c: \(=\left(4+\sqrt{3}\right)\cdot\sqrt{\left(4-\sqrt{3}\right)^2}\)

=(4+căn 3)(4-căn 3)

=16-3=13

d: \(=\sqrt{6+2\sqrt{5-2\sqrt{3}-1}}\)

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

\(=\sqrt{6+2\cdot\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)

b) \(\sqrt{\left(1-\sqrt{5}\right)^2}+\sqrt{23-8\sqrt{5}}\)

\(=\left|1-\sqrt{5}\right|+\sqrt{23-2\sqrt{60}}\)

\(=\sqrt{5}-1+\sqrt{\left(\sqrt{20}\right)^2-2.\sqrt{20}.\sqrt{3}+\left(\sqrt{3}\right)^2}\)

\(=\sqrt{5}-1+\sqrt{\left(\sqrt{20}-\sqrt{3}\right)^2}\)

\(=\sqrt{5}-1+\sqrt{20}-\sqrt{3}=\sqrt{5}+2\sqrt{5}-1-\sqrt{3}\)

\(=3\sqrt{5}-1-\sqrt{3}\)

b: x=72

d: Ta có: \(\left(2x-1\right)\left(3x+1\right)+\left(3x-4\right)\left(3-2x\right)=5\)

\(\Leftrightarrow6x^2+2x-3x-3+9x-6x^2-12+8x=5\)

\(\Leftrightarrow16x=5+15=20\)

hay \(x=\dfrac{5}{4}\)

b: \(=\dfrac{x^2-x+1-3+1-x^2}{\left(x+1\right)\cdot\left(x^2-x+1\right)}=\dfrac{-x-1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{-1}{x^2-x+1}\)

a: \(\left(x-1.2\right)^2=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1.2=2\\x-1.2=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3.2\\x=-0.8\end{matrix}\right.\)

b: Ta có: \(\left(x+1\right)^3=-125\)

\(\Leftrightarrow x+1=-5\)

hay x=-6

c) 3^(4-x)=27

3^(4-x) = 3^3

4-x = 3

x = 1

Đề bài không rõ ràng, em liên hệ người ra đề xem vẽ đồ thị đường thẳng nào? Vì đường thẳng đề cho có a chưa biết