Nhận xét : \(\sqrt{\left(5-2\sqrt{6}\right)^x}.\sqrt{\left(5+2\sqrt{6}\right)^x}=1\)

Ta đặt \(\sqrt{\left(5-2\sqrt{6}\right)^x}=a\Rightarrow\sqrt{\left(5+2\sqrt{6}\right)^x}=\frac{1}{a}\)

Khi đó phương trình ban đầu trở thành :

\(a+\frac{1}{a}=10\Rightarrow a^2-10a+1=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=5+2\sqrt{6}\\a=5-2\sqrt{6}\end{cases}}\)

+) Với \(a=5+2\sqrt{6}\Rightarrow\sqrt{\left(5-2\sqrt{6}\right)^x}=5+2\sqrt{6}\)

\(\Leftrightarrow\left(5-2\sqrt{6}\right)^x=\left(5+2\sqrt{6}\right)^2=\left(\frac{1}{5-2\sqrt{6}}\right)^2\)

\(\Leftrightarrow x=-2\)

+) Với \(a=5-2\sqrt{6}\Rightarrow\sqrt{\left(5-2\sqrt{6}\right)^x}=5-2\sqrt{6}\)

\(\Leftrightarrow\left(5-2\sqrt{6}\right)^x=\left(5-2\sqrt{6}\right)^2\)

\(\Leftrightarrow x=2\)

Vậy \(x\in\left\{-2,2\right\}\) thỏa mãn đề.

\(\left(5-2\sqrt{6}\right)^{\frac{x}{2}}+\left(5+2\sqrt{6}\right)^{\frac{x}{2}}=10\)

\(pt\Leftrightarrow\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^{2x}}+\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^{2x}}=10\)

\(\Leftrightarrow\left(\sqrt{3}-\sqrt{2}\right)^x+\left(\sqrt{3}+\sqrt{2}\right)^x=10\)

\(\Leftrightarrow\frac{1}{\left(\sqrt{3}+\sqrt{2}\right)^x}+\left(\sqrt{3}+\sqrt{2}\right)^x=10\)

\(\Leftrightarrow\frac{1}{t}+t=10\left(t=\left(\sqrt{3}+\sqrt{2}\right)^x\right)\)

\(\Leftrightarrow t^2-10t+1=0\)\(\Leftrightarrow t=5\pm2\sqrt{6}\)

\(\Rightarrow5\pm2\sqrt{6}=\left(\sqrt{3}+\sqrt{2}\right)^x\)

\(\Leftrightarrow\left(\sqrt{3}+\sqrt{2}\right)^{\pm2}=\left(\sqrt{3}+\sqrt{2}\right)^x\)

\(\Rightarrow x=\pm2\). Vậy...

1 like tức thì nào

\(\left(\sqrt{2x+3}+2\right)\left(\sqrt{x+6}-\sqrt{x+1}\right)=5\)

\(ĐKXĐ:x\ge-1\).Nhận thấy \(\sqrt{x+6}-\sqrt{x+1}>0\)

\(\Leftrightarrow\left(\sqrt{2x+3}+2\right)\frac{\left(\sqrt{x+6}+\sqrt{x+1}\right)\left(\sqrt{x+6}-\sqrt{x+1}\right)}{\sqrt{x+6}-\sqrt{x+1}}=5\)

\(\Leftrightarrow\left(\sqrt{2x+3}+2\right)\frac{5}{\sqrt{x+6}-\sqrt{x+1}}=5\)

\(\Leftrightarrow\frac{\sqrt{2x+3}+2}{\sqrt{x+6}-\sqrt{x+1}}=1\)

\(\Leftrightarrow\sqrt{2x+3}+2-\sqrt{x+6}+\sqrt{x+1}=0\)

Th1:\(\sqrt{x+1}=2\Leftrightarrow x=3\left(thoaman\right)\)

Th2:\(\sqrt{x+1}-2\ne0\Leftrightarrow x\ne3\)

\(\Leftrightarrow\left(\sqrt{2x+3}-\sqrt{x+6}\right)+\left(2+\sqrt{x+1}\right)=0\)

