= \(\frac{\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7+2\sqrt{10}}}\)

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

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

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

= \(\sqrt{3}+\sqrt{5}+3\)

Bạn Khanh đúng r chỉ sai chỗ\(\sqrt{5+2\sqrt{6}}=\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}\) mới đúng

\(1,4\sqrt{5}-5\sqrt{2}+12\sqrt{5}-8\sqrt{5}=8\sqrt{5}-5\sqrt{2}\)

\(2,\left(\sqrt{27}+\sqrt{32}-\sqrt{50}\right)\left(\sqrt{27}-\sqrt{32}+\sqrt{50}\right)\)

\(=27-\left(\sqrt{32}-\sqrt{50}\right)^2=27-2=25\)

a)+) \(A=\sqrt{2x^2-3x+1}=\sqrt{2x^2-2x-x+1}\)

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

Để A có nghĩa thì \(\hept{\begin{cases}2x-1\ge0\\x-1\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge\frac{1}{2}\\x\ge1\end{cases}}\Leftrightarrow x\ge1\)

hoặc \(\hept{\begin{cases}2x-1\le0\\x-1\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le\frac{1}{2}\\x\le1\end{cases}}\Leftrightarrow x\le\frac{1}{2}\)

A có nghĩa\(\Leftrightarrow\orbr{\begin{cases}x\ge1\\x\le\frac{1}{2}\end{cases}}\)

+) B có nghĩa\(\Leftrightarrow\hept{\begin{cases}x-1\ge0\\2x-1\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge1\\x\ge\frac{1}{2}\end{cases}}\Leftrightarrow x\ge1\)

c) \(A=B\Leftrightarrow\sqrt{\left(x-1\right)\left(2x-1\right)}=\sqrt{x-1}.\sqrt{2x-1}\)

\(\Leftrightarrow\hept{\begin{cases}x-1\ge0\\2x-1\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge1\\x\ge\frac{1}{2}\end{cases}}\Leftrightarrow x\ge1\)

Vậy \(x\ge1\)thì A = B

d) \(x\le\frac{1}{2}\)

Đề câu c ptrinh = 4 là phải riêng ra chứ

\(a,\frac{3x+2}{\sqrt{x+2}}=2\sqrt{x+2}\)

\(\Rightarrow3x+2=2\sqrt{x+2}.\sqrt{x+2}\)

\(\Rightarrow3x+2=2\left(x+2\right)\)

\(\Rightarrow3x+2=2x+4\)

\(\Rightarrow3x-2x=4-2\)

\(\Rightarrow x=2\)

\(b,\sqrt{4x^2-1}-2\sqrt{2x+1}=0\)

\(\Rightarrow\sqrt{\left(2x+1\right)\left(2x-1\right)}-2\sqrt{2x+1}=0\)

\(\Rightarrow\sqrt{2x+1}\left(\sqrt{2x-1}-2\right)=0\)

\(\Rightarrow\hept{\begin{cases}\sqrt{2x+1}=0\\\sqrt{2x-1}-2=0\end{cases}\Rightarrow\orbr{\begin{cases}2x+1=0\\\sqrt{2x-1}=2\end{cases}\Rightarrow}\orbr{\begin{cases}2x=-1\\2x-1=4\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{2}\\2x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{5}{2}\end{cases}}}\)

\(c,\sqrt{x-2}+\sqrt{4x-8}-\frac{2}{5}\sqrt{\frac{25x-50}{4}}=4\)

\(\Rightarrow\sqrt{x-2}+\sqrt{4\left(x-2\right)}-\frac{2}{5}\sqrt{\frac{25\left(x-2\right)}{4}}=4\)

\(\Rightarrow\sqrt{x-2}+2\sqrt{x-2}-\frac{2}{5}.\frac{5\sqrt{x-2}}{2}=4\)

\(\Rightarrow\sqrt{x-2}+2\sqrt{x-2}-\sqrt{x-2}=4\)

\(\Rightarrow2\sqrt{x-2}=4\)

\(\Rightarrow\sqrt{x-2}=2\)

\(\Rightarrow x-2=4\)

\(\Rightarrow x=6\)

\(d,\sqrt{x+4}-\sqrt{1-x}=\sqrt{1-2x}\)

\(\Rightarrow\sqrt{x+4}=\sqrt{1-2x}+\sqrt{1-x}\)

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

\(\Rightarrow x+4=2-3x+2\sqrt{1-3x+2x^2}\)

\(\Rightarrow x+4-2+3x=2\sqrt{1-3x+2x^2}\)

\(\Rightarrow4x+2=2\sqrt{1-3x+2x^2}\)

\(\Rightarrow2x+1=\sqrt{1-3x+2x^2}\)

\(\Rightarrow4x^2+4x+1=1-3x+2x^2\)

\(\Rightarrow4x^2-2x^2+4x+3x+1-1=0\)

\(\Rightarrow2x^2+7x=0\)

\(\Rightarrow x\left(2x+7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\2x+7=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{-7}{2}\end{cases}}}\)

\(e,\frac{2x}{\sqrt{5}-\sqrt{3}}-\frac{2x}{\sqrt{3}+1}=\sqrt{5}+1\)

