\(c25:\left(d\right)//\left(d'\right)\Leftrightarrow\left\{{}\begin{matrix}m^2=4\\m+5\ne7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m=\pm2\\m\ne2\end{matrix}\right.\)\(\Leftrightarrow m=-2\)

\(c26:\) xét tam giác ANC vuông tại N có \(sin30^o=\dfrac{AN}{AC}\Rightarrow AN=AC.sin30^o\)

xét tam giác ABN vuông tại N có 

\(tan38^o=\dfrac{AN}{BN}\Rightarrow AN=BN.tan38^o\)

\(\Rightarrow AC.sin30^o=BN.tan38^o=\left(BC-NC\right).tan38^o\)

\(\Leftrightarrow AC.sin30^o=\left(11-NC\right).tan38^o\)

\(mà:cos30^o=\dfrac{NC}{AC}\Rightarrow NC=AC.cos30^o\)

\(\Rightarrow AC.sin30^o=\left[11-AC.cos30^o\right]tan38^0\Leftrightarrow AC=7,3cm\)

\(1.A=\sqrt{\sqrt{3}-\sqrt{1-\sqrt{12-2.\sqrt{12}.\sqrt{9}+9}}}=\sqrt{\sqrt{3}-\sqrt{1-\sqrt{\left(\sqrt{12}-\sqrt{9}\right)^2}}}=\sqrt{\sqrt{3}-\sqrt{1-\sqrt{12}+\sqrt{9}}}=\sqrt{\sqrt{3}-\sqrt{3-2\sqrt{3}+1}}=\sqrt{\sqrt{3}-\sqrt{\left(\sqrt{3}-1\right)^2}}=1\)

\(2.a;đk:\forall x;đặt:x^2=t\ge0\Rightarrow\dfrac{3}{t^2+t+1}+5=3t\left(t+1\right)\)

\(\Rightarrow\dfrac{3+5t^2+5t+5}{t^2+t+1}=3t\left(t+1\right)\Leftrightarrow5t^2+5t+8=3t\left(t+1\right)\left(t^2+t+1\right)\)

\(\Leftrightarrow\left(t-1\right)\left(t+2\right)\left(3t^2+3t+4>0\right)=0\Leftrightarrow\left[{}\begin{matrix}t=1\left(tm\right)\Rightarrow x=\pm1\\t=-2\left(loại\right)\end{matrix}\right.\)

\(b;\left\{{}\begin{matrix}2x^2-3xy+y^2=0\Leftrightarrow\left(x-y\right)\left(2x-y\right)=0\\3x^2+2xy-y^2-3x+y=2\left(3\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=y\left(1\right)\\y=2x\left(2\right)\end{matrix}\right.\\3x^2+2xy-y^2-3x+y=2\left(3\right)\end{matrix}\right.\)

\(thế\left(1\right)và\left(2\right)\) \(vào\left(3\right)\Rightarrow\left(x;y\right)\)

\(B2;a;mx+y=2\Leftrightarrow y=2-mx\)

\(\Rightarrow\dfrac{x^2}{2}=2-mx\Leftrightarrow x^2+2mx-4=0\Rightarrow\Delta'=m^2+4>0\left(\forall m\right)\Rightarrowđpcm\)

\(b;\Rightarrow\left\{{}\begin{matrix}x1+x2=-2m\\x1.x2=-4\end{matrix}\right.\)  \(\Rightarrow A\left(x1;y1\right);B\left(x2;y2\right)\Leftrightarrow A\left(x1;2-mx1\right);B\left(x2;2-mx2\right)\)

\(\Rightarrow AB=\sqrt{\left(x1-x2\right)^2+\left(2-mx1-2+mx2\right)^2}=\sqrt{\left(x1-x2\right)^2+\left(mx1-mx2\right)^2}=\sqrt{\left(m^2+1\right)\left(x1-x2\right)^2}=\sqrt{\left(m^2+1\right)\left[\left(x1+x2\right)^2-4x1x2\right]}=\sqrt{\left(m^2+1\right)\left(4m^2+16\right)}\ge\sqrt{16}=4\Leftrightarrow m=0\)

