Căn bậc hai - Hỏi đáp

Lời giải:

Ta có:

\(x^2+2y+1=y^2+2z+1=z^2+2x+1=0\)

\(\Rightarrow x^2+2y+1+y^2+2z+1+z^2+2x+1=0+0+0=0\)

\(\Leftrightarrow (x^2+2x+1)+(y^2+2y+1)+(z^2+2z+1)=0\)

\(\Leftrightarrow (x+1)^2+(y+1)^2+(z+1)^2=0(*)\)

Ta thấy rằng \(\left\{\begin{matrix} (x+1)^2\geq 0\\ (y+1)^2\geq 0\\ (z+1)^2\geq 0\end{matrix}\right., \forall x,y,z\in\mathbb{R}\)

Do đó để $(*)$ xảy ra thì \((x+1)^2=(y+1)^2=(z+1)^2=0\)

\(\Leftrightarrow x=y=z=-1\)

Thử lại thấy thỏa mãn

Vậy \(x^{2017}+y^{2017}+z^{2017}=(-1)^{2017}.3=-3\)

ĐKXĐ: \(x\ge\dfrac{5}{2}\)

Đặt Vế trái là A, ta có:

\(A^2=\left(x-2\right)+\sqrt{2x-5}+\left(x-2\right)-\sqrt{2x-5}+2\sqrt{\left(x-2\right)^2-\left(2x-5\right)}=98\)\(\Leftrightarrow A^2=2x-4+2\sqrt{\left(x^2-6x+9\right)}=98\)

\(\Leftrightarrow A^2=2x-4+2\left|x-3\right|=98\)

Ta có với \(x>3\) ta có \(2x-4+2\left(x-3\right)=98\Leftrightarrow4x-10=98\Rightarrow x=27>3\)

Với \(\dfrac{5}{2}\le x\le3\) ta có: \(2x-4+2\left(3-x\right)=98\)(loại)

Thử lại với x = 27 vào phương trình ban đầu thấy thỏa mãn

Kết luận \(S=\left\{27\right\}\)

cách khác

\(\sqrt{x-2+\sqrt{2x-5}}\) + \(\sqrt{x-2-\sqrt{2x-5}}\) = 7\(\sqrt{2}\)

<=> \(\sqrt{2x-4+2\sqrt{2x-5}}\) + \(\sqrt{2x-4-2\sqrt{2x-5}}\)=14

<=> \(\sqrt{2x-5+2\sqrt{2x-5}+1}\) + \(\sqrt{2x-5-2\sqrt{2x-5}+1}\) =14

<=>\(\sqrt{\left[\left(\sqrt{2x-5}\right)+1\right]^2}\) + \(\sqrt{\left[\left(\sqrt{2x-5}\right)-1\right]^2}\)=14

<=> /\(\sqrt{2x-5}\)+1/ + /\(\sqrt{2x-5}\)-1/ =14

Vì /\(\sqrt{2x-5}\)+1/ + /\(\sqrt{2x-5}\)-1/ = /\(\sqrt{2x-5}\)+1/ + /1-\(\sqrt{2x-5}\)/ >=/\(\sqrt{2x-5}\) +1 +1 -\(\sqrt{2x-5}\)/ = 2

Dấu''=='' xảy ra <=> (​​​\(\sqrt{2x-5}\)​ + 1)(1-\(\sqrt{2x-5}\)) >=0

<=> (\(\sqrt{2x-5}\)+1)(\(\sqrt{2x-5}\)-1) <=0

đến đây bạn tự giải tiếp nhé

(ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\))

=> \(M=\left(1-\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{1}{\sqrt{x}-1}\right).\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{x-2\sqrt{x}}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-4\sqrt{x}+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{x-2\sqrt{x}}=\dfrac{x-1-4\sqrt{x}+\sqrt{x}+1}{x-2\sqrt{x}}=\dfrac{x-3\sqrt{x}}{x-2\sqrt{x}}=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\)

Sử dụng Cô-si đc ko bạn?

cái này ra rồi , nên không cần nữa nhé!

