Bài làm:

a) \(A=\sqrt{4}-2\sqrt{3}+\sqrt{7}-4\sqrt{3}\)

\(A=2+\sqrt{7}-6\sqrt{3}\)

b) \(B=\sqrt{3}+\sqrt{8}+\sqrt{3}-\sqrt{8}\)

\(B=2\sqrt{3}\)

đoạn cuối thiếu dấu"+"

\(A=\dfrac{\sqrt{4}-\sqrt{5}}{4-5}+\dfrac{\sqrt{5}-\sqrt{6}}{5-6}+....+\dfrac{\sqrt{34}-\sqrt{35}}{34-35}+\dfrac{\sqrt{35}-\sqrt{36}}{335-36}\)

\(A=\dfrac{\sqrt{4}-\sqrt{5}+\sqrt{5}-\sqrt{6}+....+\sqrt{35}-\sqrt{36}}{-1}=\dfrac{\sqrt{4}-\sqrt{36}}{-1}\)

\(A=\sqrt{36}-\sqrt{4}=6-2=4\)

\(\dfrac{1}{\sqrt{4}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{6}}+\dfrac{1}{\sqrt{6}+\sqrt{7}}+...+\dfrac{1}{\sqrt{34}+\sqrt{35}}+\dfrac{1}{\sqrt{35}+\sqrt{36}}\)

\(=-\sqrt{4}+\sqrt{5}-\sqrt{5}+\sqrt{6}-...-\sqrt{35}+\sqrt{36}\)

\(=6-2=4\)

ĐKXĐ: \(a^3+a^2-3\ge0\) (1)

Đặt \(\sqrt{a^3+a^2-3}=x\ge0\Rightarrow a^3+a^2+4=x^2+7\)

Phương trình trở thành:

\(\sqrt{x^2+7}+x=7\)

\(\Leftrightarrow\sqrt{x^2+7}=7-x\) (\(x\le7\))

\(\Leftrightarrow x^2+7=x^2-14x+49\)

\(\Rightarrow14x=42\)

\(\Rightarrow x=3\)

\(\Rightarrow a^3+a^2-3=9\)

\(\Rightarrow a^3+a^2-12=0\)

\(\Rightarrow\left(a-2\right)\left(a^2+3a+6\right)=0\)

\(\Rightarrow a=2\) (thỏa mãn (1))

a: Ta có: \(\dfrac{4}{\sqrt{7}-\sqrt{3}}+\dfrac{6}{3+\sqrt{3}}+\dfrac{\sqrt{7}-7}{\sqrt{7}-1}\)

\(=\sqrt{7}+\sqrt{3}+3-\sqrt{3}-\sqrt{7}\)

=3

a) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left(2x+1\right)^2=6^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

b) \(\sqrt{4x^2-4\sqrt{7}x+7}=\sqrt{7}\)

\(\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)

\(\Leftrightarrow\left(2x-\sqrt{7}\right)^2=\left(\sqrt{7}\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt[]{7}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)

a) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left|2x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

b) \(pt\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)

\(\Leftrightarrow\left|2x-\sqrt{7}\right|=\sqrt{7}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)

c) \(PT\Leftrightarrow\sqrt{\left(x+\sqrt{3}\right)^2}=2\sqrt{3}\)

\(\Leftrightarrow\left|x+\sqrt{3}\right|=2\sqrt{3}\)

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

d) \(pt\Leftrightarrow\left|x-3\right|=9\Leftrightarrow\left[{}\begin{matrix}x-3=-9\\x-3=9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=12\end{matrix}\right.\)

Ta có : \(A=\sqrt{4-2\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)

\(=\sqrt{3-2\sqrt{3}+1}+\sqrt{2^2-2.2\sqrt{3}+3}\)

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

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

Ta có : \(B=\sqrt{3+\sqrt{8}+\sqrt{3-\sqrt{8}}}\)

\(=\sqrt{3+\sqrt{8}+\sqrt{2-2\sqrt{2}+1}}\)\(=\sqrt{3+\sqrt{8}+\sqrt{\left(\sqrt{2}-1\right)^2}}\)

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

a: Thay \(x=7-4\sqrt{3}\) vào A, ta được:

\(A=2-\sqrt{3}-7+4\sqrt{3}=3\sqrt{3}-5\)

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

Tương tự  \(\sqrt{4-2\sqrt{3}}=\sqrt{3}-1\);   \(\sqrt{7-4\sqrt{3}}=2-\sqrt{3}\)

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

\(\sqrt{3-2\sqrt{2}}+\sqrt{4-2\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)

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

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

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

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

ok nhé

\(B=\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)

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

     \(=\sqrt{3}+2+2-\sqrt{3}\)

   \(=4\)

Còn cách nữa là bình phương

Đag làm thì ấn nhầm trả lời .V
Cách bình phương đây

\(B=\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)

\(\Rightarrow B^2=7+4\sqrt{3}+2\sqrt{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}+7-4\sqrt{3}\)

           \(=14+2\sqrt{49-48}\)

            \(=14+2\)

             \(=16\)

\(\Rightarrow B=\sqrt{16}=4\)

\(B=\sqrt{\left(2\right)^2+4\sqrt{3}+3}+\sqrt{\left(2\right)^2-4\sqrt{3}+3}\)

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

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