\(b;;2\left(x-2\right)=\sqrt{2\left(x^2-x-1\right)}\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\4\left(x-2\right)^2=2\left(x^2-x-1\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\\left[{}\begin{matrix}x=\dfrac{1}{2}\left(7-\sqrt{13}\right)\\x=\dfrac{1}{2}\left(7+\sqrt{13}\right)\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{1}{2}\left(7+\sqrt{13}\right)\Rightarrow T=5\)

\(B3..\)

\(\dfrac{4b^2}{a-1}+\dfrac{5b^2}{b-2}=4\left(a-1\right)+\dfrac{4}{a-1}+8+5\left(b-2\right)+\dfrac{20}{b-2}+20\ge2\sqrt{4^2}+8+2\sqrt{5.20}+20=56\left(đpcm\right)\)

\(a,2x^2-\left(m+2\right)x-7+m^2=0\)

\(yêu\) \(cầu\Leftrightarrow\left\{{}\begin{matrix}2\left(-7+m^2\right)< 0\left(1\right)\\\left|x1\right|=\dfrac{1}{x2}\left(2\right)\\m>0\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\left[{}\begin{matrix}m< -\sqrt{7}\\m>\sqrt{7}\end{matrix}\right.\)

\(\left(2\right)\Leftrightarrow-x1=\dfrac{1}{x2}\left(do:x1< 0\right)\Leftrightarrow x1.x2=-1\Leftrightarrow\dfrac{-7+m^2}{2}=-1\Leftrightarrow m^2=5\Leftrightarrow m=\pm\sqrt{5}\left(ktm\right)\Rightarrow m=\varnothing\)

\(c;x^2-xy+y^2=2\Leftrightarrow2x^2-2xy+2y^2-4=0\Leftrightarrow\left(x-y\right)^2+x^2+y^2-4=0\Leftrightarrow\left(x-y\right)^2=4-\left(x^2+y^2\right)\ge0\Rightarrow x^2+y^2\le4\)

\(\Rightarrow P\le4+xy=4+x^2+y^2-2=2+x^2+y^2\le2+4=6\Rightarrow maxP=6\)

\(dấu"="xayra\Leftrightarrow\left\{{}\begin{matrix}x=y\\x^2-xy+y^2=2\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)\)

\(..x^2-xy+y^2=2\Leftrightarrow\left(X+y\right)^2=2+3xy\ge0\Leftrightarrow xy\ge-\dfrac{2}{3}\Rightarrow P\ge x^2+y^2-\dfrac{2}{3}=2+xy-\dfrac{2}{3}\ge2-\dfrac{2}{3}-\dfrac{2}{3}=\dfrac{2}{3}\Rightarrow min=\dfrac{2}{3}\Leftrightarrow\left\{{}\begin{matrix}x=-y\\x^2-xy+y^2=2\end{matrix}\right.\)

\(biến\) \(đổi\) \(tương\) \(đương\)\(\Leftrightarrow cosx+sinx=sin^3x+cotx.sin^3x+sin^3x.cot^2x+sin^3x.cot^3x=sin^3x+sin^2x.cosx+sinx.cosx^2+cos^3x=\left(sinx+cosx\right)\left[sin^2x-sinx.cosx+cos^2x\right]+sinx.cosx\left(sinx+cosx\right)=\left(sinx+cosx\right)\left[1-sinx.cosx+sinx.cosx\right]=sinx+cosx\Leftrightarrow0=0\Rightarrowđpcm\)

\(tử:=\dfrac{1}{2}\left[sin\left(60^o-x+30^o-x\right)+sin\left(60^o-x-30^2+x\right)\right]+\dfrac{1}{2}\left[sin\left(30^o-x+60^o-x\right)+sin\left(30^o-x-60^o+x\right)\right]\)

