\(4\times\gamma\div17=0\)
Những câu hỏi liên quan
a) \(5\frac{8}{17}\div x+\frac{-1}{17}\div x+3\frac{1}{17}\div17\frac{1}{3}=\frac{4}{17}\)
b)\(\frac{1}{1\times4}+\frac{1}{4\times7}+\frac{1}{7\times10}+...+\frac{1}{x\times\left(x+3\right)}=\frac{6}{19}\)
a) \(5\frac{8}{17}:x+\frac{-1}{17}:x+3\frac{1}{17}:17\frac{1}{3}=\frac{4}{17}\)
\(\frac{93}{17}:x+\frac{-1}{17}:x+\frac{52}{17}:\frac{52}{3}=\frac{4}{17}\)
\(\left(\frac{93}{17}+\frac{-1}{17}\right):x+\frac{52}{17}.\frac{3}{52}=\frac{4}{17}\)
\(\frac{92}{17}:x+\frac{3}{17}=\frac{4}{17}\)
\(\frac{92}{17}:x=\frac{4}{17}-\frac{3}{17}\)
\(\frac{92}{17}:x=\frac{1}{17}\)
\(x=\frac{92}{17}:\frac{1}{17}\)
\(x=92\)
b) \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x.\left(x+3\right)}=\frac{6}{19}\)
\(\frac{1}{3}.\left(1-\frac{1}{4}\right)+\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{7}\right)+\frac{1}{3}.\left(\frac{1}{7}-\frac{1}{10}\right)+...+\frac{1}{3}.\left(\frac{1}{x}-\frac{1}{x+3}\right)=\frac{6}{19}\)
\(\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{6}{19}\)
\(\frac{1}{3}.\left(1-\frac{1}{x+3}\right)=\frac{6}{19}\)
\(1-\frac{1}{x+3}=\frac{6}{19}:\frac{1}{3}\)
\(1-\frac{1}{x+3}=\frac{18}{19}\)
\(\frac{1}{x+3}=1-\frac{18}{19}\)
\(\frac{1}{x+3}=\frac{1}{19}\)
\(\Rightarrow x+3=19\)
\(\Rightarrow x=19-3\)
\(\Rightarrow x=16\)
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4 tháng 1 2020 lúc 23:15
1. Cho a,b,c 0. Cmr: a) frac{bc}{a^2+2bc}+frac{ca}{b^2+2ca}+frac{ab}{c^2+2ab}le1
b) frac{ab^2}{a^2+2b^2+c^2}+frac{bc^2}{b^2+2c^2+a^2}+frac{ca^2}{c^2+2a^2+b^2}lefrac{a+b+c}{4}
2. Cho x,y,z0;x+frac{y}{3}+frac{z}{5}ge3;frac{y}{3}+frac{z}{5}ge2;frac{z}{5}ge1.MaxPx^2+y^2+z^2
3. Cho x0;yge2;2x+y+xyge6.MinPx^3+y^2
4. Cho 0 alpha beta gamma. Giả sử x,y,z 0 TM zgegamma;frac{x}{alpha}+frac{y}{beta}+frac{z}{gamma}+frac{xyz}{alphabetagamma}4;frac{y}{beta}+frac{z}{gamma}+frac{yz}{betagamma}3.MinPx^3+y^...
Đọc tiếp
1. Cho a,b,c > 0. Cmr: a) \(\frac{bc}{a^2+2bc}+\frac{ca}{b^2+2ca}+\frac{ab}{c^2+2ab}\le1\)
b) \(\frac{ab^2}{a^2+2b^2+c^2}+\frac{bc^2}{b^2+2c^2+a^2}+\frac{ca^2}{c^2+2a^2+b^2}\le\frac{a+b+c}{4}\)
2. Cho \(x,y,z>0;x+\frac{y}{3}+\frac{z}{5}\ge3;\frac{y}{3}+\frac{z}{5}\ge2;\frac{z}{5}\ge1.MaxP=x^2+y^2+z^2\)
3. Cho \(x>0;y\ge2;2x+y+xy\ge6.MinP=x^3+y^2\)
4. Cho \(0< \alpha< \beta< \gamma\). Giả sử x,y,z > 0 TM \(z\ge\gamma;\frac{x}{\alpha}+\frac{y}{\beta}+\frac{z}{\gamma}+\frac{xyz}{\alpha\beta\gamma}=4;\frac{y}{\beta}+\frac{z}{\gamma}+\frac{yz}{\beta\gamma}=3.MinP=x^3+y^3+z^3\)
Vì đã khuya nên não cũng không còn hoạt động tốt nữa, mình làm bài 1 thôi nhé.
