\(=\left(\dfrac{\sqrt{5}\left(\sqrt{3}-2\right)}{\sqrt{3}-2}+\dfrac{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}-\dfrac{\sqrt{6}-\sqrt{5}}{\left(\sqrt{6}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{5}\right)}\right):\sqrt{\dfrac{5}{2}}\)

\(=\left(\sqrt{5}+\sqrt{6}-\sqrt{6}+\sqrt{5}\right):\dfrac{\sqrt{5}}{\sqrt{2}}\)

\(=2\sqrt{5}.\dfrac{\sqrt{2}}{\sqrt{5}}=2\sqrt{2}\)

a) Ta có: \(\left(\dfrac{\sqrt{15}-\sqrt{20}}{\sqrt{3}-2}+\dfrac{3\sqrt{2}+2\sqrt{3}}{\sqrt{3}+\sqrt{2}}-\dfrac{1}{\sqrt{6}+\sqrt{5}}\right):\sqrt{\dfrac{5}{2}}\)

\(=\left(\sqrt{5}+\sqrt{6}-\sqrt{6}+\sqrt{5}\right):\dfrac{\sqrt{10}}{2}\)

\(=2\sqrt{5}\cdot\dfrac{2}{\sqrt{10}}=2\sqrt{2}\)

a) \(k=-5\)

b) 

Tìm x xong rồi tìm y

3 thì làm kiểu gì cũng được

Bài 3:

Hình 1:

Xét ΔABC có \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)

=>\(x+60^0+40^0=180^0\)

=>\(x=180^0-100^0=80^0\)

Hình 2:

Xét ΔABD có \(\widehat{ADC}\) là góc ngoài tại đỉnh D

nên \(\widehat{ADC}=\widehat{ABD}+\widehat{BAD}=110^0\)

=>\(y=110^0\)

ΔDAC cân tại D

=>\(\widehat{DAC}=\widehat{DCA}=\dfrac{180^0-\widehat{ADC}}{2}\)

=>\(x=\dfrac{180^0-110^0}{2}=35^0\)

Hình 3:

Ta có: \(\widehat{CAB}+60^0=180^0\)

=>\(\widehat{CAB}=180^0-60^0=120^0\)

Xét ΔCAB có \(\widehat{CAB}+\widehat{ABC}+\widehat{ACB}=180^0\)

=>\(x+2x+120^0=180^0\)

=>\(3x=60^0\)

=>\(x=20^0\)

Hình 4:

Ta có: \(\widehat{ADC}+\widehat{ADB}=180^0\)

=>\(\widehat{ADB}=180^0-80^0=100^0\)

Xét ΔADB có

\(\widehat{ADB}+\widehat{ABD}+\widehat{BAD}=180^0\)

=>\(x+60^0+100^0=180^0\)

=>\(x=20^0\)

ta có: \(\widehat{ACD}+135^0=180^0\)

=>\(\widehat{ACD}=180^0-135^0=45^0\)

Xét ΔACD có \(\widehat{ACD}+\widehat{ADC}+\widehat{DAC}=180^0\)

=>\(y+80^0+45^0=180^0\)

=>\(y=55^0\)

\(\dfrac{x}{27}=\dfrac{2}{9}-\dfrac{1}{3}\Rightarrow\dfrac{x}{27}=-\dfrac{1}{9}\Rightarrow\dfrac{x}{27}=\dfrac{-3}{27}\Rightarrow x=27\)

\(\dfrac{x}{27}=\dfrac{2}{9}-\dfrac{1}{3}=-\dfrac{1}{9}\Rightarrow x=-\dfrac{1}{9}.27=-3\).

h_trụ=\(\dfrac{S_{xq}}{2\pi.R}\)=\(\dfrac{352}{2\pi.7}\)≃8,003(cm)

có  

<=>352=2\(\pi\).7.h<=>352=14\(\pi\).h<=>h=352/(14.\(\pi\))

<=>h\(\approx\)8cm( nếu lấy\(\pi\) \(\approx\)3,14)

 làm đc câu 1 thôi

mình không biết đã làm đúng chưa nên mong bạn thông cảm nha  

1: Ta có: \(P=\left(\dfrac{x+3}{x-1}+\dfrac{\sqrt{x}}{1-\sqrt{x}}+\dfrac{1}{1+\sqrt{x}}\right)\left(\dfrac{x\sqrt{x}-1}{\sqrt{x}-1}+\sqrt{x}\right)\)

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

\(=\dfrac{x+\sqrt{x}+2-x-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\left(\sqrt{x}+1\right)^2\)

\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)