Giai pt :x^2-x/x^2-x+1 - x^2-x+2/x^2-x-2=1
giai pt |x+1|+3|x-1|=x+2+|x|+2|x-2|
1. Cho pt: x2 -2(m+1)x+m2=0 (1). Tìm m để pt có 2 nghiệm x1 ; x2 thỏa mãn (x1-m)2 + x2=m+2.
2. Giai pt: \(\left(x-1\right)\sqrt{2\left(x^2+4\right)}=x^2-x-2\)
3. Giai hệ pt: \(\left\{{}\begin{matrix}\frac{1}{\sqrt[]{x}}-\frac{\sqrt{x}}{y}=x^2+xy-2y^2\left(1\right)\\\left(\sqrt{x+3}-\sqrt{y}\right)\left(1+\sqrt{x^2+3x}\right)=3\left(2\right)\end{matrix}\right.\)
4. Giai pt trên tập số nguyên \(x^{2015}=\sqrt{y\left(y+1\right)\left(y+2\right)\left(y+3\right)}+1\)
giai pt 2(x^2+x+1)^2-7(x-1)^2=13(x^3-1)
2(x2+x+1)2-7(x-1)2=13(x3-1)
<=> 2(x2+x+1)2-7(x-1)2-13(x3-1)=0
<=>2(x2+x+1)2-14(x3-1)+(x3-1)-7(x-1)2=0
<=> 2(x2+x+1)(x2+x+1-7x+7)+(x-1)(x2+x+1-7x+7)=0
<=> (2x2+2x+2)(x2-6x+8)+(x-1)(x2-6x+8)=0
<=> (x2-6x+8)(2x2+3x+1)=0
<=> (x2-4x-2x+8)(2x2+2x+x+1)=0
<=> [x(x-4)-2(x-4)][2x(x+1)+(x+1)]=0
<=> (x-4)(x-2)(x+1)(2x+1)=0
Đến đây dễ rồi nhé bạn
Giai Pt sau | 4x + 2| - 5x + 3 = 0 nhận được nghiệm?
Giai Pt sau |-4x| = 2 ( x + 1) ta nhận được nghiệm?
Giai Pt sau |x + 2| + x^2 - ( 3 + x) x = 0 ta nhận được nghiệm?
giai pt (x+1)(x+2)(x+5)(x+10)=10x^2
giai pt : a. x^4/2x^2+1 + 2x^2+1/x^4=2
b.(x/x-1)^2+(x/x+1)^2=10/9
c. x^3+3x^2-10x-24=0
giai pt:
1)can(2(x+1)(x+3))+can((x+1)(x-1))=2(x+1)
2)can(x)-can(x+1)-can(x+4)+can(x+9)=0
giai pt:
\(x^2+x+1=\left(2-x\right)\left(\sqrt{x^2+x+2}-1\right)\)
\(\Leftrightarrow x^2+x+2-1=\left(2-x\right)\left(\sqrt{x^2+x+2}-1\right)\)
\(\Leftrightarrow\left(\sqrt{x^2+x+2}-1\right)\left(\sqrt{x^2+x+2}+1\right)-\left(2-x\right)\left(\sqrt{x^2+x+2}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x^2+x+2}-1\right)\left(\sqrt{x^2+x+2}+1-2+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+x+2}=1\left(1\right)\\\sqrt{x^2+x+2}=1-x\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x^2+x+1=0\left(vn\right)\)
\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}1-x\ge0\\x^2+x+2=\left(1-x\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\3x=-1\end{matrix}\right.\) \(\Rightarrow x=-\frac{1}{3}\)
Giai pt sau:
1/x^2-3x+2 +1/x^2-5x+6 +1/x^2-7x+12 =2(Tất cả =2 nhé!)
=>\(\dfrac{-1}{x-1}+\dfrac{1}{x-2}-\dfrac{1}{x-2}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-4}=2\)
=>\(\dfrac{1}{x-4}-\dfrac{1}{x-1}=2\)
=>\(\dfrac{x-1-x+4}{x^2-5x+4}=2\)
=>2x^2-10x+8=3
=>2x^2-10x+5=0
=>\(x=\dfrac{5\pm\sqrt{15}}{2}\)