a: Ta có: \(4\sqrt{3a}-3\sqrt{12a}+\dfrac{6\sqrt{a}}{3}-2\sqrt{20a}\)

\(=4\sqrt{3a}-6\sqrt{3a}+2\sqrt{2a}-4\sqrt{5a}\)

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

Đề bài yêu cầu gì?

Rút gọn biểu thức

gõ latex đi b=)

\(A=\sqrt{x}+1\) (đã thu gọn)

\(B=\dfrac{4\sqrt{x}}{x+4}\) (đã thu gọn)

\(A=x-\sqrt{x}+1=\sqrt{x}\cdot\sqrt{x}-\sqrt{x}+1=\sqrt{x}\left(\sqrt{x}-1\right)+1\)

\(A=\dfrac{3}{2\sqrt{x}}\) (đã thu gọn)

\(A=\dfrac{3}{\sqrt{x}+3}\) (đã thu gọn)

\(A=1-\sqrt{x}\) (đã thu gọn)

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

a) \(\sqrt[]{x^2-4x+4}=x+3\)

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

\(\Leftrightarrow\left|x-2\right|=x+3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=x+3\\x-2=-\left(x+3\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}0x=5\left(loại\right)\\x-2=-x-3\end{matrix}\right.\)

\(\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)

b) \(2x^2-\sqrt[]{9x^2-6x+1}=5\)

\(\Leftrightarrow2x^2-\sqrt[]{\left(3x-1\right)^2}=5\)

\(\Leftrightarrow2x^2-\left|3x-1\right|=5\)

\(\Leftrightarrow\left|3x-1\right|=2x^2-5\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=2x^2-5\\3x-1=-2x^2+5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x-4=0\left(1\right)\\2x^2+3x-6=0\left(2\right)\end{matrix}\right.\)

Giải pt (1)

\(\Delta=9+32=41>0\)

Pt \(\left(1\right)\) \(\Leftrightarrow x=\dfrac{3\pm\sqrt[]{41}}{4}\)

Giải pt (2)

\(\Delta=9+48=57>0\)

Pt \(\left(2\right)\) \(\Leftrightarrow x=\dfrac{-3\pm\sqrt[]{57}}{4}\)

Vậy nghiệm pt là \(\left[{}\begin{matrix}x=\dfrac{3\pm\sqrt[]{41}}{4}\\x=\dfrac{-3\pm\sqrt[]{57}}{4}\end{matrix}\right.\)

\(A=\sqrt{3+2\sqrt{2}}-\dfrac{1}{2}\sqrt{8}\)

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

=1

a:

Sửa đề: \(M=\frac{a^2+2}{a^3-1}+\frac{a+1}{a^2+a+1}-\frac{1}{a-1}\)

ĐKXĐ: a<>1

Ta có: \(M=\frac{a^2+2}{a^3-1}+\frac{a+1}{a^2+a+1}-\frac{1}{a-1}\)

\(=\frac{a^2+2}{\left(a-1\right)\left(a^2+a+1\right)}+\frac{a+1}{a^2+a+1}-\frac{1}{a-1}\)

\(=\frac{a^2+2+\left(a+1\right)\left(a-1\right)-a^2-a-1}{\left(a-1\right)\left(a^2+a+1\right)}\)

\(=\frac{-a+1+a^2-1}{\left(a-1\right)\left(a^2+a+1\right)}=\frac{a^2-a}{\left(a-1\right)\left(a^2+a+1\right)}=\frac{a}{a^2+a+1}\)

b: \(a=\sqrt{7+4\sqrt3}+\sqrt{7-4\sqrt3}\)

\(=\sqrt{\left(2+\sqrt3\right)^2\left.\right.}+\sqrt{\left(2-\sqrt3\right)^2}\)

\(=2+\sqrt3+2-\sqrt3=4\)

Thay a=4 vào M, ta được:

\(M=\frac{4}{4^2+4+1}=\frac{4}{16+5}=\frac{4}{21}\)

c: \(M=\frac15\)

=>\(\frac{a}{a^2+a+1}=\frac15\)

=>\(a^2+a+1=5a\)

=>\(a^2-4a+1=0\)

=>\(a^2-4a+4-3=0\)

=>\(\left(a-2\right)^2=3\)

=>\(\left[\begin{array}{l}a-2=\sqrt3\\ a-2=-\sqrt3\end{array}\right.\Rightarrow\left[\begin{array}{l}a=\sqrt3+2\left(nhận\right)\\ a=-\sqrt3+2\left(nhận\right)\end{array}\right.\)

a: \(\left(\sqrt3-\sqrt2\right)\left(\sqrt3+\sqrt2\right)=3-2=1\)

\(f\left(\sqrt{\left(\sqrt3-\sqrt2\right)\left(\sqrt3+\sqrt2\right)}\right)=f\left(1\right)=\sqrt{2\cdot1}=\sqrt2\)

b: \(f\left(\sqrt{3+\sqrt5}\right)=\sqrt{2\left(3+\sqrt5\right)}=\sqrt{6+2\sqrt5}\)

\(=\sqrt{\left(\sqrt5+1\right)^2}=\sqrt5+1\)