1/CMR
a/\(x^4-2x^3+2x^2-2x+1\ge0\forall x\in R\)
b/cho \(a\ge0,b\ge2,a+b+c=3\). CMR : \(a^2+b^2+c^2\le5\)
c/cho a,b,c >0 . CMR : \(\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}\ge4\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)\)
2/ cho \(x,y\ge0,x+y=1\). tìm GTLN,GTNN của A =\(x^2+y^2\)
3/ cho x,y>0 .tìm GTNN của B= \(\frac{\left(x+y\right)^2}{x^2+y^2}+\frac{\left(x+y\right)^2}{xy}\)
➤ Bài 1 : Cho đa thức :
\(f\left(x\right)=x\left(\frac{x^{2013}}{3}-\frac{x^{2014}}{5}+\frac{x^{2015}}{7}+\frac{x^2}{2}\right)-\left(\frac{x^{2014}}{3}-\frac{x^{2015}}{5}+\frac{x^{2016}}{7}+\frac{x^2}{2}\right)\).
a/ Tìm bậc của đa thức f(x).
b/ Chứng minh : Đa thức f(x) luôn nhận giá trị nguyên với \(\forall x\)\(\in \mathbb{Z}\)
➤ Bài 2 : Cho 3 số ɑ, b, c thoả mãn :
\(\frac{a}{2016}=\frac{b}{2017}=\frac{c}{2018}\)
Tính \(M=4\left(a-b\right)\left(b-c\right)\left(c-a\right)^2\).
1. \(\left\{{}\begin{matrix}a,b,c>0\\a+b+c=1\end{matrix}\right.\). Cmr: \(\frac{ab}{\sqrt{\left(1-c\right)^2\left(1+c\right)}}+\frac{bc}{\sqrt{\left(1-a\right)^2\left(1+a\right)}}+\frac{ca}{\sqrt{\left(1-b\right)^3\left(1+b\right)}}\le\frac{3\sqrt{2}}{8}\)
2. \(\left\{{}\begin{matrix}a,b,c>0\\a+b+c\le1\end{matrix}\right.\). Cmr: \(\frac{1}{a^2+b^2+c^2}+\frac{1}{ab\left(a+b\right)}+\frac{1}{bc\left(b+c\right)}+\frac{1}{ac\left(a+c\right)}\ge\frac{87}{2}\)
3. \(\left\{{}\begin{matrix}a,b,c>0\\ab+bc+ca=2abc\end{matrix}\right.\). Cmr: \(\frac{1}{a\left(2a-1\right)^2}+\frac{1}{b\left(2b-1\right)^2}+\frac{1}{c\left(2c-1\right)^2}\ge\frac{1}{2}\)
4. \(\left\{{}\begin{matrix}x,y,z>0\\x+y+z=2015\end{matrix}\right.\). Tìm min \(A=\frac{x^4+y^4}{x^3+y^3}+\frac{y^4+z^4}{y^3+z^3}+\frac{z^4+x^4}{z^2+x^2}\)
Mn giúp mk với ạ! Thanks nhiều
Mới nghĩ ra 3 câu:
a/ \(\frac{ab}{\sqrt{\left(1-c\right)^2\left(1+c\right)}}=\frac{ab}{\sqrt{\left(a+b\right)^2\left(1+c\right)}}\le\frac{ab}{2\sqrt{ab\left(1+c\right)}}=\frac{1}{2}\sqrt{\frac{ab}{1+c}}\)
\(\sum\sqrt{\frac{ab}{1+c}}\le\sqrt{2\sum\frac{ab}{1+c}}\)
\(\sum\frac{ab}{1+c}=\sum\frac{ab}{a+c+b+c}\le\frac{1}{4}\sum\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)=\frac{1}{4}\)
c/ \(ab+bc+ca=2abc\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
Đặt \(\left(x;y;z\right)=\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)\Rightarrow x+y+z=2\)
\(VT=\sum\frac{x^3}{\left(2-x\right)^2}\)
Ta có đánh giá: \(\frac{x^3}{\left(2-x\right)^2}\ge x-\frac{1}{2}\) \(\forall x\in\left(0;2\right)\)
\(\Leftrightarrow2x^3\ge\left(2x-1\right)\left(x^2-4x+4\right)\)
\(\Leftrightarrow9x^2-12x+4\ge0\Leftrightarrow\left(3x-2\right)^2\ge0\)
d/ Ta có đánh giá: \(\frac{x^4+y^4}{x^3+y^3}\ge\frac{x+y}{2}\)
\(\Leftrightarrow\left(x-y\right)^2\left(x^2+xy+y^2\right)\ge0\)
Akai Haruma, Nguyễn Ngọc Lộc , @tth_new, @Băng Băng 2k6, @Trần Thanh Phương, @Nguyễn Việt Lâm
Mn giúp e vs ạ! Thanks!
