\(B8:\)\(M=a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)\)

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

\(=1-3ab\left(1-a^2-b^2\right)+6.\left(ab\right)^2\)

\(=1-3ab\left[1-\left(a^2+2ab+b^2-2ab\right)\right]+6\left(ab\right)^2\)

\(=1-3ab\left[1-\left(1-2ab\right)\right]+6\left(ab\right)^2=1-3ab\left(2ab\right)+6\left(ab\right)^2=1-6\left(ab\right)^2+6\left(ab\right)^2=1\)

\(B7:ax+by+cz=0\Rightarrow\left(ax+by+cx\right)^2=0\)

\(\Leftrightarrow\left(ax\right)^2+\left(by\right)^2+\left(cz\right)^2+2\left(axzc+axby+bycz\right)=0\)

\(\Rightarrow-2\left(axzc+axby+bycz\right)=\left(ax\right)^2+\left(by\right)^2+\left(cz\right)^2\)

\(A=\dfrac{bc\left(y-z\right)^2+ca\left(z-x\right)^2+ab\left(x-y\right)^2}{ax^2+by^2+cz^2}=\dfrac{abx^2+aby^2+bcy^2+bcz^2+caz^2+cax^2+\left(ax\right)^2+\left(by\right)^2+\left(cz\right)^2}{ax^2+by^2+cz^2}=\dfrac{\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)}{ax^2+by^{^2}+cz^2}=a+b+c\)

bài này mình chưa giải dc triệt để ở cái cuối

\(2x^3-4x^2+3x-1=2x^3\left(2-y\right)\sqrt{3-2y}\) \(\left(y\le\dfrac{3}{2}\right)\)

\(\Leftrightarrow4x^3-8x^2+6x-2=2x^3\left(4-2y\right)\sqrt{3-2y}\left(1\right)\)

\(đặt:\sqrt{3-2y}=a\ge0\Rightarrow a^2+1=4-2y\)

\(\left(1\right)\Leftrightarrow4x^3-8x^2+6x-2=2x^3.\left(a^2+1\right)a\)

\(\Leftrightarrow4x^3-8x^2+6x-2-2x^3\left(a^2+1\right)a\)

\(\Leftrightarrow-2\left(xa-x+1\right)\left[\left(xa\right)^2+x^2a+2x^2-xa-2x+1\right]=0\)

\(\Leftrightarrow x.a-x+1=0\Leftrightarrow x\left(a-1\right)=-1\Leftrightarrow x=\dfrac{-1}{a-1}\)

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

\(=\dfrac{-a}{a-1}-\sqrt{\dfrac{-1}{a-1}}\)

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

\(\Rightarrow\left(\dfrac{-a}{a-1}-\sqrt{-\dfrac{1}{a-1}}\right)\left(\sqrt{\dfrac{-a}{a-1}+2}+\sqrt{\dfrac{a-2}{a-1}}\right)-4=0\)

\(\Leftrightarrow\left(-\dfrac{a}{a-1}-\sqrt{-\dfrac{1}{a-1}}\right).2\sqrt{\dfrac{a-2}{a-1}}=4\)

\(\Leftrightarrow\left(-\dfrac{a}{a-1}-\sqrt{-\dfrac{1}{a-1}}\right)\sqrt{\dfrac{a-2}{a-1}}=2\)

\(\Leftrightarrow\left(-1+\dfrac{-1}{a-1}-\sqrt{-\dfrac{1}{a-1}}\right)\sqrt{1-\dfrac{1}{a-1}}=2\)(3)

\(đặt:1-\dfrac{1}{a-1}=u\Rightarrow\sqrt{-\dfrac{1}{a-1}}=\sqrt{u-1}\)

\(\left(3\right)\Leftrightarrow\left(u-2-\sqrt{u-1}\right)\sqrt{u}=2\)

bình phương lên tính được u

\(\Rightarrow u=.....\Rightarrow a\Rightarrow y=...\Rightarrow x=....\)

đoạn đặt \(\sqrt{m+4}=t\left(do:m>5\right)\Rightarrow t>3\) nhé bạn sửa lại 1 đoạn đấy chút còn lại giống kết quả nhé

\(x^2-\left(m+1\right)x+m+4=0\left(1\right)\)

