Bài 1: Căn bậc hai

Lời giải:
\(A=[(\sqrt{10}+\sqrt{11})^2-(\sqrt{12})^2][(\sqrt{10}-\sqrt{11})^2-(\sqrt{12})^2]\)

\(=(9+2\sqrt{110})(9-2\sqrt{110})=9^2-(2\sqrt{110})^2=81-440=-359\)

\(B=\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)

\(B\sqrt{2}=\sqrt{8+2\sqrt{15}}+\sqrt{8-2\sqrt{15}}-2\sqrt{6-2\sqrt{5}}\)

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

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

b) Ta có: \(B=\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\cdot\sqrt{3-\sqrt{5}}\)

\(=\dfrac{\sqrt{8+2\sqrt{15}}+\sqrt{8-2\sqrt{15}}-2\cdot\sqrt{6-2\sqrt{5}}}{\sqrt{2}}\)

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

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

\(=\sqrt{2}\)

Bài 2:

a. 

\(\sqrt{8}+\sqrt{15}< \sqrt{9}+\sqrt{16}=3+4=7\)

\(\sqrt{65}-1> \sqrt{64}-1=8-1=7\)

\(\Rightarrow \sqrt{8}+\sqrt{15}< \sqrt{65}-1\)

b.

\(13-2\sqrt{3}=9+4-2\sqrt{3}=9+2\sqrt{4}-2\sqrt{3}>9\)

\(6\sqrt{2}= 6\sqrt{\frac{8}{4}}< 6\sqrt{\frac{9}{4}}=6.\frac{3}{2}=9\)

$\Rightarrow 13-2\sqrt{3}> 6\sqrt{2}$

$\Rightarrow \frac{13-2\sqrt{3}}{6}> \sqrt{2}$

Bài 3:

a. ĐKXĐ: $9-x^2\geq 0$

$\Leftrightarrow (3-x)(3+x)\geq 0$

$\Leftrightarrow 3\geq x\geq -3$

b. ĐKXĐ: $-4x^2+4x-1\geq 0$

$\Leftrightarrow 4x^2-4x+1\leq 0$

$\Leftrightarrow (2x-1)^2\leq 0$

Mà $(2x-1)^2\geq 0$ với mọi $x\in\mathbb{R}$

$\Rightarrow (2x-1)^2=0$

$\Leftrightarrow x=\frac{1}{2}$

c.

ĐKXĐ: $x^2+x-2>0$

$\Leftrightarrow (x-1)(x+2)>0$

$\Leftrightarrow x>1$ hoặc $x< -2$

Bài 3: 

a) ĐKXĐ: \(-3\le x\le3\)

b) ĐKXĐ: \(x=\dfrac{1}{2}\)

c) ĐKXĐ: \(\left[{}\begin{matrix}x>1\\x< -2\end{matrix}\right.\)

1.

Ta có: \(E=\sqrt{37-6\sqrt{30}}=\sqrt{\left(3\sqrt{3}-\sqrt{10}\right)^2}=\left|3\sqrt{3}-\sqrt{10}\right|=3\sqrt{3}-\sqrt{10}\)

\(F=\sqrt{51-6\sqrt{30}}=\sqrt{\left(3\sqrt{5}-\sqrt{6}\right)^2}=\left|3\sqrt{5}-\sqrt{6}\right|=3\sqrt{5}-\sqrt{6}\)

\(G=\sqrt{59-6\sqrt{30}}=\sqrt{\left(3\sqrt{6}-\sqrt{5}\right)^2}=\left|3\sqrt{6}-\sqrt{5}\right|=3\sqrt{6}-\sqrt{5}\)

\(H=\sqrt{17-2\sqrt{30}}=\sqrt{\left(\sqrt{15}-\sqrt{2}\right)^2}=\left|\sqrt{15}-\sqrt{2}\right|=\sqrt{15}-\sqrt{2}\)

\(E=\sqrt{37-6\sqrt{30}}\\ =\sqrt{\left(3\sqrt{3}-\sqrt{10}\right)^2}\\ =\left|3\sqrt{3}-\sqrt{10}\right|\\ =3\sqrt{3}-\sqrt{10}\)

\(F=\sqrt{51-6\sqrt{30}}\\ =\sqrt{\left(3\sqrt{5}-\sqrt{6}\right)^2}\\ =\left|3\sqrt{5}-\sqrt{6}\right|\\ =3\sqrt{5}-\sqrt{6}\)

\(G=\sqrt{59-6\sqrt{30}}\\ =\sqrt{\left(3\sqrt{6}-\sqrt{5}\right)^2}\\ =\left|3\sqrt{6}-\sqrt{5}\right|\\ =3\sqrt{6}-\sqrt{5}\)

