\(P=10^2-9^2+8^2-7^2+...+2^2-1\\=\left(10^2-9^2\right)+\left(8^2-7^2\right)+...+\left(2^2-1^2\right)\\ =\left(10+9\right)\left(10-9\right)+\left(8+7\right)\left(8-7\right)+...+\left(2+1\right)\left(2-1\right)\\ =19+15+11+7+1\\ =53\)

\(x^4+81\\ =x^4+81-18x^2+18x^2\\ =\left(x^4+18x^2+81\right)-18x^2\\ =\left(x^2+9\right)^2-\left(\sqrt{18}x\right)^2\\ =\left(x^2+9+\sqrt{18x}\right)\left(x^2+9-\sqrt{18x}\right)\)

Khi dùng Tirixo cần quan tâm đến số liệu kĩ thuật nào?

A. Dòng điện định mức

B. Điện áp định mức

C. Cả A và B đều đúng

D. Đáp án khác

\(Câu\text{ }2:\)

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

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

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

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

\(Câu\text{ }1:\)

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

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

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

30^o

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

\(\text{b) }\dfrac{xy}{x^2-y^2}-\dfrac{x^2}{y^2-x^2}\\ =\dfrac{xy}{x^2-y^2}+\dfrac{x^2}{x^2-y^2}\\ =\dfrac{xy+x^2}{x^2-y^2}\\ =\dfrac{x\left(y+x\right)}{\left(x+y\right)\left(x-y\right)}\\ =\dfrac{x}{x-y}\)