\(\Rightarrow\Delta>0\Leftrightarrow\left(m+1\right)^2-4\left(m+4\right)>0\Leftrightarrow\left[{}\begin{matrix}m< -3\\m>5\end{matrix}\right.\)\(\left(2\right)\)

\(ddkt-thỏa:\sqrt{x1}+\sqrt{x2}=2\sqrt{3}\)

\(x1=0\Rightarrow\left(1\right)\Leftrightarrow m=-4\Rightarrow\left(1\right)\Leftrightarrow x^2+3x=0\Leftrightarrow\left[{}\begin{matrix}x1=0\\x2=-3< 0\left(loại\right)\end{matrix}\right.\)

\(x1\ne0\) \(\Rightarrow0< x1< x2\)

\(\Leftrightarrow\left\{{}\begin{matrix}x1+x2>0\\x1x2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m+1>0\\m+4>0\end{matrix}\right.\)\(\Rightarrow m>-1\)\(\left(3\right)\)

\(\left(2\right)\left(3\right)\Rightarrow m>5\)

\(\Rightarrow\sqrt{x1}+\sqrt{x2}=2\sqrt{3}\)

\(\Leftrightarrow x1+x2+2\sqrt{x1x2}=12\Leftrightarrow m+1+2\sqrt{m+4}=12\)

\(\Leftrightarrow m+4+2\sqrt{m+4}-15=0\)

\(đặt:\sqrt{m+4}=t>5\Rightarrow t^2+2t-15=0\Leftrightarrow\left[{}\begin{matrix}t=-5\left(ktm\right)\\t=3\left(ktm\right)\end{matrix}\right.\)

\(\Rightarrow m\in\phi\)

\(x< -2\Rightarrow\left|x+2\right|+\left|x-3\right|=-x-2+3-x=1-2x\)

\(\Rightarrow\left\{{}\begin{matrix}x< -2\\1-2x\ge10x-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< -2\\x\le\dfrac{1}{4}\end{matrix}\right.\)\(\Leftrightarrow x< -2\left(1\right)\)

\(-2\le x< 3\Rightarrow\left|x+2\right|+\left|x-3\right|=x+2+3-x=5\Rightarrow\left\{{}\begin{matrix}-2\le x< 3\\5\ge10x-2\Leftrightarrow x\le\dfrac{7}{10}\end{matrix}\right.\)

\(\Leftrightarrow-2\le x\le\dfrac{7}{10}\left(2\right)\)

\(x\ge3\Rightarrow\left|x+2\right|+\left|x-3\right|=x+2+x-3=2x-1\Rightarrow\left\{{}\begin{matrix}x\ge3\\2x-1\ge10x-2\Leftrightarrow x\le\dfrac{1}{8}\end{matrix}\right.\)

\(\Rightarrow x\in\phi\)

\(\left(1\right)\left(2\right)\Rightarrow x\in\phi\)

\(\left\{{}\begin{matrix}9x^2-3xy+2y^2=23\\7x^2+6xy-8y^2=-37\end{matrix}\right.\)\(\left(hpt\right)\)

\(đặt:x=t.y\Rightarrow hpt\Leftrightarrow\left\{{}\begin{matrix}9\left(t.y\right)^2-3t.y^2+2y^2=23\left(1\right)\\7\left(ty\right)^2+6t.y^2-8y^2=-37\left(2\right)\end{matrix}\right.\)

\(\Rightarrow-37\left[9\left(t.y\right)^2-3ty^2+2y^2\right]=23\left[7\left(ty\right)^2+6ty^2-8y^2\right]\)

\(\Leftrightarrow494\left(ty\right)^2+27ty^2-110y^2=0\left(3\right)\)

\(x=y=0\) \(không\) \(là\) \(nghiệm\) \(hpt\)

\(y\ne0\Rightarrow\left(3\right)\Leftrightarrow494t^2+27t-110=0\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{110}{247}\Rightarrow x=\dfrac{110}{247}.y\left(4\right)\\t=-\dfrac{1}{2}\Rightarrow x=-\dfrac{1}{2}.y\left(5\right)\end{matrix}\right.\)

\(thay\left(4\right)và\left(5\right)vào-hpt\Rightarrow x,y=.....\)(đến đây dễ rồi bạn tự tìm x,y)