\(=\dfrac{1}{2}\left[2sin\left(\dfrac{\pi}{2}-2x\right)+sin\left(\dfrac{\pi}{6}\right)+sin\left(-\dfrac{\pi}{6}\right)\right]=\dfrac{1}{2}.\left[2sin\left(\dfrac{\pi}{2}-2x\right)+0\right]=sin\left(\dfrac{\pi}{2}-2x\right)=cos2x\)

\(VT=\dfrac{cos2x}{sin4x}=\dfrac{cos2x}{2sin2x.cos2x}=\dfrac{1}{2sin2x}=\dfrac{1}{4sinx.cosx}=\dfrac{\dfrac{1}{cos^2x}}{\dfrac{4sinx.cosx}{cos^2x}}=\dfrac{1+tan^2x}{\dfrac{4sĩnx}{cosx}}=\dfrac{1+tan^2x}{4tanx}=VP\)

\(\sqrt{\dfrac{b+c}{a}.1}\le\dfrac{\dfrac{b+c}{a}+1}{2}=\dfrac{a+b+c}{2a}\Rightarrow\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\)

\(tương\) \(tự\Rightarrow\sqrt{\dfrac{b}{c+a}}\ge\dfrac{2b}{a+b+c};\sqrt{\dfrac{c}{a+b}}\ge\dfrac{2c}{a+b+c}\)

\(\Rightarrow VP\ge\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)

\(dấu"="\Leftrightarrow\sqrt{\dfrac{b+c}{a}}=\sqrt{\dfrac{c+a}{b}}=\sqrt{\dfrac{a+b}{c}}=1\Leftrightarrow\left\{{}\begin{matrix}b+c=a\\c+a=b\\a+b=c\end{matrix}\right.\)

\(\Leftrightarrow a+b+c=2\left(a+b+c\right)\)\(vô\) \(lí\) \(do:a,b,c>0\)

\(\Rightarrow VP>2\)

\(VT< \dfrac{a+c}{a+b+c}+\dfrac{a+b}{a+b+c}+\dfrac{c+b}{a+b+c}=2\Rightarrow VT< 2\)

\(\Rightarrow VT< VP\left(đpcm\right)\)

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

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

\(giảibpt\Rightarrowđáp\) \(số\)

\(x^2-mx+m-1=0\)

\(\Delta\ge0\Leftrightarrow m^2-4\left(m-1\right)=m^2-4m+4=\left(m-2\right)^2\ge0\left(luôndung\right)\)

\(A=\dfrac{2x1.x2+3}{x1^2+x2^2+2\left(1+x1x2\right)}=\dfrac{2x1x2+3}{\left(x1+x2\right)^2+1}=\dfrac{2\left(m-1\right)+3}{m^2+1}\Leftrightarrow A\left(m^2+1\right)=2\left(m-1\right)+3\Leftrightarrow Am^2+A=2m+1\Leftrightarrow Am^2-2m+A-1=0\left(1\right)\Rightarrow\left[{}\begin{matrix}TH:A=0\Rightarrow\left(1\right)\Leftrightarrow m=-\dfrac{1}{2}\\TH:A\ne0\Rightarrow\Delta'\ge0\Leftrightarrow1-A\left(A-1\right)\ge0\Leftrightarrow\dfrac{1-\sqrt{5}}{2}\le A\le\dfrac{1+\sqrt{5}}{2}\end{matrix}\right.\)

\(\Rightarrow minA=\dfrac{1-\sqrt{5}}{2}\Leftrightarrow m=....\)

\(thayA=\dfrac{1-\sqrt{5}}{2}\) \(vào\left(1\right)\Rightarrow m=....\)

\(2x^2+30xy=5\left(x+5y\right)\sqrt{5xy}-50y^2\)\(\left(đk:x;y\ge0\right)\)

\(\Leftrightarrow2x^2+30xy-5\left(x+5y\right)\sqrt{5xy}+50y^2=0\left(1\right)\)

\(đặt:\sqrt{5xy}=b\ge0\Rightarrow5xy=b^2\Rightarrow10xy=2b^2\)

\(x+5y=a\ge0\Rightarrow x^2+10xy+25y^2=â^2\)

