• Trường hợp 1: Nếu \(a\in Z\)thì phương trình có nghiệm x = 5 - a. Khi đó \(x\in Z\)thỏa mãn điều kiện bài toán.
•  Trường hợp 2: Nếu \(a\notin Z\)thì phương trình không có nghiệm nguyên.

\((3\sqrt{20}-2\sqrt{80}+\frac{2}{3}\sqrt{45}-\sqrt{5}):\sqrt{5}\)

\(=\left(3\sqrt{2^2.5}-2\sqrt{4^2.5}+\frac{2}{3}\sqrt{3^2.5}-\sqrt{5}\right):\sqrt{5}\)

\(=\left(3.2\sqrt{5}-2.4\sqrt{5}+\frac{2}{3}.3\sqrt{5}\right):\sqrt{5}\)

\(=\left(6\sqrt{5}-8\sqrt{5}+2\sqrt{5}-\sqrt{5}\right):\sqrt{5}\)

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

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

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

\(=\left(\frac{4+4\sqrt{5}+5-\left(4-4\sqrt{5}+5\right)}{4-5}\right).\frac{-\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\)

\(=\frac{9+4\sqrt{5}-9+4\sqrt{5}}{-1}.\left(-\sqrt{5}\right)\)

\(-8\sqrt{5}.\left(-\sqrt{5}\right)=40\)

\(a\text{)}\:36x^2-5=\left(6x\right)^2-\left(\sqrt{5}\right)^2\\ =\left(6x-\sqrt{5}\right)\left(6x+\sqrt{5}\right)\)

\(b\text{)}\:25-3x^2=5^2-\left(\sqrt{3}x\right)^2\\ =\left(5-\sqrt{3}x\right)\left(5+\sqrt{3}\right)\)

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

\(d\text{)}\:11+9x=9.\dfrac{11}{9}+9x\\ =9\left(\dfrac{11}{9}+x\right)\)

\(e\text{)}\:31+7x=7.\dfrac{31}{7}+7x\\ =7\left(\dfrac{31}{7}+x\right)\)

1: ĐKXĐ: x<=31

\(\sqrt{31-x}=x-1\)

=>\(\begin{cases}x-1\ge0\\ \left(x-1\right)^2=31-x\end{cases}\Rightarrow\begin{cases}x\ge1\\ x^2-2x+1-31+x=0\end{cases}\)

=>\(\begin{cases}1\le x\le31\\ x^2-x-30=0\end{cases}\Rightarrow\begin{cases}1\le x\le31\\ \left(x-6\right)\left(x+5\right)=0\end{cases}\Rightarrow x=6\)

3: ĐKXĐ: x∈R
\(\sqrt{x^2-3x+5}+x=3x+7\)

=>\(\sqrt{x^2-3x+5}=2x+7\)

=>\(\begin{cases}2x+7\ge0\\ \left(2x+7\right)^2=x^2-3x+5\end{cases}\Rightarrow\begin{cases}x\ge-\frac72\\ 4x^2+28x+49-x^2+3x-5=0\end{cases}\)

=>\(\begin{cases}x\ge-\frac72\\ 3x^2+31x+44=0\end{cases}\Rightarrow\begin{cases}x\ge-\frac72\\ x^2+\frac{31}{3}+\frac{44}{3}=0\end{cases}\)

=>\(\begin{cases}x\ge-\frac72\\ x^2+2\cdot x\cdot\frac{31}{6}+\frac{961}{36}=\frac{433}{36}\end{cases}\Rightarrow\begin{cases}x\ge-\frac72\\ \left(x+\frac{31}{6}\right)^2=\frac{433}{36}\end{cases}\)

=>\(\begin{cases}x\ge-\frac72\\ x+\frac{31}{6}\in\left\lbrace\frac{\sqrt{433}}{6}^{\prime};-\frac{\sqrt{433}}{6}\right\rbrace\end{cases}\Rightarrow\begin{cases}x\ge-\frac72\\ x\in\left\lbrace\frac{\sqrt{433}-31}{6};\frac{-\sqrt{433}-31}{6}\right\rbrace\end{cases}\)

=>\(x=\frac{\sqrt{433}-31}{6}\)

câu 1 

ta có .....

lười viết Min - cốp xki nha

DKXD của A, ta có \(x^{2\le5\Rightarrow-\sqrt{5}\le x\le\sqrt{5}}\)

mà \(3x\ge-3\sqrt{5}\)

mặt kkhác \(\sqrt{5-x^2}\ge0\Rightarrow A=3x+x\sqrt{5-x^2}\ge-3\sqrt{5}\)

min A= \(-3\sqrt{5}\)\(\Leftrightarrow x=-\sqrt{5}\)

3/7(7x+2/3) - 2(5/14 - 3x) = 15/5 

3x + 2/7*x - 5/7 + 6x = 3

65/7*x - 5/7 = 3

65/7* x = 26/7

x= 2/5

a, 3.(2-2x)=4.(x-1)

6 - 6x = 4x -4

4x + 6x = 6 + 4

10x = 10

x=1