ĐKXĐ: \(x\ge2\)

\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{\left(x-1\right)\left(x+3\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x-1}=a>0\\\sqrt{x-2}=b\ge0\\\sqrt{x+3}=c>0\end{matrix}\right.\)

\(\Leftrightarrow ab+c=b+ac\)

\(\Leftrightarrow a\left(b-c\right)-\left(b-c\right)=0\)

\(\Leftrightarrow\left(a-1\right)\left(b-c\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=1\\b=c\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x-2}=\sqrt{x+3}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\-2=3\left(vn\right)\end{matrix}\right.\)

1> Nua chu vi khu dat hinh chu nhat la:

 280:2=140(m)

 Goi chieu dai va chieu rong lan luot la x,y(x,y \(\varepsilon\)N)

Theo bai ra ta co : \(\frac{x}{4}=\frac{y}{3}\)va x+y=140

Ap dung tinh chat day ti so bang nhau ta co:

   \(\frac{x}{4}=\frac{y}{3}=\frac{x+y}{4+3}=\frac{140}{7}=20\)

Tu \(\frac{x}{4}=20=>x=80\)

       \(\frac{y}{3}=20=>y=60\)

Vaay chieu dai va chieu rong lan luot la 80,60

             Dien h khu dat hinh chu nhat la 

                          80.60=4800(\(m^2\))

                                    D/s:4800m^2

Day la cach lam lop 7 nha boi vi ban hocj lop 7 nen mk giai theo cach lop7.

Nho k cho mk nhe

a) \(2\sqrt{9}+3\sqrt{16}+2\sqrt{49}\)

\(=2.3+3.4+2.7\)

\(=6+12+14\)

\(=32\)

b) \(4\sqrt{0,25}+2\sqrt{0,36}-2\sqrt{0,16}\)

\(=4.\sqrt{\frac{25}{100}}+2\sqrt{\frac{36}{100}}-2\sqrt{\frac{16}{100}}\)

\(=4.\frac{1}{2}+2.\frac{3}{5}-2.\frac{4}{25}\)

\(=2+\frac{6}{5}-\frac{8}{25}\)

\(=\frac{72}{25}\)

\(\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}\right)^2-2\sqrt{2}+1}=\sqrt{\left(\sqrt{2}-1\right)^2}=|\sqrt{2}-1|=\sqrt{2}-1\)

Tương tự  \(\sqrt{4-2\sqrt{3}}=\sqrt{3}-1\);   \(\sqrt{7-4\sqrt{3}}=2-\sqrt{3}\)

\(\Rightarrow BTT=\sqrt{2}-1+\sqrt{3}-1+2-\sqrt{3}=\sqrt{2}\)

\(\sqrt{3-2\sqrt{2}}+\sqrt{4-2\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)

\(=\sqrt{2-2\sqrt{2}+1}+\sqrt{3-2\sqrt{3}+1}-\sqrt{4-4\sqrt{3}+3}\)

\(=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)

\(=\sqrt{2}-1+\sqrt{3}-1-2+\sqrt{3}\)

\(=2\sqrt{3}+\sqrt{2}-4\)

ok nhé

\(\sqrt{13+\sqrt{48}}=\sqrt{13+\sqrt{4.12}}=\sqrt{13+2\sqrt{12}}=\sqrt{\left(\sqrt{12}+1\right)^2}\)

\(=\sqrt{12}+1=2\sqrt{3}+1\)

\(\Rightarrow\sqrt{5-\sqrt{13+\sqrt{48}}}=\sqrt{5-2\sqrt{3}-1}=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}\)

\(=\sqrt{3}-1\)

\(\Rightarrow\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{3+\sqrt{3}-1}=\sqrt{2+\sqrt{3}}\)

\(\Rightarrow\sqrt{\dfrac{4+2\sqrt{3}}{2}}=\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{2}}=\dfrac{\sqrt{3}+1}{\sqrt{2}}\)

\(\Rightarrow2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}==2.\dfrac{\sqrt{3}+1}{\sqrt{2}}=\sqrt{6}+\sqrt{2}\)

