Căn bậc hai. Căn bậc ba - Hỏi đáp

a) \(\sqrt{12-2\sqrt{35}}=\sqrt{\left(\sqrt{5}-\sqrt{7}\right)^2}=\sqrt{7}-\sqrt{5}\)

b) \(\sqrt{4+\sqrt{15}}=...\)

c) \(\left(3-\sqrt{2}\right)\sqrt{11+6\sqrt{2}}=\left(3\sqrt{2}\right)\sqrt{\left(3+\sqrt{2}\right)^2}\\ =\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)=9-2=7\)

d) \(\left(\sqrt{5}+\sqrt{7}\right)\sqrt{12-2\sqrt{35}}=\left(\sqrt{7}+\sqrt{5}\right)\sqrt{\left(\sqrt{5}-\sqrt{7}\right)^2}\\ =\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)=7-5=2\)

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

f) \(\sqrt{13-\sqrt{160}+\sqrt{53}+4\sqrt{90}}=\sqrt{13-4\sqrt{10}+\sqrt{53}+12\sqrt{10}}\\ =\sqrt{13+8\sqrt{10}+\sqrt{53}}\)

\(A=\dfrac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\dfrac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\)

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

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

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

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

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

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

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

\(=\dfrac{6}{6}\)

\(=1\)

\(x^2-7=0\)

\(\Leftrightarrow x^2=7\)

\(\Leftrightarrow x=\pm\sqrt{7}\)

\(x^2-7=0\)

\(\Rightarrow x^2=0+7\)

\(\Rightarrow x^2=7\)

\(\Rightarrow x\in\varnothing\)

Vậy ko tìm dc giá trị của x thỏa mãn theo yêu cầu

e trả lời r đăng 1 lần thui nhé

1) \(\sqrt{17}>\sqrt{16}=4\)

\(\sqrt{26}>\sqrt{25}=5\)

Vế cộng vế ta có: \(\sqrt{17}+\sqrt{26}>9\)

2) Ta có: \(13-\sqrt{35}>13-\sqrt{36}=13-6=7\left(1\right)\)

\(\sqrt{48}< \sqrt{49}=7\left(2\right)\)

Từ (1);(2), Suy ra: \(13-\sqrt{35}>\sqrt{48}\)

\(\sqrt{5}>\sqrt{4}=2\)

\(\sqrt{57}>\sqrt{49}=7\)

\(\Rightarrow\sqrt{5}+\sqrt{57}>2+7\)

Vậy \(\sqrt{5}+\sqrt{57}>9\)

a, Ta có:

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

\(\Rightarrow-3+\sqrt{2}>-3+\sqrt{1}\)

\(\Rightarrow-3+\sqrt{2}>-3+1\)

\(\Rightarrow-3+\sqrt{2}>-2\)

Chúc bạn học tốt!!!

\(BDT\Leftrightarrow\dfrac{\left(x^2-y^2\right)^2}{x^2y^2}\ge\dfrac{3\left(x-y\right)^2}{xy}\)

\(\Leftrightarrow\dfrac{\left[\left(x-y\right)\left(x+y\right)\right]^2}{x^2y^2}-\dfrac{3\left(x-y\right)^2}{xy}\ge0\)

\(\Leftrightarrow\left(x-y\right)^2\left(\dfrac{\left(x+y\right)^2}{x^2y^2}-\dfrac{3}{xy}\right)\ge0\)

\(\Leftrightarrow\left(x-y\right)^2\left(\dfrac{\left(x+y\right)^2-3xy}{x^2y^2}\right)\ge0\)

\(\Leftrightarrow\left(x-y\right)^2\left(\dfrac{x^2+y^2-xy}{x^2y^2}\right)\ge0\) (luôn đúng)

\(D=\dfrac{1}{2a-1}\cdot\sqrt{25a^4-100a^5+100a^6}\)