a) Tìm m để pt \(\left(x^2-1\right)\left(x+3\right)\left(x+5\right)=m\) có 4 nghiệm thỏa: \(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}=-1\)
b) Tìm các số \(a,b,c\ge0\)sao cho: \(\left(a^2+b+\frac{3}{4}\right)\left(b^2+a+\frac{3}{4}\right)=\left(2a+\frac{1}{2}\right)\left(2b+\frac{1}{2}\right)\)
Bài 9. Rút gọn các phân thức sau
a) \(\frac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-bc-ca}\)
d) \(\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{a^4\left(b^2-c^2\right)+b^4\left(c^2-a^2\right)+c^4\left(a^2-b^2\right)}\)
e) \(\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{ab^2-ac^2-b^3+bc^2}\)
f) \(\frac{x^{24}+x^{20}+x^{16}+...+x^4+1}{x^{26}+x^{24}+x^{22}+...+x^2+1}\)
a: \(=\dfrac{\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc}{a^2+b^2+c^2-ab-ac-bc}\)
\(=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)}{a^2+b^2+c^2-ab-ac-bc}\)
=a+b+c
e: \(=\dfrac{a^2b-a^2c+b^2c-b^2a+c^2\left(a-b\right)}{a\left(b^2-c^2\right)-b\left(b^2-c^2\right)}\)
\(=\dfrac{ab\left(a-b\right)+c\left(b-a\right)\left(b+a\right)+c^2\left(a-b\right)}{\left(b-c\right)\left(b+c\right)\left(a-b\right)}\)
\(=\dfrac{\left(a-b\right)\left(ab-ac-bc+c^2\right)}{\left(b-c\right)\left(b+c\right)\left(a-b\right)}\)
\(=\dfrac{a\left(b-c\right)-c\left(b-c\right)}{\left(b-c\right)\left(b+c\right)}=\dfrac{a-c}{b+c}\)
ta có:
(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc\(\ge\left(a+b+c\right)\left(ab+bc+ca\right)-\frac{1}{9}\left(a+b+c\right)\left(ab+bc+ca\right)=\frac{8}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\)
\(\frac{x}{x+yz}+\frac{y}{y+zx}+\frac{z}{z+xy}=\frac{x}{\left(x+y\right)\left(x+z\right)}+\frac{y}{\left(y+x\right)\left(y+z\right)}+\frac{z}{\left(z+x\right)\left(z+y\right)}=\frac{2\left(xy+yz+zx\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\le\frac{9}{4\left(xy+yz+zx\right)}=\frac{9}{4}\)
1, Cho a + b = 2
Tính a2 + b2 + 6ab
2, Tìm a, b sao cho a2 + b2 - ab - a - b + 1 = 0
3, Cho x + y = x2 + y2 = x3 + y3
