=9\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=9\\x+1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-10\end{matrix}\right.\)

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

\(\Leftrightarrow\sqrt{\left(1-2x\right)^2}=5\)

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

\(\Leftrightarrow\left[{}\begin{matrix}1-2x=5\\1-2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)

c)\(\sqrt{x^2-2x\sqrt{2}+2}=5\)

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

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

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

Mình giải tới đây thôi

a, đkxđ: x ≥ 5

\(\sqrt{x-5}=\sqrt{2x+4}\)

\(\Leftrightarrow x-5=2x+4\)

\(\Leftrightarrow-x=9\)

\(\Leftrightarrow x=-9\) (KTM)

vậy pt vô nghiệm

b/ đkxđ: x ≥ 2

\(\sqrt{x^2-x-2}-\sqrt{x-2}=0\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(TM\right)\\x=0\left(KTM\right)\end{matrix}\right.\)

Vậy pt có nghiệm x = 2

a) điều kiện xác định : \(x\ge5\)

ta có :\(\sqrt{x-5}=\sqrt{2x+4}\Leftrightarrow x-5=2x+4\Leftrightarrow x=-9\left(loại\right)\)

vậy phương trình vô nghiệm

b) điều kiện xác định : \(x\ge2\)

ta có : \(\sqrt{x^2-x-2}-\sqrt{x-2}\Leftrightarrow\sqrt{x^2-x-2}=\sqrt{x-2}\)

\(\Leftrightarrow x^2-x-2=x-2\Leftrightarrow x^2-2x=0\Leftrightarrow x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(L\right)\\x=2\left(N\right)\end{matrix}\right.\) vậy \(x=2\)

Câu a)

Ta có: \(\sqrt{x^2-2x+1}+\sqrt{x^2-6x+9}=1\)

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

\(\Leftrightarrow |x-1|+|x-3|=1(*)\)

Xét các TH sau để phá dấu trị tuyệt đối.

Nếu \(x\geq 3\)

\((*)\Leftrightarrow x-1+x-3=1\Rightarrow 2x=5\Rightarrow x=2,5\) (vô lý)

Nếu $x< 1$

\((*)\Leftrightarrow 1-x+3-x=1\rightarrow 2x=3\Rightarrow x=1,5\) (vô lý)

Nếu $1\leq x< 3$

\((*)\Leftrightarrow x-1+3-x=1\Leftrightarrow 2=1\) (vô lý)

Vậy pt vô nghiệm

Hoặc có thể sử dụng BĐT \(|a|+|b|\geq |a+b|\) thì:

\(1=|x-1|+|x-3|=|x-1|+|3-x|\geq |x-1+3-x|=2\) (vô lý nên pt vô nghiệm)

Câu b: ĐK: \(x\geq 1\)

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

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

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

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

Áp dụng BĐT \(|a|+|b|\geq |a+b|\)

\(\Rightarrow |\sqrt{x-1}+1|+|\sqrt{x-1}-1|=|\sqrt{x-1}+1|+|1-\sqrt{x-1}|\)

\(\geq |\sqrt{x-1}+1+1-\sqrt{x-1}|=2\)

Dấu "=" xảy ra khi \((\sqrt{x-1}+1)(1-\sqrt{x-1})\geq 0\)

\(\Leftrightarrow 1-\sqrt{x-1}\geq 0\)

\(\Leftrightarrow x\leq 2\)

Vậy pt có nghiệm $x$ nằm trong đoạn \([1;2]\)

Câu c)

ĐK: \(x\geq 1\)

PT \(\Leftrightarrow \sqrt{(x-1)+4\sqrt{x-1}+4}+\sqrt{(x-1)+6\sqrt{x-1}+9}=1\)

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

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

\(\Leftrightarrow 2\sqrt{x-1}=-4< 0\) (vô lý do \(2\sqrt{x-1}\geq 0, \forall x\geq 1\))

Do đó pt vô nghiệm.

Lời giải:

Ta có:

Với mọi \(n\in\mathbb{N}>1\Rightarrow 2n+1-\sqrt{n}=(\sqrt{n}-1)^2+n+\sqrt{n}>0\)

\(\Rightarrow 2n+1>\sqrt{n}\)

\(\Rightarrow 2(n+1)> \sqrt{n}+1\)

\(\Rightarrow 2(n+1)\sqrt{n}>(\sqrt{n}+1)\sqrt{n}\)

\(\Rightarrow \frac{1}{2(n+1)\sqrt{n}}< \frac{1}{\sqrt{n}(\sqrt{n}+1)}=\frac{\sqrt{n}+1-\sqrt{n}}{\sqrt{n}(\sqrt{n}+1)}\)

\(\Rightarrow \frac{1}{2(n+1)\sqrt{n}}< \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n}+1}\)

\(\Rightarrow \frac{1}{(n+1)\sqrt{n}}< \frac{2}{\sqrt{n}}-\frac{2}{\sqrt{n}+1}\) (đpcm)

\(a.3x^2+21x+18+2\sqrt{x^2+7x+7}=2\)

\(\Leftrightarrow3\left(x^2+7x+6\right)+2\sqrt{x^2+7x+7}=2\circledast\)

Đặt : \(x^2+7x+7=t\left(t\ge0\right)\) , ta có :

\(\circledast\Leftrightarrow3\left(t-1\right)+2\sqrt{t}=2\)

\(\Leftrightarrow3t+2\sqrt{t}-5=0\)

\(\Leftrightarrow3\sqrt{t}\left(\sqrt{t}-1\right)+5\left(\sqrt{t}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{t}-1=0\\3\sqrt{t}+5=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}t=1\left(TM\right)\\vô-nghiệm\end{matrix}\right.\)

Với : \(t=1\) , thì : \(x^2+7x+7=1\Leftrightarrow x^2+x+6x+6=0\)

\(\Leftrightarrow x\left(x+1\right)+6\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\)

KL...........

