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Câu a đơn giản
b)
\(A=\frac{x^4-x^3-x+1}{x^4+x^3+3x^2+2x+2}=\frac{\left(x^4-x^3\right)-\left(x-1\right)}{\left(x^4+x^3+\frac{x^2}{4}\right)+\left(\frac{11}{4}x^2+2x+\frac{4}{11}\right)+1-\frac{4}{11}}\)
\(=\frac{\left(x-1\right)\left(x^3-1\right)}{\left(x^2+\frac{x}{2}\right)^2+\left(\frac{\sqrt{11}}{2}+\frac{2}{\sqrt{11}}\right)^2+\frac{7}{11}}\)
\(=\frac{\left(x-1\right)^2\left(x^2+x+1\right)}{\left(x^2+\frac{x}{2}\right)^2+\left(\frac{\sqrt{11}}{2}+\frac{2}{\sqrt{11}}\right)^2+\frac{7}{11}}\)
\(=\frac{\left(x-1\right)^2\left[\left(x^2+x+0,25\right)+0,75\right]}{\left(x^2+\frac{x}{2}\right)^2+\left(\frac{\sqrt{11}}{2}+\frac{2}{\sqrt{11}}\right)^2+\frac{7}{11}}\)
\(=\frac{\left(x-1\right)^2\left[\left(x+0,5\right)^2+0,75\right]}{\left(x^2+\frac{x}{2}\right)^2+\left(\frac{\sqrt{11}}{2}+\frac{2}{\sqrt{11}}\right)^2+\frac{7}{11}}\)
Vì \(\left(x-1\right)^2\left[\left(x+0,5\right)^2+0,75\right]>0\)và \(\left(x^2+\frac{x}{2}\right)^2+\left(\frac{\sqrt{11}}{2}+\frac{2}{\sqrt{11}}\right)^2+\frac{7}{11}>0\)
nên \(A>0\)hay A ko âm
Nhớ k nha !
1,Ta có
3x+7y=24
<=>3x=24-7y
Vì x là số tự nhiên
=>\(24-7y\ge0\)
<=>\(7y\le24\)
<=>\(y<4\) mà y là số tự nhiên
=>\(y=\left\{0;1;2;3\right\}\)
=>\(x=\left\{....\right\}\)
b,\(x^2-4x+2y-xy+9=0\)
<=>\(\left(x^2-4x+4\right)-y\left(x-2\right)+5=0\)
<=>\(\left(x-2\right)^2-y\left(x-2\right)=-5\)
<=>\(\left(x-2\right)\left(x-2-y\right)=5\)
Đến đây giải theo pp pt nghiệm nguyên.
Nếu mình làm đúng thì tick nha bạn,cảm ơn.
tick tui làm tiếp cho nha.
mình làm bài 2 trước nha:
a) y.(a-b)+a.(y-b)=a.y-b.y+a.y-b.y
=(a.y+a.y)-(b.y+b.y)
=2.a.y-2.b.y
=2.y.(a-b)
b)x2.(x+y)-y.(x2-y2)=x3+x2.y-x2y+y3=x3+y3
a,\(A=\dfrac{x+4}{x^2-2x}+\dfrac{2}{x}\)
\(=\dfrac{x+4}{x\left(x-2\right)}+\dfrac{2}{x}\)
\(=\dfrac{x+4}{x\left(x-2\right)}+\dfrac{2\left(x-2\right)}{x\left(x-2\right)}\)
\(=\dfrac{x+4+2x-4}{x\left(x-2\right)}\)
\(=\dfrac{3x}{x\left(x-2\right)}=\dfrac{3}{x-2}\)
b, Để A có giá trị bằng - 3
\(\Leftrightarrow\dfrac{3}{x-2}=-3\)
\(\Leftrightarrow x-2=-1\)
\(\Leftrightarrow x=1\) ( t/m )
Vậy x = 1 thì A có giá trị bằng -3
\(\frac{x^4-x^3-x+1}{x^4+x^3+3x^2+2x+2}\)
\(=\frac{x^3\left(x-1\right)-\left(x-1\right)}{x^4+x^3+x^2+2x^2+2x+2}\)
\(=\frac{\left(x-1\right)\left(x^3-1\right)}{x^2\left(x^2+x+1\right)+2\left(x^2+x+1\right)}\)
\(=\frac{\left(x-1\right)\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x^2+2\right)}\)
\(=\frac{\left(x-1\right)^2}{\left(x^2+2\right)}\)
Bài 1:
8: \(=\dfrac{x+3}{x\left(x-3\right)}\)
9: \(=\dfrac{x-2}{x-5}\cdot\dfrac{\left(x-5\right)\left(x+5\right)}{\left(x-2\right)^2}=\dfrac{x+5}{x-2}\)
10: \(=1:\dfrac{a-1}{a}=\dfrac{a}{a-1}\)
12: \(=\dfrac{6\left(x+1\right)}{3x\left(x+1\right)}=\dfrac{2}{x}\)
13: \(\dfrac{3}{x+3}-\dfrac{x-6}{x\left(x+3\right)}\)
\(=\dfrac{3x-x+6}{x\left(x+3\right)}=\dfrac{2x+6}{x\left(x+3\right)}=\dfrac{2}{x}\)
\(\text{Ta có : }\left(a+b\right)^2-\left(a+2\right)\left(a-2\right)\)
\(=\left(a^2+2ab+b^2\right)-\left(a^2-2^2\right)\)
\(=\left(a^2+2ab+b^2\right)-\left(a^2-4\right)\)
\(=a^2+2ab+b^2-a^2+4\)
\(=b^2+2ab+4\)