Cho \(\hept{\begin{cases}a;b>0\\a\ne b\end{cases}}\)thỏa mãn \(3a^2+8b^3=20\)Tìm Min:
\(A=\frac{4}{a^2}+\frac{4}{b^2}+\frac{1}{\left(a-b\right)^2}\)
1, Cho \(\hept{\begin{cases}a,b>0\\a^2+b^2=1\end{cases}.}\)Tìm min A= \(\left(1+a\right)\left(1+\frac{1}{b}\right)+\left(1+b\right)\left(1+\frac{1}{a}\right)\)
2, Cho \(\hept{\begin{cases}a^2+2b^2\le3c^2\\a,b,c>0\end{cases}}\).Chứng minh : \(\frac{1}{a}+\frac{2}{b}\ge\frac{3}{c}\)
1,
\(A=1+a+\frac{1}{b}+\frac{a}{b}+1+b+\frac{1}{a}+\frac{b}{a}\)
\(\ge1+1+2\sqrt{\frac{a}{b}.\frac{b}{a}}+a+b+\frac{a+b}{ab}=4+a+b+\frac{4\left(a+b\right)}{\left(a+b\right)^2}=4+a+b+\frac{4}{a+b}\)
lại có \(\left(1+1\right)\left(a^2+b^2\right)\ge\left(a+b\right)^2\Rightarrow a+b\le\sqrt{2}\)
\(4+a+b+\frac{4}{a+b}=4+\left(a+b+\frac{2}{a+b}\right)+\frac{2}{a+b}\ge4+2\sqrt{2}+\sqrt{2}=4+3\sqrt{2}\)
\(\Rightarrow A\ge4+3\sqrt{2}\)
câu 2
ta có:\(\left(2b^2+a^2\right)\left(2+1\right)\ge\left(2b+a\right)^2\Rightarrow3c\ge a+2b\)
\(\frac{1}{a}+\frac{2}{b}=\frac{1}{a}+\frac{4}{2b}\ge\frac{9}{a+2b}\ge\frac{9}{3c}=\frac{3}{c}\left(Q.E.D\right)\)
Cho đề \(\hept{\begin{cases}2y^2-x^2=1\\2\left(x^3-y\right)=y^3-x\end{cases}\Leftrightarrow}\)\(\hept{\begin{cases}2\left(y^2+1\right)-\left(x^2+1\right)=2\\x\left(2x^2+1\right)-y\left(y^2+2\right)=0\end{cases}}\)
đặt \(a=y^2+1,b=x^2+1\)
\(\Leftrightarrow\hept{\begin{cases}2a-b=2\\x\left(2b-1\right)-y\left(a+1\right)=0\end{cases}\Leftrightarrow\hept{\begin{cases}b=2a-2\\x\left(4a-5\right)-ya-y=0\end{cases}}}\Leftrightarrow\hept{\begin{cases}b=2a-2\\a=\frac{5x+y}{4x-y}\end{cases}\Leftrightarrow\hept{\begin{cases}b=\frac{2x+4y}{4x-y}\\a=\frac{5x+y}{4x-y}\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}y^2+1=\frac{5x+y}{4x-y}\left(1\right)\\x^2+1=\frac{2x+4y}{4x-y}\left(2\right)\end{cases}}\)
pt(1)-pt(2),ta dc:\(\left(x-y\right)\left(\frac{3}{4x-y}+x+y\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=y\left(3\right)\\\frac{3}{4x-y}+x+y=0\left(4\right)\end{cases}}\)
CM:PT (4) vô nghiệm giúp mình nha!Và xem lại nếu mình có lm sai hay thiếu đk j đó hãy chỉ giúp mình nha!!!Hoặc pt(4) có nghiệm thì hãy giải giúp mình luôn nha!Thanks
Cho\(\hept{\begin{cases}a,b,c>0\\abc>1\end{cases}CMR:}2\left(a^2+b^2+c^2\right)+4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge7\left(a+b+c\right)-3\)
\(P\left(x\right)=x^2+bx+c=\left(x+\frac{b}{2}\right)^2-\frac{b^2}{4}+c\ge c-\frac{b^2}{4}\)
Có \(P\left(x\right)_{min}=-1\) tại x=2 => \(\hept{\begin{cases}2+\frac{b}{2}=0\\c-\frac{b^2}{4}=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}b=-4\\c=3\end{cases}}\)
lớp 1 ????
mà đây cũng đâu phải câu hỏi đâu ??
