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vuông cân tại suy ra 

suy ra 

Gọi \(DE\cap CF=H\)

ΔADE=ΔDCF(c.g.c)

\(\Rightarrow\widehat{ADE}=\widehat{DCF}\)

\(\Rightarrow\widehat{ADE}+\widehat{DFH}=\widehat{DCF}+\widehat{DFH}=90\)

\(\Rightarrow\Delta FHD\) vuông tại H

\(\Rightarrow CF\perp DE\)

1.D

2.B

9.D

\(x^2+xy+x+y\\ =x\left(x+y\right)+\left(x+y\right)\\ =\left(x+1\right)\left(x+y\right)\)

5.

a)\(A=\sqrt{8+2\sqrt{15}}+\sqrt{8-2\sqrt{15}}\)

\(\Leftrightarrow A=\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}\\ \Leftrightarrow A=\sqrt{3}+\sqrt{5}+\sqrt{3}-\sqrt{5}=2\sqrt{3}\)

\(B=\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{2-\sqrt{x}}\left(x\ge0;x\ne4\right)\\ \Leftrightarrow B=\dfrac{2-\sqrt{x}+\sqrt{x}+2}{4-x}=\dfrac{4}{4-x}\)

b) Để \(A=B\Leftrightarrow\dfrac{4}{4-x}=2\sqrt{3}\)

\(\Leftrightarrow2\sqrt{3}\left(4-x\right)=4\\ \Leftrightarrow8\sqrt{3}-2\sqrt{3}x=4\\ \Leftrightarrow2\sqrt{3}x=8\sqrt{3}-4\\ \Leftrightarrow\sqrt{3}x=4\sqrt{3}-2\\ \Leftrightarrow x=\dfrac{4\sqrt{3}-2}{\sqrt{3}}=\dfrac{12-2\sqrt{3}}{3}\)

Tick nha

4.

a)\(A=\left(2\sqrt{75}-5\sqrt{27}-\sqrt{192}+4\sqrt{48}\right):\sqrt{3}\)

\(\Leftrightarrow A=\left(10\sqrt{3}-15\sqrt{3}-8\sqrt{3}+16\sqrt{3}\right):\sqrt{3}\\ \Leftrightarrow A=10-15-8+16=3\)

\(P=\left(\dfrac{\sqrt{x}}{2+\sqrt{x}}+\dfrac{\sqrt{x}}{2-\sqrt{x}}\right):\dfrac{\sqrt{x}}{2+\sqrt{x}}\left(x>0;x\ne4\right)\\ \Leftrightarrow P=\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)+\sqrt{x}\left(2+\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\cdot\dfrac{2+\sqrt{x}}{\sqrt{x}}\\ \Leftrightarrow P=\dfrac{2\sqrt{x}-x+2\sqrt{x}+x}{\sqrt{x}\left(2-\sqrt{x}\right)}=\dfrac{4}{2-\sqrt{x}}\)

b) Để \(A=P\)

\(\Leftrightarrow\dfrac{4}{2-\sqrt{x}}=3\\ \Leftrightarrow6-3\sqrt{x}=4\\ \Leftrightarrow3\sqrt{x}=2\\ \Leftrightarrow\sqrt{x}=\dfrac{2}{3}\\ \Leftrightarrow x=\dfrac{4}{9}\)

1.

a)\(A=\sqrt{3}\left(2\sqrt{27}-6\sqrt{\dfrac{1}{3}}+\dfrac{3}{2}\sqrt{12}\right)\)

\(\Leftrightarrow A=\sqrt{3}\left(6\sqrt{3}-2\sqrt{3}+3\sqrt{3}\right)=\sqrt{3}\cdot7\sqrt{3}\)

\(\Leftrightarrow A=21\)

\(B=\dfrac{x+\sqrt{x}}{\sqrt{x}}+\dfrac{x-4}{\sqrt{x}+2}\left(x>0\right)\\ \Leftrightarrow B=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\sqrt{x}+2}\\ \Leftrightarrow B=\sqrt{x}+1+\sqrt{x}-2=2\sqrt{x}-1\)

b) Để \(A=B\)

\(\Leftrightarrow2\sqrt{x}-1=21\\ \Leftrightarrow2\sqrt{x}=22\\ \Leftrightarrow\sqrt{x}=11\\ \Leftrightarrow x=121\)

3.

a)\(A=\left(\sqrt{5}-\sqrt{2}\right)^2+\sqrt{40}\)

\(\Leftrightarrow A=7-2\sqrt{10}+2\sqrt{10}\\ \Leftrightarrow A=7\)

\(B=\left(\dfrac{x-\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}}\left(x>0;x\ne1\right)\\ \Leftrightarrow B=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{\sqrt{x}}{\sqrt{x}+1}\\ \Leftrightarrow B=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\cdot\dfrac{\sqrt{x}}{\sqrt{x}+1}\\ \Leftrightarrow B=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)

b) Để \(A=B\)

\(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=7\Leftrightarrow\sqrt{x}-1=7\sqrt{x}+7\\ \Leftrightarrow6\sqrt{x}=-8\\ \Leftrightarrow\sqrt{x}=-\dfrac{4}{3}\\ \Leftrightarrow x=\dfrac{16}{9}\)

Kẻ AG⊥CD, BH⊥CD, IK⊥CD kẻ thêm nha

??

Ta có \(\left(1-2\sqrt{11}\right)^2=45-4\sqrt{11}\)

và \(\left(-5\right)^2=25=45-20\)

Ta có \(\left(4\sqrt{11}\right)^2=88;20^2=400\)

\(\Rightarrow4\sqrt{11}< 20\Rightarrow45-4\sqrt{11}>45-20\)

\(\Leftrightarrow1-2\sqrt{11}>-5\)

Tick nha

3.

a)\(4\sqrt{x^2}\left(x< 0\right)\)

\(=4\left|x\right|=-4x\)

b)\(\sqrt{9x^2}-4x\left(x\ge0\right)\)

\(=\left|3x\right|-4x=3x-4x=-x\)

c)\(\sqrt{x^2-6x+9}+x+3\left(x\ge3\right)\)

\(=\sqrt{\left(x-3\right)^2}+x+3\\ =\left|x-3\right|+x+3\\ =x-3+x+3=2x\)

d)\(\dfrac{3+\sqrt{a}}{a-9}\left(a\ge0;a\ne9\right)\)

\(=\dfrac{\sqrt{a}+3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}=\dfrac{1}{\sqrt{a}-3}\)

Tick nha