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a) x2 - 5x - y2 -5y
= ( x2 - y2 ) + ( -5x - 5y)
= ( x - y ) ( x + y) - 5( x + y )
= ( x + y ) ( x - y -5)
b) x3 + 2x2 - 4x - 8
= x2 ( x + 2 ) - 4 ( x + 2 )
= ( x +2 ) ( x2 -4 )
= ( x+2)2 ( x-2)
Bai 2 :
a, \(A=\left(x+3\right)^2+\left(x-2\right)^2-2\left(x+3\right)\left(x-2\right)\)
\(=x^2+6x+9+x^2-4x+4-2\left(x^2-2x+3x-6\right)\)
\(=2x^2+2x+13-2x^2-2x+12=25\)
b, \(B=\left(x-2\right)^2-x\left(x-1\right)\left(x-3\right)+3x^2-9x+8\)
\(=x^2-4x+4-x\left(x^2-3x-x+3\right)+3x^2-9x+8\)
\(=4x^2-13x+12-x^3+4x^2-3x=-16x+12-x^3\)
\(\dfrac{x^2+2x}{2\left(x+5\right)}+\dfrac{5x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
a ) ĐKXĐ : \(x\ne0,x\ne-5\)
b ) Rút gọn trước cái đã
\(\dfrac{x^2+2x}{2\left(x+5\right)}+\dfrac{5x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+2x^2+10x^2+50x-10x-50+50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+12x^2+35x}{2x\left(x+5\right)}\)
\(=\dfrac{x\left(x+5\right)\left(x+7\right)}{2x\left(x+5\right)}=\dfrac{x+7}{2x}\)
Khi \(A=1\), thì :
\(\dfrac{x+7}{2x}=1\Leftrightarrow x=7\)
Khi A = 3, thì :
\(\dfrac{x+7}{2x}=3\Leftrightarrow x=-1.\)
Bài 2 :
a ) ĐKXĐ : x\(\ne-3;2\)
b ) \(\dfrac{x-2}{x+3}-\dfrac{5}{\left(x-2\right)\left(x+3\right)}+\dfrac{1}{2-x}\)
\(=\dfrac{\left(x+2\right)\left(x-2\right)-5-\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}=\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\dfrac{x-4}{x-2}\)
c ) Khi \(A=-\dfrac{3}{4}\), thì :
\(\dfrac{x-4}{x-2}=-\dfrac{3}{4}\)
\(\Leftrightarrow4x-16=-3x+6\)
\(\Leftrightarrow x=\dfrac{22}{7}\)
d ) Ta có :
\(A=\dfrac{x-4}{x-2}=\dfrac{x-2-2}{x-2}=1-\dfrac{2}{x-2}\)
Để A nguyên thi \(x-2\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
Thay vào rồi tìm ra nếu x có trong đkxđ thì loại .
e ) \(x^2-9=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
Thay từng x vào A là tìm ra
\(1.a,Q=\frac{x+3}{2x+1}-\frac{x-7}{2x+1}=\frac{x+3}{2x+1}+\frac{7-x}{2x+1}\)
\(=\frac{x+3+7-x}{2x+1}=\frac{10}{2x+1}\)
\(b,\) Vì \(x\inℤ\Rightarrow\left(2x+1\right)\inℤ\)
Q nhận giá trị nguyên \(\Leftrightarrow\frac{10}{2x+1}\) nhận giá trị nguyên
\(\Leftrightarrow10⋮2x+1\)
\(\Leftrightarrow2x+1\inƯ\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
Mà \(\left(2x+1\right):2\) dư 1 nên \(2x+1=\pm1;\pm5\)
\(\Rightarrow x=-1;0;-3;2\)
Vậy.......................
