#)Góp ý :

Tham khảo ở : Câu hỏi của Nguyễn Ánh Tuyền - Toán lớp 9 - Học toán với OnlineMath

Link : https://olm.vn/hoi-dap/detail/54093991490.html

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

\(A=\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}\)

\(A=\frac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}+\frac{\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)}\)

\(A=\frac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\)

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

\(A=\frac{6}{6}=1\)

\(A=1\)

ĐKXĐ : x > 2

Ta có \(\left(\sqrt{x+3}-\sqrt{x-2}\right)\left(1+\sqrt{x^2+x-6}\right)=5\)

\(\Leftrightarrow\left(\sqrt{x+3}-\sqrt{x-2}\right)\left(1+\sqrt{\left(x+3\right)\left(x-2\right)}\right)=5\)

Đặt \(\hept{\begin{cases}\sqrt{x+3}=a\left(a>0\right)\\\sqrt{x-2}=b\left(b\ge0\right)\end{cases}}\)

\(\Rightarrow a^2-b^2=x+3-x+2=5\) và \(a\ne b\)

Pt trở thành \(\left(a-b\right)\left(1+ab\right)=a^2-b^2\)

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

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

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

       \(\Leftrightarrow a=b\left(h\right)a=1\left(h\right)b=1\)                                     ^{(h) là hoặc nhé}

*Với a = b (Loại do a khác b)

*Với \(a=1\Rightarrow\sqrt{x+3}=1\)

                    \(\Leftrightarrow x+3=1\)

                    \(\Leftrightarrow x=-2\)(Loại do ko thỏa mãn ĐKXĐ)

*Với \(b=1\Rightarrow\sqrt{x-2}=1\)

                    \(\Leftrightarrow x-2=1\)

                    \(\Leftrightarrow x=3\left(Tm\cdotĐKXĐ\right)\)

Vậy pt có nghiệm duy nhất x = 3

hehe chiều mình cũng thế

https://diendantoanhoc.net/topic/74052-cho-xyz0-xyz1-tim-gtnn-c%E1%BB%A7a-p-fracx2yzyzfracy2zxzxfracz2xyxy/

vào là có ok

\(a,19^2=\left(18+1\right)^2=18^2+2.18.1+1^2=324+36+1=361\)

\(28^2=\left(27+1\right)^2=27^2+2.27.1+1^2=729+54+1=784\)

\(81^2=\left(80+1\right)^2=80^2+2.80.1+1^2=6400+160+1=6561\)

\(91^2=\left(90+1\right)^2=90^2+2.90.1+1^2=8100+180+1=8281\)

\(b,19.21=\left(20-1\right)\left(20+1\right)=20^2-1^2=400-1=399\)

\(29.31=\left(30-1\right)\left(30+1\right)=30^2-1^2=900-1=899\)

\(39.41=\left(40-1\right)\left(40+1\right)=40^2-1^2=1600-1=1599\)

\(c,28^2-8^2=\left(28-8\right)\left(28+8\right)=20.36=720\)

\(56^2-46^2=\left(56-46\right)\left(56+46\right)=10.102=1020\)

\(67^2-57^2=\left(67-57\right)\left(67+57\right)=10.124=1240\)

a) \(19^2=\left(20-1\right)^2=20^2-2.20.1+1^2=400-40+1=361\)

\(28^2=\left(30-2\right)^2=30^2-2.30.2+2^2=900-120+4=784\)

\(81^2=\left(80+1\right)^2=80^2+2.80.1+1^2=6400+160+1=6561\)

\(91^2=\left(90+1\right)^2=90^2+2.90.1+1^2=8100+180+1=8281\)

Áp dụng bđt Cô si ta có: 

\(\sqrt{x-2}\le\frac{x-2+1}{2}=\frac{x-1}{2}\)

\(\sqrt{y+2014}\le\frac{y+2014+1}{2}=\frac{y+2015}{2}\)

\(\sqrt{z-2015}\le\frac{z-2015+1}{2}=\frac{z-2014}{2}\)

Cộng theo vế: \(\sqrt{x-2}+\sqrt{y+2014}+\sqrt{z-2015}\le\frac{x-1+y+2015+z-2014}{2}=\frac{1}{2}\left(x+y+z\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}x=3\\y=-2013\\z=2016\end{cases}}\)

Áp dụng bất đẳng thức Cô si ta có

\(\sqrt{x-2}\le\frac{x-2+1}{2}=\frac{x-1}{2}\)

\(\sqrt{y+2014}\le\frac{y+2014+1}{2}=\frac{y+2015}{2}\)

\(\sqrt{z-2015}\le\frac{z-2015+1}{2}=\frac{z-2014}{2}\)

Cộng theo vế

\(\sqrt{x-2}+\sqrt{y+2014}+\sqrt{z-2015}\le\)\(\frac{x-1+y+2015+z-2014}{2}=\frac{1}{2}\left(x+y+z\right)\)

Dấu = xảy ra khi

\(\hept{\begin{cases}x=3\\y=-2013\\z=2016\end{cases}}\)

rút gọn hay là sao

\(\sqrt{x^3}-\sqrt{y^3}+\sqrt{x^2y}-\sqrt{xy^2}\)

\(=\left(\sqrt{x}-\sqrt{y}\right)\left(x-\sqrt{xy}-y\right)+\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\)

\(=\left(\sqrt{x}-\sqrt{y}\right)\left(x-\sqrt{xy}-y+\sqrt{xy}\right)\)

\(=\left(\sqrt{x}-\sqrt{y}\right)\left(x-y\right)\)

Ta có:\(x^2-4xy+6y^2+2x+4\)

\(=\left(x-2y\right)^2+\left(x+x+\frac{8}{x^2}\right)+\left(2y^2+\frac{2}{y^2}\right)\)

\(\ge0+6+4=10\)

\(\Rightarrow x^2-4xy+6y^2+2x\ge10-4=6\)

Dấu bằng xảy ra khi x=2 và y=1.

\(\sqrt{6-x}+\sqrt{x+2}=\sqrt{\left(1.\sqrt{6-x}+1.\sqrt{x+2}\right)^2}\) \(\le\left(1^2+1^2\right)\left(6-x+x+2\right)=2.8=16\)

bạn tìm điều kiện xác định r dùng bunhiacopxki là ra nhé