Bài 2

\(a,x+x+x\times\dfrac{1}{2}=\dfrac{7}{2}\\ \Rightarrow x\left(1+1+\dfrac{1}{2}\right)=\dfrac{7}{2}\\ \Rightarrow x.\left(\dfrac{4+1}{2}\right)=\dfrac{7}{2}\\ \Rightarrow x=\dfrac{7}{2}:\dfrac{5}{2}\\ \Rightarrow x=\dfrac{7}{5}\)

\(b,\dfrac{3}{8}-\dfrac{3}{8}\times\left(x-1\right)=\dfrac{1}{5}\\ \Rightarrow\dfrac{3}{8}\times\left(x-1\right)=\dfrac{3}{8}-\dfrac{1}{5}\\ \Rightarrow\dfrac{3}{8}\times\left(x-1\right)=\dfrac{3.5-8}{40}\\ \Rightarrow\dfrac{3}{8}\times\left(x-1\right)=\dfrac{7}{40}\\ \Rightarrow x-1=\dfrac{7}{40}:\dfrac{3}{8}\\ \Rightarrow x-1=\dfrac{7}{15}\\ \Rightarrow x=\dfrac{7}{15}+1\\ \Rightarrow x=\dfrac{22}{15}\)

\(R=d\left(I;\Delta\right)=\dfrac{\left|3.3-4.\left(-1\right)+2\right|}{\sqrt{3^3+\left(-4\right)}^2}=3\)

Phương trình đường tròn có tâm \(I\left(3;-1\right)\) và \(R=3\)

\(\Rightarrow\left(x-3\right)^2+\left(y+1\right)^2=9\)

\(\dfrac{HB}{HC}=\dfrac{2}{5}\\ \Rightarrow HB=\dfrac{2}{5}HC\)

Xét tam giác ABC vuông tại A
\(AH^2=BH.CH\\ \Rightarrow16^2=\dfrac{2}{5}HC.HC\\ \Rightarrow HC^2=640\\ \Rightarrow HC=8\sqrt{10}\)

\(\Rightarrow HB=\dfrac{2}{5}.8\sqrt{10}=\dfrac{16\sqrt{10}}{5}\)

\(BC=HC+HB=8\sqrt{10}+\dfrac{16\sqrt{10}}{5}=\dfrac{56\sqrt{10}}{5}\)

\(AB^2=BH.BC\\ \Rightarrow AB=\sqrt{\dfrac{16\sqrt{10}}{5}.\dfrac{56\sqrt{10}}{5}}=\dfrac{16\sqrt{35}}{5}\)

\(AC^2=CH.BC\\ \Rightarrow AC=\sqrt{8\sqrt{10}.\dfrac{56\sqrt{10}}{5}}=8\sqrt{14}\)

Chu vi : \(AB+AC+BC==8\sqrt{14}+\dfrac{56\sqrt{10}}{5}+\dfrac{16\sqrt{35}}{5}=84,28\)

Xét tam giác ABC : \(AB^2+AC^2=3^2+4^2=5^2=BC^2\)

\(\Rightarrow\Delta ABC\) vuông tại A \(\Rightarrow\widehat{A}=90^o\)
\(sinB=\dfrac{AC}{BC}=\dfrac{4}{5}\\ \Rightarrow\widehat{B}=53^o8'\)

\(sinC=\dfrac{AB}{BC}=\dfrac{3}{5}\\ \Rightarrow\widehat{C}=36^o52'\)

Bài 3

\(a,\sqrt{\left(\sqrt{15}-4\right)^2}=\left|\sqrt{15}-4\right|=-\sqrt{15}+4\\ b,=\sqrt{6-2.\sqrt{6}.1+1}\\ =\sqrt{\left(\sqrt{6}-1\right)^2}\\ =\left|\sqrt{6}-1\right|\\ =\sqrt{6}-1\)

Bài 4

\(a,=\left|2\sqrt{3}-3\sqrt{2}\right|+\sqrt{12-2.2\sqrt{3}.1+1}-\sqrt{18+2.3\sqrt{2}.2+4}\\ =3\sqrt{2}-2\sqrt{3}+\sqrt{\left(2\sqrt{3}-1\right)^2}-\sqrt{\left(3\sqrt{2}+2\right)^2}\\ =3\sqrt{2}-2\sqrt{3}+\left|2\sqrt{3}-1\right|-\left|3\sqrt{2}+2\right|\\ =3\sqrt{2}-2\sqrt{3}+2\sqrt{3}-1-3\sqrt{2}-2\\ =-3\)

\(b,=\sqrt{17-6\sqrt{2+\sqrt{8+2.2\sqrt{2}.1+1}}}\\ =\sqrt{17-6\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}\\ =\sqrt{17-6\sqrt{2+\left|2\sqrt{2}+1\right|}}\\ =\sqrt{17-6\sqrt{2+2\sqrt{2}+1}}\\ =\sqrt{17-6\sqrt{\left(\sqrt{2}+1\right)^2}}\\ =\sqrt{17-6\left|\sqrt{2}+1\right|}\\ =\sqrt{17-6\sqrt{2}-6}\\ =\sqrt{11-6\sqrt{2}}\\ =\sqrt{9-2.3.\sqrt{2}+2}\\ =\sqrt{\left(3-\sqrt{2}\right)^2}\\ =\left|3-\sqrt{2}\right|\\ =3-\sqrt{2}\)

\(A=2x^2+2\sqrt{2}x+3\\ =2\left(x^2+\sqrt{2}x+\dfrac{3}{2}\right)\\ =2.\left(x^2+2.\dfrac{1}{\sqrt{2}}x+\dfrac{1}{2}+1\right)\\ =2.\left(x^2+2.\dfrac{1}{\sqrt{2}}x+\dfrac{1}{2}\right)+2\\ =2.\left(x+\dfrac{1}{\sqrt{2}}\right)^2+2\)

Ta có \(2.\left(x+\dfrac{1}{\sqrt{2}}\right)^2\ge0\forall x\)

\(2.\left(x+\dfrac{1}{\sqrt{2}}\right)^2+2\ge2\forall x\)

Dấu bằng xảy ra khi : \(x+\dfrac{1}{\sqrt{2}}=0\\ \Rightarrow x=\dfrac{-\sqrt{2}}{2}\)

Vậy \(Min_A=2\) khi \(x=\dfrac{-\sqrt{2}}{2}\)

1- C

2-B

3- A

4- B