a)xét \(\Delta\)ABC vuông tại A có

\(\widehat{B}+\widehat{C}=90'\Rightarrow\widehat{C}=90'-30'=60'\)

\(\sin C=\frac{AB}{BC}\Rightarrow BC=\frac{AB}{\sin B}=\frac{6}{\sin30'}=12\left(cm\right)\)

\(\tan B=\frac{AC}{AB}\Rightarrow AC=AB.\tan B=6.\tan30'=2\sqrt{3}\left(cm\right)\)

b)Xét \(\Delta ABC\left(\widehat{BAC}=90'\right)AHvuôngócBC\)

\(AB^2=BC.HB\Rightarrow HB=\frac{AB^2}{BC}=\frac{6^2}{12}=3cm\)

\(AH.BC=AB.AC\)

\(\Rightarrow AH=\frac{AB.AC}{BC}=6.2\sqrt{3}=12\sqrt{3}cm\)(1)

VÌ AM LÀ ĐƯỜNG TRUNG TUYẾN CỦA TG ABC NÊN

\(MB=MC=\frac{BC}{2}=\frac{12}{2}=6cm\)

MÀ\(MB=MH+HB\)

\(\Rightarrow MH=MB-HB=6-3=3cm\)(2)

TỪ (1)và (2) SUY RA

\(S\Delta AHM=\frac{1}{2}AH.HM=\frac{1}{2}.12\sqrt{3}.3=18\sqrt{3}\approx31.18\left(cm^2\right)\left(do\Delta AHMvuôngtạiH\right)\)

Từ pt 1 ta có thể biến đổi : \(ax+y+z=a^2\)

\(< =>a=\frac{ax+y+z}{a}\)

\(< =>x+y+z=a\)

\(< =>3x+3y+3z=x+ay+z\)

\(< =>2x+y\left(3-a\right)+2z=0\)

\(< =>2a+y-ay=0\)

\(< =>2a+y-ay-2=-2\)

\(< =>a\left(2-y\right)-\left(2-y\right)=-2\)

\(< =>\left(a-1\right)\left(2-y\right)=2.\left(-1\right)=-1.2=-2.1=1.\left(-2\right)\)

\(< =>\left(a;y\right)=\left(3;3\right)=\left(0;0\right)=\left(-1;1\right)=\left(2;4\right)\)

Bạn thay vào là đc :)) giải sai hay đúng cg ko bt nx :(

ĐKXĐ:.............

1.\(\sqrt{x^2-6x+9}=2x-1\)

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

\(\Leftrightarrow\left|x-3\right|=2x-1\)

................

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

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

\(\Leftrightarrow\left|\sqrt{x}+2\right|=5x+2\)

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

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

\(\Leftrightarrow\left|x-1\right|+\left|x+2\right|=4\)

\(P=\frac{2005x+2006\sqrt{1-x^2}+2007}{\sqrt{1-x^2}}\)

\(=\frac{2006\left(1+x\right)+\left(1-x\right)}{\sqrt{1-x^2}}+2006\)

\(\ge\frac{2\sqrt{2006\left(1+x\right)\left(1-x\right)}}{\sqrt{1-x^2}}+2006=2\sqrt{2006}+2006\)

Dấu = xảy ra khi:

\(2006\left(1+x\right)=1-x\)

\(\Leftrightarrow x=-\frac{2005}{2007}\)