1)Thu gọn
a; \(\sqrt{12-6\sqrt{3}}-\sqrt{21-12\sqrt{3}}\)
b; \(\sqrt{12}-\sqrt{27}\)
2) \(A=\left(\dfrac{1}{\sqrt{x+2}}+\dfrac{1}{\sqrt{x-2}}\right)\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
a, Tìm tập xác định và rút gọn A
b, x= bao nhiêu để A\(>\dfrac{1}{2}\)
3) Rút gọn C
\(C=\left(\dfrac{2x-10}{x}+\dfrac{5x+50}{x^2+5x}+\dfrac{x^2}{5x+25}\right):\dfrac{3x+15}{7}\)
4) Rút gọn B
\(B=\left(\dfrac{\sqrt{x}+1}{x-4}-\dfrac{\sqrt{x}-1}{x+4\sqrt{x}+4}\right):\dfrac{\sqrt{x}}{x\sqrt{x}+2x-4\sqrt{x}-8}\)
5) Tam giác ABC gọi D, E trung điểm AB, AC. Trên tia đối tia DC lấy M trên tia đối tia EB lấy N sao cho DM= DC; EN= Be.
b, Chứng minh BC song song và bằng MA
b, Chứng minh AN song song và bằng BC
c, Chứng minh A trung điểm MN
6) \(\widehat{xOy}\) , Oz phân giác. Từ A\(\in\)Oz kẻ các đường song song, với Ox cắt Oy ở B, Oy cắt Ox ở C
a, Chừng minh OB = OC, AB=AC
b, Kẻ AH vuông góc với Ox, AK vuông góc với Oy. Chứng minh, AH=AK
1/
a/ \(\sqrt{12-6\sqrt{3}}-\sqrt{21-12\sqrt{3}}\)
\(\sqrt{\left(3+\sqrt{3}\right)^2}-\sqrt{\left(3+2\sqrt{3}\right)^2}=3+\sqrt{3}-3-2\sqrt{3}=\sqrt{3}-2\sqrt{3}=-\sqrt{3}\)
b/ \(\sqrt{12}-\sqrt{27}=2\sqrt{3}-3\sqrt{3}=-\sqrt{3}\)
3/ \(C=\left(\dfrac{2x-10}{x}+\dfrac{5x+50}{x^2+5x}+\dfrac{x^2}{5x+25}\right):\dfrac{3x+15}{7}\)
\(=\left(\dfrac{2\left(x-5\right)}{x}+\dfrac{5\left(x+10\right)}{x\left(x+5\right)}+\dfrac{x^2}{5\left(x+5\right)}\right)\cdot\dfrac{7}{3\left(x+5\right)}\)
\(=\left(\dfrac{10\left(x+5\right)\left(x-5\right)}{5x\left(x+5\right)}+\dfrac{25\left(x+10\right)}{5x\left(x+5\right)}+\dfrac{x^3}{5x\left(x+5\right)}\right)\cdot\dfrac{7}{3\left(x+5\right)}\)
\(=\dfrac{10x^2-250+25x+250+x^3}{5x\left(x+5\right)}\cdot\dfrac{7}{3\left(x+5\right)}\)
\(=\dfrac{x^3+10x^2+25x}{5x\left(x+5\right)}\cdot\dfrac{7}{3\left(x+5\right)}\)
\(=\dfrac{x\left(x^2+10x+25\right)}{5x\left(x+5\right)}\cdot\dfrac{7}{3\left(x+5\right)}\)
\(=\dfrac{7\left(x+5\right)^2}{5\left(x+5\right)\cdot3\left(x+5\right)}=\dfrac{7}{15}\)
3) \(C=\left(\dfrac{2x-10}{x}+\dfrac{5x+50}{x^2+5x}+\dfrac{x^2}{5x+25}\right):\dfrac{3x+15}{7}\)
\(C=\left(\dfrac{2x-10}{x}+\dfrac{5x+50}{x\left(x+5\right)}+\dfrac{x^2}{5\left(x+5\right)}\right):\dfrac{3x+15}{7}\)
\(C=\left[\dfrac{10\left(x+5\right)\left(x-5\right)}{5x\left(x+5\right)}+\dfrac{25\left(x+10\right)}{5x\left(x+5\right)}+\dfrac{x^3}{5x\left(x+5\right)}\right]:\dfrac{3x+15}{7}\)
\(C=\left[\dfrac{10\left(x^2-25\right)+25x+250+x^3}{5x\left(x+5\right)}\right]:\dfrac{3x+15}{7}\)
\(C=\left(\dfrac{10x^2-250+25x+250-x^3}{5x\left(x+5\right)}\right).\dfrac{7}{3\left(x+5\right)}\)
\(C=\dfrac{x\left(x+2.x.5+25\right)}{5x\left(x+5\right)}.\dfrac{7}{3\left(x+5\right)}=\dfrac{x\left(x+5\right)^2}{5x\left(x+5\right)}.\dfrac{7}{3\left(x+5\right)}=\dfrac{x+5}{5}.\dfrac{7}{3\left(x+5\right)}=\dfrac{7}{15}\)
5)
Hình vẽ :
a) Xét \(\Delta DBC,\Delta DAM\) có :
\(\left\{{}\begin{matrix}DA=DB\left(gt\right)\\\widehat{ADM}=\widehat{BDC}\left(đ.đỉnh\right)\\DC=DM\left(gt\right)\end{matrix}\right.\)
=> \(\Delta DBC=\Delta DAM\left(c.g.c\right)\)
=> \(\left\{{}\begin{matrix}AM=BC\left(\text{2 cạnh tương ứng}\right)\\\widehat{AMD}=\widehat{BCD}\left(\text{2 góc tương ứng - so le trong}\right)\Rightarrow AM//BC\end{matrix}\right.\)
b) Xét \(\Delta ANE,\Delta BCE\) có :
\(\left\{{}\begin{matrix}AE=EC\left(gt\right)\\\widehat{AEN}=\widehat{CEB}\left(đ.đỉnh\right)\\BE=EN\left(gt\right)\end{matrix}\right.\)
=> \(\Delta ANE=\Delta BCE\left(c.g.c\right)\)
=> \(\left\{{}\begin{matrix}AN=BC\left(\text{2 cạnh tương ứng}\right)\\\widehat{ANE}=\widehat{CBE}\left(\text{2 góc tương ứng - so le trong}\right)\Rightarrow AN//BC\end{matrix}\right.\)
c) Ta có :
Từ câu a : AM = BC (2 cạnh tương ứng)
Từ câu b : AN = BC (2 cạnh tương ứng)
Suy ra: AM = AN (=BC)
Vậy : A là trung điểm của MN.
Mk Chua lam nen co j said tho Thời nha
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