\(\left(\sqrt{a}+\sqrt{b}\right)^4\le\left(\sqrt{a}+\sqrt{b}\right)^4+\left(\sqrt{a}-\sqrt{b}\right)^4=2a^2+2b^2+6ab\)

\(tt\Rightarrow B\le6\left(a^2+b^2+c^2+d^2+2ab+2ac+2ad+2bc+2bd+2cd\right)=6\left(a+b+c+d\right)^2\le6\)

\(dấu"='\Leftrightarrow a=b=c=d=\dfrac{1}{4}\)

a) \(\sqrt{\left|1-\sqrt{3}\right|^2}\cdot\sqrt{4+2\sqrt{3}}\)

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

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

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

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

b) \(2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)

\(=2\sqrt{40\cdot2\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{5\cdot4\sqrt{3}}\)

\(=2\sqrt{80\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)

\(=8\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}=0\)

a) $\sqrt{{\left|1-\sqrt{3}\right|}^2}\cdot \sqrt{4+2\sqrt{3}}$

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

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

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

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

b) $2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}$

$=2\sqrt{40\cdot 2\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{5\cdot 4\sqrt{3}}$

$=2\sqrt{80\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}$

$=8\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}=0$

Lời giải:

Bổ sung đk $m$ nguyên
Để pt có 2 nghiệm nguyên thì:

\(\Delta=m^2-4(m+2)=t^2\) với $t\in\mathbb{N}^*$

$\Leftrightarrow m^2-4m-8=t^2$

$\Leftrightarrow (m-2)^2-12=t^2$
$\Leftrightarrow 12=(m-2)^2-t^2=(m-2-t)(m-2+t)$

Vì $m-2-t, m-2+t$ có cùng tính chẵn lẻ nên $(m-2-t, m-2+t)=(2,6), (6,2), (-2,-6), (-6,-2)$

$\Rightarrow m=-2$ hoặc $m=6$

Thử lại thấy tm

Lời giải:

a. $\sqrt{x^2}-x+1=|x|-x+1=x-x+1=1$

b. $\sqrt{x^2}+2x=|x|+2x=-x+2x=x$

c. $\sqrt{(\frac{x}{y})^2}=|\frac{x}{y}|=\frac{x}{y}$ do $\frac{x}{y}\geq 0$ với $x\geq 0$ và $y>0$

d.

$\sqrt{(\frac{x}{y})^2}=|\frac{x}{y}|=\frac{-x}{y}$ do $\frac{x}{y}<0$ với $x>0; y<0$

Equation of the intersection of (P) and (d) is:

\(x^2=\left(m+1\right)x-m\) \(\Leftrightarrow x^2-\left(m+1\right)x+m=0\)  (1)\(a=1;b=-\left(m+1\right);c=m\)

We can see that \(a+b+c=1-\left(m+1\right)+m=0\) so the equation (1) has 2 roots: \(x_1=1;x_2=\dfrac{c}{a}=m\)

We have \(y_1=x_1^2=1^2=1\); \(y_2=x_2^2=m^2\)

Thus, \(y_1+y_2=1+m^2\)

Because \(m^2\ge0\Leftrightarrow m^2+1\ge1\) or \(y_1+y_2\ge1\). "=" happens when \(m=0\)

In conclusion, in order to minimize the value of \(y_1+y_2\), m must be equal to 0.

vậy em sẽ nhận được sự giải thích như sau em nhé:

khi ta đổi chỗ các hạng tử của một tổng thì tổng đó không đổi nên:

5\(\sqrt{23}\) +   5 - \(\sqrt{23}\)   =  5 + 5\(\sqrt{23}\)- \(\sqrt{23}\) 

                               = 5 + 5 x \(\sqrt{23}\) - 1 x \(\sqrt{23}\)

                               = 5 + ( 5 - 1) \(\sqrt{23}\)

                                 = 5 +   4\(\sqrt{23}\)

Nó sẽ là 5/23 + 5 - 1/23 = 5 + 4/23

Dấu / là dấu căn nhé vì đt không có dấu căn ấy 😅😅

\(P=\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}-\dfrac{5}{ab+bc+ca}=\dfrac{a^2}{ab}+\dfrac{b^2}{bc}+\dfrac{c^2}{ca}-\dfrac{5}{ab+bc+ca}\ge\dfrac{\left(a+b+c\right)^2-5}{ab+bc+ca}=\dfrac{4}{ab+bc+ca}\ge\dfrac{4.3}{\left(a+b+c\right)^2}=\dfrac{12}{3^2}=\dfrac{4}{3}\left(đpcm\right)\)

\(dấu"="\Leftrightarrow a=b=c=1\)

\(x^2+1=2x+\sqrt{3x-1}\) (không cần điệu kiện)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-\sqrt{3x-1}=0\left(1\right)\\x+1+\sqrt{3x-1}=0\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow x=\sqrt{3x-1}\Leftrightarrow\left\{{}\begin{matrix}x^2=3x-1\\x\ge0\end{matrix}\right.\Leftrightarrow x=\dfrac{3\pm\sqrt{5}}{2}\)

\(\left(2\right)\Leftrightarrow-x-1=\sqrt{3x-1}\Leftrightarrow\left\{{}\begin{matrix}\left(-x-1\right)^2=3x-1\\-x-1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2-x+2=0\\x\le-1\end{matrix}\right.\)(vô nghiệm)

Vậy phương trình có hai nghiệm \(x_{1,2}=\dfrac{3\pm\sqrt{5}}{2}\)