Bài 1:
a)
\(2\text{VT}=\sum \frac{2bc}{a^2+2bc}=\sum (1-\frac{a^2}{a^2+2bc})=3-\sum \frac{a^2}{a^2+2bc}\)
Áp dụng BĐT Cauchy-Schwarz:
\(\sum \frac{a^2}{a^2+2bc}\geq \frac{(a+b+c)^2}{a^2+2bc+b^2+2ac+c^2+2ab}=\frac{(a+b+c)^2}{(a+b+c)^2}=1\)
Do đó: \(2\text{VT}\leq 3-1\Rightarrow \text{VT}\leq 1\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
b)
Áp dụng BĐT Cauchy-Schwarz:
\(\text{VT}=\sum \frac{ab^2}{a^2+2b^2+c^2}=\sum \frac{ab^2}{\frac{a^2+b^2+c^2}{3}+\frac{a^2+b^2+c^2}{3}+\frac{a^2+b^2+c^2}{3}+b^2}\leq \sum \frac{1}{16}\left(\frac{9ab^2}{a^2+b^2+c^2}+\frac{ab^2}{b^2}\right)\)
\(=\frac{1}{16}.\frac{9(ab^2+bc^2+ca^2)}{a^2+b^2+c^2}+\frac{a+b+c}{16}(1)\)
Áp dụng BĐT AM-GM:
\(3(ab^2+bc^2+ca^2)\leq (a^2+b^2+c^2)(a+b+c)\)
\(\Rightarrow \frac{1}{16}.\frac{9(ab^2+bc^2+ca^2)}{a^2+b^2+c^2)}\leq \frac{3}{16}(a+b+c)(2)\)
Từ $(1);(2)\Rightarrow \text{VT}\leq \frac{a+b+c}{4}$ (đpcm)
Dấu "=" xảy ra khi $a=b=c$
Bài 2/Áp dụng BĐT Bunyakovski:
\(\left(x^2+y^2+z^2\right)\left(1^2+3^2+5^2\right)\ge\left(x+3y+5z\right)^2\)
\(\Rightarrow P\ge\frac{\left(x+3y+5z\right)^2}{35}\) (*)
Ta có: \(x+3y+5z=x.1+\frac{y}{3}.9+\frac{z}{5}.25\)
\(=\frac{16z}{5}+8\left(\frac{y}{3}+\frac{z}{5}\right)+1\left(\frac{z}{5}+\frac{y}{3}+x\right)\)
\(\ge16+8.2+1.3=35\). Thay vào (*) là xong.
Đẳng thức xảy ra khi x = 1; y =3; z = 5
No choice teen, Akai Haruma, Arakawa Whiter, Phạm Lan Hương, soyeon_Tiểubàng giải, tth, Nguyễn Văn Đạt
giúp em với ạ! Cần gấp lắm! Thanks nhiều!
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Chứng minh đẳng thức:
\(\dfrac{sin\left(\alpha-\beta\right)}{sin\alpha sin\beta}+\dfrac{sin\left(\beta-\gamma\right)}{sin\beta sin\gamma}+\dfrac{sin\left(\gamma-\alpha\right)}{sin\gamma sin\alpha}=0\)
\(\dfrac{sin\left(a-b\right)}{sina.sinb}+\dfrac{sin\left(b-c\right)}{sinb.sinc}+\dfrac{sin\left(c-a\right)}{sinc.sina}\)
\(=\dfrac{sina.cosb-cosa.sinb}{sina.sinb}+\dfrac{sinb.cosc-cosb.sinc}{sinb.sinc}+\dfrac{sinc.cosa-cosc.sina}{sina.sinc}\)
\(=\dfrac{cosb}{sinb}-\dfrac{cosa}{sina}+\dfrac{cosc}{sincc}-\dfrac{cosb}{sinb}+\dfrac{cosa}{sina}-\dfrac{cosc}{sincc}\)
\(=0\)
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cho \(\hept{\begin{cases}x;y;z>0\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{4}\end{cases}}\)Tìm \(Min_P=\frac{1}{\alpha a+\beta b+\gamma c}+\frac{1}{\beta a+\gamma b+\alpha c}+\frac{1}{\gamma a+\alpha b+\beta c}\)với \(\alpha;\beta;\gamma\in\)N*
Cho \(\left\{{}\begin{matrix}\text{x, y, z > 0}\\\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{4}\end{matrix}\right.\). Tìm \(\min\limits_P=\dfrac{1}{\alpha\text{a}+\beta b+\gamma c}+\dfrac{1}{\beta\text{a}+\gamma b+\alpha c}+\dfrac{1}{\gamma\text{a}+\alpha b+\beta c} v\text{ới} \alpha; \beta;\text{ \gamma}\in\) \(\mathbb{N}^*\)
Cho \(\hept{\begin{cases}x,y,z>0\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{4}\end{cases}}\). Tìm \(max_p=\frac{1}{\alpha\text{a}+\beta b+\gamma c}=\frac{1}{\beta\text{a}+\gamma b+\alpha c}=\frac{1}{\gamma\text{a}+\alpha b+\beta c}\) với \(\alpha,\beta,\gamma\inℕ^∗\).
