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a, \(\sqrt{3+2\sqrt{2}}-\sqrt{6-4\sqrt{2}}\)
= \(\sqrt{2+2\sqrt{2}+1}-\sqrt{4-4\sqrt{2}+2}\)
= \(\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(2-\sqrt{2}\right)^2}\)
= \(\sqrt{2}\) + 1 - 2 + \(\sqrt{2}\)
= 2\(\sqrt{2}\) - 1
b, \(\sqrt{9-4\sqrt{5}}-\sqrt{5}\)
= \(\sqrt{5-4\sqrt{5}+4}-\sqrt{5}\)
= \(\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}\)
= \(\sqrt{5}-2-\sqrt{5}\)
= - 2
c, \(\sqrt{28+8\sqrt{7}}-\sqrt{7}\)
= \(\sqrt{16+8\sqrt{7}+7}-\sqrt{7}\)
= \(\sqrt{\left(4+\sqrt{7}\right)^2}-\sqrt{7}\)
= 4 + \(\sqrt{7}\) - \(\sqrt{7}\)
= 4
b) \(\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{2}+2}-\sqrt{5+2\cdot\sqrt{5}\cdot\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}\)
\(=\left|\sqrt{5}-\sqrt{2}\right|-\left|\sqrt{5}+\sqrt{2}\right|\)
\(=\sqrt{5}-\sqrt{2}-\sqrt{5}-\sqrt{2}\) (vì \(\sqrt{5}\ge\sqrt{2}\)
=0
c) \(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{3-2\sqrt{3}+1}+\sqrt{3+2\sqrt{3}+1}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left|\sqrt{3}-1\right|+\left|\sqrt{3}+1\right|\)
\(=\sqrt{3}-1+\sqrt{3+1}\) (vì \(\sqrt{3}\ge1\))
\(=2\sqrt{3}\)
a)\(\sqrt{5+2\sqrt{6}}-\sqrt{5+2\sqrt{6}}\)
\(=\sqrt{3+2\cdot\sqrt{3}\cdot\sqrt{2}+2}-\sqrt{3-2\cdot\sqrt{3}\cdot\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\left|\sqrt{3}+\sqrt{2}\right|-\left|\sqrt{3}-\sqrt{2}\right|\)
\(=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}\) (vì \(\sqrt{3}\ge\sqrt{2}\))
=0
\(A=\sqrt[3]{2-\sqrt{5}}\left(\sqrt[6]{\left(2+\sqrt{5}\right)^2}+\sqrt[3]{2+\sqrt{5}}\right)\)
\(=\sqrt[3]{2-\sqrt{5}}.2\sqrt[3]{2+\sqrt{5}}=2\sqrt[3]{4-5}=-2\)
\(B=\sqrt[4]{\left(3-2\sqrt{2}\right)^2}-\sqrt{2}=\sqrt{3-2\sqrt{2}}-\sqrt{2}\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{2}=\sqrt{2}-1-\sqrt{2}=-1\)
\(C=\sqrt[4]{\left(6-2\sqrt{5}\right)^2}=\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-1\)
\(D=1+\sqrt[4]{\left(4-2\sqrt{3}\right)^2}=1+\sqrt{4-2\sqrt{3}}\)
\(=1+\sqrt{\left(\sqrt{3}-1\right)^2}=1+\sqrt{3}-1=\sqrt{3}\)
Câu e lấy nguyên văn từ sách thầy Vũ Hữu Bình:
Đặt \(x=\sqrt[4]{5}\Rightarrow x^4=5\Rightarrow5-x^4=0\)
\(E=\frac{2}{\sqrt{4-3x+2x^2-x^3}}=\frac{2\left(x+1\right)}{\sqrt{\left(x+1\right)^2\left(4-3x+2x^2-x^3\right)}}=\frac{2\left(x+1\right)}{\sqrt{-x^5+5x+4}}\)
\(E=\frac{2\left(x+1\right)}{\sqrt{x\left(5-x^4\right)+4}}=\frac{2\left(x+1\right)}{\sqrt{4}}=x+1=\sqrt[4]{5}+1\)
Không hiểu ý tưởng nhân cả tử và mẫu với \(x+1\) từ đâu ra luôn
a/ \(\Leftrightarrow\sqrt{x^2+x+3}-\sqrt{x^2+2}+\sqrt{x^2+x+8}-\sqrt{x^2+7}=0\)
\(\Leftrightarrow\frac{x+1}{\sqrt{x^2+x+3}+\sqrt{x^2+2}}+\frac{x+1}{\sqrt{x^2+x+8}+\sqrt{x^2+7}}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{\sqrt{x^2+x+3}+\sqrt{x^2+2}}+\frac{1}{\sqrt{x^2+x+8}+\sqrt{x^2+7}}\right)=0\)