\(-\dfrac{1}{3}\le x\le6\)

\(\sqrt{3x+1}-4-\left(\sqrt{6-x}-1\right)+3x^2-14x-5=0\)

\(\Leftrightarrow\dfrac{3x-15}{\sqrt{3x+1}+4}+\dfrac{x-5}{\sqrt{6-x}+1}+\left(x-5\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(\dfrac{3}{\sqrt{3x+1}}+\dfrac{1}{\sqrt{6-x}+1}+3x-1\right)=0\)

do \(x\ge\dfrac{-1}{3}\Rightarrow3x+1\ge0\)

\(\dfrac{3}{\sqrt{3x+1}}+\dfrac{1}{\sqrt{6-x}+1}+3x-1>0\)

\(\Rightarrow x=5\)

ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)

Câu a :

\(P=\left(\dfrac{4\sqrt{x}}{2+\sqrt{x}}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)

\(=\left(\dfrac{4\sqrt{x}\left(2-\sqrt{x}\right)+8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right):\left(\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)

\(=\dfrac{8\sqrt{x}+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}:\dfrac{3-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\dfrac{4\sqrt{x}}{2-\sqrt{x}}\times\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)

\(=\dfrac{-4x}{3-\sqrt{x}}=\dfrac{4x}{\sqrt{x}-3}\)

Câu b :

\(P=-1\)

\(\Leftrightarrow\) \(\dfrac{4x}{\sqrt{x}-3}=-1\)

\(\Leftrightarrow4x=-\sqrt{x}+3\)

\(\Leftrightarrow4x+\sqrt{x}-3=0\)

\(\Leftrightarrow4x+4\sqrt{x}-3\sqrt{x}-3=0\)

\(\Leftrightarrow4\sqrt{x}\left(\sqrt{x}+1\right)-3\left(\sqrt{x}+1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(4\sqrt{x}-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+1=0\\4\sqrt{x}-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\x=\dfrac{9}{16}\end{matrix}\right.\)

Vậy \(x=\dfrac{9}{16}\)

Chúc bạn học tốt !!

Lời giải:

ĐKXĐ của biểu thức là:

\(\left\{\begin{matrix} x\neq 0\\ x+\frac{3}{x}\geq 0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\neq 0\\ \frac{x^2+3}{x}\geq 0\end{matrix}\right.\). Vì \(x^2+3>0, \forall x\in\mathbb{R}\neq 0\) do đó điều kiện tương đương với:

\(x>0\)

Vậy ĐKXĐ là \(x>0\)

để 2x\(\sqrt{1-9x^2}\) có nghĩa thì:

\(1-9x^2\ge0\Rightarrow-9x^2\ge-1\Rightarrow9x^2\le1\Rightarrow x^2\le\dfrac{1}{9}\)

\(\Rightarrow\left[{}\begin{matrix}x\le\dfrac{1}{3}\\x\ge-\dfrac{1}{3}\end{matrix}\right.\)

Để : \(2x\sqrt{1-9x^2}\) xác đình , thì :

\(1-9x^2\) ≥ 0 ⇔ 9x² ≤ 1 ⇔ x² ≤ \(\dfrac{1}{9}\) ⇔ \(x\) ≤ \(\dfrac{1}{3}\) hoặc x ≥ \(\dfrac{-1}{3}\)

KL.........

\(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)

\(=\sqrt{\dfrac{2\left(4+\sqrt{7}\right)}{2}}-\sqrt{\dfrac{2\left(4-2\sqrt{7}\right)}{2}}-\sqrt{2}\)

\(=\sqrt{\dfrac{8+2\sqrt{7}}{2}}-\sqrt{\dfrac{8-2\sqrt{7}}{2}}-\sqrt{2}\)

\(=\sqrt{\dfrac{\left(\sqrt{7}\right)^2+2\sqrt{7}+1}{2}}-\sqrt{\dfrac{\left(\sqrt{7}\right)^2-2\sqrt{7}+1}{2}}-\sqrt{2}\)

\(=\sqrt{\dfrac{\left(\sqrt{7}+1\right)^2}{2}}-\sqrt{\dfrac{\left(\sqrt{7}-1\right)^2}{2}}-\sqrt{2}\)

\(=\dfrac{|\sqrt{7}+1|}{\sqrt{2}}-\dfrac{|\sqrt{7}-1|}{\sqrt{2}}-\sqrt{2}\)

\(=\dfrac{\sqrt{7}+1}{\sqrt{2}}-\dfrac{\sqrt{7}-1}{\sqrt{2}}-\dfrac{2}{\sqrt{2}}\)

\(=\dfrac{0}{\sqrt{2}}=0\)

\(\sqrt{4}+\sqrt{\sqrt{7}}-\sqrt{4}-\sqrt{\sqrt{7}}-\sqrt{2}\) ​\(=-\sqrt{2}\)​

\(\sqrt{1-9x^2}\) có nghĩa ⇔ 1 - 9x² ≥ 0 ⇔ 9x² ≤ 1 ⇔ x² ≤ \(\dfrac{1}{9}\) ⇔ \(\dfrac{-1}{3}\le x\le\dfrac{1}{3}\)

