Chương I - Căn bậc hai. Căn bậc ba

a: \(=\sqrt{\dfrac{16}{9}\cdot\dfrac{4}{100}}=\dfrac{4}{3}\cdot\dfrac{2}{10}=\dfrac{4}{3}\cdot\dfrac{1}{5}=\dfrac{4}{15}\)

b: \(=\sqrt{0.09\cdot0.09}\cdot\sqrt{1.21\cdot0.4}\)

\(=0.09\cdot\dfrac{11\sqrt{10}}{50}=\dfrac{99\sqrt{10}}{5000}\)

c: \(=\dfrac{9\sqrt{2}-14\sqrt{2}+6\sqrt{2}}{\sqrt{2}}=9+6-14=1\)

Bạn thông cảm..trình bày ko đc đẹp

a)\(\dfrac{\sqrt{243a}}{\sqrt{3a}}=\dfrac{\sqrt{24}.\sqrt{3a}}{\sqrt{3a}}=2\sqrt{6}\)

b)\(\dfrac{3\sqrt{18a^2b^4}}{\sqrt{2a^2b^2}}=3\sqrt{9b^2}=\left[{}\begin{matrix}9b\\-9b\end{matrix}\right.\)

éo biết

We have \(\dfrac{s}{\dfrac{t}{2}}+\dfrac{s}{\dfrac{t}{2}}=40+60=100\)

After that. we have:

\(\dfrac{s}{\dfrac{t}{2}}=\dfrac{s}{\dfrac{t}{2}}=\dfrac{s+s}{\dfrac{t}{2}}=\dfrac{2s}{\dfrac{t}{2}}=\dfrac{4s}{t}\)

\(\Rightarrow\dfrac{s}{t}.4=100\)

\(\Rightarrow\dfrac{s}{t}=\dfrac{100}{4}=25\left(\dfrac{km}{h}\right)\)

Ơ nhầm...

1.do cắt tại trục tung có tung độ bằng 1+\(\sqrt{2}\) nên x= 0 y=1+\(\sqrt{2}\)thay vào hàm số là xong 2.tìm x y theo m thay vào biểu thức x\(^2\) +xy=30 ta tìm được m

Đặt \(\left\{{}\begin{matrix}a^2-1=x\\b^2-1=y\\c^2-1=z\end{matrix}\right.\)(x,y,z>0)thì giả thiết trở thành \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\)

chứng minh \(\dfrac{1}{\sqrt{x+1}+1}+\dfrac{1}{\sqrt{y+1}+1}+\dfrac{1}{\sqrt{z+1}+1}\le1\)

Áp dụng BĐT cauchy:(dạng \(\dfrac{1}{a+b+c}\le\dfrac{1}{9}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\))(

\(\sum\dfrac{1}{\sqrt{x+1}+1}\le\sum\dfrac{1}{9}\left(\dfrac{4}{\sqrt{x+1}}+1\right)=\dfrac{4}{9}\left(\dfrac{1}{\sqrt{x+1}}+\dfrac{1}{\sqrt{y+1}}+\dfrac{1}{\sqrt{z+1}}\right)+\dfrac{1}{3}\)(*)

mà theo BĐT bunyakovsky:\(\left(\sum\dfrac{1}{\sqrt{x+1}}\right)^2\le3\left(\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}\right)\le\dfrac{3}{16}\left(\dfrac{9}{x}+\dfrac{9}{y}+\dfrac{9}{z}+3\right)=\dfrac{3}{16}\left(9+3\right)=\dfrac{9}{4}\)

\(\Leftrightarrow\sum\dfrac{1}{\sqrt{x+1}}\le\dfrac{3}{2}\)kết hợp với (*), ta có

\(VT\le\dfrac{4}{9}.\dfrac{3}{2}+\dfrac{1}{3}=\dfrac{2}{3}+\dfrac{1}{3}=1\)

Dấu = xảy ra khi x=y=z=3 hay a=b=c=2 (a,b,c>1)

Sửa đề: C/m: \(\dfrac{1}{a+1}+\dfrac{1}{b+1}-\dfrac{1}{c+1}< 1\)

Ta có: \(a,b,c>1:\)

\(\Rightarrow\left\{{}\begin{matrix}a-1>0\\b-1>0\\c-1>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a-1}>0\\\dfrac{1}{b-1}>0\\\dfrac{1}{c-1}>0\end{matrix}\right.\)