\(b.2x^2-8x-3\sqrt{x^2-4x-5}=12\circledast\)

ĐKXĐ : \(\left[{}\begin{matrix}x\ge5\\x\le-1\end{matrix}\right.\)

\(\circledast\Leftrightarrow2x^2-8x-12-3\sqrt{x^2-4x-5}=0\)

\(\Leftrightarrow2\left(x^2-4x-3\right)-3\sqrt{x^2-4x-5}=0\)

Đặt : \(x^2-4x-5=t\left(t\ge0\right)\) , ta có :

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

\(\Leftrightarrow2t-3\sqrt{t}+4=0\)

\(\Leftrightarrow2\left(t-2.\dfrac{3}{4}\sqrt{t}+\dfrac{9}{16}\right)+4-\dfrac{9}{8}=0\)

\(\Leftrightarrow\left(\sqrt{t}-\dfrac{3}{4}\right)^2=\dfrac{23}{16}\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{t}-\dfrac{3}{4}=\dfrac{\sqrt{23}}{4}\\\sqrt{t}-\dfrac{3}{4}=-\dfrac{\sqrt{23}}{4}\end{matrix}\right.\)

Tới đây dễ rồi , bạn tự làm nốt nhé...:)

☛ Câu hỏi của Tô Thu Huyền - Toán lớp 9 | Học trực tuyến

a: \(=\dfrac{1}{x-y}\cdot x^2\cdot\left(x-y\right)=x^2\)

b: \(=\sqrt{27\cdot48}\cdot\left|a-2\right|=36\left(a-2\right)\)

c: \(=\left(\sqrt{2012}+\sqrt{2011}\right)^2\)

d: \(=\dfrac{8}{7}\cdot\dfrac{-x}{y+1}\)

e: \(=\dfrac{11}{12}\cdot\dfrac{x}{-y-2}=\dfrac{-11x}{12\left(y+2\right)}\)

Câu a : \(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)

\(=\sqrt{2+2.\sqrt{2}.\sqrt{3}+3}-\sqrt{2-2.\sqrt{2}.\sqrt{3}+3}\)

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

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

\(=\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{2}\)

\(=2\sqrt{2}\)

Câu b : \(\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)

\(=\sqrt{2-2.\sqrt{2}.\sqrt{5}+5}-\sqrt{2+2.\sqrt{2}.\sqrt{5}+5}\)

\(=\sqrt{\left(\sqrt{2}-\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{2}+\sqrt{5}\right)^2}\)

\(=\left|\sqrt{2}-\sqrt{5}\right|-\left|\sqrt{2}+\sqrt{5}\right|\)

\(=\sqrt{5}-\sqrt{2}-\sqrt{2}-\sqrt{5}\)

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

a, \(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}=\sqrt{2+2.\sqrt{2}.\sqrt{3}+3}-\sqrt{2-2\sqrt{2}.\sqrt{3}+3}\)

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

\(b,\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}=\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\)

\(=\sqrt{5}+\sqrt{2}-\sqrt{5}+\sqrt{2}=2\sqrt{2}\)

a: \(\Leftrightarrow\left|x-1\right|+\left|x-3\right|=1\)

TH1: x<1

Pt sẽ là 1-x+3-x=1

=>4-2x=1

=>x=3/2(loại)

TH2: 1<=x<3

Pt sẽ là x-1+3-x=1

=>2=1(loại)

TH3: x>=3

Pt sẽ là x-1+x-3=1

=>2x=5

hay x=5/2(loại)

b: \(\Leftrightarrow\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|=2\)

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

TH1: x>=2

Pt sẽ là \(\sqrt{x-1}+1+\sqrt{x-1}-1=2\)

=>căn x-1=1

=>x-1=1

hay x=2(nhận)

TH2: 1<=x<2

Pt sẽ là \(\sqrt{x-1}+1-\sqrt{x-1}+1=2\)

=>2=2(luôn đúng)

a/ \(\sqrt{x^2-14x+49}+4x-7=0\)

\(\Leftrightarrow\sqrt{\left(x-7\right)^2}=7-4x\)

\(\Leftrightarrow\left|x-7\right|=7-4x\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}\left(KTM\right)\\x=0\left(TM\right)\end{matrix}\right.\)

Vậy pt có 1 nghiệm x = 0

b/ đkxđ: x ≥2

\(\sqrt{x+2+4\sqrt{x-2}}=4\sqrt{x-2}-5\)

Đặt \(\sqrt{x-2}\) = t (t ≥ 0)

PT \(\Leftrightarrow\sqrt{t^2+4t+4}=4t-5\)

\(\Leftrightarrow\sqrt{\left(t+2\right)^2}=4t-5\)

\(\Leftrightarrow\left|t+2\right|=4t-5\)

Vì t ≥ 0 => t + 2 > 0

=> \(t+2=4t-5\)

\(\Leftrightarrow-3t=-7\Leftrightarrow t=\dfrac{7}{3}\left(TM\right)\)

\(\Rightarrow\sqrt{x-2}=\dfrac{7}{3}\Rightarrow x-2=\dfrac{49}{9}\)

\(\Leftrightarrow x=\dfrac{67}{9}\)(TM)

Vậy pt có nghiệm \(x=\dfrac{67}{9}\)