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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3:
a) Đặt f(x)=0
\(\Leftrightarrow x^2-4x+3=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
b) Đặt f(x)=0
\(\Leftrightarrow x^2-7x+12=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
Bài 3:
c) Đặt f(x)=0
\(\Leftrightarrow x^2+2x+1=0\)
\(\Leftrightarrow\left(x+1\right)^2=0\)
\(\Leftrightarrow x+1=0\)
hay x=-1
d) Đặt f(x)=0
\(\Leftrightarrow x^4+2=0\)
\(\Leftrightarrow x^4=-2\)(Vô lý)
![](https://rs.olm.vn/images/avt/0.png?1311)
d. Áp dụng BĐT Caushy Schwartz ta có:
\(x+y+\dfrac{1}{x}+\dfrac{1}{y}\le x+y+\dfrac{\left(1+1\right)^2}{x+y}=x+y+\dfrac{4}{x+y}\le1+\dfrac{4}{1}=5\)
-Dấu bằng xảy ra \(\Leftrightarrow x=y=\dfrac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1.
$x(x+2)(x+4)(x+6)+8$
$=x(x+6)(x+2)(x+4)+8=(x^2+6x)(x^2+6x+8)+8$
$=a(a+8)+8$ (đặt $x^2+6x=a$)
$=a^2+8a+8=(a+4)^2-8=(x^2+6x+4)^2-8\geq -8$
Vậy $A_{\min}=-8$ khi $x^2+6x+4=0\Leftrightarrow x=-3\pm \sqrt{5}$
2.
$B=5+(1-x)(x+2)(x+3)(x+6)=5-(x-1)(x+6)(x+2)(x+3)$
$=5-(x^2+5x-6)(x^2+5x+6)$
$=5-[(x^2+5x)^2-6^2]$
$=41-(x^2+5x)^2\leq 41$
Vậy $B_{\max}=41$. Giá trị này đạt tại $x^2+5x=0\Leftrightarrow x=0$ hoặc $x=-5$
![](https://rs.olm.vn/images/avt/0.png?1311)
a:
Sửa đề: \(P=\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{4x^2}{x^2-9}\right):\left(\dfrac{5}{3-x}-\dfrac{4x+2}{3x-x^2}\right)\)\(P=\left(\dfrac{-\left(x+3\right)}{x-3}+\dfrac{x-3}{x+3}-\dfrac{4x^2}{\left(x-3\right)\left(x+3\right)}\right):\dfrac{5x-4x-2}{x\left(3-x\right)}\)
\(=\dfrac{-x^2-6x-9+x^2-6x+9-4x^2}{\left(x-3\right)\left(x+3\right)}:\dfrac{x-2}{x\left(3-x\right)}\)
\(=\dfrac{-4x^2-12x}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x\left(3-x\right)}{x-2}\)
\(=\dfrac{-4x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{-x\left(x-3\right)}{x-2}=\dfrac{4x^2}{x-2}\)
b: x^2-4x+3=0
=>x=1(nhận) hoặc x=3(loại)
Khi x=1 thì \(P=\dfrac{4\cdot1^2}{1-2}=-4\)
c: P>0
=>x-2>0
=>x>2
d: P nguyên
=>4x^2 chia hết cho x-2
=>4x^2-16+16 chia hết cho x-2
=>x-2 thuộc {1;-1;2;-2;4;-4;8;-8;16;-16}
=>x thuộc {1;4;6;-2;10;-6;18;-14}
![](https://rs.olm.vn/images/avt/0.png?1311)
1)
a) 40% x X + X x 1/2 = 72
0,4 x X + X x 0,5 = 72
X x (0,4 + 0,5) = 72
X x 0,9 = 72
X = 72 : 0,9
X = 80
b) X x 2 + X : 0,125 = 55
X x 2 + X x 8 = 55
X x (2 + 8) = 55
X x 10 = 55
X = 55 : 10
X = 5,5
2.
a) 160 x (X - 17,4) = 2912
X - 17,4 = 2912 : 160
X - 17,4 = 18,2
X = 18,2 + 17,4
X = 35,6
b) X x 4,8 + 5,2 x X = 25,6
X x (4,8 + 5,2) = 25,6
X x 10 = 25,6
X = 25,6 : 10
X = 2,56
`@` `\text {Ans}`
`\downarrow`
`x - 1,78 = 11,69 - 4,57`
`x - 1,78 = 7,12`
`x = 7,12 + 11,78`
`x = 8,9`
Vậy, `x = 8,9.`
\(x\) - 1,78 = 11,69 - 4,57
\(x\) - 1,78 = 7,12
\(x\) = 7,12 + 1,78
\(x\) = 8,9