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C7, \(\dfrac{\left(b+c\right)\left(a^2+bc\right)}{b^2+bc+c^2}\ge\dfrac{\left(2\sqrt{bc}\right).\left(2a\sqrt{bc}\right)}{3\sqrt[3]{b^2.bc.c^2}}=\dfrac{4abc}{3abc}=\dfrac{4}{3}\left(1\right)\)
tương tự \(=>\dfrac{\left(a+c\right)\left(b^2+Ac\right)}{a^2+ac+c^2}\ge\dfrac{4}{3}\left(2\right)\)
\(=>\dfrac{\left(b+a\right)\left(c^2+ba\right)}{a^2+ab+b^2}\ge\dfrac{4}{3}\left(3\right)\)
cộng vế (1)(2)(3) \(=>P\ge4\)
dấu"=" xảy ra<=>a=b=c=1
\(\left\{{}\begin{matrix}x+2y=2\\mx-y=m\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+2y=2\\2mx-2y=2m\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2mx+x=2+2m\\x+2y=2\end{matrix}\right.\\ \left\{{}\begin{matrix}x\left(2m+1\right)=2\left(m+1\right)\\x+2y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2\left(m+1\right)}{2m+1}\\\dfrac{2\left(m+1\right)}{2m+1}+2y=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2\left(m+1\right)}{2m+1}\\2m+2+4my+2y=4m+2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2\left(m+1\right)}{2m+1}\\y\left(4m+2\right)=2m\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2\left(m+1\right)}{2m+1}\\y=\dfrac{2m}{4m+2}\end{matrix}\right.\\ thay.....x,y....vào....ta.....được\\ \dfrac{2\left(m+1\right)}{2m+1}+\dfrac{2m}{4m+2}=1\\ \Leftrightarrow\dfrac{4\left(m+1\right)}{4m+2}+\dfrac{2m}{4m+2}=\dfrac{4m+2}{4m+2}\\ \Rightarrow4m+4+2m=4m+2\\ \Leftrightarrow2m=-2\\ \Leftrightarrow m=-1\\ vậy...m=-1...thì...tm\) \(thay....m=3...vào...ta...có...hpt:\\ \left\{{}\begin{matrix}x+2y=2\\3x-y=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+2y=2\\6x-2y=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}7x=8\\x+2y=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{8}{7}\\y=\dfrac{3}{7}\end{matrix}\right.\)
\(thay...m=3....ta...có:\\ \left\{{}\begin{matrix}x+2y=2\\3x-y=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+2y=2\\6x-2y=6\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}7x=8\\x+2y=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{8}{7}\\y=\dfrac{3}{7}\end{matrix}\right.\\ vậy...với..m=3...thì...hệ....phương....trình....có...nghiệm...duy...nhất\left\{x=\dfrac{8}{7};y=\dfrac{3}{7}\right\}\)
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Anh đừng buồn bởi đây là những câu hỏi 0.5 đ ở cuối đề thi và có thể mấy bạn học sinh khá hay giỏi mới làm được đó là lớp 9 còn anh lớp 10 thì .... chắc quyên thôi ...
Câu 1:
PT \(\Leftrightarrow\left(x-3\right)\left(x-2\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
Vậy \(S=\left\{2;3\right\}\)
Câu 2:
a) HPT \(\Leftrightarrow\left\{{}\begin{matrix}2x+4y=10\\3x+4y=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=\dfrac{5-x}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=5\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(-5;5\right)\)
b) HPT \(\Leftrightarrow\left\{{}\begin{matrix}5x=10\\y=2x-7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-3\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(2;-3\right)\)
Câu 5:
Đặt \(P=\dfrac{1}{x^2+y^2}+\dfrac{1}{xy}=\left(\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\right)+\dfrac{1}{2xy}\)
Áp dụng bất đẳng thức Bunhiacopxki dạng phân thức ta có:
\(\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\ge\dfrac{4}{x^2+y^2+2xy}=\dfrac{4}{\left(x+y\right)^2}\ge4\)
Áp dụng bất đẳng thức Cosi ta có:
\(2xy\le\dfrac{\left(x+y\right)^2}{2}\le\dfrac{1}{2}\Rightarrow\dfrac{1}{2xy}\ge2\)
\(\Rightarrow P\ge6\)
Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)
5.
