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a. A=\(1+\left(\frac{x+1}{x^3+1}-\frac{1}{x-x^2-1}-\frac{2}{x+1}\right):\frac{x^3-2x^2}{x^3-x^2+x}\)
\(=1+\left(\frac{x+1+x+1-2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\right).\frac{x\left(x^2-x+1\right)}{x^2\left(x-2\right)}\)
\(=1+\frac{-2x^2+4x}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{x^2-x+1}{x\left(x-2\right)}\)
\(=1+\frac{-2x\left(x-2\right)}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{x^2-x+1}{x\left(x-2\right)}\)
\(=1-\frac{2}{x+1}=\frac{x-1}{x+1}\)
b.\(\left|x-\frac{3}{4}\right|=\frac{5}{4}\Rightarrow\orbr{\begin{cases}x-\frac{3}{4}=\frac{5}{4}\\x-\frac{3}{4}=-\frac{5}{4}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=-\frac{1}{2}\end{cases}}\)
Với \(x=2\Rightarrow A=\frac{2-1}{2+1}=\frac{1}{3}\)
Với \(x=-\frac{1}{2}\Rightarrow A=\frac{-\frac{1}{2}-1}{-\frac{1}{2}+1}=-3\)
a) \(\left(x+3\right)\left(2x-1\right)-2x\left(x-4\right)\)
\(=2x^2-x+6x-3-2x^2+8x\)
\(=13x-3\)
b) \(\left(3x-2\right)\left(2x-3\right)-5\left(x^2+4\right)\)
\(=6x^2-9x-4x+6-5x^2-20\)
\(=x^2-13x-14\)
c) \(2x\left(x^2+1\right)-\left(2x-5\right)\left(x^2-6\right)\)
\(=2x^3+2x-\left(2x^3-12x-5x^2+30\right)\)
\(=2x^3+2x-2x^3+12x+5x^2-30\)
\(=5x^2+14x-30\)
\(A=\frac{2x}{x^2-25}+\frac{5}{5-x}-\frac{1}{x+5}\)
\(=\frac{2x}{\left(x-5\right)\left(x+5\right)}-\frac{5}{x-5}-\frac{1}{x+5}\)
\(=\frac{2x}{\left(x-5\right)\left(x+5\right)}-\frac{5\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{x-5}{\left(x-5\right)\left(x+5\right)}\)
\(=\frac{2x-5\left(x+5\right)-x+5}{\left(x-5\right)\left(x+5\right)}\)
\(=\frac{-6x+5}{x^2-25}\)
B1:dài quá :vv
B2:\(Q=\frac{x^2}{x^4+x^2+1}=\frac{x^2}{x^4+2x^2+1-x^2}=\frac{x^2}{\left(x^2+1\right)-x^2}=\frac{x^2}{\left(x^2-x+1\right)\left(x^2+x+1\right)}\)
\(=\frac{x}{x^2-x+1}.\frac{x}{x^2+x+1}=\frac{2}{3}.\frac{x}{x^2+x+1}\)
\(\frac{x}{x^2-x+1}=\frac{2}{3}\Rightarrow\frac{x^2-x+1}{x}=\frac{3}{2}\Rightarrow\frac{x^2-x+1}{x}+2=\frac{3}{2}+2\Rightarrow\frac{x^2+x+1}{x}=\frac{7}{2}\)
\(\Rightarrow\frac{x}{x^2+x+1}=\frac{2}{7}\Rightarrow Q=\frac{2}{3}.\frac{2}{7}=\frac{4}{21}\)
3.
