*Gọi D là giao của BM và EF.

△ABC vuông tại B có: BM là trung tuyến.

\(\Rightarrow BM=CM=\dfrac{BC}{2}\)

\(\Rightarrow\)△BMC cân tại M.

\(\Rightarrow\widehat{MCB}=\widehat{MBC}\)

Tứ giác BEHF có: \(\widehat{EBF}=\widehat{BFH}=\widehat{BEH}=90^0\)

\(\Rightarrow\)BEHF là hình chữ nhật.

\(\Rightarrow\widehat{HBE}=\widehat{DEB}\).

Ta có: \(\widehat{MCB}+\widehat{HBE}=90^0\) (△BHC vuông tại H).

\(\Rightarrow\widehat{MBC}+\widehat{DEB}=90^0\)

\(\Rightarrow180^0-\widehat{BDE}=90^0\)

\(\Rightarrow\widehat{BDE}=90^0\)

\(\Rightarrow\)BM⊥EF tại D.

Suppose I is the intersection of BM and EF.

Consider the right triangle ABC, which has \(\widehat{B}=90^o\), has the median BM, so, \(BM=\dfrac{AC}{2}\)  (1)

On the other hand, M is the midpoint of AC, therefore, \(AM=\dfrac{AC}{2}\)  (2)

From, (1) and (2), we have \(BM=AM\left(=\dfrac{AC}{2}\right)\), which means MAB is an isosceles triangle, and this leads to \(\widehat{A}=\widehat{ABM}\) or \(\widehat{A}=\widehat{FBI}\)

Consider the right triangle ABH (right at H), has the height HF, thus, \(BH^2=BF.BA\)

Similarly, we have \(BH^2=BE.BC\)

From these, we get \(BF.BA=BE.BC\left(=BH^2\right)\) or \(\dfrac{BF}{BC}=\dfrac{BE}{BA}\)

Consider the 2 right triangles (which are both right at B), we have \(\dfrac{BF}{BC}=\dfrac{BE}{BA}\). Therefore, \(\Delta BEF~\Delta BAC\left(s.a.s\right)\), which means \(\widehat{BFE}=\widehat{C}\) or \(\widehat{BFI}=\widehat{C}\)

Also, \(\widehat{A}+\widehat{C}=90^o\) due to the right triangle ABC (right at B). Because \(\widehat{FBI}=\widehat{A};\widehat{BFI}=\widehat{C}\), we have \(\widehat{FBI}+\widehat{BFI}=90^o\). This means FBI is a right triangle (whose right angle is I). Thus, \(BM\perp EF\), and that is what we must prove!

       `{(ax+by=c\text{   (1)}),(a'x+b'y=c'):}`

`<=>{(ab'x+bb'y=cb'),(a'bx+bb'y=c'b):}`

`<=>ab'x-a'bx=cb'-c'b`

`<=>(ab'-a'b)x=cb'-c'b`

`<=>x=[cb'-c'b]/[ab'-a'b]`      `(2)`

`<=>x=[D_x]/D`

Thế `(2)` vào `(1)` có: `a . [cb'-c'b]/[ab'-a'b]+by=c`

                             `<=>[acb'-ac'b]/[ab'-a'b]+by=c`

                             `<=>by=c-[acb'-ac'b]/[ab'-a'b]`

                             `<=>by=[cab'-ca'b-acb'+ac'b]/[ab'-a'b]`

                             `<=>y=[ac'b-ca'b]/[b(ab'-a'b)]`

                             `<=>y=[ac'-a'c]/[ab'-a'b]=[D_y]/D`

 `->` Đpcm

\(\sqrt{3-x}=\sqrt{3}-2\)          ĐK: \(x \le 3\)

Vì \(\sqrt{3-x} >= 0\)

 Mà \(\sqrt{3}-2 < 0\)

  => Ptr vô nghiệm

  \(\sqrt{3-x}\) = \(\sqrt{3}\)  - 2 

\(\sqrt{3-x}\) ≥ 0

\(\sqrt{3}\) - 2 < \(\sqrt{4}\) -2 = 0

vậy  pt vô nghiệm 

b,√11-4 và √12-5 ạ

 3 - \(\sqrt{7}\) > 3 - \(\sqrt{7,29}\) = 3- 2,7 = 0,3

2 - \(\sqrt{3}\) < 2 - \(\sqrt{2,89}\) = 2 - 1,7 = 0,3

vậy 3 - \(\sqrt{7}\) > 2 - \(\sqrt{3}\)

