giải pt :\(\frac{2x-5}{2-x}>-1\)
Câu 1 : Giải pt: \(8x^2+\sqrt{\frac{1}{x}}=\frac{5}{2}\)
Câu 2: Giải pt: \(\frac{2x^2}{\left(3-\sqrt{9+2x}\right)^2}=x+21\\\)
\(\frac{2x}{x+1}+\frac{18}{x^2+2x-3}=\frac{2x-5}{x+3}\)
giải pt
\(\frac{2x}{x+1}+\frac{18}{x^2+2x-3}=\frac{2x-5}{x+3}ĐKXĐ:x\ne-1;-3\)
\(\frac{2x}{x+1}+\frac{18}{\left(x-1\right)\left(x+3\right)}=\frac{2x-5}{x+3}\)
\(2x\left(x-1\right)\left(x+3\right)+18\left(x+1\right)=\left(2x-5\right)\left(x+1\right)\left(x-1\right)\)
\(4x^2+12x+18=-2x-5x^2+5\)
\(4x^2+12x+18+2x+5x^2-5=0\)
\(9x^2-14x+13=0\)
=> vô nghiệm
giải PT \(\frac{3x-1}{x-1}-\frac{2x+5}{x+3}=1-\frac{4}{x^2+2x-3}\)
\(\frac{3\text{x}-1}{x-1}-\frac{2\text{x}+5}{x+3}=1-\)\(\frac{4}{x^2+2\text{x}-3}\) \(\left(\text{Đ}K\text{X}\text{Đ}:x\ne1;x\ne-3\right)\)
\(\Leftrightarrow\frac{\left(3\text{x}-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{\left(2\text{x}+5\right)\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}=\frac{\left(x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{4}{\left(x-1\right)\left(x+3\right)}\)
\(\Rightarrow\left(3\text{x}-1\right)\left(x+3\right)-\left(2\text{x}+5\right)\left(x-1\right)=\left(x-1\right)\left(x+3\right)-4\)
\(\Leftrightarrow3\text{x}^2+8\text{x}-3-2\text{x}^2-3\text{x}+5=x^2+2\text{x}-3-4\)
\(\Leftrightarrow3\text{x}^2-2\text{x}^2-x^2+8\text{x}-3\text{x}-2\text{x}=-3-4+3-5\Leftrightarrow3\text{x}=-9\Leftrightarrow x=-3\)(không thỏa mãn ĐKXĐ)
Vậy pt vô nghiệm
giải pt: a. (x - 2)(x+1)(x+3) = (x+3)(x+1)(2x-5)
b. \(5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}+\frac{3x-1}{x-4}\)
a.\(\Leftrightarrow\left(x+3\right)\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(-x^2+2x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=3\\x=-1\end{matrix}\right.\)
(x-2)(x+1)(x+3)=(x+3)(x+1)(2x-58)
\(x^3+2x^2-5x-6\)=\(2x^3+3x^2-14x-15\)
\(-x^3-x^2+9x+9=0\)
\(-x^2\left(x+1\right)+9\left(x+1\right)=0\)
\(\left(x+1\right)\left(9-x^2\right)\)=0
(x+1)(3-x)(3+x)=0
*x+1=0 =>x=-1
*3-x=0=>x=3
*3+x=0=>x=-3
Giải PT:
\(\frac{5}{x^2}+\frac{2x}{\sqrt{x^2+5}}=1\)
Giải PT:
\(\frac{5}{x^2}+\frac{2x}{\sqrt{x^2+5}}=1\)
giải pt
\(\frac{x-2}{x+1}+1=\frac{5}{2x-2}\)
giải pt
\(\frac{1}{\left(x^2+2x+2\right)^2}+\frac{1}{\left(x^2+2x+3\right)^2}=\frac{5}{4}\)
Giải pt : \(\frac{3x-1}{x-1}-\frac{2x+5}{x+3}+\frac{4}{x^2+2x-3}=1\)
\(\frac{3x-1}{x-1}-\frac{2x-5}{x+3}+\frac{4}{x^2+2x-3}=1\)
\(\frac{3x-1}{x-1}-\frac{2x-5}{x+3}+\frac{4}{\left(x+1\right)^2-4}=1\)
\(\frac{3x-1}{x-1}-\frac{2x-5}{x+3}+\frac{4}{\left(x+1+2\right)\left(x+1-2\right)}=1\)
\(\frac{3x-1}{x-1}-\frac{2x-5}{x+3}+\frac{4}{\left(x+3\right)\left(x-1\right)}=1\)
ĐKXĐ: x \(\ne\) 1 và x \(\ne\) - 3
\(\left(3x-1\right)\left(x+3\right)-\left(2x-5\right)\left(x-1\right)+4=\left(x+3\right)\left(x-1\right)\)
3x2 + 9x - x - 3 - 2x2 + 2x + 5x - 5 + 4 = x2 - x + 3x - 3
3x2 + 9x - x - 3 - 2x2 + 2x + 5x - 5 + 4 - x2 + x - 3x + 3 = 0
13x - 1 = 0
x = \(\frac{1}{13}\)
giải pt \(\frac{x^2+x+1}{x^2+2x+1}+\frac{x^2+3x+1}{x^2+4x+1}=\frac{5}{6}\)
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)