\(\Rightarrow2a^2=2x^2+20xy+50y^2\)

\(\Leftrightarrow\left(1\right)\Leftrightarrow2a^2+2b^2-5ab=0\Leftrightarrow\left(2a-b\right)\left(a-2b\right)=0\Leftrightarrow\left[{}\begin{matrix}b=2a\left(2\right)\\a=2b\left(3\right)\end{matrix}\right.\)

\(\left(2\right)\Rightarrow\sqrt{5xy}=2x+10y\Leftrightarrow4x^2+35xy+100y^2=0\left(4\right)\)

\(với:y=0\) \(ko\) \(là\) \(nghiệm\)

\(với:y\ne0\Rightarrow\left(4\right)\Leftrightarrow4\left(\dfrac{x}{y}\right)^2+35\left(\dfrac{x}{y}\right)+100=0\)\(\left(vô-lí\right)\)

\(do:4\left(\dfrac{x}{y}\right)^2+35\left(\dfrac{x}{y}\right)+100>0\)

\(\left(3\right)\Rightarrow x+5y=2\sqrt{5xy}\Leftrightarrow x^2+10xy+25y^2=20xy\Leftrightarrow x^2-10xy+25y^2=0\Leftrightarrow\left(x-5y\right)^2=0\Leftrightarrow x=5y\)

\(thay:x=5y\) \(vào:2x^2+y^2=51\Rightarrow2\left(5y\right)^2+y^2-51=0\Leftrightarrow51y^2-51=0\Leftrightarrow\left[{}\begin{matrix}y=1\left(tm\right)\Rightarrow x=5\left(tm\right)\\y=-1\left(loại\right)\end{matrix}\right.\)

\(f\left(x\right)=x^2+4x+\left|x+2\right|-m< 0\) 

\(\Leftrightarrow f\left(x\right)=x^2+4x+4+\left|x+2\right|-4-m< 0\)

\(\Leftrightarrow f\left(x\right)=\left(x+2\right)^2+\left|x+2\right|-4-m< 0\)

\(đặt:\left|x+2\right|=t\ge0\Rightarrow f\left(t\right)=t^2+t-4-m< 0\)

\(có\) \(f\left(x\right)nghiệm\Leftrightarrow f\left(t\right)có\)  \(nghiệm\) \(t\ge0\)

\(f\left(t\right)=t^2+t-4< m\)\(có\) \(nghiệm\) \(t\ge0\)

\(\Leftrightarrow m>minf\left(t\right)\left(trên[0;+\infty\right)\)\(\Leftrightarrow m>-4\)

\(\left\{{}\begin{matrix}\sqrt{x^2+3}+2\sqrt{x}=3+\sqrt{y}\left(1\right)\\\sqrt{y^2+3}+2\sqrt{y}=3+\sqrt{x}\left(2\right)\end{matrix}\right.\)\(\left(đk;x;y\ge0\right)\)

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

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

\(với:x=y=0\Rightarrow ko\) \(là\) \(nghiệm\)

\(vỡi:x=y\ne0\Rightarrow x;y>0\)

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

\(\Leftrightarrow\left(x-y\right)\left[\dfrac{x+y}{\sqrt{x^2+3}+\sqrt{y^2+3}}+\dfrac{4}{2\sqrt{x}+2\sqrt{y}}+\dfrac{1}{\sqrt{x}+\sqrt{y}}>0\left(\forall x;y>0\right)\right]=0\)

\(\Rightarrow x=y\left(4\right)\)

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

\(\Leftrightarrow\sqrt{x^2+3}-2+\sqrt{x}-1=0\Leftrightarrow\dfrac{x^2+3-4}{\sqrt{x^2+3}+2}+\dfrac{x-1}{\sqrt{x}+1}=0\Leftrightarrow\left(x-1\right)\left[\dfrac{x+1}{\sqrt{x^2+3}+2}+\dfrac{1}{\sqrt{x}+1}>0\left(\forall x>1\right)\right]=0\Leftrightarrow x=y=1\)