để A xác định
\(\Rightarrow\hept{\begin{cases}x+2\ne0\\x-2\ne0\\x^2\ne4\end{cases}}\Rightarrow x\ne\pm2\)
\(A=\frac{4}{x+2}+\frac{3}{x-2}-\frac{5x-6}{x^2-4}\)
\(A=\frac{4.x-8}{\left(x+2\right).\left(x-2\right)}+\frac{3.x+6}{\left(x-2\right).\left(x+2\right)}-\frac{5x-6}{\left(x-2\right).\left(x+2\right)}\)
\(A=\frac{4x-8+3x+6-5x+6}{\left(x+2\right).\left(x-2\right)}=\frac{2.\left(x+2\right)}{\left(x+2\right).\left(x-2\right)}=\frac{2}{x-2}\)
\(\frac{4}{x+2}+\frac{3}{x-2}-\frac{5x-6}{x^2-4}=\frac{4}{x+2}+\frac{3}{x-2}-\frac{5x-6}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{4x-8}{\left(x+2\right)\left(x-2\right)}+\frac{3x+4}{\left(x-2\right)\left(x+2\right)}-\frac{5x-6}{\left(x-2\right)\left(x+2\right)}=\frac{4x-8+3x+4-5x+6}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{2x+2}{\left(x+2\right)\left(x-2\right)}=\frac{2x+2}{x^2-4}\)
C, \(x=4\Rightarrow A=\frac{2x+2}{x^2-4}=\frac{-6}{12}=\frac{-1}{2}\)
d, \(A\inℤ\Leftrightarrow2x+2⋮x^2-4\Leftrightarrow2x^2+2x-2x^2+8⋮x^2-4\Leftrightarrow2x+8⋮x^2-4\)
\(\Leftrightarrow2x^2+8x⋮x^2-4\Leftrightarrow16⋮x^2-4\)
\(x^2-4\inℕ\)
\(\Rightarrow x^2\in\left\{0;4;12\right\}\)
Thử lại thì 12 ko là số chính phương vậy x=0 hoặc x=2 thỏa mãn
mk học lớp 6 mong mn thông cảm nếu có sai sót
a) Phân thức B xác định \(\Leftrightarrow\hept{\begin{cases}2x-2\ne0\\x^2-1\ne0\\2x+2\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne\left\{\pm1\right\}\\x\ne-1\end{cases}\Leftrightarrow}x\ne\left\{\pm1\right\}}\)
b) \(B=\left(\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right)\cdot\frac{4x^2-4}{5}\)
\(B=\left[\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}+\frac{3\cdot2}{2\left(x-1\right)\left(x+1\right)}-\frac{\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\right]\cdot\frac{\left(2x\right)^2-2^2}{5}\)
\(B=\frac{x^2+2x+1+6-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}\cdot\frac{\left(2x-2\right)\left(2x+2\right)}{5}\)
\(B=\frac{10\cdot2\left(x-1\right)\cdot2\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)\cdot5}\)
\(B=\frac{40\left(x-1\right)\left(x+1\right)}{10\left(x-1\right)\left(x+1\right)}\)
\(B=4\)
Vậy với mọi giá trị của x thì B luôn bằng 4
Vậy giá trị của B không phụ thuộc vào biến ( đpcm )
\(Giải:\)
\(ĐKXĐ:x\ne\pm1\)
\(B=\left[\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right]=\left[\frac{x+1}{2x-2}+\frac{12}{4x^2-4}-\frac{x+3}{2x+2}\right]\)
\(=\left[\frac{x+1}{2x-2}+\frac{12}{\left(2x+2\right)\left(2x-2\right)}-\frac{x+3}{2x+2}\right]\)
\(=\left[\frac{\left(x+1\right)\left(2x+2\right)}{\left(2x+2\right)\left(2x-2\right)}+\frac{12}{\left(2x+2\right)\left(2x-2\right)}-\frac{\left(x+3\right)\left(2x-2\right)}{\left(2x-2\right)\left(2x+2\right)}\right]\)
\(=\frac{2x^2+4x+14-2x^2+2x-6x+6}{\left(2x-2\right)\left(2x+2\right)}\)
\(=\frac{6}{\left(2x-2\right)\left(2x+2\right)}\)
a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne-5\end{cases}}\)