Ta có: \(a^5-a=a\left(a^4-1\right)=a\left(a^2-1\right)\left(a^2+1\right)=a\left(a+1\right)\left(a-1\right)\left(a^2+1\right)\)
\(=a\left(a-1\right)\left(a+1\right)\left(a^2-4+5\right)=a\left(a-1\right)\left(a+1\right)\left(a^2-4\right)+5a\left(a-1\right)\left(a+1\right)\)
\(=a\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)+5a\left(a-1\right)\left(a+1\right)\)
Do a(a-1)(a+1)(a-2)(a+2) là tích của 5 số hạng liên tiếp nên chia hết cho 2,3 và 5
Lại có a(a-1)(a+1) là tích của 3 số hạng liên tiếp nên chia hết cho 2,3 suy ra 5a(a-1)(a+1) chia hết cho 2,3,5
Từ đó:a(a-1)(a+1)(a-1)(a+2)+5a(a-1)(a+1) chia hết cho 2,3,5 hay a(a-1)(a+1)(a-2)(a+2)+5a(a-1)(a+1) chia hết cho 30 \(\Leftrightarrow a^5-a\) chia hết cho 30
Tương tự ta có\(b^5-b\) chia hết cho 30, \(c^5-c\) chia hết cho 30
Do đó:\(a^5-a+b^5-b+c^5-c⋮30\)
\(\Leftrightarrow a^5+b^5+c^5-\left(a+b+c\right)⋮30\)
Mà a+b+c=0 nên;
\(a^5+b^5+c^5⋮30\left(ĐCCM\right)\)
Với \(x\ne1\)ta có
\(P=\left(\frac{4}{x-1}-\frac{7x+5}{x^3-1}\right):\left(1-\frac{x-4}{x^2+x+1}\right)\)
\(=\left[\frac{4x^2+4x+4-7x-5}{\left(x-1\right)\left(x^2+x+1\right)}\right]:\left(\frac{x^2+x+1-x-4}{x^2+x+1}\right)\)
\(=\frac{4x^2-3x-1}{\left(x-1\right)\left(x^2+x+1\right)}:\frac{x^2-3}{x^2+x+1}=\frac{4x+1}{x^2-3}\)
\(Câu\text{ }1:\)
\(\text{ a) }A=\dfrac{4}{x^2+2}+\dfrac{3}{2-x^2}-\dfrac{12}{4-x^4}\\ A=\dfrac{4\left(2-x^2\right)}{\left(x^2+2\right)\left(2-x^2\right)}+\dfrac{3\left(2+x^2\right)}{\left(2-x^2\right)\left(2+x^2\right)}-\dfrac{12}{\left(2+x^2\right)\left(2-x^2\right)}\\ A=\dfrac{4\left(2-x^2\right)+3\left(2+x^2\right)-12}{\left(x^2+2\right)\left(2-x^2\right)}\\ A=\dfrac{8-4x^2+6+3x^2-12}{\left(x^2+2\right)\left(2-x^2\right)}\\ A=\dfrac{-x^2-2}{\left(x^2+2\right)\left(2-x^2\right)}\\ A=\dfrac{-\left(x^2+2\right)}{\left(x^2+2\right)\left(2-x^2\right)}\\ A=\dfrac{-1}{2-x^2}\)
\(\text{b) }Để\text{ }A=-3\\ thì\Rightarrow\dfrac{-1}{2-x^2}=-3\\ \Leftrightarrow2-x^2=3\\ \Leftrightarrow x^2=-1\\ \Leftrightarrow x\text{ }không\text{ }có\text{ }giá\text{ }trị\left(vì\text{ }x^2\ge0\forall x\right)\\ \text{ }Vậy\text{ }để\text{ }A=-3\text{ }thì\text{ }x\text{ }không\text{ }có\text{ }giá\text{ }trị.\)
\(\text{c) }Ta\text{ }có:\text{ }A=\dfrac{-1}{2-x^2}\\ A=\dfrac{1}{x^2-2}\\ x^2\ge0\forall x\\ \Rightarrow x^2-2\ge-2\forall x\\ \Rightarrow A=\dfrac{1}{x^2-2}\le-\dfrac{1}{2}\\ Dấu\text{ }"="\text{ }xảy\text{ }khi:\\ x^2=0\\ \Leftrightarrow x=0\\\text{ }Vậy\text{ }A_{\left(Max\right)}=-\dfrac{1}{2}\text{ }khi\text{ }x=0\)
\(Câu\text{ }2:\)
\(\text{a) }B=\dfrac{1}{x}+\dfrac{1}{x+5}+\dfrac{x-5}{x\left(x+5\right)}\\ B=\dfrac{x+5}{x\left(x+5\right)}+\dfrac{x}{\left(x+5\right)x}+\dfrac{x-5}{x\left(x+5\right)}\\ B=\dfrac{x+5+x+x-5}{x\left(x+5\right)}\\ B=\dfrac{3x}{x\left(x+5\right)}\\ B=\dfrac{3}{x+5}\left(\text{*}\right)\)