e,

\(\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\\ =\sqrt{17-6\cdot\sqrt{4}\cdot\sqrt{2}}+\sqrt{9+2\cdot\sqrt{4}\cdot\sqrt{2}}\\ =\sqrt{17-6\cdot\sqrt{8}}+\sqrt{9+2\cdot\sqrt{8}}\\ =\sqrt{9-2\cdot3\cdot\sqrt{8}+8}+\sqrt{8+2\cdot\sqrt{8}\cdot1+1}\\ =\sqrt{3^2-2\cdot3\cdot\sqrt{8}+\sqrt{8}^2}+\sqrt{\sqrt{8}^2+2\cdot\sqrt{8}\cdot1+1^2}\\ =\sqrt{\left(3-\sqrt{8}\right)^2}+\sqrt{\left(\sqrt{8}+1\right)^2}\\ =\left|3-\sqrt{8}\right|+\left|\sqrt{8}+1\right|\\ =3-\sqrt{8}+\sqrt{8}+1\\ =4\)

f,

\(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\\ =\sqrt{6-4\sqrt{2}}+\sqrt{22-4\cdot\sqrt{9}\cdot\sqrt{2}}\\ =\sqrt{6-4\sqrt{2}}+\sqrt{22-4\cdot\sqrt{18}}\\ =\sqrt{4-2\cdot2\cdot\sqrt{2}+2}+\sqrt{18-2\cdot2\cdot\sqrt{18}+4}\\ =\sqrt{2^2-2\cdot2\cdot\sqrt{2}+\sqrt{2}^2}+\sqrt{\sqrt{18}^2-2\cdot2\cdot\sqrt{18}+2^2}\\ =\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{18}-2\right)^2}\\ =\left|2-\sqrt{2}\right|+\left|\sqrt{18}-2\right|\\ =2-\sqrt{2}+\sqrt{18}-2\\ =\sqrt{2}+\sqrt{18}\\ =\sqrt{2}+\sqrt{9}\cdot\sqrt{2}\\ =\sqrt{2}+3\sqrt{2}\\ =4\sqrt{2}\)

d) \(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}=2\sqrt{6+2\sqrt{5}}+\sqrt{\left(\sqrt{5}-2\right)^2}\)

\(=2\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{5}-2=2\sqrt{5}+2+\sqrt{5}-2=3\sqrt{5}\)

\(\dfrac{3}{\sqrt{20}}\)= \(\dfrac{3}{2\sqrt{5}}\)=\(\dfrac{3.\sqrt{5}}{2(\sqrt{5})\sqrt{5}}\)=\(\dfrac{3\sqrt{5}}{10}\)

Mk làm đc có 3 câu thôi.

D = (4\(\sqrt{10}\) - 4\(\sqrt{6}\) + 5\(\sqrt{6}\) - 3\(\sqrt{10}\) )\(\sqrt{4-\sqrt{15}}\)

D = (\(\sqrt{10}\) + \(\sqrt{6}\) )\(\sqrt{4-\sqrt{15}}\)

D = \(\sqrt{\left(4-\sqrt{15}\right)10}\) + \(\sqrt{\left(4-\sqrt{15}\right)6}\)

D = \(\sqrt{40-10\sqrt{15}}\) + \(\sqrt{24-6\sqrt{15}}\)

D = \(\sqrt{\left(\sqrt{15}\right)^2-2.5.\sqrt{5}+5^2}\) + \(\sqrt{\left(\sqrt{15}\right)^2-2.3.\sqrt{15}+3^2}\)

D = \(\sqrt{\left(\sqrt{15}-5\right)^2}\) + \(\sqrt{\left(\sqrt{15}-3\right)^2}\)

D = 5 - \(\sqrt{15}\) + \(\sqrt{15}\) - 3 = 2

a)\(\sqrt{13-4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)

\(=\sqrt{12-2.2\sqrt{3}.1+1}+\sqrt{4-2.2.\sqrt{3}+3}\)

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

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

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

b)\(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)

\(=\sqrt{5+2\sqrt{5}.1+1}+\sqrt{5-2\sqrt{5}.1+1}\)

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

\(=\left(\sqrt{5}+1\right)+\left(\sqrt{5}-1\right)=2\sqrt{5}\)

c)\(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)

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

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

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

d)\(\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)

\(=\sqrt{4+2.2\sqrt{3}+3}+\sqrt{4-2.2.\sqrt{3}+3}\)

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

\(=\left(2+\sqrt{3}\right)+\left(2-\sqrt{3}\right)=4\)

e)\(\sqrt{9+4\sqrt{5}}=\sqrt{5+2.\sqrt{5}.2+4}=\sqrt{\left(\sqrt{5}+2\right)^2}=\sqrt{5}+2\)

f)\(\sqrt{23+8\sqrt{7}}=\sqrt{16+2.4.\sqrt{7}+7}=\sqrt{\left(4+\sqrt{7}\right)^2}=4+\sqrt{7}\)