Không mất tính tổng quát, giả sử \(c=min\left\{a;b;c\right\}\Rightarrow0\le c\le1\Rightarrow1-\dfrac{c}{2}>0\)
\(P=bc+ca+ab\left(1-\dfrac{c}{2}\right)\ge0\)
\(P_{min}=0\) khi \(\left(a;b;c\right)=\left(0;0;3\right)\) và các hoán vị
\(P=c\left(a+b\right)+ab\left(1-\dfrac{c}{2}\right)\le c\left(3-c\right)+\dfrac{\left(a+b\right)^2}{4}\left(1-\dfrac{c}{2}\right)\)
\(P\le3c-c^2+\dfrac{\left(3-c\right)^2}{4}\left(1-\dfrac{c}{2}\right)\)
\(P\le\dfrac{5}{2}-\dfrac{c^3}{8}+\dfrac{3c}{8}-\dfrac{1}{4}=\dfrac{5}{2}-\dfrac{1}{8}\left(c-1\right)^2\left(c+2\right)\le\dfrac{5}{2}\)
\(P_{max}=\dfrac{5}{2}\) khi \(a=b=c=1\)
Cách 2 phần tìm max bài 5:
Áp dụng BĐT: \(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\)
\(\Leftrightarrow abc\ge\left(3-2a\right)\left(3-2b\right)\left(3-2c\right)\)
\(\Leftrightarrow abc\ge-8abc+12\left(ab+bc+ca\right)-27\)
\(\Leftrightarrow3abc+27\ge12\left(ab+bc+ca\right)-6abc\)
\(\Leftrightarrow ab+bc+ca-\dfrac{1}{2}abc\le\dfrac{abc}{4}+\dfrac{9}{4}\le\dfrac{1}{4}.\left(\dfrac{a+b+c}{3}\right)^3+\dfrac{9}{4}=\dfrac{5}{2}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Bài 2.
Ta có:a2+b2+c2+2abc+1≥2(ab+bc+ca)
⇔ (a2-2ab+b2)+(c2-2c+1)+(2c+2abc-2bc-2ca)≥0
⇔ (a-b)2+(c-1)2+2c(a-1)(b-1)≥0
Vì a,b,c≥0 ⇒ 2c(a-1)(b-1)≥0
Dấu "=" xảy ra ⇔ a=b=c=1
C25: b5: Sử dụng kĩ thuật Côsi ngược dấu:
Ta có: \(\dfrac{1}{2bc^2+1}=1-\dfrac{2bc^2}{2bc^2+1}\ge1-\dfrac{2bc^2}{3\sqrt[3]{b^2c^4}}=1-\dfrac{2\sqrt[3]{bc^2}}{3}\)
Cmtt ta được: \(\dfrac{1}{2ca^2+1}\ge1-\dfrac{2\sqrt[3]{ca^2}}{3};\dfrac{1}{2ab^2+1}\ge1-\dfrac{2\sqrt[3]{ab^2}}{3}\)
\(\Rightarrow VT\ge1-\dfrac{2\sqrt[3]{bc^2}}{3}+1-\dfrac{2\sqrt[3]{ca^2}}{3}+1-\dfrac{2\sqrt[3]{ab^2}}{3}=3-2\left(\dfrac{\sqrt[3]{bc^2}+\sqrt[3]{ca^2}+\sqrt[3]{ab^2}}{3}\right)\)
Ta có: Theo bđt Côsi:
\(\sqrt[3]{bc^2}=\sqrt[3]{b.c.c}\le\dfrac{b+c+c}{3}=\dfrac{b+2c}{3}\)
\(\sqrt[3]{ca^2}=\sqrt[3]{c.a.a}\le\dfrac{c+a+a}{3}=\dfrac{c+2a}{3}\)
\(\sqrt[3]{ab^2}=\sqrt[3]{a.b.b}\le\dfrac{a+b+c}{3}=\dfrac{a+2b}{3}\)
\(\Rightarrow\sqrt[3]{bc^2}+\sqrt[3]{ca^2}+\sqrt[3]{ab^2}\le\dfrac{b+2c+c+2a+a+2b}{3}=a+b+c=3\)
\(\Rightarrow3-2\left(\dfrac{\sqrt[3]{bc^2}+\sqrt[3]{ca^2}+\sqrt[3]{ab^2}}{3}\right)=1\)
\(\Rightarrow VT\ge1\)
Dấu ''='' xảy ra khi a=b=c=1
các bạn khác k làm thì đừng cmt vô đây mấy bài của các bạn giải bị trôi
1, \(\)BDT AM-GM
\(=>\sqrt{a^2+b^2}\ge\sqrt{2ab}\left(1\right)\)
tương tuqj \(=>\sqrt{b^2+c^2}\ge\sqrt{2bc}\left(2\right)\)
\(=>\sqrt{c^2+a^2}\ge\sqrt{2ac}\left(3\right)\)
cộng vế (1)(2)(3)
\(=>Vt=\sqrt{2}\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)=\sqrt{2021}\)
\(=>\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=\dfrac{\sqrt{2021}}{\sqrt{2}}\)
\(=>\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\le a+b+c\)\(=>a+b+C\ge\dfrac{\sqrt{2021}}{\sqrt{2}}\)
đặt \(P=\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\)
\(=>P\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}=\dfrac{1}{2}.\dfrac{\sqrt{2021}}{\sqrt{2}}\)
dấu"=" xảy ra<=>\(a=b=c=\dfrac{\sqrt{2021}}{3\sqrt{2}}\)