Tìm x, y
4, Cho ab + bc + ca = 1
Rút gọn: P = \(\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}-\frac{2\left(a+b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
5, Cho P = x3 + y3 + 3xy là số nguyên tố, x và y \(\in N\). Tìm x,y
Bài 1: Chưa đủ dữ kiện để tính. Từ $a+b=2$ bạn chỉ có thể tính $a^2+b^2+2ab$
Bài 2:
\(a^2+b^2-ab-a-b+1=0\)
\(\Leftrightarrow 2a^2+2b^2-2ab-2a-2b+2=0\)
\(\Leftrightarrow (a^2-2ab+b^2)+(a^2-2a+1)+(b^2-2b+1)=0\)
\(\Leftrightarrow (a-b)^2+(a-1)^2+(b-1)^2=0\)
Vì \((a-b)^2\geq 0; (a-1)^2\geq 0;(b-1)^2\geq 0, \forall a,b\in\mathbb{R}\)
\(\Rightarrow (a-b)^2+(a-1)^2+(b-1)^2\geq 0\)
Dấu "=" xảy ra khi \((a-b)^2=(a-1)^2=(b-1)^2=0\Leftrightarrow a=b=1\)
Bài 3:
\(x+y=x^3+y^3=(x+y)(x^2-xy+y^2)\)
\(\Leftrightarrow (x+y)(x^2-xy+y^2-1)=0\)
\(\Rightarrow \left[\begin{matrix} x+y=0\\ x^2-xy+y^2-1=0\end{matrix}\right.\).
Nếu $x+y=0$ \(\Rightarrow x^2+y^2=x+y=0\)
Mà \(x^2\geq 0, y^2\geq 0, \forall x,y\) nên để tổng của chúng bằng $0$ thì \(x^2=y^2=0\Leftrightarrow x=y=0\) (thỏa mãn)
Nếu \(x^2-xy+y^2-1=0\)
\(\Leftrightarrow (x^2+y^2)-xy-1=0\)
\(\Leftrightarrow x+y-xy-1=0\)
\(\Leftrightarrow (x-1)(1-y)=0\) \(\Rightarrow \left[\begin{matrix} x=1\\ y=1\end{matrix}\right.\)
\(x=1\Rightarrow 1+y=1+y^2=1+y^3\)
\(\Leftrightarrow y=y^2=y^3\Rightarrow y=0\) hoặc $y=1$
\(y=1\Rightarrow x+1=x^2+1=x^3+1\)
\(\Leftrightarrow x=x^2=x^3\Rightarrow x=0\) hoặc $x=1$.
Vậy $(x,y)=(0,0); (1,0), (0,1), (1,1)$
Bài 4:
\(ab+bc+ac=1\)
\(\Rightarrow a^2+1=a^2+ab+bc+ac=a(a+b)+c(b+a)=(a+c)(a+b)\)
Hoàn toàn tương tự:
\(b^2+1=(b+c)(b+a); c^2+1=(c+a)(c+b)\)
Do đó:
\(P=\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}-\frac{2(a+b+c)}{(a+b)(b+c)(c+a)}=\frac{1}{(a+b)(a+c)}+\frac{1}{(b+c)(b+a)}+\frac{1}{(c+a)(c+b)}-\frac{2(a+b+c)}{(a+b)(b+c)(c+a)}\)
\(=\frac{(b+c)+(a+c)+(a+b)}{(a+b)(b+c)(c+a)}-\frac{2(a+b+c)}{(a+b)(b+c)(c+a)}\)
\(=\frac{2(a+b+c)}{(a+b)(b+c)(c+a)}-\frac{2(a+b+c)}{(a+b)(b+c)(c+a)}=0\)
Vậy $P=0$
