Bài 2: Căn thức bậc hai và hằng đẳng thức căn bậc hai của bình phương

b) ta có pt \(\sqrt{25-x^2}-\sqrt{9-x^2}=2\)

Đặt \(\sqrt{25-x^2}=a;\sqrt{9-x^2}=b\left(a,b\ge0\right)\Rightarrow a-b=2\)

Mà \(a^2-b^2=25-x^2-9+x^2=16\Leftrightarrow\left(a-b\right)\left(a+b\right)=16\Leftrightarrow a+b=8\)

ta có a-b=2;a+b=8=> a=5;b=3

a) ta có pt \(\dfrac{4}{x}+\sqrt{x-\dfrac{1}{x}}=x+\sqrt{2x-\dfrac{5}{x}}\Leftrightarrow x-\dfrac{4}{x}+\sqrt{2x-\dfrac{5}{x}}-\sqrt{x-\dfrac{1}{x}}=0\)

đặt \(\sqrt{2x-\dfrac{5}{x}}=a;\sqrt{x-\dfrac{1}{x}}=b\Rightarrow a^2-b^2=2x-\dfrac{5}{x}-x+\dfrac{1}{x}=x-\dfrac{4}{x}\)

nên pt \(\Leftrightarrow a^2-b^2+a-b=0\Leftrightarrow\left(a-b\right)\left(a+b+1\right)=0\)

\(A=\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)=\sqrt{3-\sqrt{5}}.\sqrt{2}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)=\sqrt{5-2\sqrt{5}+1}\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)=\left(\sqrt{5}-1\right)^2\left(3+\sqrt{5}\right)=2\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)=2\left(9-5\right)=8\)

\(B=\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}=\sqrt{x-1-2\sqrt{x-1}+1}+\sqrt{x-1+2\sqrt{x-1}+1}=|\sqrt{x-1}-1|+\left|\sqrt{x-1}+1\right|\left(x\ge1\right)\)

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

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

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

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

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

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

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

điều kiện xác định : \(x>0;x\ne1\)

ta có : \(\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right)\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

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

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

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

\(6\pm2\sqrt{5}=5\pm2.\sqrt{5}.1+1=\left(\sqrt{5}\pm1\right)^2\)

\(\Rightarrow M=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5-1}\right)^2}=\sqrt{5}+1-\sqrt{5}+1=2\)

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

ĐKXĐ:\(-1\le x\le\dfrac{1}{2}\)

Bình phương cả hai vế ta có:

\(x+1=\left(1-2x\right)^2\Leftrightarrow x+1=4x^2-4x+1\)\(\Leftrightarrow4x^2-5x=0\Leftrightarrow4x\left(x-\dfrac{5}{4}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{5}{4}\left(loại\right)\end{matrix}\right.\)

Vậy \(S=\left\{0\right\}\)

a.b. \(M=\dfrac{x\sqrt{x}-3}{x-2\sqrt{x}-3}-\dfrac{2\sqrt{x}-6}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{3-\sqrt{x}}\) \(\left(x\ne9,x\ge0\right)\)

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

\(M=\dfrac{x\sqrt{x}-3-2x+12\sqrt{x}-18-x-4\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)

\(M=\dfrac{x\sqrt{x}-3x+8\sqrt{x}-24}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)

\(M=\dfrac{x\left(\sqrt{x}-3\right)+8\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-3\right)\left(x+8\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)

\(M=\dfrac{x+8}{\sqrt{x}+1}=\dfrac{36+8}{\sqrt{36}+1}=\dfrac{44}{7}\)

c. Mk nghĩ là GTLN :(

\(M=\dfrac{x+8}{\sqrt{x}+1}\le\dfrac{0+8}{0+1}=8\)

\("="\Leftrightarrow x=0\)

Lời giải:

Đặt \(1+\sqrt{3}=m\).

Ta phân tích đa thức ra như sau:

\(3x^3+ax^2+bx+12=(x+m)(3x^2+nx+p)\)

\(=3x^3+x^2n+xp+3mx^2+mnx+mp\)

\(=3x^3+x^2(n+3m)+x(p+mn)+mp\)

Đồng nhất hệ số:

\(\Rightarrow \left\{\begin{matrix} n+3m=a\\ p+mn=b\\ mp=12\end{matrix}\right.\). Thay $m=\sqrt{3}+1$ vào hệ trên:

\(\Rightarrow p=6\sqrt{3}-6\); \(n=a-3(1+\sqrt{3})\)

\(\Rightarrow 6\sqrt{3}-6+(1+\sqrt{3})[a-3(1+\sqrt{3})=b\)

\(\Rightarrow -18+(1+\sqrt{3})a=b\)

\(\Rightarrow (1+\sqrt{3})a=b+18\in\mathbb{Z}\)

Mà \(1+\sqrt{3}\not\in\mathbb{Q}\) nên suy ra $a=0$

\(\Rightarrow b=-18\)

Vậy $(a,b)=(0,-18)$

Làm sai rồi làm lại đi bác Akai Haruma

Lời giải:

Đặt $\sqrt{x^2+2}=a$. Khi đó ta có hệ sau:

\(\left\{\begin{matrix} x^2+(3-a)x=1+2a\\ a^2-x^2=2\end{matrix}\right.\)

\(\Rightarrow a^2-2+(3-a)x=1+2a\)

\(\Leftrightarrow a^2-2a-3-(a-3)x=0\)

\(\Leftrightarrow (a-3)(a+1)-(a-3)x=0\)

\(\Leftrightarrow (a-3)(a+1-x)=0\)

\(\Rightarrow \left[\begin{matrix} a=3\\ a+1=x\end{matrix}\right.\)

Nếu $a=3$ suy ra $x^2=a^2-2=7$ \(\Rightarrow x=\pm \sqrt{7}\)

Nếu \(a+1=x\Rightarrow (a+1)^2=x^2=a^2-2\)

\(\Rightarrow 1+2a=-2\Rightarrow a=-\frac{3}{2}\) (vô lý vì $a>0$)

\(A=\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}=\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{2}}-\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}}=\dfrac{\sqrt{3}+1-\sqrt{3}+1}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)