\(\text{b) }Ta\text{ }có:\text{ }\left|x-1\right|=6\\ \Leftrightarrow\left[{}\begin{matrix}x-1=6\\x-1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-5\end{matrix}\right.\\ Ta\text{ }lại\text{ }có:\text{ }B=\dfrac{3}{x+5}\\ \RightarrowĐKCĐ:x+5\ne0\\ \Rightarrow x\ne-5\\ \Rightarrow x=7\text{ }thỏa\text{ }mãn\text{ }với\text{ }điều\text{ }kiện\text{ }của\text{ }biến.\\ x=-5\text{ }không\text{ }thỏa\text{ }mãn\text{ }với\text{ }điều\text{ }kiện\text{ }của\text{ }biến.\\ Thay\text{ }x=7\text{ }vào\text{ }\left(\text{*}\right),ta\text{ }được:\text{ }B=\dfrac{3}{7+5}=\dfrac{3}{12}=\dfrac{1}{4}\\ \text{ }Vậy\text{ }với\text{ }x=7\text{ }thì\text{ }B=\dfrac{1}{4}\\ với\text{ }x=-5\text{ }thì\text{ }B\text{ }không\text{ }có\text{ }giá\text{ }trị.\)
\(\text{c) }Ta\text{ }có:B=\dfrac{3}{x+5}\\ \RightarrowĐể\text{ }B\in Z\\ thì\Rightarrow3⋮x+5\\ \Rightarrow x+5\inƯ_{\left(3\right)}\\ Mà\text{ }Ư_{\left(3\right)}=\left\{\pm1;\pm3\right\}\\ Ta\text{ }lập\text{ }bảng\text{ }xét\text{ }giá\text{ }trị:\)
| \(x+5\) | \(-3\) | \(-1\) | \(1\) | \(3\) |
| \(x\) | \(-8\) | \(-6\) | \(-4\) | \(-2\) |
\(\Rightarrow x\in\left\{-8;-6;-4;-2\right\}\\ Vậy\text{ }để\text{ }B\in Z\\ thì x\in\left\{-8;-6;-4;-2\right\}\)
a. \(2x\left(x-5\right)-\left(x-2\right)^2-\left(x+3\right)\left(x-3\right)\)
\(=2x^2-10x-x^2+4x-4-x^2+9\)
\(=-6x+5\)
b. \(\left(x+1\right)^2+3\left(x-5\right)\left(x+5\right)-\left(2x-1\right)^2\)
\(=x^2+2x+1+3x^2-75-4x^2+4x-1\)
\(=6x-75\)
c. \(2x\left(x-7\right)-\left(x+3\right)\left(x-2\right)-\left(x+4\right)\left(x-4\right)\)
\(=2x^2-14x-x^2-x+6-x^2+16\)
\(=-15x+22\)
d. \(\left(x+3\right)\left(x-3\right)-\left(x+5\right)\left(x-1\right)-\left(x-4\right)^2\)
\(=x^2-9-x^2-4x+5-x^2+8x-16\)
\(=-x^2+4x-20\)
Bài làm:
a) \(2x\left(x-5\right)-\left(x-2\right)^2-\left(x+3\right)\left(x-3\right)\)
\(=2x^2-10x-x^2+4x-4-x^2+9\)
\(=-6x+5\)
b) \(\left(x+1\right)^2+3\left(x-5\right)\left(x+5\right)-\left(2x-1\right)^2\)
\(=x^2+2x+1+3x^2-75-4x^2+4x-1\)
\(=6x-75\)
c) \(2x\left(x-7\right)-\left(x+3\right)\left(x-2\right)-\left(x+4\right)\left(x-4\right)\)
\(=2x^2-14x-x^2-x+6-x^2+16\)
\(=-15x+22\)
d) \(\left(x+3\right)\left(x-3\right)-\left(x+5\right)\left(x-1\right)-\left(x-4\right)^2\)
\(=x^2-9-x^2-4x+5-x^2+8x-16\)
\(=-x^2-4x-20\)
Bài 1. Rút gọn:
\(a, x\left(1-x\right)+6\left(x+3\right)\left(x+3\right)\)
\(=x-x^2+6\left(x^2+6x+9\right)\)
\(=x-x^2+6x^2+36x+54\)
\(=5x^2+37x+54\)
\(b, \left(2-3x\right)\left(2+3x\right)-\left(x+5\right)\left(x-5\right)\)
\(=\left(4-9x^2\right)-\left(x^2-25\right)\)
\(=-10x^2+29\)
\(c, \left(3x+1\right)\left(x+5\right)-\left(x-1\right)\left(x+1\right)\)
\(=3x^2+15x+x+5-x^2+1\)
\(=2x^2+16x+6\)
\(d,\left(2-3x\right)\left(2x+3\right)+6\left(x-1\right)^2\)
\(=\left(4x+6-6x^2-9x\right)+6\left(x^2-2x+1\right)\)
\(=4x+6-6x^2-9x+6x^2-12x+6\)
\(=-17x+12\)
\(e, x\left(5-x\right)-\left(2x+2\right)\left(3x+2\right)-\left(x-2\right)\left(x+2\right)\)
\(=5x-x^2-\left(6x^2+4x+6x+4\right)-\left(x^2-4\right)\)
\(=5x-x^2-6x^2-4x-6x-4-x^2+4\)
\(=-8x^2-5x\)
Bài 2:
a: VT\(=x^3-xy+x^2y^2-y^3-x^3+y^3-x^2y^2\)
=-xy
b: \(VT=x^2+6xy+9y^2-x^2+9y^2-6xy=18y^2=VP\)
Chọn đáp án B