a) Ta có: \(2\sqrt{9x-27}-\dfrac{1}{5}\sqrt{25x-75}-\dfrac{1}{7}\sqrt{49x-147}=20\)

\(\Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)

\(\Leftrightarrow4\sqrt{x-3}=20\)

\(\Leftrightarrow x-3=25\)

hay x=28

b) Ta có: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)

\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)

\(\Leftrightarrow2\sqrt{x+2}=6\)

\(\Leftrightarrow x+2=9\)

hay x=7

Giải câu d thôi mấy câu còn lại đơn giản lắm nên bạn tự làm.

d/ \(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\)

Điều kiện \(x\ge1\)

\(\Leftrightarrow\sqrt{x-1-4\sqrt{x-1}+4}+\sqrt{x-1-6\sqrt{x-1}+9}=1\)

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

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

Đây chỉ là phương trình cơ bản của trị tuyệt đối lớp 6, 7 học rồi nên bạn tự làm nhé.

Lời giải:

\(x=\sqrt{4+\sqrt{8}}.\sqrt{(2+\sqrt{2+\sqrt{2}})(2-\sqrt{2+\sqrt{2}})}\)

\(=\sqrt{4+2\sqrt{2}}.\sqrt{2^2-(2+\sqrt{2})}=\sqrt{2(2+\sqrt{2})}.\sqrt{2-\sqrt{2}}\)

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

\(y=\frac{6\sqrt{2}-4\sqrt{3}+2\sqrt{5}}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}}=\frac{\frac{2}{3}(9\sqrt{2}-6\sqrt{3}+3\sqrt{5})}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}}=\frac{2}{3}\)

Do đó:

\(E=\frac{1+xy}{x+y}-\frac{1-xy}{x-y}=\frac{1+\frac{4}{3}}{2+\frac{2}{3}}-\frac{1-\frac{4}{3}}{2-\frac{2}{3}}=\frac{9}{8}\)

a) Ta có: \(\sqrt{25x+75}+2\sqrt{9x+27}=5\sqrt{x+3}+18\)

\(\Leftrightarrow5\sqrt{x+3}+6\sqrt{x+3}-5\sqrt{x+3}=18\)

\(\Leftrightarrow\sqrt{x+3}=3\)

\(\Leftrightarrow x+3=9\)

hay x=6

b) Ta có: \(\sqrt{4x-8}-14\sqrt{\dfrac{x-2}{49}}=\sqrt{9x-18}+8\)

\(\Leftrightarrow2\sqrt{x-2}-2\sqrt{x-2}-3\sqrt{x-2}=8\)

\(\Leftrightarrow-3\sqrt{x-2}=8\)(Vô lý)

a) \(\sqrt{x-3}=2\)

\(\Leftrightarrow\) \(x-3=4\)

\(\Leftrightarrow\) \(x=7\)

b) \(\sqrt{x^2-6x+9}=5\) (ĐKXĐ: \(x\ne0\) , \(x\ge3\) )

\(\Leftrightarrow\) \(\sqrt{\left(x-3\right)^2}=5\)

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

\(\Leftrightarrow\) \(x-3=5\) với x > 0

\(x-3=-5\) với x < 0

\(\Leftrightarrow\) \(x=8\) (thỏa mãn)

\(x=-2\) (loại) | NOTE: cũng có thể ghi là không thỏa mãn)

c) \(x\sqrt{12}+\sqrt{18}=x\sqrt{8}+\sqrt{27}\) (ĐKXĐ: \(x\ne0\) )

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

\(\Leftrightarrow\) \(2x\sqrt{3}-2x\sqrt{2}=3\sqrt{3}-3\sqrt{2}\)

\(\Leftrightarrow\) \(2x\left(\sqrt{3}+\sqrt{2}\right)=3\left(\sqrt{3}-\sqrt{2}\right)\) | Có lẽ không nên làm theo cách này vì nó khá dài dòng|

\(\Leftrightarrow\) \(2x\left(\sqrt{3}+\sqrt{2}\right)-3\left(\sqrt{3}+\sqrt{2}\right)=0\)

\(\Leftrightarrow\) \(\left(2x-3\right)\left(\sqrt{3}+\sqrt{2}\right)=0\)

\(\Leftrightarrow\) \(2x-3=0\) hoặc \(\sqrt{3}+\sqrt{2}=0\) (luôn đúng)

\(\Leftrightarrow\) \(2x=3\)

\(\Leftrightarrow\) \(x=\dfrac{3}{2}\) (thỏa mãn)

\(\sqrt{x-3}=2\\ \Rightarrow x-3=4\\ \Rightarrow x=7\)

\(\sqrt{x^2-6x+9}=5\\ \Rightarrow\sqrt{\left(x-3\right)^2}=5\\ \Rightarrow x-3=5\\ \Rightarrow x=8\)

\(x\sqrt{12}+\sqrt{18}=x\sqrt{8}+\sqrt{27}\\ \Rightarrow2\sqrt{3}x+3\sqrt{2}=2\sqrt{2}x+3\sqrt{3}\\ \Rightarrow2x\left(\sqrt{3}-\sqrt{2}\right)=3\left(\sqrt{3}-\sqrt{2}\right)\\ \Rightarrow2x=3\\ \Rightarrow x=\dfrac{3}{2}\)

b, \(\sqrt{x^2-6x+9}=5\)

\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=5\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

c, \(x\sqrt{12}+\sqrt{18}=x\sqrt{8}+\sqrt{27}\)

\(\Leftrightarrow x2\sqrt{3}+3\sqrt{2}-2x\sqrt{2}-3\sqrt{3}=0\)

\(\Leftrightarrow2x\left(\sqrt{3}-\sqrt{2}\right)-3\left(\sqrt{3}-\sqrt{2}\right)=0\)

\(\Leftrightarrow\left(\sqrt{3}-\sqrt{2}\right)\left(2x-3\right)=0\)

\(\Leftrightarrow2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\dfrac{3}{2}\)

Lời giải:

\(x=\sqrt{4+\sqrt{8}}.\sqrt{(2+\sqrt{2+\sqrt{2}})(2-\sqrt{2+\sqrt{2}})}\)

\(=\sqrt{4+\sqrt{8}}.\sqrt{2^2-(2+\sqrt{2})}=\sqrt{4+\sqrt{8}}.\sqrt{2-\sqrt{2}}\)

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

\(=\sqrt{2}.\sqrt{2^2-2}=\sqrt{2}.\sqrt{2}=2\)

\(y=\frac{3.2\sqrt{2}-2.2\sqrt{3}+2\sqrt{5}}{3.3\sqrt{2}-2.3\sqrt{3}+3\sqrt{5}}=\frac{6\sqrt{2}-4\sqrt{3}+2\sqrt{5}}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}}\)

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