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

\(\Rightarrow x\left(\sqrt{5}+\sqrt{3}\right)-x\left(\sqrt{3}-1\right)=\sqrt{5}+1\)

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

\(\Rightarrow\sqrt{5}x+x=\sqrt{5}+1\)

\(\Rightarrow x\left(\sqrt{5}+1\right)=\sqrt{5}+1\)

\(\Rightarrow x=1\)

Câu 1

a)

Để biểu thức A có nghĩa thì \(2x^2-3x+1\ge0\Leftrightarrow\left(x-1\right)\left(2x-1\right)\ge0\)

\(\Leftrightarrow x\ge1\)

b)

Để biểu thức B có nghĩa thì \(x-1\ge0;2x-1\ge0\Rightarrow x\ge1\)

c)

Với \(x\ge1\) thì biểu thức A luôn luôn bằng biểu thức B

d)

Vô lý vcl

Câu 2

Xài BĐT Bunhiacopski:

\(A^2=\left(2x+3y\right)^2=\left(2\cdot x+3\cdot y\right)^2\le13\left(x^2+y^2\right)=1521\)

\(\Rightarrow A\le39\)

Câu 1:

a) A=\(\sqrt{2x^2-3x+1}\)

ĐKXĐ: \(\orbr{\begin{cases}x\le\frac{1}{2}\\x\ge1\end{cases}}\)

b) B=\(\sqrt{x-1}\cdot\sqrt{2x-1}\)

ĐKXĐ:\(\orbr{\begin{cases}x\ge1\\x\ge\frac{1}{2}\end{cases}}\)

=>\(x\ge1\)

c) Với \(x\ge1\)thì A=B đc xác định

d) Với \(x\le\frac{1}{2}\)thì A có nghĩa,B không có nghĩa

a) \(\sqrt{x-1-2\sqrt{x-2}}=\sqrt{\left(x-2\right)-2\sqrt{x-2}+1}=\sqrt{\left(\sqrt{x-2}-1\right)^2}=\left|\sqrt{x-2}-1\right|\)

b) \(\sqrt{x-2+2\sqrt{x-3}}-\sqrt{x-3}\)

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

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

a.\(A=x^2+2x+16=x^2+2x+1+15=\left(x+1\right)^2+15\)

Với : \(x=\sqrt{2}-1\)ta có:

\(A=\left(\sqrt{2}-1+1\right)^2+15=2+15=17\)

b. \(B=x^2+12x-14=x^2+2.x.6+36-36-14=\left(x+6\right)^2-50\)

Với \(x=5\sqrt{2}-6\)

Ta có: \(B=\left(5\sqrt{2}+6-6\right)^2-50=50-50=0\)

\(PT\Leftrightarrow\sqrt{8x+1}-3+\sqrt{46x-10}-6=-x^3+5x^2+4x+1-3-6\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{8}{\sqrt{8x+1}+3}-5+x^2-4x-3-\frac{10}{\sqrt{46-10x}+6}\right)=0\)

Xét \(\left(\frac{8}{\sqrt{8x+1}+3}-5+x^2-4x-3-\frac{10}{\sqrt{46-10x}+6}\right)\)(*) (đk\(\frac{23}{5}\ge x\ge-\frac{1}{8}\))

(*)\(=\frac{8-5\left(\sqrt{8x+1}+3\right)}{\sqrt{8x+1}+3}+\left(x^2-4x-3\right)-\frac{10}{\sqrt{46-10x}+6}\)

\(=\frac{-7-5\left(\sqrt{8x+1}\right)}{\sqrt{8x+1}+3}+\left(x^2-4x-3\right)-\frac{10}{\sqrt{46-10x}+6}< 0\)

\(\Rightarrow x-1=0\Leftrightarrow x=1\)

Vậy..................

Đề thi thuyển sinh lớp 10 môn Toán Chuyên, TP HCM năm 2012-2013

ĐK \(\frac{-1}{8}\le x\le\frac{23}{5}\)(*) Ta có:

\(\sqrt{8x+1}+\sqrt{46-10x}=-x^3+5x^2+4x+1\)

\(\Leftrightarrow\sqrt{8x+1}-3+\sqrt{46-10x}-6+x^3-x^2-4x^2+4x-8x+8=0\)

\(\Leftrightarrow\frac{8x-1}{\sqrt{8x+1}+3}+\frac{10-10x}{\sqrt{46-10x}+6}+x^2\left(x-1\right)-4x\left(x-1\right)-8\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{8}{\sqrt{8x+1}+3}+\frac{10}{\sqrt{46-10x}+6}+x^2-4x-8\right)=0\)(**)

(*) \(\Rightarrow-1< x< 5\Rightarrow\left(x+1\right)\left(x+5\right)< 0\Rightarrow x^2-4x-5< 0\)

Và \(\frac{8}{\sqrt{8x+1}+3}< \frac{9}{3}=3\Rightarrow\frac{8}{\sqrt{8x+1}+3}-3< 0\) Do vậy:

\(\frac{8}{\sqrt{8x+1}+3}-\frac{10}{\sqrt{46-10x}+6}+x^2-4x-8< 0\)Do đó:

(**)\(\Leftrightarrow x=1\)

Vậy S={1}