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bài này dễ ẹt ak
nhưng giúp mình bài này đi
chotam giac abc . co canh bc=12cm, duong cao ah=8cm
a> tinh s tam giac abc
b> tren canh bc lay diem e sao cho be=3/4bc. tinh s tam giac abe va s tam giac ace ( bằng nhiều cách )
c> lay diem chinh giua cua canh ac va m . tinh s tam giac ame
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: 3x + y = 1 => y = 1 - 3x
a, Thay y = 1 - 3x vào M, ta có:
\(\Rightarrow M=3x^2+\left(1-3x\right)^2=3x^2+1-6x+9x^2=12x^2-6x+1=3\left(4x^2-2x+\frac{1}{3}\right)\)
\(=3\left(4x^2-2x+\frac{1}{4}+\frac{1}{12}\right)=3\left(2x-\frac{1}{2}\right)^2+\frac{3}{12}=3\left(2x-\frac{1}{2}\right)^2+\frac{1}{4}\)
Vì \(\left(2x-\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow3\left(2x-\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow3\left(2x-\frac{1}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}\forall x\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}2x-\frac{1}{2}=0\\3x+y=1\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}\\y=1-3x=1-3.\frac{1}{4}=\frac{1}{4}\end{cases}}\)\(\Leftrightarrow x=y=\frac{1}{4}\)
Vậy GTNN M = 1/4 khi x = y = 1/4
b, Thay y = 1 - 3x vào N
\(\Rightarrow N=x\left(1-3x\right)=x-3x^2=-3\left(x^2-\frac{x}{3}+\frac{1}{36}-\frac{1}{36}\right)\)
\(=-3\left(x-\frac{1}{6}\right)^2-3.\left(-\frac{1}{36}\right)=-3\left(x-\frac{1}{6}\right)^2+\frac{1}{12}\)
Vì \(\left(x-\frac{1}{6}\right)^2\ge0\forall x\)
\(\Rightarrow-3\left(x-\frac{1}{6}\right)^2\le0\forall x\)
\(\Rightarrow-3\left(x-\frac{1}{6}\right)^2+\frac{1}{12}\le\frac{1}{12}\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x-\frac{1}{6}=0\\3x+y=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{6}\\y=1-3x=1-3.\frac{1}{6}=\frac{1}{2}\end{cases}}\)
Vậy GTLN N = 1/12 khi x = 1/6 và y = 1/2
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Bài 3:
\(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{4}{xy}\)
\(\Leftrightarrow x^2y^2\left(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\ge\dfrac{4}{xy}.x^2y^2\)
\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2+y^2\ge4xy\)
\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2-2xy+y^2\ge2xy\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2+\left(x-y\right)^2\ge2xy\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2-2xy+\left(x-y\right)^2\ge0\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}-x+y\right)^2=0\) (luôn đúng)
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a) x2 + 1 ≤ (x - 2)2 ⇔ x2 + 1 ≤ x2 - 4x + 4 ⇔ 4x ≤ 3
⇔ x ≤ 3/4
Vậy: x ≤ 3/4
b) a, b > 0
Ta có: a + b = 1 suy ra: (a + b)2 = 1 ⇒ a2 + 2ab + b2 = 1 (1)
Mặt khác (a - b)2 ≥ 0 với mọi a, b ⇒ a2 - 2ab + b2 ≥ 0 (2)
Cộng (1) và (2) vế theo vế, ta được:
2a2 + 2b2 ≥ 1 ⇒ 2(a2 + b2) ≥ 1 ⇒ a2 + b2 ≥ 1/2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Rightarrow x^2+y^2\ge2\sqrt{x^2y^2}=2xy\)
\(\Rightarrow1\ge2xy\)
\(\Rightarrow\frac{1}{2}\ge xy\)
Có \(x+y\ge2\sqrt{xy}\ge2\sqrt{\frac{1}{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)
Vậy \(Min_{x+y}=\sqrt{2}\)
Làm tương tự với max
Thêm đk: x,y>0
Tìm max:
Áp dụng BĐT bunhiacopxki ta có:
\(\left(1+1\right)\left(x^2+y^2\right)\ge\left(x+y\right)^2\)
\(\Leftrightarrow2\ge\left(x+y\right)^2\)
\(\Leftrightarrow\sqrt{2}\ge x+y\)
Dấu " = " xảy ra <=> x=y
KL:...............................
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b1:
ĐKXĐ: \(x\ne0;x\ne\pm2\)
Ta có : \(A=\left(\frac{4x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{8x^2}{x^2-4}\right)\left(\frac{x-1}{x\left(x-2\right)}-\frac{2\left(x-2\right)}{x\left(x-2\right)}\right)\)
\(=\left(\frac{4x^2-8x-8x^2}{\left(x-2\right)\left(x+2\right)}\right)\left(\frac{x-1-2x+4}{x\left(x-2\right)}\right)\)
\(=\left(\frac{4x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\right)\left(\frac{3-3x}{x\left(x-2\right)}\right)\)
\(=\frac{12\left(x-1\right)}{x-2}\)
Vậy ....
Ta có : \(A< 0\Rightarrow\frac{12\left(x-1\right)}{x-2}< 0\)
Đến đây xét 2 TH 12(x-1)<0 & (x-2)>0 hoặc 12(x-1)>0 & (x-2)<0
a)Có \(a^2+1\ge2a\) với mọi a; \(b^2+1\ge2b\) với mọi b
Cộng vế với vế \(\Rightarrow a^2+b^2+2\ge2\left(a+b\right)\)
Dấu = xảy ra <=> a=b=1
b) Áp dụng BĐT bunhiacopxki có:
\(\left(x+y\right)^2\le\left(1+1\right)\left(x^2+y^2\right)\Leftrightarrow\left(x+y\right)^2\le2\)
\(\Leftrightarrow-\sqrt{2}\le x+y\le\sqrt{2}\)
\(\Rightarrow\left(x+y\right)_{max}=\sqrt{2}\Leftrightarrow\left\{{}\begin{matrix}x+y=\sqrt{2}\\x=y\end{matrix}\right.\)\(\Leftrightarrow x=y=\dfrac{\sqrt{2}}{2}\)
\(\left(x+y\right)_{min}=-\sqrt{2}\Leftrightarrow\left\{{}\begin{matrix}x+y=-\sqrt{2}\\x=y\end{matrix}\right.\)\(\Leftrightarrow x=y=-\dfrac{\sqrt{2}}{2}\)
c) \(S=\dfrac{1}{ab}+\dfrac{1}{a^2+b^2}=\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}+\dfrac{1}{2ab}\)
Với x,y>0, ta có: \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) (1)
Thật vậy (1) \(\Leftrightarrow\dfrac{y+x}{xy}\ge\dfrac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\)\(\Leftrightarrow\left(x-y\right)^2\ge0\) (lđ)
Áp dụng (1) vào S ta được:
\(S\ge\dfrac{4}{a^2+b^2+2ab}+\dfrac{1}{2ab}\)
Lại có: \(ab\le\dfrac{\left(a+b\right)^2}{4}\) \(\Leftrightarrow2ab\le\dfrac{\left(a+b\right)^2}{2}\Leftrightarrow2ab\le\dfrac{1}{2}\)\(\Rightarrow\dfrac{1}{2ab}\ge2\)
\(\Rightarrow S\ge\dfrac{4}{\left(a+b\right)^2}+2=6\)
\(\Rightarrow S_{min}=6\Leftrightarrow a=b=\dfrac{1}{2}\)