\(H=\sqrt{17-2\sqrt{30}}\\ =\sqrt{\left(\sqrt{15}-\sqrt{2}\right)^2}\\ =\left|\sqrt{15}-\sqrt{2}\right|=\sqrt{15}-\sqrt{2}\)

\(E=\sqrt{37-6\sqrt{30}}=3\sqrt{3}-\sqrt{10}\)

\(F=\sqrt{51-6\sqrt{30}}=3\sqrt{5}-\sqrt{6}\)

\(G=\sqrt{59-6\sqrt{30}}=3\sqrt{6}-\sqrt{5}\)

\(H=\sqrt{17-2\sqrt{30}}=\sqrt{15}-\sqrt{2}\)

Đề là: \(P=x^3+y^3-\dfrac{x^2+y^2}{\left(x-1\right)\left(y-1\right)}\)

Hay \(P=\dfrac{x^3+y^3-\left(x^2+y^2\right)}{\left(x-1\right)\left(y-1\right)}\) 

Cái nào em nhỉ?

\(P=\dfrac{x^3+y^3-\left(x^2+y^2\right)}{\left(x-1\right)\left(y-1\right)}=\dfrac{x^3-x^2+y^3-y^2}{\left(x-1\right)\left(y-1\right)}=\dfrac{x^2\left(x-1\right)+y^2\left(y-1\right)}{\left(x-1\right)\left(y-1\right)}\)

\(P=\dfrac{x^2}{y-1}+\dfrac{y^2}{x-1}\)

Ta có:

\(\dfrac{x^2}{y-1}+4\left(y-1\right)\ge2\sqrt{\dfrac{4x^2\left(y-1\right)}{y-1}}=4x\)

Tương tự: \(\dfrac{y^2}{x-1}+4\left(x-1\right)\ge4y\)

Cộng vế:

\(P+4\left(x+y\right)-8\ge4\left(x+y\right)\)

\(\Rightarrow P\ge8\)

\(P_{min}=8\) khi \(x=y=2\)

a) Ta có: \(\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}\)

\(=2\sqrt{5}-3\sqrt{5}+3\cdot3\sqrt{2}+6\sqrt{2}\)

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

b) Ta có: \(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\cdot\sqrt{7}+\sqrt{84}\)

\(=14-2\sqrt{21}+7+2\sqrt{21}\)

=21

c) Ta có: \(\left(\sqrt{6}+\sqrt{5}\right)^2-\sqrt{120}\)

\(=11+2\sqrt{30}-2\sqrt{30}\)

=11

3.

a. $2\sqrt{2}(\sqrt{3}-2)+(1+2\sqrt{2})^2-2\sqrt{6}$

$=2\sqrt{6}-4\sqrt{2}+(1+8+4\sqrt{2})-2\sqrt{6}$

$=2\sqrt{6}-2\sqrt{6}-4\sqrt{2}+4\sqrt{2}+9$

$=9$

b.

$\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\sqrt{\frac{4+2\sqrt{3}}{2}}+\sqrt{\frac{4-2\sqrt{3}}{2}}$

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

$=\frac{\sqrt{3}+1}{\sqrt{2}}+\frac{\sqrt{3}-1}{\sqrt{2}}$

$=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}$

c.

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

\(=|\frac{2}{2-\sqrt{5}}|-|\frac{2}{2+\sqrt{5}}|=\frac{2}{\sqrt{5}-2}-\frac{2}{\sqrt{5}+2}=\frac{2(\sqrt{5}+2)-2(\sqrt{5}-2)}{(\sqrt{5}-2)(\sqrt{5}+2)}=\frac{8}{5-2^2}=8\)

Bài 3:

a) Ta có: \(2\sqrt{2}\left(\sqrt{3}-2\right)+\left(2\sqrt{2}+1\right)^2-2\sqrt{6}\)

\(=2\sqrt{6}-4\sqrt{2}-2\sqrt{6}+9+4\sqrt{2}\)

=9

b) Ta có: \(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)

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

\(=\dfrac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\sqrt{6}\)

Bài 2: 

a) Ta có: \(A=\dfrac{1}{\sqrt{5}+\sqrt{3}}-\dfrac{1}{\sqrt{5}-\sqrt{3}}\)

\(=\dfrac{\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}}{2}\)

\(=-\sqrt{3}\)

Vì là trắc nghiệm nên mình làm tắt thôi nkaaa.

Thay `x=1/4` vào từng ý:

a: `0=0 =>` Đúng.

b. `23/4 = 5` => Sai.

`n`có căn bậc 2 `<=> n>=0`

`=>` Các số thỏa mãn là: `0,5,(x^2+1),7`.

a. `A>=0`.

A.