1. a) \(\left\{{}\begin{matrix}x,y,z>0\\xyz=1\end{matrix}\right.\). Tìm max \(P=\frac{1}{\sqrt{x^5-x^2+3xy+6}}+\frac{1}{\sqrt{y^5-y^2+3yz+6}}+\frac{1}{\sqrt{z^5-z^2+zx+6}}\)
b) \(\left\{{}\begin{matrix}x,y,z>0\\xyz=8\end{matrix}\right.\). Min \(P=\frac{x^2}{\sqrt{\left(1+x^3\right)\left(1+y^3\right)}}+\frac{y^2}{\sqrt{\left(1+y^3\right)\left(1+z^3\right)}}+\frac{z^2}{\sqrt{\left(1+z^3\right)\left(1+x^3\right)}}\)
c) \(x,y,z>0.\) Min \(P=\sqrt{\frac{x^3}{x^3+\left(y+z\right)^3}}+\sqrt{\frac{y^3}{y^3+\left(z+x\right)^3}}+\sqrt{\frac{z^3}{z^3+\left(x+y\right)^3}}\)
d) \(a,b,c>0;a^2+b^2+c^2+abc=4.Cmr:2a+b+c\le\frac{9}{2}\)
e) \(\left\{{}\begin{matrix}a,b,c>0\\a+b+c=3\end{matrix}\right.\). Cmr: \(\frac{a}{b^3+ab}+\frac{b}{c^3+bc}+\frac{c}{a^3+ca}\ge\frac{3}{2}\)
f) \(\left\{{}\begin{matrix}a,b,c>0\\ab+bc+ca+abc=4\end{matrix}\right.\) Cmr: \(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le3\)
g) \(\left\{{}\begin{matrix}a,b,c>0\\ab+bc+ca+abc=2\end{matrix}\right.\) Max : \(Q=\frac{a+1}{a^2+2a+2}+\frac{b+1}{b^2+2b+2}+\frac{c+1}{c^2+2c+2}\)
Câu c quen thuộc, chém trước:
Ta có BĐT phụ: \(\frac{x^3}{x^3+\left(y+z\right)^3}\ge\frac{x^4}{\left(x^2+y^2+z^2\right)^2}\) \((\ast)\)
Hay là: \(\frac{1}{x^3+\left(y+z\right)^3}\ge\frac{x}{\left(x^2+y^2+z^2\right)^2}\)
Có: \(8(y^2+z^2) \Big[(x^2 +y^2 +z^2)^2 -x\left\{x^3 +(y+z)^3 \right\}\Big]\)
\(= \left( 4\,x{y}^{2}+4\,x{z}^{2}-{y}^{3}-3\,{y}^{2}z-3\,y{z}^{2}-{z}^{3 } \right) ^{2}+ \left( 7\,{y}^{4}+8\,{y}^{3}z+18\,{y}^{2}{z}^{2}+8\,{z }^{3}y+7\,{z}^{4} \right) \left( y-z \right) ^{2} \)
Từ đó BĐT \((\ast)\) là đúng. Do đó: \(\sqrt{\frac{x^3}{x^3+\left(y+z\right)^3}}\ge\frac{x^2}{x^2+y^2+z^2}\)
\(\therefore VT=\sum\sqrt{\frac{x^3}{x^3+\left(y+z\right)^3}}\ge\sum\frac{x^2}{x^2+y^2+z^2}=1\)
Done.
Câu 1 chuyên phan bội châu
câu c hà nội
câu g khoa học tự nhiên
câu b am-gm dựa vào hằng đẳng thử rồi đặt ẩn phụ
câu f đặt \(a=\frac{2m}{n+p};b=\frac{2n}{p+m};c=\frac{2p}{m+n}\)
Gà như mình mấy câu còn lại ko bt nha ! để bạn tth_pro full cho nhé !