Với \(\alpha\ge\beta\ge\gamma>0\) , \(a\ge\alpha\) , \(ab\ge\alpha\beta\) , \(abc\ge\alpha\beta\gamma\)
Chứng minh rằng \(a+b+c\ge\alpha+\beta+\gamma\)
\(VT=a+b+c=\alpha.\frac{a}{\alpha}+\beta.\frac{b}{\beta}+\gamma.\frac{c}{\gamma}\)
Áp dụng phương pháp nhóm ABEL
\(\Rightarrow VT=\left(\alpha-\beta\right)\frac{a}{\alpha}+\left(\beta-\gamma\right)\left(\frac{a}{\alpha}+\frac{b}{\beta}\right)+\gamma\left(\frac{a}{\alpha}+\frac{b}{\beta}+\frac{c}{\gamma}\right)\)
Áp dụng bất đẳng thức Cauchy
\(\Rightarrow\left\{\begin{matrix}\frac{a}{\alpha}+\frac{b}{\beta}\ge2\sqrt{\frac{ab}{\alpha\beta}}\left(1\right)\\\frac{a}{\alpha}+\frac{b}{\beta}+\frac{c}{\gamma}\ge3\sqrt[3]{\frac{abc}{\alpha\beta\gamma}}\left(3\right)\end{matrix}\right.\)
Ta có \(ab\ge\alpha\beta\Rightarrow\frac{ab}{\alpha\beta}\ge1\) \(\Rightarrow2\sqrt{\frac{ab}{\alpha\beta}}\ge2\left(2\right)\)
Ta có \(abc\ge\alpha\beta\gamma\Rightarrow\frac{abc}{\alpha\beta\gamma}\ge1\Rightarrow3\sqrt[3]{\frac{abc}{\alpha\beta\gamma}}\ge3\left(4\right)\)
Từ ( 1 ) và ( 2 )
\(\Rightarrow\frac{a}{\alpha}+\frac{b}{\beta}\ge2\)
\(\Rightarrow\left(\beta-\gamma\right)\left(\frac{a}{\alpha}+\frac{b}{\beta}\right)\ge2\left(\beta-\gamma\right)\) ( 5 )
Từ ( 3 ) và ( 4 )
\(\Rightarrow\frac{a}{\alpha}+\frac{b}{\beta}+\frac{c}{\gamma}\ge3\)
\(\Rightarrow\gamma\left(\frac{a}{\alpha}+\frac{b}{\beta}+\frac{c}{\gamma}\right)\ge3\gamma\) ( 6 )
Theo đề bài ta có \(a\ge\alpha\Rightarrow\frac{a}{\alpha}\ge1\)\(\Rightarrow\left(\alpha-\beta\right)\frac{a}{\alpha}\ge\alpha-\beta\) ( 7 )
Từ ( 5 ) , ( 6 ) , ( 7 ) cộng theo từng vế
\(\Rightarrow VT=\left(\alpha-\beta\right)\frac{a}{\alpha}+\left(\beta-\gamma\right)\left(\frac{a}{\alpha}+\frac{b}{\beta}\right)+\gamma\left(\frac{a}{\alpha}+\frac{b}{\beta}+\frac{c}{\gamma}\right)\ge2\left(\beta-\gamma\right)+3\gamma+\alpha-\beta\)
\(\Rightarrow VT\ge2\beta-2\gamma+3\gamma+\alpha-\beta\)
\(\Rightarrow VT\ge\alpha+\beta+\gamma\)
\(\Leftrightarrow a+b+c\ge\alpha+\beta+\gamma\) ( đpcm )
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Cho \(\Delta ABC.M,N,P\in BC,CA,AB.\)CM: AM,BN,CP đồng quy tại tâm tỉ cự của hệ điểm{A;B;C} với hệ số \(\left\{\alpha,\beta,\gamma\right\}\Leftrightarrow\hept{\begin{cases}\alpha+\beta+\gamma\ne0\\\beta\overrightarrow{MB}+\gamma\overrightarrow{MC}=\gamma\overrightarrow{NC}+\alpha\overrightarrow{NA}=\alpha\overrightarrow{PA}+\beta\overrightarrow{PB}=\overrightarrow{0}\end{cases}}\)
2 tháng 11 2023 lúc 19:20
Chứng minh rằng đa thức \(f\left(x\right)\) bậc chẵn có ít nhất 2 nghiệm khi \(\exists\alpha,\beta,\gamma\) phân biệt sao cho \(f\left(\alpha\right)+f\left(\beta\right)+f\left(\gamma\right)=0\)