Question

cho biểu thức sau:

A=\(\dfrac{4}{x^2+2}+\dfrac{3}{2-x^2}-\dfrac{12}{4-x^4}\)biết x \(\ne\pm2\)

a) Rút gọn các biểu thức

b) Tính giá trị của A khi x= -3

c) Tìm GTLN của A

B= \(\dfrac{1}{x}+\dfrac{1}{x+5}+\dfrac{x-5}{x\left(x+5\right)}\) biết x \(\ne\) 0;-5

a) Rút gọn các biểu thức

b) Tính giá trị của A khi \(|x-1|\)=6

c) Tìm x nguyên để A có giá trị nguyên

Answer 1

\(Câu\text{ }1:\)

\(\text{ a) }A=\dfrac{4}{x^2+2}+\dfrac{3}{2-x^2}-\dfrac{12}{4-x^4}\\ A=\dfrac{4\left(2-x^2\right)}{\left(x^2+2\right)\left(2-x^2\right)}+\dfrac{3\left(2+x^2\right)}{\left(2-x^2\right)\left(2+x^2\right)}-\dfrac{12}{\left(2+x^2\right)\left(2-x^2\right)}\\ A=\dfrac{4\left(2-x^2\right)+3\left(2+x^2\right)-12}{\left(x^2+2\right)\left(2-x^2\right)}\\ A=\dfrac{8-4x^2+6+3x^2-12}{\left(x^2+2\right)\left(2-x^2\right)}\\ A=\dfrac{-x^2-2}{\left(x^2+2\right)\left(2-x^2\right)}\\ A=\dfrac{-\left(x^2+2\right)}{\left(x^2+2\right)\left(2-x^2\right)}\\ A=\dfrac{-1}{2-x^2}\)

\(\text{b) }Để\text{ }A=-3\\ thì\Rightarrow\dfrac{-1}{2-x^2}=-3\\ \Leftrightarrow2-x^2=3\\ \Leftrightarrow x^2=-1\\ \Leftrightarrow x\text{ }không\text{ }có\text{ }giá\text{ }trị\left(vì\text{ }x^2\ge0\forall x\right)\\ \text{ }Vậy\text{ }để\text{ }A=-3\text{ }thì\text{ }x\text{ }không\text{ }có\text{ }giá\text{ }trị.\)

\(\text{c) }Ta\text{ }có:\text{ }A=\dfrac{-1}{2-x^2}\\ A=\dfrac{1}{x^2-2}\\ x^2\ge0\forall x\\ \Rightarrow x^2-2\ge-2\forall x\\ \Rightarrow A=\dfrac{1}{x^2-2}\le-\dfrac{1}{2}\\ Dấu\text{ }"="\text{ }xảy\text{ }khi:\\ x^2=0\\ \Leftrightarrow x=0\\\text{ }Vậy\text{ }A_{\left(Max\right)}=-\dfrac{1}{2}\text{ }khi\text{ }x=0\)

Answer 2

\(Câu\text{ }2:\)

\(\text{a) }B=\dfrac{1}{x}+\dfrac{1}{x+5}+\dfrac{x-5}{x\left(x+5\right)}\\ B=\dfrac{x+5}{x\left(x+5\right)}+\dfrac{x}{\left(x+5\right)x}+\dfrac{x-5}{x\left(x+5\right)}\\ B=\dfrac{x+5+x+x-5}{x\left(x+5\right)}\\ B=\dfrac{3x}{x\left(x+5\right)}\\ B=\dfrac{3}{x+5}\left(\text{*}\right)\)

\(\text{b) }Ta\text{ }có:\text{ }\left|x-1\right|=6\\ \Leftrightarrow\left[{}\begin{matrix}x-1=6\\x-1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-5\end{matrix}\right.\\ Ta\text{ }lại\text{ }có:\text{ }B=\dfrac{3}{x+5}\\ \RightarrowĐKCĐ:x+5\ne0\\ \Rightarrow x\ne-5\\ \Rightarrow x=7\text{ }thỏa\text{ }mãn\text{ }với\text{ }điều\text{ }kiện\text{ }của\text{ }biến.\\ x=-5\text{ }không\text{ }thỏa\text{ }mãn\text{ }với\text{ }điều\text{ }kiện\text{ }của\text{ }biến.\\ Thay\text{ }x=7\text{ }vào\text{ }\left(\text{*}\right),ta\text{ }được:\text{ }B=\dfrac{3}{7+5}=\dfrac{3}{12}=\dfrac{1}{4}\\ \text{ }Vậy\text{ }với\text{ }x=7\text{ }thì\text{ }B=\dfrac{1}{4}\\ với\text{ }x=-5\text{ }thì\text{ }B\text{ }không\text{ }có\text{ }giá\text{ }trị.\)

\(\text{c) }Ta\text{ }có:B=\dfrac{3}{x+5}\\ \RightarrowĐể\text{ }B\in Z\\ thì\Rightarrow3⋮x+5\\ \Rightarrow x+5\inƯ_{\left(3\right)}\\ Mà\text{ }Ư_{\left(3\right)}=\left\{\pm1;\pm3\right\}\\ Ta\text{ }lập\text{ }bảng\text{ }xét\text{ }giá\text{ }trị:\)

\(x+5\)	\(-3\)	\(-1\)	\(1\)	\(3\)
\(x\)	\(-8\)	\(-6\)	\(-4\)	\(-2\)

\(\Rightarrow x\in\left\{-8;-6;-4;-2\right\}\\ Vậy\text{ }để\text{ }B\in Z\\ thì x\in\left\{-8;-6;-4;-2\right\}\)