\(\frac{4}{x+1}+\frac{x^2+x-2}{\left(x+1\right)\left(x+2\right)}\)
Những câu hỏi liên quan
left(frac{1}{2+2sqrt{x}}+frac{1}{2-2sqrt{x}}-frac{x^2+1}{1-x^2}right)left(1+frac{1}{x}right)left(frac{2-2sqrt{x}+2+2sqrt{x}}{left(2+2sqrt{x}right)left(2-2sqrt{x}right)}-frac{x^2+1}{1-x^2}right)left(1+frac{1}{x}right)left(frac{4}{4-4x}-frac{x^2+1}{left(1-xright)left(1+xright)}right)left(1+frac{1}{x}right)left(frac{1+x-x^2-1}{left(1-xright)left(1+xright)}right)left(1+frac{1}{x}right)frac{xleft(1-xright)}{left(1-xright)left(1+xright)}.frac{x+1}{x}1
Đọc tiếp
\(\left(\frac{1}{2+2\sqrt{x}}+\frac{1}{2-2\sqrt{x}}-\frac{x^2+1}{1-x^2}\right)\left(1+\frac{1}{x}\right)\)
\(=\left(\frac{2-2\sqrt{x}+2+2\sqrt{x}}{\left(2+2\sqrt{x}\right)\left(2-2\sqrt{x}\right)}-\frac{x^2+1}{1-x^2}\right)\left(1+\frac{1}{x}\right)\)
\(=\left(\frac{4}{4-4x}-\frac{x^2+1}{\left(1-x\right)\left(1+x\right)}\right)\left(1+\frac{1}{x}\right)\)
\(=\left(\frac{1+x-x^2-1}{\left(1-x\right)\left(1+x\right)}\right)\left(1+\frac{1}{x}\right)=\frac{x\left(1-x\right)}{\left(1-x\right)\left(1+x\right)}.\frac{x+1}{x}=1\)
Nếu bạn bảo kiểm tra thì lời giải đúng rồi nhé!
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Giải phương trình:
\(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{^{x^2}}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2=\left(x+4\right)^2\)
\(\frac{\left(2009-x\right)^2+\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}{\left(2009-x\right)^2-\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}=\frac{19}{49}\)
Giải phương trình \(\frac{1}{\left(x^2+5\right)\left(x^2+4\right)}+\frac{1}{\left(x^2+4\right)\left(x^2+3\right)}+\frac{1}{\left(x^2+3\right)\left(x^2+2\right)}+\frac{1}{\left(x^2+2\right)}+\frac{1}{\left(x^2+1\right)}\)
AYUASGSHXHFSGDB HAGGAHAJF
Tìm x biết: \(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2\)=\(\left(x+4\right)^2\)
Đặt \(t=\left(x+\frac{1}{x}\right)^2\)\(\Rightarrow\)\(x^2+\frac{1}{x^2}=t-2\)điều kiện t>=0,x # 0
Phương trình trở thành
8t +4(t-2)2 - 4(t-2)2t =(x+4)2
8t + 4t2 - 16t + 16 -4t3 + 16t2 - 16t=(x+4)2
-4t3 + 20t2 -24t=x2 +8x
-4t(t2 -5t +6)=x(x+8)
-4t(t-2)(t-3)=x(x+8)
Mình chỉ giúp dược tới đó
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Tính A = \(\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{2}{\left(x+3\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+5\right)}+\frac{8}{\left(x+5\right)\left(x+6\right)}\)
A= \(\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{2}{x+3}-...+\frac{8}{x+5}-\frac{8}{x+6}\)
A=\(\frac{1}{x+1}+\frac{1}{x+3}+\frac{2}{x+4}+\frac{4}{x+5}-\frac{8}{x+6}\)
Rồi tiếp tục làm nhé bạn.
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Tính:
\(\frac{1}{x.\left(x+1\right)}+\frac{1}{\left(x+1\right).\left(x+2\right)}+\frac{1}{\left(x+2\right).\left(x+3\right)}+\frac{1}{\left(x+3\right).\left(x+4\right)}+\frac{1}{\left(x+4\right).\left(x+5\right)}+\frac{1}{x+5}\)
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{x+5}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}\)
\(=\frac{1}{x}\)
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ta có: \(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{x+5}\)
=\(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}\)
= \(\frac{1}{x}\)
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tim x
\(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2=\left(x+4\right)^2\)
Nhân hết ra,giải phương trình bậc cao đi
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Giải phương trình :\(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x^2}\right)^2=\left(x+4\right)^2\)
\(\Leftrightarrow8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)\left[\left(x^2+\frac{1}{x^2}\right)-\left(x+\frac{1}{x}\right)^2\right]=\left(x+4\right)^2.ĐKXĐ:x\ne0\)
\(\Leftrightarrow8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)\left(x^2+\frac{1}{x^2}-x^2-2-\frac{1}{x^2}\right)=\left(x+4\right)^2\)
\(\Leftrightarrow8\left(x+\frac{1}{x}\right)^2-8\left(x^2+\frac{1}{x^2}\right)=\left(x+4\right)^2\)
\(\Leftrightarrow8\left[\left(x+\frac{1}{x}\right)^2-\left(x^2+\frac{1}{x^2}\right)\right]=\left(x+4\right)^2\)
\(\Leftrightarrow8\left(x^2+2+\frac{1}{x^2}-x^2+\frac{1}{x^2}\right)=\left(x+4\right)^2\)
\(\Leftrightarrow16=\left(x+4\right)^2\)
\(\Leftrightarrow x^2+8x+16=16\)
\(\Leftrightarrow x^2+8x=0\)
\(\Leftrightarrow x\left(x+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\left(l\right)\\x=-8\left(n\right)\end{cases}}\)
V...\(S=\left\{-8\right\}\)
^^
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bạn ghi sai đề ở chỗ \(\left(x+\frac{1}{x}\right)^2\)chứ ko phải \(\left(x+\frac{1}{x^2}\right)^2\)nhé
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\(0=-\frac{\left(x+2\right)^2+12}{\left(x+2\right)^2}+\frac{\left(x+1\right)^2+1}{\left(x+1\right)^2}-\frac{\left(x+3\right)^2+3}{\left(x+3\right)^2}+\frac{\left(x+4\right)^2+4}{\left(x+4\right)^2}\)
Tính D= \(\frac{x^4-\left(x-1\right)^2}{\left(x+1\right)^2-x^2}+\frac{\left(x-1\right)^2-x^2}{1-\left(x^2+x\right)^2}+\frac{\left(x^2-x\right)^2-1}{x^4-\left(x+1^2\right)}\)