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The pollution of water is an important problem today. Today, throughout moutain to city, almost rivers are not pure. They are dark, black, and foul and make the bad trouble with our lives. We need fresh air, we need fresh water. We need water to drink, to bath, to clean and wash. We very need water. Land Dad, and water Mom. Water smooth, transparent, clear, and it has 70% of all our body. We can not don't need water. Water help tree grow. Water can refresh and clear all things. But today, we only have water from water city company. The water we drink, is not fresh. And it must fire to hot as 100 degree. And then, we make it cool, so we can drink. In once upon a time, all rivers are fresh. And we can swim in that, it's very funny. In that rivers, we have fishes, too, and shrimp, and very much see animals. But now, with our rivers, there have no fish, or shrimp. Because there're too dirty. If we want to make our rivers revival, we need learn the way to not pour the pollution things to our rivers. It's very difficult, I know. But if we truely want our rivers have the blue, not black, or grown color, we need do all things from the begin. Nowadays, we have many polluted company, we always try to make pure water. And to have pure water, we must pay money. It's not cheap. And the last thing I want to talk with you, that we need to learn the way to not pour the dirty things to rivers. And we need save our water.
làm câu a là làm đc tất cả nha:
((2x2 - x - 1)2 - (x2 - 7x + 6)2 = 0
(2x^2-x-1-x^2+7x-6)(2x^2-x-1+x^2-7x+6)=0
(x^2+6x-7)(3x^2-8x+5)=0
x^2+6x-7=0 hoặc 3x^2-8x+5=0
TH1: x^2+6x-7=0
<=>x^2-x+7x-7=0
<=> (x-1)(x+7)=0
<=> x=1 hoặc x=-7
TH2: 3x^2-8x+5=0
<=>3x^2-3x-5x+5=0
<=>(x-1)(3x-5)=0
<=> x=1 hoặc x=5/3
Câu hỏi của Đoàn Thanh Kim Kim - Toán lớp 7 - Học toán với OnlineMath
Em tham khảo ở link này nhé :)
Do a,b,c là độ dài các cạnh của tam giác nên luôn dương.
Do đó: \(VP>0\)
Nhân 2 vào mỗi vễ,điều cần c/m tương đương với:
\(2a^2+2b^2+2c^2\ge2ab+2bc+2ca\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)(Chuyển vế)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\) (luôn đúng) (đpcm)
Dấu "=" xảy ra khi a = b = c
Áp dụng BĐT AM-GM ta có:
\(1=\frac{3}{x}+\frac{2}{y}\ge2.\sqrt{\frac{6}{xy}}\)
\(\Leftrightarrow1^2\ge4.\frac{6}{xy}\)
\(\Leftrightarrow1\ge\frac{24}{xy}\)
\(\Leftrightarrow xy\ge24\)
Dấu " = " xảy ra \(\Leftrightarrow\frac{3}{x}=\frac{2}{y}=\frac{1}{2}\Leftrightarrow\hept{\begin{cases}x=6\\y=4\end{cases}}\)
Vậy \(xy_{min}=24\Leftrightarrow\hept{\begin{cases}x=6\\y=4\end{cases}}\)
T nghĩ ra câu b rồi nhé Pain,bớt xạo lz!
b) Từ \(\frac{3}{x}+\frac{2}{y}=1\),ta có: \(x+y=1\left(x+y\right)=\left(\frac{3}{x}+\frac{2}{y}\right)\left(x+y\right)\)
Áp dụng BĐT Bunhiacopxki,ta có: \(\left(\frac{3}{x}+\frac{2}{y}\right)\left(x+y\right)\ge\left(\sqrt{\frac{3}{x}.x}+\sqrt{\frac{2}{y}.y}\right)\)
\(=\left(\sqrt{3}+\sqrt{2}\right)^2=5+2\sqrt{6}\)
Vậy \(Min_{x+y}=5+2\sqrt{6}\Leftrightarrow\hept{\begin{cases}x=3+\sqrt{6}\\y=2+\sqrt{6}\end{cases}}\)
Với [x>0x<−1] [x>0x<−1] ta có:
x3<x3+x2+x+1<(x+1)3⇒x3<y3<(x+1)3x3<x3+x2+x+1<(x+1)3⇒x3<y3<(x+1)3 (không thỏa mãn)
Suy ra −1≤x≤0−1≤x≤0. Mà x∈Z⇒x∈{−1;0}x∈Z⇒x∈{−1;0}
⋆⋆ Với x=−1x=−1 ta có: y=0
⋆⋆ Với x=0x=0 ta có: y=1
x^3+x^2+x+1=y^3 => y^3 - x^3 = x^2 + x + 1 = (x + 1/2)^2 + 3/4 > 0
=> y^3 > x^3 (1)
mặt khác:
5x^2 +11x+5 =5(x+11/10)^2 +19/20 > 0
y^3 = x^3 + x^2 + x +1 < x^3 + x^2 + x +1 + 5x^2 + 11x +5 = x^3 +6x^2 +12x +8 = (x + 2)^3 (2)
(1) và (2) => y^3 = (x + 1)^3 => y = x +1
=> x^3+x^2 +x +1 = x^3 +3x^2 +3x +1 = y^3
<=> 2x^2 + 2x =0
<=> 2x(x+1)=0
=> x = 0 và y=1
hoặc x = -1 và y = 0
D E A B C
Theo định lí Ta let
\(\frac{AD}{AB}=\frac{AE}{AC}\Rightarrow\frac{AE}{AC}+\frac{CE}{AC}=\frac{AE+CE}{AC}=\frac{AC}{AC}=1\)
