1) 2x^3 - 8x = 0 2)2x (x - 15) - 4 (x - 15) = 0
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32 . 53 - 20{300 - [540 - 23 (78 : 76 + 70)]}
= 9 . 125 - 20{300 - [540 - 8(72 + 70)]}
= 9 . 125 - 20{300 - [540 - 8(49 + 1)]
= 9 . 125 - 20{300 - [540 - 8.50]
= 9 . 125 - 20{300 - [540 - 400]}
= 9 . 125 - 20{300 - 140}
= 9 . 125 - 20. 160
= 1125 - 3200 = -2075
\(=9\cdot125-20\cdot\left\{300-\left[540-8\cdot\left(7^2+1\right)\right]\right\}\)
\(=1125-20\cdot\left\{300-\left[540-8\cdot\left(49+1\right)\right]\right\}\)
\(=1125-20\cdot\left[300-\left(540-8\cdot50\right)\right]\)
\(=1125-20\cdot\left[300-\left(540-400\right)\right]\)
\(=1125-20\cdot\left(300-140\right)\)
\(=1125-20\cdot160\)
\(=1125-3200\)
\(=-2075\)
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mk bận đi ch nên chỉ tạm câu a nha
vẽ 3 đường trung tuyến AD ; BE ; CF
VT =
\(GA+GB+GC\) ( nhớ thêm dấu vec tơ nha )
\(=-\frac{2}{3}AD-\frac{2}{3}BE-\frac{2}{3}CF\)
\(=-\frac{2}{3}\cdot\frac{1}{2}\left(AB+BC\right)-\frac{2}{3}\cdot\frac{1}{2}\left(BA+BC\right)-\frac{2}{3}\cdot\frac{1}{2}\left(CA+CB\right)\) ( quy tắc hình bình hành )
\(=-\frac{1}{3}\left(AB+AC\right)-\frac{1}{3}\left(BA+BC\right)-\frac{1}{3}\left(CA+CB\right)\)
\(=-\frac{1}{3}AB-\frac{1}{3}AC-\frac{1}{3}BA-\frac{1}{3}BC-\frac{1}{3}CA-\frac{1}{3}CB\)
\(=0=VP\)
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Bài 1:
Ta có:
\(P=\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)+1\)
\(P=\left[\left(a+1\right)\left(a+4\right)\right]\cdot\left[\left(a+2\right)\left(a+3\right)\right]+1\)
\(P=\left(a^2+5a+4\right)\left(a^2+5a+6\right)+1\)
Đặt \(x=a^2+5a+5\) , khi đó:
\(P=\left(a-1\right)\left(a+1\right)+1\)
\(P=a^2-1+1\)
\(P=a^2=\left(x^2-5x+5\right)^2\)
Mà \(a\inℤ\Rightarrow x^2-5x+5\inℤ\)
=> P là số chính phương
\(\left(xy+yz+zx\right)^2+\left(x^2-yz\right)^2+\left(y^2-zx\right)^2+\left(z^2-xy\right)^2=x^2y^2+y^2z^2+z^2x^2+2xyz\left(x+y+z\right)+x^4-2x^2yz+y^2z^2+y^4-2y^2zx+z^2x^2+z^4-2z^2xy+x^2y^2=x^4+y^4+z^4+2\left(x^2y^2+y^2z^2+z^2x^2\right)=\left(x^2+y^2+z^2\right)^2=100^2=10000\)
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Mình xem phép làm câu 1 ạ.
Đề là?