\(=\dfrac{1}{2a-1}\cdot\sqrt{\left(5a^2-10a^3\right)^2}\)

\(=\dfrac{1}{2a-1}\cdot\left(5a^2-10a^3\right)\)

\(=\dfrac{5a^2-10a^3}{2a-1}\)

D = \(\dfrac{1}{2a-1}\sqrt{25a^4-100a^5+100a^6}\) = \(\dfrac{1}{2a-1}\sqrt{a^4\left(25-100a+100a^2\right)}\)

D = \(\dfrac{1}{2a-1}\sqrt{a^4\left(10a-5\right)^2}\) (1)

th1: \(x< \dfrac{1}{2}\) thì D \(\Leftrightarrow\) \(\dfrac{a^2\left(5-10a\right)}{2a-1}\) = \(\dfrac{-5a^2\left(2a-1\right)}{2a-1}\) = \(-5a^2\)

th2 : \(x>\dfrac{1}{2}\) thì D \(\Leftrightarrow\) \(\dfrac{a^2\left(10a-5\right)}{2a-1}\) = \(\dfrac{5a^2\left(2a-1\right)}{2a-1}\) = \(5a^2\)

vậy D = \(\pm5a^2\)

Ta có: \(\sqrt{3\left(a^2+6\right)}\ge\left(a+b\right)\sqrt{2}\)

<=> \(3\left(a^2+6\right)\ge2\left(a+b\right)^2\)

<=> \(3\left(a^2+b^2+a^2\right)\ge2a^2+2b^2+4ab\)

<=> \(6a^2+3b^2\ge2a^2+2b^2+4ab\)

<=> \(4a^2-4ab+b^2\ge0\)

<=> \(\left(2a-b\right)^2\ge0\) ( Luôn đúng) => đpcm

=> Dấu = xảy ra <=> \(\left\{{}\begin{matrix}2a=b\\a^2+b^2=6\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}a=\sqrt{\dfrac{6}{5}}=\dfrac{\sqrt{30}}{5}\\b=\dfrac{2\sqrt{30}}{5}\end{matrix}\right.\)

\(A=\sqrt{9a^2-12a+4}-9a+1\)

\(A=\sqrt{9a^2-6a-6a+4}-9a+1\)

\(A=\sqrt{\left(9a^2-6a\right)-\left(6a-4\right)}-9a+1\)

\(A=\sqrt{3a.\left(3a-2\right)-2.\left(3a-2\right)}-9a+1\)

\(A=\sqrt{\left(3a-2\right)^2}-9a+1\)

\(A=3a-2-9a+1\)

\(A=-6a-1\)

\(A=-\left(6a+1\right)\)

Chúc bạn học tốt!!!

a) B = \(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{x+5}{x-\sqrt{x}+2}\) ( đk: x \(\ge\) 0; x\(\ne\)4)

<=> B = \(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

<=> B = \(\dfrac{x-3\sqrt{x}+2-x-4\sqrt{x}-3-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

<=> B = \(\dfrac{x+7\sqrt{x}+6}{-\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}=\dfrac{-\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

=> B = \(\dfrac{-\sqrt{x}-6}{\sqrt{x}-2}\)

b) Để B > -1 => \(\dfrac{-\sqrt{x}-6}{\sqrt{x}-2}>-1\)

=> \(\dfrac{-\sqrt{x}-6}{\sqrt{x}-2}+1>0\)

<=> \(\dfrac{-8}{\sqrt{x}-2}\) > 0 => \(\sqrt{x}-2< 0\) => \(x< 4\)

Đối chiếu với điều kiện ta được: \(0\le x< 4\)

c) Để B \(\in\) Z => \(\dfrac{-\sqrt{x}-6}{\sqrt{x}-2}\) \(\in\) Z

Mà \(\dfrac{-\sqrt{x}-6}{\sqrt{x}-2}\) = \(\dfrac{-\left(\sqrt{x}-2\right)-8}{\sqrt{x}-2}=-1-\dfrac{8}{\sqrt{x}-2}\)

