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cho \(x,y,z\ge0\) thỏa mãn \(x y z=6\). tìm GTLN và GTNN của biểu thức \(A=x^2 y^2 z^2\) - Hoc24
\(a,\dfrac{-5}{x+6}\ge0\\ mà\left(-5< 0\right)\\ \Rightarrow x+6< 0\\ \Rightarrow x< -6\\ b,\dfrac{2}{6-x}\ge0\\ mà\left(2>0\right)\\ \Rightarrow6-x>0\\ \Rightarrow x< 6\\ c,\dfrac{-x+3}{-6}\ge0\\ mà-6< 0\\ \Rightarrow-x+3< 0\\ \Rightarrow x>3\\\)
\(d,\dfrac{7x-1}{-9}\ge0\\mà-9< 0\\ \Rightarrow 7x-1\le0\\ \Rightarrow x\le\dfrac{1}{7}\\ e,\dfrac{x+2}{x^2+2x+1}\ge0\\ mà\left(x^2+2x+1\right)>0\forall x\\ \Rightarrow x+2\ge0\\ \Rightarrow x\ge-2\\ f,\dfrac{x-2}{x^2-2x+4}\ge0\\ mà\left(x^2-2x+4\right)>0\forall x\\ \Rightarrow x-2\ge0\\ \Rightarrow x\ge2\)
Chứng minh : \(x^2-2x+4>0\\ x^2-2x+1+3=\left(x-1\right)^2+3\ge3>0\)
a: ĐKXĐ: \(\dfrac{-5}{x+6}>=0\)
=>x+6<0
=>x<-6
b: ĐKXĐ: (-2)/(6-x)>=0
=>6-x<0
=>x>6
c: ĐKXĐ: (-x+3)/(-6)>=0
=>-x+3<=0
=>-x<=-3
=>x>=3
d: ĐKXĐ: (7x-1)/-9>=0
=>7x-1<=0
=>x<=1/7
e: ĐKXĐ: (x+2)/(x^2+2x+1)>=0
=>x+2>=0
=>x>=-1
f: ĐKXĐ: (x-2)/(x^2-2x+4)>=0
=>x-2>=0
=>x>=2
\(\left[3\left(x-1\right)^2+6\right]\left(3+6\right)\ge\left[3\left(x-1\right)+6\right]^2\)
\(\Leftrightarrow3x^2-6x+9\ge x+5\)
\(\Rightarrow A\ge x^4-8x^2+2024=\left(x^2-4\right)^2+2008\ge2008\)
Dấu "=" xảy ra khi \(x=2\)
Có phát hiện ra lỗi sai trong bài làm trên ko? :D
Bài 1:
ĐKXĐ: $3-2x\geq 0\Leftrightarrow x\leq \frac{3}{2}$
Bài 2:
a. ĐKXĐ: $x\geq \frac{1}{3}$
PT $\Leftrightarrow 3x-1=2^2=4$
$\Leftrightarrow x=\frac{5}{3}$ (tm)
b. ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow \sqrt{x-2}+2\sqrt{x-2}=6$
$\Leftrightarrow 3\sqrt{x-2}=6$
$\Leftrightarrow \sqrt{x-2}=2$
$\Leftrightarrow x-2=4$
$\Leftrightarrow x=6$ (tm)
\(M=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\left(\text{đ}k\text{x}\text{đ}:x\ge3\right)\\ =\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\\ =\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{2\sqrt{x}-9-\left(x-9\right)-\left(2x-4\sqrt{x}+\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2\sqrt{x}-9-x+9-2x+4\sqrt{x}-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ =\dfrac{5\sqrt{x}-3x+2}{x-5\sqrt{x}+6}\)
__
Để \(M\in Z\) thì \(x-5\sqrt{x}+6\) thuộc ước của \(5\sqrt{x}-3x+2\)
\(\Rightarrow x-5\sqrt{x}+6=-5\sqrt{x}-3x+2\\ \Leftrightarrow x-5\sqrt{x}+6+5\sqrt{x}+3x-2=0\\ \Leftrightarrow4x-4=0\\ \Leftrightarrow4x=4\\ \Leftrightarrow x=1\)
a) ĐKXĐ: \(x\ge0;x\ne9;x\ne4\)
\(M=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)
\(M=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\)
\(M=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(M=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(M=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(M=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(M=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
b) Ta có M ϵ Z thì \(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3+4}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3}{\sqrt{x}-3}+\dfrac{4}{\sqrt{x}-3}=1+\dfrac{4}{\sqrt{x}-3}\)
Phải thuộc Z vậy:
4 ⋮ \(\sqrt{x}-3\)
\(\Rightarrow\sqrt{x}-3\inƯ\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
Mà: \(x\ge0,x\ne4,x\ne9\) nên \(\sqrt{x}-3\in\left\{1;2;-2;4\right\}\)
\(\Rightarrow x\in\left\{16;25;1;49\right\}\)
a, Thay m = -2 ta được :
x^2 + 6x + 3 = 0
\(\Leftrightarrow x=-3+\sqrt{6};x=-3-\sqrt{6}\)
b, Để pt có 2 nghiệm
\(\Delta'=\left(m-1\right)^2-\left(-m+1\right)=m^2-2m+1+m-1=m^2-m\)> 0
Theo Viet : \(\left\{{}\begin{matrix}x_1+x_2=2m-2\\x_1x_2=-m+1\end{matrix}\right.\)
Ta có : \(\left(x_1+x_2\right)^2+5x_1x_2=9\)
\(\Leftrightarrow4\left(m-1\right)^2+5\left(-m+1\right)=9\)
\(\Leftrightarrow4m^2-8m+4-5m+5=9\Leftrightarrow4m^2-13m=0\)
\(\Leftrightarrow m\left(4m-13\right)=0\Leftrightarrow m=0\left(ktm\right);m=\dfrac{13}{4}\)(tm)
a, Thay m=-2 vào pt ta có:
\(x^2-2\left(m-1\right)x-m+1=0\\ \Leftrightarrow x^2-2\left(-2-1\right)x-\left(-2\right)+1=0\\ \Leftrightarrow x^2+6x+3=0\\ \Leftrightarrow\left(x^2+6x+9\right)-6=0\\ \Leftrightarrow\left(x+3\right)^2-\sqrt{6^2}=0\\ \Leftrightarrow\left(x+3-\sqrt{6}\right)\left(x+3+\sqrt{6}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3+\sqrt{6}\\x=-3-\sqrt{6}\end{matrix}\right.\)
\(b,\Delta'=\left[-\left(m-1\right)\right]^2-\left(-m+1\right)\\ =m^2-2m+1+m-1\\ =m^2-m\)
Để pt có 2 nghiệm thì \(\) \(\Delta'\ge0\Leftrightarrow m^2-m\ge0\Leftrightarrow\left[{}\begin{matrix}m\ge1\\m\le0\end{matrix}\right.\)
Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=2m-2\\x_1x_2=-m+1\end{matrix}\right.\)
\(x_1^2+x_2^2+7x_1x_2=9\\ \Leftrightarrow\left(x_1+x_2\right)^2+5x_1x_2=9\\ \Leftrightarrow\left(2m-2\right)^2+5\left(-m+1\right)=9\\ \Leftrightarrow4m^2-8m+4-5m+5-9=0\\ \Leftrightarrow4m^2-13m=0\\ \Leftrightarrow m\left(4m-13\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}m=0\left(tm\right)\\m=\dfrac{13}{4}\left(tm\right)\end{matrix}\right.\)