Question

Answer 1

Bài 1:

a.

$(3x-4)^2-x(3-x)=(9x^2-24x+16)-(3x-x^2)$

$=10x^2-27x+16$

b.

$(x-2)(x+2)(x^2+4)=(x^2-4)(x^2+4)=(x^2)^2-4^2=x^4-16$

 

 

Answer 2

Bài 2:

a. $(x+3)^2-x(x-3)=5$

$\Leftrightarrow (x^2+6x+9)-(x^2-3x)=5$

$\Leftrightarrow 9x+9=5$

$\Leftrightarrow x=\frac{-4}{9}$

b.

$x^2-3x-4=0$

$\Leftrightarrow (x^2+x)-(4x+4)=0$

$\Leftrightarrow x(x+1)-4(x+1)=0$

$\Leftrightarrow (x-4)(x+1)=0$

$\Leftrightarrow x-4=0$ hoặc $x+1=0$

$\Leftrightarrow x=4$ hoặc $x=-1$

c.

$(2x-1)^2=25$

$\Leftrightarrow (2x-1)^2-5^2=0$

$\Leftrightarrow (2x-1-5)(2x-1+5)=0$

$\Leftrightarrow (2x-6)(2x+4)=0$

$\Leftrightarrow 2x-6=0$ hoặc $2x+4=0$

$\Leftrightarrow x=3$ hoặc $x=-2$

Answer 3

Bài 3:

a. ĐKXĐ: \(\left\{\begin{matrix} x-x^2\neq 0\\ 1-x\neq 0\\ x+1\neq 0\\ (1-x)^2\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 1-x\neq 0\\ 1+x\neq 0\end{matrix}\right.\Leftrightarrow x\neq \pm 1\)

\(P=\left[\frac{1}{x(1-x)}+\frac{x}{x(1-x)}\right].\frac{(1-x)^2}{x+1}=\frac{x+1}{x(1-x)}.\frac{(1-x)^2}{x+1}\)

\(=\frac{1-x}{x}\)

b. $P>0\Leftrightarrow \frac{1-x}{x}>0$

$\Leftrightarrow 1-x>0; x>0$ hoặc $1-x< 0; x< 0$

$\Leftrightarrow 1> x>0$ (chọn) hoặc $1< x< 0$ (vô lý- loại)

Kết hợp đkxđ suy ra $P>0$ khi $1> x>0$

Answer 4

Bài 5. ĐKXĐ: $x\neq 10$

$A=\frac{x}{(x-10)^2}=\frac{x-10+10}{(x-10)^2}$

$=\frac{1}{x-10}+\frac{10}{(x-10)^2}$

Đặt $\frac{1}{x-10}=a$

Khi đó:
$A=a+10a^2=10(a^2+0,1a+0,05^2)-0,025$

$=10(a+0,05)^2-0,025\geq -0,025$
Vậy $A_{\min}=-0,025$ khi $a=-0,05$ hay $x=-10$

 

Answer 5

Bài 4.

Với $x\in\mathbb{Z}$, để $A\in\mathbb{Z}$ thì $3x-1\vdots 2x+1$

$\Rightarrow 2(3x-1)\vdots 2x+1$

$\Leftrightarrow 6x-2\vdots 2x+1$

$\Leftrightarrow 3(2x+1)-5\vdots 2x+1$

$\Leftrightarrow 5\vdots 2x+1$ hay $2x+1\in \text{Ư(5)}$

$\Leftrightarrow 2x+1\in\left\{\pm 1;\pm 5\right\}$

$\Leftrightarrow x\in\left\{0;-1; -3; 2\right\}$ 

Thử lại thấy đều tm. Vậy...............

 

Answer 6

Bài 6: ĐKXĐ: $x\neq -1$

Ta thấy:

$x^2+x+1=\frac{3}{4}(x^2+2x+1)+\frac{1}{4}(x^2-2x+1)$

$=\frac{3}{4}(x+1)^2+\frac{1}{4}(x-1)^2\geq \frac{3}{4}(x+1)^2$

$\Rightarrow B\geq \frac{\frac{3}{4}(x+1)^2}{x^2+2x+1}=\frac{\frac{3}{4}(x+1)^2}{(x+1)^2}=\frac{3}{4}$

Vậy $B_{\min}=\frac{3}{4}$ khi $x=1$

Answer 7

Bài 4: 

Để A nguyên thì \(3x-1⋮2x+1\)

\(\Leftrightarrow6x+3-5⋮2x+1\)

\(\Leftrightarrow2x+1\in\left\{1;-1;5;-5\right\}\)

\(\Leftrightarrow2x\in\left\{0;-2;4;-6\right\}\)

hay \(x\in\left\{0;-1;2;-3\right\}\)