Đặt \(\sqrt{x^2+7x+8}=t\left(t\ge0\right)\)

\(\sqrt{7x+7}+\sqrt{7x-6}=t\ge0\)

\(bpt\Leftrightarrow t+t^2< 182\Leftrightarrow-14< t< 13\Leftrightarrow t< 13\Leftrightarrow\sqrt{7x+7}+\sqrt{7x-6}< 13\left(đk:x\ge\dfrac{6}{7}\right)\Leftrightarrow14x+1+2\sqrt{\left(7x+7\right)\left(7x-6\right)}< 169\Leftrightarrow2\sqrt{\left(7x+7\right)\left(7x-6\right)}< 168-14x\Leftrightarrow\left\{{}\begin{matrix}\left(7x+7\right)\left(7x-6\right)\ge0\\168-14x\ge0\\4\left(7x+7\right)\left(7x-6\right)< \left(168-14x\right)^2\end{matrix}\right.\)

\(giảibpt\Rightarrowđáp\) \(số\)

a,\(x^2-7x+\sqrt{x^2-7x+8}=12\)

ĐKXĐ: .....

Đặt \(x^2-7x=t\)

Phương trình trở thành

\(t+\sqrt{t+8}=12\)

\(\Leftrightarrow\sqrt{t+8}=12-t\)

\(\Leftrightarrow t+8=\left(12-t\right)^2\)

\(\Leftrightarrow t+8=144-24t+t^2\)

\(\Leftrightarrow t^2-25t+136=0\)

\(\Leftrightarrow\left(t-17\right)\left(t-8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t-17=0\\t-8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}t=17\\t=8\end{cases}}}\)

tại t = 17 , ta có

\(x^2-7x=17\Leftrightarrow x^2-7x-17=0\)

\(\Leftrightarrow.......\)

Tại t = 8 ta có

\(x^2-7x=8\Leftrightarrow x^2-7x-8=0\)

\(\Leftrightarrow\left(x-8\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-8=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=8\\x=-1\end{cases}}}\)

b, \(x^2+4x+5=2\sqrt{2x+3}\)

mik ko bt :)

a,đkxđ:\(x^2-7x+8\ge0\Leftrightarrow x^2-2\cdot\frac{7}{2}x+\frac{49}{4}-\frac{17}{4}\ge0\Leftrightarrow\left(x-\frac{7}{2}\right)^2\ge\frac{17}{4}\Leftrightarrow\hept{\begin{cases}x-\frac{7}{2}\ge\frac{\sqrt{17}}{2}\approx2,06\\x-\frac{7}{2}\le-\frac{\sqrt{17}}{2}\approx-2,06\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge5,56\\x\le1,44\end{cases}}\)

\(\Leftrightarrow\left(x^2-7x+8\right)+\sqrt{x^2-7x+8}=12+8=20\)

\(\Leftrightarrow4\left(x^2-7x+8\right)+4\sqrt{x^2-7x+8}+1=20\cdot4+1=81\)

\(\Leftrightarrow\left(2\sqrt{x^2-7x+8}+1\right)^2=81\)

\(\Leftrightarrow2\sqrt{x^2-7x+8}+1=\pm9\)

Mà vế trái >0 nên \(2\sqrt{x^2-7x+8}+1=9\)

\(\Leftrightarrow\sqrt{x^2-7x+8}=\frac{9-1}{2}=4\)

\(\Leftrightarrow x^2-7x+8=16\)

\(\Leftrightarrow x^2-7x-8=0\Leftrightarrow\left(x-8\right)\left(x+1\right)=0\Leftrightarrow\orbr{\begin{cases}x=8\\x=-1\end{cases}}\)

\(\sqrt[3]{7x-8}+5\sqrt{x-1}=x\sqrt{2x-1}-2\)

\(\Leftrightarrow\sqrt[3]{7x-8}-3+5\sqrt{x-1}-10=x\sqrt{2x-1}-15\)

\(\Leftrightarrow\frac{7x-8-27}{\sqrt[3]{7x-8}^2+3\sqrt[3]{7x-8}+9}+5\frac{x-1-4}{\sqrt{x-1}-2}-\frac{x^2\left(2x-1\right)-225}{x\sqrt{2x-1}+15}=0\)

\(\Leftrightarrow\frac{7\left(x-5\right)}{\sqrt[3]{7x-8}^2+3\sqrt[3]{7x-8}+9}+5\frac{x-5}{\sqrt{x-1}-2}-\frac{\left(x-5\right)\left(2x^2+9x+45\right)}{x\sqrt{2x-1}+15}=0\)

\(\Leftrightarrow\left(x-5\right)\left(\frac{7}{\sqrt[3]{7x-8}^2+3\sqrt[3]{7x-8}+9}+\frac{5}{\sqrt{x-1}-2}-\frac{2x^2+9x+45}{x\sqrt{2x-1}+15}\right)=0\)

Suy ra x=5

Bài này có 2 nghiệm là x = 1 và x = 5 nhưng không biết giải thế nào. 

