tìm Min của A=\(\dfrac{a^4}{\left(b-1\right)^3}+\dfrac{b^4}{\left(a-1\right)^3}\) biết a,b >1 và a+b≤4
+) Tìm min
\(E=\dfrac{1+\sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}}{xy+yz+zx}\)
+) Tìm max và min
\(F=\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\)
Trong đó a,b,c>0 và \(min\left\{a,b,c\right\}\ge\dfrac{1}{4}max\left\{a,b,c\right\}\)
Tìm Min:
a) A=\(\left|x-\dfrac{1}{2}\right|\)
b) B=\(\left|x+\dfrac{3}{4}\right|\) +2
`a)A=|x-1/2|>=0`
Dấu "=" xảy ra khi `x-1/2=0<=>x=1/2`
`b)B=|x+3/4|+2`
`|x+3/4|>=0`
`=>|x+3/4|+2>=2`
Hay `A>=2`
Dấu "=" xảy ra khi `x+3/4=0<=>x=-3/4`.
Cho a,b > 1 và a + b \(\ge\) 4 . Tìm GTNN của P = \(\dfrac{a^4}{\left(b-1\right)^3}+\dfrac{b^4}{\left(a-1\right)^3}\)
\(\dfrac{a^4}{\left(b-1\right)^3}+\dfrac{256}{81}\left(b-1\right)+\dfrac{256}{81}\left(b-1\right)+\dfrac{256}{81}\left(b-1\right)\ge4\sqrt[4]{\dfrac{a^4.256^3.\left(b-1\right)^3}{81^3\left(b-1\right)^3}}=\dfrac{256a}{27}\)
\(\dfrac{b^4}{\left(a-1\right)^3}+\dfrac{256}{81}\left(a-1\right)+\dfrac{256}{81}\left(a-1\right)+\dfrac{256}{81}\left(a-1\right)\ge\dfrac{256b}{27}\)
Cộng vế với vế:
\(P+\dfrac{256}{27}\left(a+b\right)-\dfrac{512}{27}\ge\dfrac{256}{27}\left(a+b\right)\)
\(\Rightarrow P\ge\dfrac{512}{27}\)
Dấu "=" xảy ra khi \(a=b=4\)
Cho a,b >0 và \(a+b\le3\). Tìm min
\(K=\dfrac{1}{a^2+b^2-2\left(a+b\right)+2}+\dfrac{1}{ab-\left(a+b\right)+1}+4\left(ab-a-b\right)\)
Đề bài sai, bạn có thể thử kiểm tra với \(a=1.0001\) và \(b=0.9999\)
Cho:
\(A=\left(\dfrac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}\right):\left(\dfrac{1}{a}-\dfrac{1}{b}\right)\)
Biết \(2\sqrt{a}-\sqrt{b}=4\sqrt{ab}\). Tìm min A
1. Cho a,b,c t/m: \(\left\{{}\begin{matrix}a\ge\dfrac{4}{3}\\b\ge\dfrac{4}{3}\\c\ge\dfrac{4}{3}\end{matrix}\right.\) và \(a+b+c=6\)
\(CMR:\dfrac{a}{a^2+1}+\dfrac{b}{b^2+1}+\dfrac{c}{c^2+1}\ge\dfrac{6}{5}\)
2. Cho x,y >0 t/m: \(2x+3y-13\ge0\)
Tìm min \(P=x^2+3x+\dfrac{4}{x}+y^2+\dfrac{9}{y}\)
Xét \(\dfrac{a}{a^2+1}+\dfrac{3\left(a-2\right)}{25}-\dfrac{2}{5}=\dfrac{a}{a^2+1}+\dfrac{3a-16}{25}=\dfrac{\left(3a-4\right)\left(a-2\right)^2}{25\left(a^2+1\right)}\ge0\)
\(\Rightarrow\dfrac{a}{a^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(a-2\right)}{25}\)
CMTT \(\Rightarrow\left\{{}\begin{matrix}\dfrac{b}{b^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(b-2\right)}{25}\\\dfrac{c}{c^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(c-2\right)}{25}\end{matrix}\right.\)
Cộng vế theo vế:
\(\Rightarrow VT\ge\dfrac{2}{5}+\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{3\left(a-2\right)+3\left(b-2\right)+3\left(c-2\right)}{25}\ge\dfrac{6}{5}-\dfrac{3\left(a+b+c-6\right)}{25}=\dfrac{6}{5}\)
Dấu \("="\Leftrightarrow a=b=c=2\)
Cho 3 số thực \(a,b,c\ge0\), \(a^2+b^2+c^2=4\left(a+b+c\right)-2bc\).