\(\frac{1}{a}+\frac{1}{c}=\frac{2}{b}\)(1)
Chứng minh tương đương
\(\frac{a+b}{2a-b}+\frac{c+b}{2c-b}\ge4\)<=> 12ac - 9bc - 9ab + 6b2 \(\le\)0 ( quy đồng ) (2)
Từ (1) <=> 2ac = ab + bc Thay vào (2) <=> 6ab + 6bc - 9bc - 9ab + 6b2 \(\le\)0
<=> a + c \(\ge\)2b
Từ (1) => \(\frac{2}{b}=\frac{1}{a}+\frac{1}{c}\ge\frac{4}{a+c}\)
=> a + c \(\ge\)2b đúng => BĐT ban đầu đúng
Dấu "=" xảy ra <=> a = c = b
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1. BĐT tương đương với \(6\left(a^2+b^2\right)-2ab+8-4\left(a\sqrt{b^2+1}+b\sqrt{a^2+1}\right)\ge0\)
\(\Leftrightarrow\left[a^2-4a\sqrt{b^2+1}+4\left(b^2+1\right)\right]+\left[b^2-4b\sqrt{a^2+1}+4\left(a^2+1\right)\right]\)\(+\left(a^2-2ab+b^2\right)\ge0\)
\(\Leftrightarrow\left(a-2\sqrt{b^2+1}\right)^2+\left(b-2\sqrt{a^2+1}\right)^2+\left(a-b\right)^2\ge0\)(đúng)
=> Đẳng thức không xảy ra
2. \(a^4+b^4+c^2+1\ge2a\left(ab^2-a+c+1\right)\)
\(\Leftrightarrow a^4+b^4+c^2+1\ge2a^2b^2-2a^2+2ac+2a\)
\(\Leftrightarrow\left(a^4-2a^2b^2+b^4\right)+\left(c^2-2ac+a^2\right)+\left(a^2-2a+1\right)\ge0\)
\(\Leftrightarrow\left(a^2-b^2\right)^2+\left(c-a\right)^2+\left(a-1\right)^2\ge0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sin^25x+1=\cos^23x\)
<=> \(\sin^25x+1-\cos^23x=0\)
<=> \(\frac{1-\cos10x}{2}+1-\frac{\cos6x+1}{2}=0\)
<=> \(\cos10x+\cos6x=2\)
Mà \(\cos10x;\cos6x\ge1\)=> \(\cos10x+\cos6x\ge2\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}\cos10x=1\\\cos6x=1\end{cases}}\Leftrightarrow\hept{\begin{cases}10x=k2\pi\\6x=l2\pi\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{k\pi}{5}\\x=\frac{l\pi}{3}\end{cases}};k,l\in Z\Leftrightarrow x=m\pi;m\in\)
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Đk: x \(\ge\)2013; y \(\ge\)2013
Ta có: A = \(\frac{\sqrt{x-2013}}{x}+\frac{\sqrt{y-2012}}{y+1}\ge0\forall x;y\)
(vì \(\sqrt{x-2013}\ge0\); x > 0; \(\sqrt{y-2012}\ge0\); y + 1 > 0)
Dấu "=" xảy ra <=> x - 2013 = 0 và y - 2012 = 0 <=> x = 2013 và y = 2012
Vậy MinA = 0 khi x =2013 và y = 2012
Ta lại có: A = \(\frac{\sqrt{x-2013}}{x}+\frac{\sqrt{y-2012}}{y+1}\le\frac{\frac{x-2013+1}{2}}{x}+\frac{\frac{y-2012+1}{2}}{y+1}\)(bđt cosi)
<=> A \(\le\frac{x-2012}{2x}+\frac{y-2011}{2\left(y+1\right)}=\frac{1}{2}-\frac{1006}{x}+\frac{y+1-2012}{2\left(y+1\right)}\)
<= > A \(\le\frac{1}{2}-\frac{1006}{x}+\frac{1}{2}-\frac{1006}{y+1}=1-1006\left(\frac{1}{x}+\frac{1}{y+1}\right)\)
Áp dụng bđt \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) (1)
CM bđt đúng: Từ (1) => (a + b)2 > = 4ab <=> (a - b)2 > = 0 (luôn đúng với mọi a,b > 0)
Khi đó: A \(\le1-\frac{1006.4}{x+y+1}=1-\frac{4024}{x+y+1}\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}\sqrt{x-2013}=1\\\sqrt{y-2012}=1\\x=y+1\end{cases}}\) <=> \(\hept{\begin{cases}x=2014\\y=2013\end{cases}}\)
Vậy MaxA = \(1-\frac{4024}{2014+2013+1}=1-\frac{1006}{1007}=\frac{1}{1007}\) <=> x = 2014 và y = 2013
1) \(2x^3-8x=0\)
\(\Leftrightarrow2x\left(x^2-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm2\end{cases}}\)
Vậy \(x\in\left\{0;\pm2\right\}\)
2) \(2x\left(x-15\right)-4\left(x-15\right)=0\)
\(\Leftrightarrow\left(2x-4\right)\left(x-15\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-4=0\\x-15=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=15\end{cases}}\)
Vậy \(x\in\left\{2;15\right\}\)
1
\(2x^3-8x=0\)
\(2x\left(x^2-4\right)=0\)
\(\orbr{\begin{cases}2x=0\\x^2-4=0\end{cases}}\)
\(\orbr{\begin{cases}x=0\\x^2=4\end{cases}}\)
\(\orbr{\begin{cases}x=0\\x=\pm2\end{cases}}\)
2
\(2x\left(x-15\right)-4\left(x-15\right)=0\)
\(\left(2x-4\right)\left(x-15\right)=0\)
\(\orbr{\begin{cases}2x-4=0\\x-15=0\end{cases}}\)
\(\orbr{\begin{cases}2x=4\\x=0+15\end{cases}}\)
\(\orbr{\begin{cases}x=2\\x=15\end{cases}}\)