\(\sqrt[3]{7x-8}+5\sqrt{x-1}=x\sqrt{2x-1}-2\)\(\Leftrightarrow\left[\sqrt[3]{7x-8}-\left(x-2\right)\right]+5\left(\sqrt{x-1}-\frac{x-1}{2}\right)+x\left(\frac{x+1}{2}-\sqrt{2x-1}\right)\)\(+\left(x-2\right)-\frac{x\left(x+1\right)}{2}+\frac{5}{2}\left(x-1\right)+2\)

\(\Leftrightarrow2\left[\sqrt[3]{7x-8}-\left(x-2\right)\right]+x\left(x+1-2\sqrt{2x-1}\right)+\)\(5\left[2\sqrt{x-1}-\left(x-1\right)\right]-x^2+6x-5=0\)

\(\Leftrightarrow2\left[\left(x-2\right)-\sqrt[3]{7x-8}\right]+x\left[2\sqrt{2x-1}-\left(x-1\right)\right]+\)\(5\sqrt{x-1}\left(\sqrt{x-1}-2\right)+x^2-6x+5=0\)

\(\Leftrightarrow\left(x-5\right)\sqrt{x-1}\left[\frac{2x\sqrt{x-1}}{A}+\frac{-x\sqrt{x-1}}{2\sqrt{2x-1}+x+1}+\frac{5}{\sqrt{x-1}+2}+\sqrt{x-1}\right]=0\)

\(\Leftrightarrow\left(x-5\right)\sqrt{x-1}=0\Leftrightarrow\orbr{\begin{cases}x=5\\x=1\end{cases}}\).

\(x^2-2\sqrt{x^2-7x+10}< 7x-2\)

\(ĐK:x\ge5\)

BPT \(\Leftrightarrow x^2-7x+2-2\sqrt{x^2-7x+10}< 0\)

\(\Leftrightarrow t^2-8-2t< 0\left(t=\sqrt{x^2-7x+10}\ge0\right)\)

\(\Leftrightarrow\left(t+2\right)\left(t-4\right)< 0\)

\(\Leftrightarrow-2< t< 4\Leftrightarrow-2< \sqrt{x^2-7x+10}< 4\)

\(\Leftrightarrow\sqrt{x^2-7x+10}< 4\Leftrightarrow x^2-7x-6< 0\)

\(\Leftrightarrow\orbr{\begin{cases}5\le x< \frac{7+\sqrt{73}}{2}\\\frac{7-\sqrt{73}}{2}< x\le2\end{cases}}\)

Chúc bạn học tốt !!!

\(ĐK:x\ge-8\)

\(\left(4x+2\right)\sqrt{x+8}=3x^2+7x+8\)

\(\Leftrightarrow x+8-3x\sqrt{x+8}-\left(x+2\right)\sqrt{x+8}+3x\left(x+2\right)=0\)

\(\Leftrightarrow\sqrt{x+8}\left(\sqrt{x+8}-3x\right)-\left(x+2\right)\left(\sqrt{x+8}-3x\right)=0\)

\(\Leftrightarrow\left(\sqrt{x+8}-x-2\right)\left(\sqrt{x+8}-3x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x+8}=x+2\left(1\right)\\\sqrt{x+8}=3x\left(2\right)\end{cases}}\)

\(\left(1\right)\Leftrightarrow x+8=x^2+4x+4\Leftrightarrow x^2+3x-4=0\Leftrightarrow\orbr{\begin{cases}x=1\left(tm\right)\\x=-4\left(L\right)\end{cases}}\)

\(\left(2\right)\Leftrightarrow9x^2-x-8=0\Leftrightarrow\orbr{\begin{cases}x=1\left(tm\right)\\x=\frac{-8}{9}\left(L\right)\end{cases}}\)

Vậy nghiệm duy nhất của phương trình là 1

ĐKXĐ : x \(\ge\)-8

PT đã cho tương đương với :

\(2\left(2x+1\right)\sqrt{x+8}=4x^2+4x+1+x+8-\left(x^2-2x+1\right)\)

\(\Leftrightarrow\left(2x+1\right)^2-2\left(2x+1\right)\sqrt{x+8}+x+8-\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(2x+1-\sqrt{x+8}\right)^2-\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x+2-\sqrt{x+8}\right)\left(3x-\sqrt{x+8}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+2-\sqrt{x+8}=0\\3x-\sqrt{x+8}=0\end{cases}}\)

Từ đó giải ra x = 1 thỏa mãn đề bài

giúp mình giải câu đấy nữa

Đặt:

\(a=\sqrt[3]{x^2-x-8};b=\sqrt[3]{x^2-8x-1}\)

Để ý thấy rằng: \(a^3-b^3=7x-7=\left(7x+1\right)+8\)nên PT trở thành:

\(b-a+\sqrt[3]{a^3-b^3+8}=2\)

\(\Leftrightarrow a^3-b^3+8=\left(2+a-b\right)^3\)

\(\Leftrightarrow\left(a-b\right)\left(a^2+b^2+ab\right)=\left(a-b\right)^3+6\left(a-b\right)\left[2+\left(a-b\right)\right]\)

\(\Leftrightarrow\orbr{\begin{cases}a-b=0\\\left(a-b\right)^2+3ab=\left(a-b\right)^2+12+6\left(a-b\right)\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}a=b\\\left(a+2\right)\left(2-b\right)=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=b\\a=-2\\b=2\end{cases}}\)

\(\left(+\right)a=b\Leftrightarrow x^2-x-8=x^2-8x-1\Leftrightarrow x=1\)

\(\left(+\right)a=-2\Leftrightarrow x^2-x-8=-8\Leftrightarrow\orbr{\begin{cases}a=0\\x=1\end{cases}}\)

\(\left(+\right)b=2\Leftrightarrow x^2-8x-1=8\Leftrightarrow\orbr{\begin{cases}x=8\\x=-1\end{cases}}\)

\(\Rightarrow x\in\left\{\pm1;0;9\right\}\)