\(\left|x+1\right|+\left|x+2\right|+...+\left|x+100\right|=10+x\)
Chứng minh: \(\frac{2}{x^2-1}+\frac{4}{x^2-4}+...+\frac{20}{x^2-100}=\frac{11}{\left(x-10\right)\left(x+1\right)}+\frac{11}{\left(x-9\right)\left(x+2\right)}+...+\frac{11}{\left(x-1\right)\left(x+10\right)}\)
Vê trái:
\(=\frac{2}{\left(x-1\right)\left(x+1\right)}+\frac{4}{\left(x-2\right)\left(x+2\right)}+...+\frac{20}{\left(x-10\right)\left(x+10\right)}\)
\(=\frac{\left(x+1\right)-\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{\left(x+2\right)-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+...+\frac{\left(x+10\right)-\left(x-10\right)}{\left(x+10\right)\left(x-10\right)}\)
\(=\frac{1}{x-1}-\frac{1}{x+1}+\frac{1}{x-2}-\frac{1}{x+2}+...+\frac{1}{x-10}-\frac{1}{x+10}\)
\(=\left(\frac{1}{x-1}+\frac{1}{x-2}+...+\frac{1}{x-10}\right)-\left(\frac{1}{x+1}+\frac{1}{x+2}+...+\frac{1}{x+10}\right)\)
Vế phải:
\(=\frac{\left(x+1\right)-\left(x-10\right)}{\left(x-10\right)\left(x+1\right)}+\frac{\left(x+2\right)-\left(x-9\right)}{\left(x-9\right)\left(x+2\right)}+...+\frac{\left(x+10\right)-\left(x-1\right)}{\left(x-1\right)\left(x+10\right)}\)
\(=\frac{1}{x-10}-\frac{1}{x+1}+\frac{1}{x-9}-\frac{1}{x+2}+...+\frac{1}{x-1}-\frac{1}{x+10}\)
\(=\left(\frac{1}{x-1}+\frac{1}{x-2}+...+\frac{1}{x-10}\right)-\left(\frac{1}{x+1}+\frac{1}{x+2}+...+\frac{1}{x+10}\right)\) = vế phải
=> đpcm
\(\left|x+y+\frac{1}{100}\right|\)\(=-\left|x\right|-\left|y\right|-\left|\frac{1}{10}\right|\)
\(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+...+\left|x+200\right|=201\)
5. Tìm x biết:
a, \(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+...+\left|x+10\right|=11x+1\)
b, \(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+...+\left|x+\frac{100}{101}\right|=101x\)
cho hai đa thức khác nhau sau
\(f\left(x\right)=x^2+ax+b\)
\(G\left(x\right)=x^2+mx+n\)
biết \(f\left(1\right)+f\left(10\right)+f\left(100\right)=G\left(1\right)+G\left(10\right)+G\left(100\right)\)
với giá trị nào của x thì \(f\left(x\right)=G\left(x\right)\)
Ta có \(f\left(1\right)+f\left(10\right)+f\left(100\right)=1+a+b+100+10a+b+10000+100a+b\)
\(=10101+111a+3b\)
Tương tự \(G\left(1\right)+G\left(10\right)+G\left(100\right)=10101+111m+3n\)
Từ đây ta có \(111a-3b=111m-3n\Rightarrow111\left(a-m\right)-3\left(b-n\right)=0\)
Xét \(h\left(x\right)=f\left(x\right)-G\left(x\right)\) , khi đó \(h\left(x_0\right)=f\left(x_0\right)-G\left(x_0\right)\)
\(=ax_0+b-mx_0-n=\left(a-m\right)x_0+\left(b-n\right)\)
Để \(h\left(x_0\right)=0\Rightarrow\left(a-m\right)x_0+\left(b-n\right)=0\Rightarrow3\left(a-m\right)x_0+3\left(b-n\right)=0\)
Ta đã có \(111a-3b=111m-3n\Rightarrow111\left(a-m\right)-3\left(b-n\right)=0\)
Vậy nên \(3x_0=111\Rightarrow x_0=37\)
Tóm lại \(f\left(37\right)=G\left(37\right)\)
tìm x biết
a)\(x+2x+3x+4x+...+2015x=2016\times2017\)
b)\(1-3+3^2-3^3+...+\left(-3\right)^x=\frac{9^{1008}-1}{4}\)
c)\(\left|x+1\right|+\left|x+2\right|+...+\left|x+100\right|=605x\)
d)tìm x nguyên biết \(\left|x-1\right|+\left|x-2\right|+...+\left|x-100\right|=2500\)
e) tìm x nguyên biết \(2004=\left|x-4\right|+\left|x-10\right|+\left|x+101\right|+\left|x+99x\right|+\left|x+1000\right|\)