2) biến đổi khúc sau như câu 1:

\(\Rightarrow\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{6+2\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}\)

\(=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)

4) Ta có: \(\sqrt{30-2\sqrt{16+6\sqrt{11+4\sqrt{4-2\sqrt{3}}}}}\)

\(=\sqrt{30-2\sqrt{16+6\sqrt{11+4\left(\sqrt{3}-1\right)}}}\)

\(=\sqrt{30-2\sqrt{16+6\sqrt{7+4\sqrt{3}}}}\)

\(=\sqrt{30-2\sqrt{16+6\left(2+\sqrt{3}\right)}}\)

\(=\sqrt{30-2\sqrt{28+6\sqrt{3}}}\)

\(=\sqrt{30-2\left(3\sqrt{3}+1\right)}\)

\(=\sqrt{28-6\sqrt{3}}=3\sqrt{3}-1\)

5) Ta có: \(\dfrac{\left(5\sqrt{3}+\sqrt{50}\right)\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}\)

\(=\dfrac{5\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)^2}{\sqrt{75}-5\sqrt{2}}\)

\(=\dfrac{5\left(\sqrt{3}-\sqrt{2}\right)}{5\left(\sqrt{3}-\sqrt{2}\right)}=1\)

3.

Đặt \(f\left(x\right)=x^4-3x^3+x-\dfrac{1}{8}\)

Hàm \(f\left(x\right)\) liên tục trên R

Do \(f\left(x\right)\) là đa thức bậc 4 nên có tối đa 4 nghiệm

Ta có: \(f\left(-1\right)=\dfrac{23}{8}>0\)

\(f\left(0\right)=-\dfrac{1}{8}< 0\Rightarrow f\left(-1\right).f\left(0\right)< 0\)

\(\Rightarrow f\left(x\right)\) luôn có ít nhất 1 nghiệm thuộc \(\left(-1;0\right)\)

\(f\left(\dfrac{1}{2}\right)=\dfrac{1}{16}>0\Rightarrow f\left(0\right).f\left(\dfrac{1}{2}\right)< 0\)

\(\Rightarrow f\left(x\right)\) luôn có ít nhất 1 nghiệm thuộc \(\left(0;\dfrac{1}{2}\right)\)

\(f\left(1\right)=-\dfrac{9}{8}< 0\Rightarrow f\left(\dfrac{1}{2}\right).f\left(1\right)< 0\)

\(\Rightarrow f\left(x\right)\) luôn có ít nhất 1 nghiệm thuộc \(\left(\dfrac{1}{2};1\right)\)

\(f\left(3\right)=\dfrac{23}{8}>0\Rightarrow f\left(1\right).f\left(3\right)< 0\)

\(\Rightarrow f\left(x\right)\) luôn có ít nhất 1 nghiệm thuộc \(\left(1;3\right)\)

Vậy pt có 4 nghiệm thuộc các khoảng nói trên

4.

\(\lim\limits_{x\rightarrow-\infty}\left(\sqrt{x^2+ax+2017}+x\right)=\lim\limits_{x\rightarrow-\infty}\dfrac{ax+2017}{\sqrt{x^2+ax+2017}-x}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{a+\dfrac{2017}{x}}{-\sqrt{1+\dfrac{a}{x}+\dfrac{2017}{x^2}}-1}=-\dfrac{a}{2}\)

\(\Rightarrow-\dfrac{a}{2}=6\Rightarrow a=-12\)

1) ĐKXĐ: \(x\ge5\)

2) ĐKXĐ: \(\left[{}\begin{matrix}x< -2\\x>2\end{matrix}\right.\)

5) ĐKXĐ: \(\left[{}\begin{matrix}x\le2\\x\ge3\end{matrix}\right.\)

 đọc là căn 3 đó bạn

Bình phương  lên là 9

~Chúc bạn học giỏi~

= 1,732.......

nó tương ứng với 

17320508*...

   đó là kết quả ai thấy đúng thì tk nha