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Ta có:\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}=\frac{1}{x}-\frac{1}{x+6}=\frac{x+6}{x\left(x+6\right)}-\frac{x}{x\left(x+6\right)}=\frac{6}{x\left(x+6\right)}\)k mik nha
ĐKXĐ : \(x\ne0;-1;-2;-3;-4;-5;-6\)
Giá trị của của tổng trên rất dễ
Giá trị của nó là:
\(\frac{1}{x}-\frac{1}{x+6}\)
\(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)
\(\Rightarrow\frac{x^2+x+1}{x^3-1}+\frac{2x^2-5}{x^3-1}=\frac{4\left(x-1\right)}{x^3-1}\)
\(\Rightarrow x^2+x+1+2x^2-5=4x-4\)
\(\Rightarrow3x^2-3x=0\)
\(\Rightarrow3x\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
a/ \(\left(x^4+\frac{1}{x^4}\right)\left(x^3+\frac{1}{x^3}\right)-\left(x+\frac{1}{x}\right)\)
\(=x^7+x+\frac{1}{x}+\frac{1}{x^7}-\left(x+\frac{1}{x}\right)=x^7+\frac{1}{x^7}\)
b/ Ta có:
\(\left(x+\frac{1}{x}\right)^2=49\)
\(\Leftrightarrow x^2+\frac{1}{x^2}=49-2=47\)
\(\left(x+\frac{1}{x}\right)^3=343\)
\(\Leftrightarrow x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}\right)=343\)
\(\Leftrightarrow x^3+\frac{1}{x^3}=343-3.7=322\)
\(\Rightarrow\left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)=47.322=15134\)
\(\Leftrightarrow x^5+\frac{1}{x}+x+\frac{1}{x^5}=15134\)
\(\Leftrightarrow x^5+\frac{1}{x^5}=15134-7=15127\)
a)\(\left(x^4+\frac{1}{x^4}\right)\left(x^3+\frac{1}{x^3}\right)-\left(x+\frac{1}{x}\right)=x^7+x+\frac{1}{x}+\frac{1}{x^7}-x-\frac{1}{x}\)
=\(x^7+\frac{1}{x^7}\)
\(x+\frac{1}{x}=7\)
=>\(x\left(x+\frac{1}{x}\right)=7x\)
=>\(^{x^2-7x+1=0}\)
=>\(x=\frac{7+3\sqrt{5}}{2};x=\frac{7-3\sqrt{5}}{2}loại\)
=>\(x^5+\frac{1}{x^5}=15127\)
\(P=\left(\frac{x}{x^2-25}-\frac{x-5}{x^2+5x}\right):\frac{2x-5}{x^2+5x}+\frac{x}{5-x}\)
\(P=\left[\frac{x}{\left(x-5\right)\left(x+5\right)}-\frac{x-5}{x\left(x+5\right)}\right]:\frac{2x-5}{x\left(x+5\right)}+\frac{x}{5-x}\)
\(P=\frac{x^2-\left(x-5\right)\left(x-5\right)}{\left(x-5\right)\left(x+5\right)x}.\frac{x\left(x+5\right)}{2x-5}+\frac{x}{5-x}\)
\(P=\frac{x^2-x^2+10x-25}{x\left(x-5\right)\left(x+5\right)}.\frac{x\left(x+5\right)}{2x-5}+\frac{x}{5-x}\)
\(P=\frac{10x-25}{x\left(x-5\right)\left(x+5\right)}.\frac{x\left(x+5\right)}{2x-5}+\frac{x}{5-x}\)
\(P=\frac{5\left(2x-5\right).x\left(x+5\right)}{x\left(x-5\right)\left(x+5\right)\left(2x-5\right)}+\frac{x}{5-x}\)
\(P=\frac{5}{x-5}+\frac{x}{5-x}\)
\(P=\frac{5}{x-5}-\frac{x}{x-5}\)
\(P=\frac{5-x}{x-5}\)
\(P=\frac{-\left(x-5\right)}{x-5}\)
\(P=-1\)
=> Giá trị của biểu thức P không phụ thuộc vào biến
đpcm
\(\frac{2}{x-3}+\frac{x-5}{x-1}=1\)
\(ĐKXĐ:x\ne1;x\ne3\)
\(pt\Leftrightarrow\frac{2x-2}{x^2-4x+3}+\frac{x^2-8x+15}{x^2-4x+3}=1\)
\(\Leftrightarrow\frac{x^2-6x+13}{x^2-4x+3}=1\)
\(\Leftrightarrow x^2-6x+13=x^2-4x+3\)
\(\Leftrightarrow-2x+10=0\Leftrightarrow x=-5\left(t/mđkxđ\right)\)
Vậy pt có 1 nghiệm là -5
2/x - 3 + x - 5/x - 1 = 1
2(x - 1) + (x - 5)(x - 3) = (x - 3)(x - 1)
-6x + 13 + x^2 = x^2 - 4x + 3
-6x + 13 = -4x + 3
13 = -4x + 3 + 6x
13 = 2x + 3
13 - 3 = 3x
10 = 2x
5 = x
=> x = 5