Tính:
a) \(\dfrac{x^2}{\left(x-y\right)\left(x-z\right)}+\dfrac{y^2}{\left(y-z\right)\left(y-x\right)}+\dfrac{z^2}{\left(z-x\right)\left(z-y\right)}\)
b) \(\dfrac{x^2-yz}{\left(x+y\right)\left(x+z\right)}+\dfrac{y^2-zx}{\left(y+z\right)\left(y+x\right)}+\dfrac{z^2-xy}{\left(z+x\right)\left(z+y\right)}\)
c) \(\dfrac{1}{x\left(x-y\right)\left(x-z\right)}+\dfrac{1}{y\left(y-x\right)\left(y-z\right)}+\dfrac{1}{z\left(z-x\right)\left(z-y\right)}\)
d) \(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+...+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)
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d)
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+.....+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)=\(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+.....-\dfrac{1}{x+99}+\dfrac{1}{x+100}\)=\(\dfrac{1}{x}-\dfrac{1}{x+100}\)
=\(\dfrac{x+100}{x\left(x+100\right)}-\dfrac{x}{x\left(x+100\right)}\)
=\(\dfrac{x+100-x}{x\left(x+100\right)}=\dfrac{100}{x\left(x+100\right)}\)
A=\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+...+\frac{1}{\left(x+9\right)\left(x+100\right)}\)
thực hiện phép tính
Phân thức cuối hình như mẫu sai rồi bạn
Phải là (x+9)(x+10) mới đúng chứ
Nếu đề bài đúng thì sẽ làm như sau:
A = \(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+...+\frac{1}{\left(x+9\right)\left(x+100\right)}\) (ĐKXĐ tự tìm nhé, chứ viết dài lắm)
A = \(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+...+\frac{1}{x+9}-\frac{1}{x+100}\)
A = \(\frac{1}{x}-\frac{1}{x+100}\)
A = \(\frac{x+100-x}{x\left(x+100\right)}\)
A = \(\frac{100}{x\left(x+100\right)}\)
Vậy A = \(\frac{100}{x\left(x+100\right)}\)
Chúc bn học tốt!!
c)\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+\left(x+4\right)+\left(x+5\right)=90\)
d)\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+.....+\left(x+99\right)+\left(x+100\right)+=20150\)
c) (x+1) + (x+2) + ... + (x+5) = 90
=> 5x + ( 1 + 2 + ... + 5 ) = 90
5x + 15 = 90
5x = 90 - 15
5x = 75
x = 75 : 5
x = 15
d) (x+1) + (x+2) + .... + (x+100) = 20150
=> 100x + ( 1+2+...+100 ) = 20150
100x + 5050 = 20150
100x = 20150 - 5050
100x = 15100
x = 15100 : 100
x = 151
Ta có : (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) = 90
<=> x + x + x+ x + x + (1 + 2 + 3 + 4 + 5) = 90
<=> 5x + 15 = 90
=> 5x = 75
=> x = 15
c) \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+\left(x+4\right)+\left(x+5\right)=90\)
\(\Leftrightarrow x+1+x+2+x+3+x+4+x+5=90\)
\(\Leftrightarrow5x+\left(1+2+3+4+5\right)=90\)
\(\Leftrightarrow5x+15=90\)
\(\Leftrightarrow5x=75\)
\(\Leftrightarrow x=15\)
d) \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+......+\left(x+99\right)+\left(x+100\right)=20150\)
\(\Leftrightarrow x+1+x+2+x+3+......+x+99+x+100=20150\)
\(\Leftrightarrow100x+\left(1+2+3+.....+99+100\right)=20150\)
\(\Leftrightarrow100x+5050=20150\)
\(\Leftrightarrow100x=15100\)
\(\Leftrightarrow x=151\)
Tìm x :
\(a,x.\left(5x+10\right)+5.\left(x23+30\right)-9.\left(3x+5\right)=100\)
\(b,10.\left(8x+9x\right)+8.\left(2x-1\right)-2.\left(5-6x\right)=20\)
\(c,\left(x.5^2\right).\left(194+x.20\right)=50\)