\(\Rightarrow\dfrac{1}{a-1}+\dfrac{1}{b-1}+\dfrac{1}{c-1}>0\)

Quay lại bài toán, theo giả thiết ta có:

\(\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}-\dfrac{1}{a-1}-\dfrac{1}{b-1}-\dfrac{1}{c-1}=1\)

\(\Leftrightarrow\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c-1}=1+\dfrac{1}{a-1}+\dfrac{1}{b-1}+\dfrac{1}{c-1}\)

\(\Leftrightarrow\dfrac{1}{a+1}+\dfrac{1}{b+1}-\dfrac{1}{c+1}< 1\)(đpcm)

\(A=\dfrac{3a}{b+c}+\dfrac{4b}{c+a}+\dfrac{3c}{a+b}\)

Đặt \(\left\{{}\begin{matrix}a+b=x\\b+c=y\\c+a=z\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{x+z-y}{2}\\b=\dfrac{x+y-z}{2}\\c=\dfrac{y+z-x}{2}\end{matrix}\right.\)

\(\Rightarrow2A=\dfrac{3\left(x+z-y\right)}{y}+\dfrac{4\left(x+y-z\right)}{z}+\dfrac{5\left(y+z-x\right)}{x}\)

\(=-12+\left(\dfrac{3x}{y}+\dfrac{5y}{x}\right)+\left(\dfrac{4y}{z}+\dfrac{3z}{y}\right)+\left(\dfrac{4x}{z}+\dfrac{5z}{x}\right)\ge-12+2\sqrt{15}+4\sqrt{3}+4\sqrt{5}\)

Dấu = xảy ra khi \(\left\{{}\begin{matrix}\dfrac{3x}{y}=\dfrac{5y}{x}\\\dfrac{4y}{z}=\dfrac{3z}{y}\\\dfrac{4x}{z}=\dfrac{5z}{x}\end{matrix}\right.\)

Ace Legona , Xuân Tuấn Trịnh , Nguyễn Huy Tú ,Hung nguyen,..... làm giúp đi , còn nhiều lắm luôn ......

thôi bí

từ giả thiết: \(x^2y+x+1\le y\Leftrightarrow y-x^2y-x-1\ge0\)

\(\Leftrightarrow y\left(1-x\right)\left(1+x\right)-\left(1+x\right)\ge0\)

\(\left(1+x\right)\left(y-xy-1\right)\ge0\)

x>0 ,\(\Rightarrow y-xy-1\ge0\Leftrightarrow1-x-\dfrac{1}{y}\ge0\)

\(1\ge x+\dfrac{1}{y}\ge2\sqrt{\dfrac{x}{y}}\Leftrightarrow\dfrac{x}{y}\le\dfrac{1}{4}\)

\(P=\dfrac{xy}{\left(x+y\right)^2}=\dfrac{1}{\dfrac{x}{y}+\dfrac{y}{x}+2}=\dfrac{1}{\dfrac{y}{x}+\dfrac{16x}{y}-\dfrac{15x}{y}+2}\le\dfrac{1}{10-\dfrac{15x}{y}}\le\dfrac{1}{10-\dfrac{15}{4}}=\dfrac{4}{25}\)dấu = xảy ra khi \(x=\dfrac{1}{y}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{2};y=2\)

2) Do \(\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}=2\\\)\(\Rightarrow\dfrac{1}{a+1}=2-\left(\dfrac{1}{b+1}+\dfrac{1}{c+1}\right)\)

=\(\dfrac{b}{b+1}+\dfrac{c}{c+1}\)

Áp dụng BĐT AM-GM ta có

\(\dfrac{1}{a+1}=\dfrac{b}{b+1}+\dfrac{c}{c+1}\) \(\ge\)\(2\sqrt{\dfrac{bc}{\left(b+1\right)\left(c+1\right)}}\)

Tương tự ta được

\(\dfrac{1}{b+1}\ge2\sqrt{\dfrac{ca}{\left(c+1\right)\left(a+1\right)}}\)

\(\dfrac{1}{c+1}\ge2\sqrt{\dfrac{ab}{\left(a+1\right)\left(b+1\right)}}\)

Nhân vế theo vế của 3 BĐT cùng chiều ta được

\(\dfrac{1}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\)\(\ge\dfrac{8abc}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\)

\(\Rightarrow abc\le\dfrac{1}{8}\)

Đẳng thức xảy ra\(\Leftrightarrow a=b=c=\dfrac{1}{2}\)