=> \(\dfrac{8}{\sqrt{x}-2}\in Z\) => 8 \(⋮\) \(\sqrt{x}-2\)

=> \(\sqrt{x}\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

=> \(\sqrt{x}\in\left\{3;1;4;0;6;-4;10;-6\right\}\)

Mà \(x\ge0\) => \(x\in\left\{9;0;1;16;36;100\right\}\)

Vậy .......................................................

a) đk : \(x\ne4;x\ge0\)

B = \(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{x+5}{x-\sqrt{x}-2}\)

B = \(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)-\left(x+5\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

B = \(\dfrac{x-2\sqrt{x}-\sqrt{x}+2-\left(x+\sqrt{x}+3\sqrt{x}+3\right)-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

B = \(\dfrac{x-2\sqrt{x}-\sqrt{x}+2-x-\sqrt{x}-3\sqrt{x}-3-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

B = \(\dfrac{-x-7\sqrt{x}-6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\) = \(\dfrac{\left(-\sqrt{x}-6\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

B = \(\dfrac{-\sqrt{x}-6}{\sqrt{x}-2}\)

a) \(A=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+4\sqrt{x}\right)\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\)

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

\(=\left(\dfrac{2\cdot2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+4\sqrt{x}\right)\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\)

\(=\left(\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+4\sqrt{x}\right)\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\)

\(=\dfrac{4\sqrt{x}+4\sqrt{x}\cdot\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\)

\(=4\sqrt{x}\cdot\left[1+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\right]\cdot\dfrac{1}{\sqrt{x}}\)

\(=4\left(1+x-1\right)\)

\(=4x\)

b) Thay \(x=\dfrac{\sqrt{6}}{2+\sqrt{6}}\) vào biểu thức A.

Ta có:

\(4\cdot\dfrac{\sqrt{6}}{2+\sqrt{6}}=\dfrac{4\sqrt{6}}{2+\sqrt{6}}=\dfrac{4\sqrt{6}\cdot\left(2-\sqrt{6}\right)}{-2}\\ =-2\sqrt{6}\cdot\left(2-\sqrt{6}\right)=-4\sqrt{6}+12\)

Vậy giá trị biểu thức A tại \(x=\dfrac{\sqrt{6}}{2+\sqrt{6}}\) là \(-4\sqrt{6}+12\)

c) Để \(\sqrt{A}>A\)

\(\Rightarrow\sqrt{4x}>4x\)

\(\Leftrightarrow\sqrt{4x}>4x\left(đk:x\ge0\right)\)

ĐKXĐ : \(x\ne1;x\ne0;x>0\)

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

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

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

b) \(x=\dfrac{\sqrt{6}}{2+\sqrt{6}}=\dfrac{\sqrt{6}\cdot\left(2-\sqrt{6}\right)}{-2}=\dfrac{2\sqrt{6}-6}{-2}=3-\sqrt{6}\)

Suy ra : \(A=4\cdot\left(3-\sqrt{6}\right)=12-4\sqrt{6}\)

c) \(\sqrt{A}>A\Leftrightarrow\sqrt{4x}>4x\)

\(\Leftrightarrow2\sqrt{x}>4x\)

\(\Leftrightarrow2\sqrt{x}-4x>0\)

\(\Leftrightarrow2\sqrt{x}\cdot\left(1-2\sqrt{x}\right)>0\)

\(\Rightarrow1-2\sqrt{x}>0\) (do x > 0 nên \(2\sqrt{x}>0\))

\(\Leftrightarrow1>2\sqrt{x}\)

\(\Leftrightarrow\dfrac{1}{2}>\sqrt{x}\Rightarrow x< \dfrac{1}{4}\)

Theo điều kiện suy ra giá trị của x để \(\sqrt{A}>A\) là \(0< x< \dfrac{1}{4}\)