Giải phương trình \(\frac{1}{\left(x^2+5\right)\left(x^2+4\right)}+\frac{1}{\left(x^2+4\right)\left(x^2+3\right)}+\frac{1}{\left(x^2+3\right)\left(x^2+2\right)}+\frac{1}{\left(x^2+2\right)}+\frac{1}{\left(x^2+1\right)}\)
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 :\(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x^2}\right)^2=\left(x+4\right)^2\)
\(\Leftrightarrow8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)\left[\left(x^2+\frac{1}{x^2}\right)-\left(x+\frac{1}{x}\right)^2\right]=\left(x+4\right)^2.ĐKXĐ:x\ne0\)
\(\Leftrightarrow8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)\left(x^2+\frac{1}{x^2}-x^2-2-\frac{1}{x^2}\right)=\left(x+4\right)^2\)
\(\Leftrightarrow8\left(x+\frac{1}{x}\right)^2-8\left(x^2+\frac{1}{x^2}\right)=\left(x+4\right)^2\)
\(\Leftrightarrow8\left[\left(x+\frac{1}{x}\right)^2-\left(x^2+\frac{1}{x^2}\right)\right]=\left(x+4\right)^2\)
\(\Leftrightarrow8\left(x^2+2+\frac{1}{x^2}-x^2+\frac{1}{x^2}\right)=\left(x+4\right)^2\)
\(\Leftrightarrow16=\left(x+4\right)^2\)
\(\Leftrightarrow x^2+8x+16=16\)
\(\Leftrightarrow x^2+8x=0\)
\(\Leftrightarrow x\left(x+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\left(l\right)\\x=-8\left(n\right)\end{cases}}\)
V...\(S=\left\{-8\right\}\)
^^
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
Giải phương trình:
\(\frac{2\left(x-\sqrt{3}\right)\left(x-\sqrt{2}\right)}{\left(1-\sqrt{2}\right)\left(1-\sqrt{3}\right)}+\frac{3\left(x-1\right)\left(x-\sqrt{3}\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}-\sqrt{3}\right)}+\frac{4\left(x-1\right)\left(x-\sqrt{2}\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}-\sqrt{2}\right)}=3x-1\)
giải hộ mk bài này nha????
giải phương trình :
1)\(5\left(\frac{x^2-4}{x^2-1}\right)-\left(\frac{x+2}{x-1}\right)^2-\left(\frac{x-2}{x+1}\right)^2=0\)
2)\(x^2+\left(\frac{x}{x-1}\right)^2=8\)
3)\(x^2+\left(\frac{81x^2}{\left(x+9\right)^2}\right)=40\)
4)\(\frac{\left(x-1\right)^2}{x^2}+\frac{\left(x-1\right)^2}{\left(x-2\right)^2}=\frac{40}{49}\)
5)\(\left(\frac{x}{x+1}\right)^2+\left(\frac{x}{x-1}\right)^2=90\)
giúp nha!!!!
Giải phương trình sau:
a) \(\frac{10}{\left(x+5\right)\left(x-1\right)}+\frac{3}{1-x}=\frac{5}{x+5}\)
b) \(\frac{x-1}{x+2}+\frac{x+3}{x-4}=\frac{2}{\left(x-2\right)\left(4-x\right)}\)
c) \(\frac{7x-3}{x-x^3}=\frac{1}{x-1}-\frac{5}{x\left(x-1\right)}\)
d) \(\frac{1}{\left(x+2\right)}+\frac{1}{\left(x+3\right)}=\frac{1}{\left(x+2\right)\left(x+3\right)}\)
Giai phương trình sau :
a) \(\frac{10}{\left(x+5\right)\left(x-1\right)}+\frac{3}{1-x}=\frac{5}{x+5}\)
ĐKXĐ : \(x\ne1;x\ne-5\)
Với điều kiện trên ta có :
\(\Leftrightarrow\)\(\frac{10}{\left(x+5\right)\left(x-1\right)}+\frac{-3}{x-1}=\frac{5}{x+5}\)
\(\Leftrightarrow10-3\left(x+5\right)=5\left(x-1\right)\)
\(\Leftrightarrow10-3x-15=5x-5\)
\(\Leftrightarrow-8x=0\)
\(\Leftrightarrow x=0\) (nhận)
Vậy : \(S=\left\{0\right\}\)
d) \(\frac{1}{\left(x+2\right)}+\frac{1}{\left(x+3\right)}=\frac{1}{\left(x+2\right)\left(x+3\right)}\)
ĐKXĐ : \(x\ne-2;x\ne-3\)
\(\Leftrightarrow x+3+x+2=1\)
\(\Leftrightarrow2x=-4\)
\(\Leftrightarrow x=-2\) (không nhận)
Vậy : \(S=\varnothing\)
giải phương trình: \(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2+4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2=\left(x+4\right)^2\)
Giải phương trình:
\(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2=\left(x+4\right)^2\)
Giải phương trình:
\(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2=\left(x-4\right)^2\)
ĐK: x khác 0
Đặt \(x+\frac{1}{x}=a\)\(\Rightarrow\left(x+\frac{1}{x}\right)^2=a^2\Leftrightarrow a^2=x^2+\frac{1}{x^2}+2\cdot x\cdot\frac{1}{x}\Leftrightarrow a^2-2=x^2+\frac{1}{x^2}\)
Có:
\(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2\)
\(=8a^2+4\left(a^2-2\right)^2-4\left(a^2-2\right)a^2\)
\(=8a^2+4\left(a^4-4a^2+4\right)-4\left(a^4-2a^2\right)\)
\(=8a^2+4a^4-16a^2+16-4a^4+8a^2=16\)
Thay \(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2=16\)
vào phương trình, ta có: \(\left(x-4\right)^2=16\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=-4\\x-4=4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=8\end{cases}}\)Mà điều kiện x khác 0 nên x=8
Vậy phương trình có nghiệm x=8
Giải phương trình
\(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2=\left(x+4\right)^2\)