\(\Leftrightarrow\frac{x-3}{\sqrt{2x+3}+\sqrt{x+6}}+\frac{x-3}{\sqrt{x+1}-2}=0\)

\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{\sqrt{2x+3}+\sqrt{x+6}}+\frac{1}{\sqrt{x+1}-2}\right)=0\)

Tự lm tiếp nha

đề tào lao, nhìn VP biết có vấn đề

Xét thấy : \(\sqrt{25-125+6}\)<0

mà: \(\sqrt{a}\)\(\ge\)0

\(\Rightarrow\)Vế phải ko hợp lí\(\Rightarrow\)X vô nghiệm

nhiều thế giải ko đổi đâu bạn

vậy trả lời câu a thôi

đkxđ : \(\frac{1}{2}\le x\le7\)

\(x^2-5x+3\sqrt{2x-1}=2\sqrt{14-2x}+5\)

\(\Leftrightarrow\left(x^2-5x\right)+3\left(\sqrt{2x-1}-3\right)=2\left(\sqrt{14-2x}-2\right)\)

\(\Leftrightarrow x\left(x-5\right)+\frac{3.\left(2x-10\right)}{\sqrt{2x-1}+3}+\frac{2.\left(2x-10\right)}{\sqrt{14-2x}+2}=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+\frac{6}{\sqrt{2x-1}+3}+\frac{4}{\sqrt{14-2x}+2}\right)=0\)

\(\Leftrightarrow x=5\)

còn bài a,c lười đánh lắm

Điều kiện \(x\ge-1\)

Phương trình đã cho tương đương với

\(\left(x+1\right)\sqrt{x+1}+4\sqrt{x+1}+1=\sqrt[3]{3x+4}\)

\(\Leftrightarrow\left(x+1\right)\sqrt{x+1}+4\sqrt{x+1}+1+3\left(x+1\right)+1=\sqrt[3]{3x+4}+\left(\sqrt[3]{3x+4}\right)^3\)

\(\Leftrightarrow\left(\sqrt{x+1}+1\right)^2+\left(\sqrt{x+1}+1\right)=\left(\sqrt[3]{3x+4}\right)^3+\sqrt[3]{3x+4}\) (*)

Xét hàm số f(t) =t³+t trên R

                   f'(t)=3t²+1>0 với mọi x \(\in\)R

Nên (*) \(\Leftrightarrow f\left(\sqrt{x+1}+1\right)=f\left(\sqrt[3]{3x+4}\right)\Leftrightarrow\sqrt{x+1}+1=\sqrt[3]{3x+4}\)

Đặt \(\left\{{}\begin{matrix}u=\sqrt{x+1}\\y=\sqrt[3]{3x+4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}u+1=v\\3u^2+1=v^3\end{matrix}\right.\)

\(\Rightarrow v^3=3\left(v-1\right)^2+1\Leftrightarrow v^3-1-3\left(v-1\right)^2=0\Leftrightarrow v=1\)

Với v=1 => x=-1

Vậy x=-1 là nghiệm của phương trình

ĐKXĐ: \(x\ge1;x\le-3;x=-1\)

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

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

\(\left(1\right)\Leftrightarrow x+1=0\Rightarrow x=-1\)

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

\(\Leftrightarrow2x+6=x-1+4\sqrt{\left(x-1\right)\left(x+1\right)}+4x+4\)

\(\Leftrightarrow4\sqrt{x^2-1}=3-3x\) \(\Leftrightarrow\left\{{}\begin{matrix}3-3x\ge0\\16\left(x^2-1\right)=\left(3-3x\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\7x^2+18x-25=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-25}{7}\end{matrix}\right.\)

Vậy pt có 3 nghiệm: \(x=-1;1;\dfrac{-25}{7}\)

ĐK : \(x\ge\dfrac{-5}{2}\) PT tương đương

\(\Leftrightarrow\sqrt{2x+5}-3+\sqrt{x^2+5}-3=0\)

\(\Leftrightarrow\dfrac{2\left(x-2\right)}{\sqrt{2x+5}+3}+\dfrac{\left(x-2\right)\left(x+2\right)}{\sqrt{x^2+5}+3}=0\)

đến đây thì ez rồi