\(\Leftrightarrow x+1=0\) (ngoặc to phía sau luôn dương)
\(\Rightarrow x=-1\)
b/
\(\sqrt{7-x^2+x\sqrt{x+5}}=\sqrt{3-2x-x^2}\) (1)
\(\Rightarrow7-x^2+x\sqrt{x+5}=3-2x-x^2\)
\(\Leftrightarrow x\sqrt{x+5}=-2x-4\)
\(\Rightarrow x^2\left(x+5\right)=4x^2+16x+16\)
\(\Rightarrow x^3+x^2-16\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x^2-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\\x=-2\end{matrix}\right.\)
Do các phép biến đổi ko tương đương nên cần thay nghiệm vào (1) để kiểm tra
c/ ĐKXĐ: \(x\ge\frac{5}{3}\)
\(\Leftrightarrow\sqrt{10x+1}-\sqrt{9x+4}+\sqrt{3x-5}-\sqrt{2x-2}=0\)
\(\Leftrightarrow\frac{x-3}{\sqrt{10x+1}+\sqrt{9x+4}}+\frac{x-3}{\sqrt{3x-5}+\sqrt{2x-2}}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{\sqrt{10x+1}+\sqrt{9x+4}}+\frac{1}{\sqrt{3x-5}+\sqrt{2x-2}}\right)=0\)
\(\Leftrightarrow x-3=0\) (ngoặc phía sau luôn dương)
d/ Đề bài là \(2\sqrt{2x+3}\) hay \(2\sqrt{2x-3}\) bạn?
e/ ĐKXĐ: \(x\ge-3\)
\(\Leftrightarrow\sqrt{x+3+2\sqrt{x+3}+1}=x+4\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x+3}+1\right)^2}=x+4\)
\(\Leftrightarrow\sqrt{x+3}+1=x+4\)
\(\Leftrightarrow x+3-\sqrt{x+3}=0\)
\(\Leftrightarrow\sqrt{x+3}\left(\sqrt{x+3}-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x+3=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)
\(A=\left(\frac{2}{\sqrt{x}-2}+\frac{3}{2\sqrt{x}+1}-\frac{5\sqrt{x}-7}{2x-3\sqrt{x}-2}\right):\)\(\frac{2\sqrt{x}+3}{5x-10\sqrt{x}}\)
\(=\left(\frac{2}{\sqrt{x}-2}+\frac{3}{2\sqrt{x}+1}-\frac{5\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\right)\)\(:\frac{2\sqrt{x}+3}{5x-10\sqrt{x}}\)
\(=\frac{2\left(2\sqrt{x}+1\right)+3\left(\sqrt{x}-2\right)-5\sqrt{x}+7}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\)\(:\frac{2\sqrt{x}+3}{5\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\)\(.\frac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\frac{2\sqrt{x}+3}{2\sqrt{x}+1}.\frac{5\sqrt{x}}{2\sqrt{x}+3}=\frac{5\sqrt{x}}{2\sqrt{x}+1}\)
\(A\in Z\Leftrightarrow\frac{5\sqrt{x}}{2\sqrt{x}+1}\in Z\Leftrightarrow\frac{10\sqrt{x}}{2\sqrt{x}+1}\in Z\)
\(\Rightarrow\frac{10\sqrt{x}+5-5}{2\sqrt{x}+1}\in Z\Leftrightarrow5-\frac{5}{2\sqrt{x}+1}\in Z\)
\(\Rightarrow\frac{5}{2\sqrt{x}+1}\in Z\Rightarrow2\sqrt{x}+1\inƯ_5\)
Mà \(Ư_5=\left\{\pm1;\pm5\right\}\)
Nhưng \(2\sqrt{x}+1\ge1\)
\(\Rightarrow\orbr{\begin{cases}2\sqrt{x}+1=1\\2\sqrt{x}+1=5\end{cases}\Rightarrow\orbr{\begin{cases}2\sqrt{x}=0\\2\sqrt{x}=4\end{cases}}}\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}=2\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}}\)
Vậy \(x\in\left\{0;4\right\}\)
a/ ĐKXĐ: \(x\ge-1\)
\(\Leftrightarrow\sqrt{x+1}-1+\sqrt{x+4}-2>0\)