Đây có phải là câu hỏi đâu bạn
Phùng Minh Quân làm đúng rồi CHÚC MỪNG
giúp mình với ạ , mình đang cần gấp !!!
a,\(\hept{\begin{cases}3\left(x+1\right)+2\left(x+2y\right)=4\\4\left(x+1\right)-\left(x+2y\right)=9\end{cases}}\)
b, \(\hept{\begin{cases}x+\frac{1}{y}=\frac{-1}{2}\\2x-\frac{3}{y}=\frac{-7}{2}\end{cases}}\)
c,\(\hept{\begin{cases}\frac{x+2}{x+1}+\frac{2}{y-2}=6\\\frac{5}{x+1}-\frac{1}{y-2}=3\end{cases}}\)
c) Ta có: \(\left\{{}\begin{matrix}\dfrac{x+2}{x+1}+\dfrac{2}{y-2}=6\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x+1}+\dfrac{10}{y-2}=25\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{11}{y-2}=22\\\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-2=\dfrac{1}{2}\\\dfrac{1}{x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=1\\y-2=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{5}{2}\end{matrix}\right.\)
Tìm a, b, c thỏa mãn:
\(\hept{\begin{cases}a^4-2b=\frac{-1}{2}\\b^4-2c=\frac{-1}{2}\\c^4-2a=\frac{-1}{2}\end{cases}}\)
GIẢI hpt:
\(a,\hept{\begin{cases}\frac{1}{\sqrt{x}}+\sqrt{2.\frac{1}{y}}=2\\\frac{1}{\sqrt{y}}+\sqrt{2.\frac{1}{x}}=2\end{cases}}\)
\(b,\hept{\begin{cases}x+y+2=4\\2xy-x^2=16\end{cases}}\)
\(c,\hept{\begin{cases}x\left(x-1\right)\left(x-2y\right)=0\\\frac{1}{x}-\frac{1}{y}=\frac{4}{3}\end{cases}}\)
cho a,b,c>0 thỏa mãn \(\hept{\begin{cases}a+b+c=2\\a^2+b^2+c^2=2\end{cases}}\)
Tính \(A=a\sqrt{\frac{\left(1+b^2\right)\left(1+c^2\right)}{1+a^2}}+b\sqrt{\frac{\left(1+a^2\right)\left(1+c^2\right)}{1+b^2}}+c\sqrt{\frac{\left(1+a^2\right)\left(1+b^2\right)}{1+c^2}}\)
Ta có\(ab+bc+ca=\frac{\left(a+b+c\right)^2-\left(a^2+b^2+c^2\right)}{2}=1\)
Thay 1=ab+bc+ca vào, ta có
\(a\sqrt{\frac{\left(1+b^2\right)\left(1+c^2\right)}{1+a^2}}=a\sqrt{\frac{\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(c+b\right)}{\left(a+b\right)\left(a+c\right)}}=a\left(b+c\right)\)
Tương tự rồi cộng lại, ta có
A=2(ab+bc+ca)=2
^_^
Cho \(\hept{\begin{cases}a,b,c>0\\a+b+c=3\end{cases}}\)Tìm min \(A=3\left(a+b+c\right)+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
\(A\ge3\left(a+b+c\right)+\frac{9}{a+b+c}=3.3+\frac{9}{3}=12\)
\(A_{min}=12\) khi \(a=b=c=1\)
Ta cần chứng minh: \(3a+\frac{1}{a}\ge2a+2\Leftrightarrow3a+\frac{1}{a}-4\ge2\left(a-1\right)\)
\(\Leftrightarrow\frac{3a^2-4a+1}{a}-2\left(a-1\right)\ge0\Leftrightarrow\left(a-1\right)\left(\frac{3a-1}{a}-2\right)\ge0\Leftrightarrow\frac{\left(a-1\right)^2}{a}\)(đúng)
Tương tự: \(3b+\frac{1}{b}\ge2b+2;3c+\frac{1}{c}\ge2c+2\)
Cộng theo vế: \(A\ge2\left(a+b+c\right)+6=12\)
Dấu bằng xảy ra khi a=b=c=1
\(A=3\left(a+b+c\right)+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3^2+\frac{9}{a+b+c}=12.\)(BĐT Bunhia)
Dấu "=" xra khi a=b=c=1