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ĐKXĐ: x ≠ \(\pm\) 1
Từ phương trình ban đầu suy ra:
\(x^2\left(x+1\right)^2+x^2\left(x-1\right)^2=\frac{10}{9}.\left(x^2-1\right)^2\)
⇒ \(x^4+2x^3+x^2+x^4-2x^3+x^2=\frac{10}{9}\left(x^4-2x^2+1\right)\)
⇒ \(18\left(x^4+x^2\right)=10\left(x^4-2x^2+1\right)\)
⇒ \(4x^4+19x^2-5=0\Leftrightarrow\left(x^2+5\right)\left(4x^2-1\right)=0\)
⇔ \(x^2=\frac{1}{4}\Leftrightarrow x=\pm\frac{1}{2}\)( thỏa mãn ĐKXĐ)
Vậy ...
\(\left(x^2-6x+9\right)+15\left(x^2-6x+10\right)=1\)
\(\Leftrightarrow\left(x-3\right)^2+15\left[\left(x-3\right)^2+1\right]=1\)
\(\Leftrightarrow16\left(x-3\right)^2+15=1\)
\(\Leftrightarrow16\left(x-3\right)^2=-14\)
=> Phương trình vô nghiệm
\(\left(x^2-6x+9\right)-15\left(x^2-6x+10\right)=1\)
Đặt : \(x^2-6x+9=\left(x-3\right)^2=t\) thay vào pt ta được :
\(t^2-15\left(t+1\right)=1\)
\(\Leftrightarrow t^2-15t-16=0\)
\(\Leftrightarrow\left(t+1\right)\left(t-16\right)=0\)
\(\Leftrightarrow t=\left\{{}\begin{matrix}16\\-1\end{matrix}\right.\)
với : \(t=-1\) thì \(\left(x-3\right)^2=-1\)
\(\Rightarrow ptvonghiem\)
Với : \(t=16\) thì \(\left(x-3\right)^2=16\)
\(\Leftrightarrow x\in\left\{{}\begin{matrix}7\\-1\end{matrix}\right.\)
\(vay...\)
ĐK x khác +- 1 :
\(pt\Leftrightarrow\left(\frac{x}{x-1}\right)^2+\left(\frac{x}{x+1}\right)^2-\frac{2x^2}{\left(x-1\right)\left(x+1\right)}+\frac{2x^2}{\left(x-1\right)\left(x+1\right)}=\frac{10}{9}\)
<=> \(\left(\frac{x}{x-1}+\frac{x}{x+1}\right)^2-\frac{2x^2}{\left(x-1\right)\left(x+1\right)}=\frac{10}{9}\)
<=> \(\left[\frac{2x^2}{\left(x-1\right)\left(x+1\right)}\right]^2-\frac{2x^2}{\left(x-1\right)\left(x+1\right)}=\frac{10}{9}\)
Đặt \(\frac{2x^2}{\left(x-1\right)\left(x+1\right)}=t\)
pt <=> \(t^2-t=\frac{10}{9}\Leftrightarrow9t^2-9t-10=0\)
Đến đây tự làm tiếp nha
\(\left(\frac{x}{x-1}\right)^2+\left(\frac{x}{x+1}\right)^2=\frac{10}{9}\Leftrightarrow\frac{x^2}{\left(x-1\right)^2}+\frac{x^2}{\left(x+1\right)^2}=\frac{10}{9}\)
\(\Leftrightarrow\frac{x^2\left(x+1\right)^2+x^2\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)^2}=\frac{10}{9}\Leftrightarrow\frac{x^2\left[\left(x+1\right)^2-\left(x-1\right)^2\right]}{\left[\left(x-1\right)\left(x+1\right)\right]^2}=\frac{10}{9}\)
\(\Leftrightarrow\frac{x^2\left(x+1-x+1\right)\left(x+1+x-1\right)}{\left(x^2-1\right)^2}=\frac{10}{9}\Leftrightarrow\frac{x^2.2.2x}{x^4-2x^2+1}=\frac{10}{9}\)
\(\Leftrightarrow36x^3=10x^4-20x^2+10\Leftrightarrow18x^3=5x^4-10x^2+5\Leftrightarrow5x^4-18x^3-10x^2\)+5=0
đến đây tự giải tiếp
ĐK:\(x\ne1;x\ne-1\)
\(pt\Leftrightarrow\frac{x^2}{\left(x-1\right)^2}+\frac{x^2}{\left(x+1\right)^2}=\frac{10}{9}\)
\(\Leftrightarrow\frac{9x^2\left(x+1\right)^2+9x^2\left(x-1\right)^2-10\left(x-1\right)^2\left(x+1\right)^2}{9\left(x-1\right)^2\left(x+1\right)^2}=0\)
\(\Leftrightarrow9x^2\left(x+1\right)^2+9x^2\left(x-1\right)^2-10\left(x-1\right)^2\left(x+1\right)^2=0\)
\(\Leftrightarrow9x^4+18x^3+9x^2+9x^4-18x^3+9x^2-10x^4+20x^2-10=0\)
\(\Leftrightarrow8x^4+38x^2-10=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=\frac{1}{4}\\x^2=5\left(l\right)\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)
a) x≠0
x^2+4/x^2=(x-2/x)^2+4
Pt<=>(x-2/x)^2+4-4(x-2/x)-9=0
<=>(x-2/x)^2-4(x-2/x)-5=0
Đặt t=x-2/x
Pt<=> t^2-4t-5=0
Đến đây tìm t rồi quy đồng lên tìm ra x nhé!
b)x>=-2
(√(x+5)-√(x+2))(1+√(x^2+7x+10))=3
<=> (√(x+5)-√(x+2))(1+√(x+5)(x+2))=3
Đặt √(x+5)=a;√(x+2)=b (a>b>=0)
=> a^2-b^2=3
Pt<=>(a-b)(1+ab)=a^2-b^2
<=>(a-b)(1+ab)=(a-b)(a+b)
Mà a>b=>a-b>0
=>ab+1=a+b
<=>(a-1)(b-1)=0
a=1=>x+5=1<=>x=-4(loại)
b=1=>x+2=1<=>x=-1 (thoả mãn)
Vậy x=-1
a. 1/2*(-x^5)
b. (10-x)^5
c. x-4+(4-x) = 0
d. 6-2x-(3-x) = 3-x