Tìm min \(P=8\left(c+b\right)+a^2+\dfrac{2025}{\sqrt{2a+2b+1}}+\dfrac{2025}{\sqrt{2c+1}}\)
\(4\left(a+b+c\right)=a^2+\left(b+c\right)^2\ge\dfrac{1}{2}\left(a+b+c\right)^2\)
\(\Rightarrow a+b+c\le8\)
\(a^2+16-16\ge8a-16\)
\(\Rightarrow P\ge8\left(a+b+c\right)-16+\dfrac{8100}{\sqrt{2a+2b+1}+\sqrt{2c+1}}\)
\(\Rightarrow P\ge8\left(a+b+c\right)-16+\dfrac{48600}{6\sqrt{2a+2b+1}+6\sqrt{2c+1}}\)
\(\Rightarrow P\ge8\left(a+b+c\right)-16+\dfrac{24300}{a+b+c+10}\)
\(\Rightarrow P\ge8\left(a+b+c+10+\dfrac{324}{a+b+c+10}\right)+\dfrac{21708}{a+b+c+10}-96\)
\(\Rightarrow P\ge16.\sqrt{324}+\dfrac{21708}{18}-96=1398\)
Dấu "=" xảy ra tại \(\left(a;b;c\right)=\left(4;0;4\right)\)
Cho a và b là các số dương. Đơn giản các biểu thức sau :
a) \(\dfrac{a^{\dfrac{4}{3}}\left(a^{-\dfrac{1}{3}}+a^{\dfrac{2}{3}}\right)}{a^{\dfrac{1}{4}}\left(a^{\dfrac{3}{4}}+a^{-\dfrac{1}{4}}\right)}\)
b) \(\dfrac{a^{\dfrac{1}{3}}\sqrt{b}+b^{\dfrac{1}{3}}\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}}\)
c) \(\left(\sqrt[3]{a}+\sqrt[3]{b}\right)\left(a^{\dfrac{2}{3}}+b^{\dfrac{2}{3}}-\sqrt[3]{ab}\right)\)
d) \(\left(a^{\dfrac{1}{3}}+b^{\dfrac{1}{3}}\right):\left(2+\sqrt[3]{\dfrac{a}{b}}+\sqrt[3]{\dfrac{b}{a}}\right)\)
a)
\(A=\dfrac{a^{\dfrac{4}{3}}\left(a^{-\dfrac{1}{3}}+a^{\dfrac{2}{3}}\right)}{a^{\dfrac{1}{4}}\left(a^{\dfrac{3}{4}}+a^{-\dfrac{1}{4}}\right)}=\dfrac{a^{\left(\dfrac{4}{3}-\dfrac{1}{3}\right)+}a^{\left(\dfrac{4}{3}+\dfrac{2}{3}\right)}}{a^{\left(\dfrac{1}{4}+\dfrac{3}{4}\right)}+a^{\left(\dfrac{1}{4}-\dfrac{1}{4}\right)}}=\dfrac{a+a^2}{a+1}=\dfrac{a\left(a+1\right)}{a+1}\)
\(a>0\Rightarrow a+1\ne0\) \(\Rightarrow A=a\)
rút gọn \(\dfrac{1}{2\left(a+b\right)^3}\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}\right)+\dfrac{3}{2\left(a+b\right)^4}\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}\right)+\dfrac{3}{\left(a+b\right)^5}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
cái này tương tự này, do dài quá nên ngại làm, bn tham khảo nhé Câu hỏi của Thiên An - Toán lớp 9 - Học toán với OnlineMath