\(\Leftrightarrow\frac{x}{\sqrt{x+1}+1}+\frac{x}{\sqrt{x+4}+2}>0\)
\(\Leftrightarrow x>0\)
b/
Chắc bạn ghi nhầm đề, thấy đề hơi kì lạ
c/ ĐKXĐ: \(\left[{}\begin{matrix}-\frac{3}{2}\le x\le\frac{3-\sqrt{57}}{8}\\x\ge\frac{3+\sqrt{57}}{8}\end{matrix}\right.\)
\(\Leftrightarrow2x+3>4x^2-3x-3\)
\(\Leftrightarrow4x^2-5x-6< 0\) \(\Rightarrow-\frac{3}{4}< x< 2\)
Kết hợp ĐKXĐ ta được nghiệm của BPT: \(\left[{}\begin{matrix}-\frac{3}{4}< x\le\frac{3-\sqrt{57}}{8}\\\frac{3+\sqrt{57}}{8}\le x< 2\end{matrix}\right.\)
d/
\(\Leftrightarrow x^2+5x+28-5\sqrt{x^2+5x+28}-24< 0\)
Đặt \(\sqrt{x^2+5x+28}=t>0\)
\(\Leftrightarrow t^2-5t-24< 0\) \(\Rightarrow-3< t< 8\)
\(\Rightarrow t< 8\Rightarrow\sqrt{x^2+5x+28}< 8\)
\(\Leftrightarrow x^2+5x-36< 0\Rightarrow-9< x< 4\)
a/ ĐKXĐ: ...
\(\Leftrightarrow3\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)-7\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow a^2=x+\frac{1}{4x}+1\)
\(\Rightarrow x+\frac{1}{4x}=a^2-1\)
Pt trở thành:
\(3a=2\left(a^2-1\right)-7\)
\(\Leftrightarrow2a^2-3a-9=9\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}=3\)
\(\Leftrightarrow2x-6\sqrt{x}+1=0\)
\(\Rightarrow\sqrt{x}=\frac{3+\sqrt{7}}{2}\Rightarrow x=\frac{8+3\sqrt{7}}{2}\)
b/ ĐKXĐ:
\(\Leftrightarrow5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow x+\frac{1}{4x}=a^2-1\)
\(\Rightarrow5a=2\left(a^2-1\right)+4\Leftrightarrow2a^2-5a+2=0\)
\(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+\frac{1}{2\sqrt{x}}=2\\\sqrt{x}+\frac{1}{2\sqrt{x}}=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-4\sqrt{x}+1=0\\2x-\sqrt{x}+1=0\left(vn\right)\end{matrix}\right.\)
c/ ĐKXĐ: ...
\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)
\(\Leftrightarrow\frac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)
\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\frac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)
\(\Leftrightarrow2x^2-8x+5=0\)
d/ ĐKXĐ: ...
\(\Leftrightarrow x+1-\frac{15}{6}\sqrt{x}+\sqrt{x^2-4x+1}-\frac{1}{2}\sqrt{x}=0\)
\(\Leftrightarrow\frac{x^2-\frac{17}{4}x+1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{x^2-\frac{17}{4}x+1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}=0\)
\(\Leftrightarrow\left(x^2-\frac{17}{4}x+1\right)\left(\frac{1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}\right)=0\)
\(\Leftrightarrow x^2-\frac{17}{4}x+1=0\)
\(\Leftrightarrow4x^2-17x+4=0\)
\(A=\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\)
\(\Rightarrow A^3=14+3\sqrt[3]{\left(7+5\sqrt{2}\right)\left(7-5\sqrt{2}\right)}\left(\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\right)\)
<=>A3=14-3A
<=>A3+3A-14=0
<=>A3-4A+7A-14=0
<=>A.(A-2)(A+2)+7.(A-2)=0
<=>(A-2)(A2+9A-14)=0
<=>A=2(nhận)
Vậy A=2
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