\(P=\frac{x^2}{5x+25}+\frac{2x-10}{x}+\frac{50+5x}{x^2+5x}\)\(=\frac{x^2}{5\left(x+5\right)}+\frac{2\left(x-5\right)}{x}+\frac{5\left(x+10\right)}{x\left(x+5\right)}\)
\(=\frac{x^3}{5x\left(x+5\right)}+\frac{10\left(x-5\right)\left(x+5\right)}{5x\left(x+5\right)}+\frac{25\left(x+10\right)}{5x\left(x+5\right)}\)
\(=\frac{x^3+10\left(x-5\right)\left(x+5\right)+25\left(x+10\right)}{5x\left(x+5\right)}=\frac{x^3+10\left(x^2-25\right)+25x+250}{5x\left(x+5\right)}\)
\(=\frac{x^3+10x^2-250+25x+250}{5x\left(x+5\right)}=\frac{x^3+10x^2+25x}{5x\left(x+5\right)}\)\(=\frac{x\left(x^2+10x+25\right)}{5x\left(x+5\right)}\)\(=\frac{\left(x+5\right)^2}{5\left(x+5\right)}=\frac{x+5}{5}\)
b) \(x^2-3x=0\)\(\Leftrightarrow x\left(x-3\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
So sánh với ĐKXĐ, ta thấy \(x=0\)không thoả mãn
Thay \(x=3\)vào biểu thức ta được: \(P=\frac{3+5}{5}=\frac{8}{5}\)
c) Để \(P=-4\)thì \(\frac{x+5}{5}=-4\)\(\Leftrightarrow x+5=-20\)\(\Leftrightarrow x=-25\)( thoả mãn ĐKXĐ )
Vậy \(P=-4\)\(\Leftrightarrow x=-25\)
d) Để \(P\ge0\)thì \(\frac{x+5}{5}\ge0\)\(\Leftrightarrow x+5\ge0\)( vì \(5>0\))\(\Leftrightarrow x\ge-5\)
So sánh với ĐKXĐ, ta thấy x phải thoả mãn \(x>-5\)và \(x\ne0\)
Vậy \(P\ge0\)\(\Leftrightarrow\)\(x>-5\)và \(x\ne0\)
\(1)\)
\(a,2x^3-6x^2=2x^2.\left(x-3\right)\)
\(b,6x-6y-x^2+xy=6.\left(x-y\right)-x.\left(x-y\right)=\left(x-y\right).\left(6-x\right)\)
\(2)\)
\(a,ĐKXĐ:x\ne0;x\ne1\)
\(b,B=\left(\frac{2}{x\left(1-x\right)}-\frac{1}{1-x}\right):\frac{2-x}{1-2x+x^2}\)
\(B=\left(\frac{2}{x\left(1-x\right)}-\frac{x}{x\left(1-x\right)}\right):\frac{2-x}{\left(1-x\right)^2}\)
\(B=\frac{2-x}{x\left(1-x\right)}.\frac{\left(1-x\right)^2}{2-x}\)
\(B=\frac{1-x}{x}\)
\(c,\)Thay x=-2014 vào B ta có :
\(B=\frac{-2014-1}{-2014}=\frac{2015}{2014}\)
1, a, để A có giá trị xác định <=> 5x-5y \(\ne\) 0 => 5x\(\ne\)5y =>x\(\ne\)y b, A=\(\dfrac{x^2-y^2}{5x-5y}=\dfrac{\left(x+y\right)\left(x-y\right)}{5\left(x-y\right)}=\dfrac{\left(x+y\right)}{5}\) 2, a,
A=\(\dfrac{2x^3+4x}{x^3-4x}+\dfrac{x^2-4}{x^2+2x}+\dfrac{2}{2-x}\) =\(\dfrac{2x\left(x+2\right)}{x\left(x^2-4\right)}+\dfrac{\left(x+2\right)\left(x-2\right)}{x\left(x+2\right)}-\dfrac{2}{x-2}\) =\(\dfrac{2x\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{x}-\dfrac{2}{x-2}\) =\(\dfrac{2x}{x\left(x-2\right)}+\dfrac{\left(x-2\right)^2}{x\left(x-2\right)}-\dfrac{2x}{x\left(x-2\right)}\) =\(\dfrac{2x+\left(x-2\right)^2-2x}{x\left(x-2\right)}\) =\(\dfrac{\left(x-2\right)^2}{x\left(x-2\right)}\) =\(\dfrac{\left(x-2\right)}{x}\)
b, thay x=4 vào A ta có : A=\(\dfrac{4-2}{4}\) =\(\dfrac{2}{4}=\dfrac{1}{2}\)
c, để A \(\in\) Z => (x-2)\(⋮\)x mà x\(⋮\)x =>-2\(⋮\)x => x\(\in\){ \(\pm1;\pm2\)} mà x\(\ne\)\(\pm2\) => x\(\in\left\{-1,+1\right\}\)
Bài 3 : a, Ta có B= 2.(-1)2+-(-1)+1 =2+1+1=4 b, Ta có A=2x3 +5x2 -2x +a =(2x3 -x2 +x )+(6x2-3x +3) +(a-3) \(⋮\) 2x2-x+1 => x(2x2-x+1)+3(2x2-x+1) +(a-3)\(⋮\) 2x2-x+1
=>a-3=0 (vì a-3 là số dư )=>a-3 Vậy a=3 thì A\(⋮\)B c,B=1 => 2x2 -x+1=1 =>x(2x-1)=0 => x=0 hoặc 2x-1 =0 => x=0 hoặc x=\(\dfrac{1}{2}\)