Nguyễn Ngọc Lộc , ?Amanda?, Phạm Lan Hương, Akai Haruma, @Trần Thanh Phương, @Nguyễn Việt Lâm,
Giúp em vs ạ! Thanks nhiều ạ
Bài 1: Cho a,b,c dương
a) Tìm Max \(P=\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}\)
b) Tìm Max \(Q=\frac{a^2}{5a^2+\left(b+c\right)^2}+\frac{b^2}{5b^2+\left(c+a\right)^2}+\frac{c^2}{5c^2+\left(a+b\right)^2}\)
Bài 2: Cho x,y,z là các số thực không âm thỏa mãn \(x+y+z=\frac{3}{2}\).Chứng minh rằng \(x+2xy+4xyz\le2\)
Bài 3: Cho a,b thỏa mãn \(\left(x+y\right)^3+4xy\ge2\). Tìm Min \(P=3\left(x^4+y^4+x^2y^2\right)-2\left(x^2+y^2\right)+1\)
Bài 4: Cho x,y,z >0: \(x\left(x+y+z\right)=3yz\). Chứng minh: \(\left(x+y\right)^3+\left(x+z\right)^3+3\left(x+y\right)\left(y+z\right)\left(z+x\right)\le5\left(y+z\right)^3\)
Bài 5:Cho a,b,c không âm thỏa mãn \(a^2+b^2+c^2+abc=4\). CMR: \(a+b+c\le3\)
1a.
\(2P=1-\frac{bc}{2a^2+bc}+1-\frac{ca}{2b^2+ca}+1-\frac{ab}{2c^2+ab}\)
\(\Rightarrow2P=3-\left(\frac{bc}{2a^2+bc}+\frac{ca}{2b^2+ca}+\frac{ab}{2c^2+ab}\right)\)
\(\Rightarrow2P=3-\left(\frac{b^2c^2}{2a^2bc+b^2c^2}+\frac{c^2a^2}{2b^2ca+c^2a^2}+\frac{a^2b^2}{2c^2ab+a^2b^2}\right)\)
\(\Rightarrow2P\le3-\frac{\left(ab+bc+ca\right)^2}{a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)}=3-1=2\)
\(\Rightarrow P\le1\)
\(P_{max}=1\) khi \(a=b=c\)
1b.
\(Q=\frac{a^2}{5a^2+b^2+c^2+2bc}+\frac{b^2}{5b^2+a^2+c^2+2ca}+\frac{c^2}{5c^2+a^2+b^2+2ab}\)
\(Q=\frac{a^2}{a^2+b^2+c^2+\left(2a^2+bc\right)+\left(2a^2+bc\right)}+\frac{b^2}{a^2+b^2+c^2+\left(2b^2+ca\right)+\left(2b^2+ca\right)}+\frac{c^2}{a^2+b^2+c^2+\left(2c^2+ab\right)+\left(2c^2+ab\right)}\)
\(\Rightarrow Q\le\frac{1}{9}\left(\frac{a^2}{a^2+b^2+c^2}+\frac{b^2}{a^2+b^2+c^2}+\frac{c^2}{a^2+b^2+c^2}+2\left(\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}\right)\right)\)
\(\Rightarrow Q\le\frac{1}{9}\left(1+2\left(\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}\right)\right)\)
Theo kết quả câu a ta có:
\(\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}\le1\)
\(\Rightarrow Q\le\frac{1}{9}\left(1+2\right)=\frac{1}{3}\)
\(Q_{max}=\frac{1}{3}\) khi \(a=b=c\)
2.
Do \(x+y+z=\frac{3}{2}\Rightarrow x< \frac{3}{2}< 2\)
Ta có:
\(VT=x+x.4y\left(\frac{1}{2}+z\right)\le x+x\left(\frac{1}{2}+z+y\right)^2=x+x\left(\frac{1}{2}+\frac{3}{2}-x\right)^2\)
\(\Leftrightarrow VT\le x+x\left(2-x\right)^2=x^3-4x^2+5x\)
\(\Leftrightarrow VT\le x^3-4x^2+5x-2+2=\left(x-1\right)^2\left(x-2\right)+2\le2\) (đpcm)
Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(1;\frac{1}{2};0\right)\)