\(dkd:x\ge1\)

\(pt\Leftrightarrow\sqrt{x+4}-3-\sqrt{x-1}+2=0\)

\(\Leftrightarrow\dfrac{x+4-9}{\sqrt{x+4}+3}-\dfrac{x-1-4}{\sqrt{x-1}+2}=0\)

\(\Leftrightarrow\dfrac{x-5}{\sqrt{x+4}+3}-\dfrac{x-5}{\sqrt{x-1}+2}=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\\dfrac{1}{\sqrt{x+4}+3}-\dfrac{1}{\sqrt{x-1}+2}=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\dfrac{\sqrt{x-1}+2-\sqrt{x+4}-3}{\left(\sqrt{x+4}+3\right)\left(\sqrt{x-1}+2\right)}=0\)

\(\Rightarrow\sqrt{x-1}+2-\sqrt{x+4}-3=0\)

\(\Leftrightarrow\sqrt{x-1}-\sqrt{x+4}-1=0\)

\(\Leftrightarrow\sqrt{x+4}-\sqrt{x-1}+1=0\)

\(do:x\ge1\Rightarrow\sqrt{x+4}-\sqrt{x-1}+1>0\Rightarrow vô-nghiệm\)

\(\Rightarrow S=\left\{5\right\}\)

\(pt\) \(hoành\) \(độ\) \(giao\) \(điểm:-x^2=2\left(2m+1\right)x-1\)

\(\Leftrightarrow-x^2-2\left(2m+1\right)x+1=0\)

\(\Leftrightarrow x^2+2\left(2m+1\right)x-1=0\)

\(\Delta'>0\Leftrightarrow\left(2m+1\right)^2+4>0\left(đúng\forall m\right)\)

\(vi-ét\Rightarrow\left\{{}\begin{matrix}x1+x2=-2\left(2m+1\right)\\x1x2=-1\end{matrix}\right.\)

\(\Rightarrow A\left(x1;x1^2+2\left(2m+1\right)x1-1\right)\)

\(B\left(x2;x2^2+2\left(2m+1\right)x2-1\right)\)

\(\Rightarrow y1+y2+2\left(x1+x2\right)-16m^2=10\)

\(\Leftrightarrow x1^2+2\left(2m+1\right)x1-1+x2^2+2\left(2m+1\right)x2-1+2\left(x1+x2\right)-16m^2-10=0\)

\(\Leftrightarrow\left(x1+x2\right)^2-2x1x2+2\left(2m+1\right)\left(x1+x2\right)+2\left(x1+x2\right)-16m^2-12=0\)

\(\Leftrightarrow4\left(2m+1\right)^2-4\left(2m+1\right)^2+10-16m^2=0\Leftrightarrow16m^2=10\Leftrightarrow m=\pm\sqrt{\dfrac{10}{16}}\)

(kiểm tra lại biến đổi với tính toán giùm nhé)

\(đặt:\sqrt{x^2+1}=t>0\Rightarrow\left(x+3\right)t^2+4\left(x+2\right)t-16=0\)

\(\Leftrightarrow\left(t+4\right)\left(tx+3t-4\right)=0\Leftrightarrow\left[{}\begin{matrix}t=-4\left(loại\right)\\tx+3t-4=0\Leftrightarrow t=\dfrac{4}{x+3}\end{matrix}\right.\)

\(\Leftrightarrow\sqrt{x^2+1}=\dfrac{4}{x+3}\left(x>-3\right)\Leftrightarrow x^2+1=\dfrac{16}{\left(x+3\right)^2}\)

\(\Leftrightarrow\left(x^2+1\right)\left(x+3\right)^2-16=0\Leftrightarrow x^4+6x^3+10x^2+6x-7=0\Rightarrow x=....\)

bài này nghiệm xấu quá

\(P=\dfrac{3}{x}+\dfrac{1}{3y}=\dfrac{3}{x}+\dfrac{\dfrac{1}{3}}{y}\ge\dfrac{\left(\sqrt{3}+\dfrac{1}{\sqrt{3}}\right)^2}{x+y}=\dfrac{\dfrac{16}{3}}{\dfrac{4}{3}}=4\)

\(min_P=4\Leftrightarrow x=1;y=\dfrac{1}{3}\)