1.09866399

Làm ơn ns cách giải 

Ta có \(\tan50=\frac{AC}{AB}\Rightarrow AB=\frac{AC}{\tan50}\approx12.5\left(cm\right)\)

Theo định lí Pitago ta có \(BC=\sqrt{AB^2+AC^2}=\sqrt{15^2+12,5^2}\approx19,6\left(cm\right)\)

Có \(\widehat{C}=180^0-\widehat{A}-\widehat{B}=40^0\)

Vì CD là phân giác trong của góc C \(\Rightarrow\widehat{ACD}=20^0\)

\(\Rightarrow CD=\frac{AC}{\cos20}\approx16\left(cm\right)\)

\(\Delta ABC\left(\widehat{A}=90^0\right)\)

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

\(BC^2=\left(3\sqrt{3}\right)^2+\left(2\sqrt{5}\right)^2=47\)

\(\Rightarrow BC=\sqrt{47}\left(cm\right)\)

\(\sin\widehat{C}=\frac{3\sqrt{3}}{\sqrt{47}}\Rightarrow\widehat{C}\approx55^0\)

\(\widehat{B}=90^0-\widehat{C}\)(2 góc phụ nhau)

\(\widehat{B}=90^0-55^0=35^0\)

Chúc bạn học tốt.

cảm ơn bạn 

a)\(y^4+4(2x-3)y^2-48x-48y+155=0\)

\(\Leftrightarrow y^4+8y^2x+16(9-3y)-12(y^2+4x)+11=0\)

\(\Leftrightarrow(y^2+4x)^2-12(y^2+4x)+11=0\)

<=>....

b)\(y^2-5x^2-4xy+16x-8y+16=0\)

\(\Leftrightarrow-\left(5x-y+4\right)\left(x+y-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}y=4-x\\y=5x+4\end{cases}}\)

tới đây nhìn vào pt thứ 1 là thấy 1 sự dễ ko hề nhẹ

c)\(pt\left(1\right)\Leftrightarrow2x\left(x+y\right)+2y^2=8x-2\)

cộng theo vế pt(1) vừa tương đương vs pt 2

\(\Leftrightarrow x\left(\left(x+y\right)^2+2\left(x+y\right)-15\right)=0\)

....

Hướng dẫn thui nhé sắp bão to nên phải off r` ko lm dc tiếp thì ib :333

câu 1 có vấn đề , (2x+3) , ko phải (2x-3) 

Giúp với mn~~

a) \(P=\frac{\left(x\sqrt{x}-3\right)-2\left(\sqrt{x}-3\right)^2-\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)

\(P=\frac{x\sqrt{x}-3-2\left(x-6\sqrt{x}+9\right)-\left(x+4\sqrt{x}+3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)

\(P=\frac{x\sqrt{x}-3x+8\sqrt{x}-24}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)

\(P=\frac{\left(\sqrt{x}-3\right)\left(x+8\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)

\(P=\frac{x+8}{\sqrt{x}+1}\)

b) Ta có \(x=14-6\sqrt{5}=9-2.3.\sqrt{5}+5=\left(3-\sqrt{5}\right)^2\)

Vậy nên \(\sqrt{x}=3-\sqrt{5}\)

Suy ra \(P=\frac{\left(3-\sqrt{5}\right)^2+8}{3-\sqrt{5}+1}=\frac{58-2\sqrt{5}}{11}\)

c) \(P=\frac{x+8}{\sqrt{x}+1}=\frac{\left(x-1\right)+9}{\sqrt{x}+1}=\left(\sqrt{x}-1\right)+\frac{9}{\sqrt{x}+1}\)

\(=\left(\sqrt{x}+1\right)+\frac{9}{\sqrt{x}+1}-2\ge2\sqrt{\left(\sqrt{x}+1\right).\frac{9}{\sqrt{x}+1}}-2=4\)

minP = 4 khi \(\sqrt{x}+1=\frac{9}{\sqrt{x}+1}\Rightarrow\sqrt{x}+1=3\Rightarrow x=4.\)

Ta có \(\tan B=\tan30^0=\frac{AC}{AB}\Rightarrow\frac{\sqrt{3}}{3}=\frac{AC}{4}\Rightarrow AC=\frac{4\sqrt{3}}{3}\)

\(\Rightarrow BC=\sqrt{AB^2+AC^2}=\sqrt{4^2+\left(\frac{4\sqrt{3}}{3}\right)^2}=\frac{8\sqrt{3}}{3}\)

Lại có \(AH.BC=AB.AC\Rightarrow AH=\frac{4.\frac{4\sqrt{3}}{3}}{\frac{8\sqrt{3}}{3}}=2\) 

a/ \(x^2-2x-1< 0\)

\(\Leftrightarrow\left(x-1\right)^2< 2\)

\(\Leftrightarrow-\sqrt{2}< x-1< \sqrt{2}\)

\(\Leftrightarrow1-\sqrt{2}< x< 1+\sqrt{2}\)

b/ \(2x^2-6x+5=\left(2x^2-\frac{2.\sqrt{2}.x.3}{\sqrt{2}}+\frac{9}{2}\right)+\frac{1}{2}=\left(\sqrt{2}x-\frac{3}{\sqrt{2}}\right)^2+\frac{1}{2}>0\)

Câu 2 tự làm nhé.

\(x^2-2x-1< 0\)

\(\left(x-2\right)x-1< 0\)

\(\left(x-2\right)x\le1\)

\(\Leftrightarrow1-\sqrt{2}< x< 1+\sqrt{2}\)