We must have \(x\ge0\) and \(y\ge0\)

We have: \(A=x+6y-4\sqrt{xy}+2\sqrt{x}-16\sqrt{y}+20\)\(A=\left(x+4y+1-4\sqrt{xy}+2\sqrt{x}-4\sqrt{y}\right)+2y-12\sqrt{y}+19\)\(A=\left(\sqrt{x}-2\sqrt{y}+1\right)^2+2\left(y-6\sqrt{y}+9\right)+1\)\(A=\left(\sqrt{x}-2\sqrt{y}+1\right)^2+2\left(\sqrt{y}-3\right)^2+1\)

Because \(\left(\sqrt{x}-2\sqrt{y}+1\right)^2\ge0;2\left(\sqrt{y}-3\right)^2\ge0;1>0\), we must have \(A>0\), and that is what we must prove.

Xét tam giác ABC vuông tại A, đường cao AH 

Ta có sinB = AC/BC -> \(\dfrac{1}{2}=\dfrac{AC}{8}\Leftrightarrow AC=4cm\)

Theo định lí Pytago ta có \(AB=\sqrt{BC^2-AC^2}=4\sqrt{3}cm\)

Áp dụng hệ thức \(AB^2=BH.BC\Rightarrow BH=\dfrac{AB^2}{BC}=6cm\)

Áp dụng hệ thức \(AH.BC=AB.AC\Rightarrow AH=\dfrac{AB.AC}{BC}=\dfrac{16\sqrt{3}}{8}=2\sqrt{3}cm\)

Ta có \(S_{AHB}=\dfrac{1}{2}.BH.AH=\dfrac{1}{2}.6.2\sqrt{3}=6\sqrt{3}cm^2\)

Vì tam giác ABC vuông tại A, M là trung điểm nên AM là đường trung tuyến 

AM= BM = BC/2 = 4 cm 

HM = BH - BM = 6 - 4 = 2 cm 

\(S_{AHM}=\dfrac{1}{2}.HM.AH=\dfrac{1}{2}.2.2\sqrt{3}=2\sqrt{3}cm^2\)

\(S_{AMB}=S_{ABH}-S_{AHM}=6\sqrt{3}-2\sqrt{3}=4\sqrt{3}cm^2\)

\(P=\sqrt{\left(2x-1\right)^2}+\sqrt{\left(2x+1\right)^2}=\left|2x-1\right|+\left|2x+1\right|\)

Theo BĐT Cosi ta được 

\(P\ge\left|1-2x+2x+1\right|=2\)

Dấu ''='' xảy ra khi \(\left\{{}\begin{matrix}2x-1\le0\\2x+1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\x\ge-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow-\dfrac{1}{2}\le x\le\dfrac{1}{2}\)

khó quá rồi

a) Ta có : \(AB^2+AC^2=6^2+8^2=100\)

               \(BC^2=10^2=100\)

\(\Rightarrow AB^2+AC^2=BC^2\)

\(\Rightarrow\bigtriangleup ABC\) vuông tại \(A\)  (đpcm)

b) Từ \(AB\cdot AC=AH\cdot BC\)

\(\Rightarrow6\cdot8=AH\cdot10\)

\(\Rightarrow AH=4,8\)

c) Từ \(AB^2=BC\cdot BH\)

\(\Rightarrow6^2=10\cdot HB\)

\(\Rightarrow HB=3,6\)

Từ \(HB+HC=BC\)

\(\Rightarrow3,6+HC=10\)

\(\Rightarrow HC=6,4\)

\(S_{\bigtriangleup ABC}=\dfrac{1}{2}AB\cdot AC\)  .

Xét tam giác ABC vuông tại A, đường cao AH 

Áp dụng hệ thức \(AB^2=HB.BC=HB\left(HC+HB\right)=HB\left(16+HB\right)\Leftrightarrow225=16HB+HB^2\)

\(\Leftrightarrow HB^2+16BH-225=0\Leftrightarrow HB=9cm\)

BC = HC + HB = 9 + 16 = 25 cm

Áp dụng hệ thức \(AH^2=HB.HC=144\Leftrightarrow AH=12cm\)

.