a: Trường hợp 1: x<-1
Pt sẽ là:
-x+2-3(-x-1)=5
=>-x+2+3x+3=5
=>2x+5=5
=>x=0(vô lý)
Trường hợp 2: -1<=x<2
Pt sẽ là -x+2-3(x+1)=5
=>-x+2-3x-3=5
=>-4x-1=5
=>-4x=6
hay x=-3/2(loại)
Trường hợp 3: x>=2
Phương trình sẽ là:
x-2-3(x+1)=5
=>x-2-3x-3=5
=>-2x=10
hay x=-5(loại)
b: \(\Leftrightarrow x^2+\left|x-1\right|=x^2+3\)
\(\Leftrightarrow\left|x-1\right|=3\)
=>x-1=3 hoặc x-1=-3
=>x=4 hoặc x=-2
Tim x
a) \(\left(4x+3\right).\left(4x-3\right)-\left(4x-5\right)^2=46\)
b) \(\left(x+1\right)^3+2x-\left(x-1\right)^3-3.\left[\left(x+1\right)^2+\left(x-1\right)^2\right]+5=0\)
Tìm x
a) \(\left(2x-1\right).\left(2x+1\right)-4x^2=3\)
b) \(5x.\left(x-3\right)^2-5.\left(x-1\right)^3+15.\left(x+2\right).\left(x-2\right)=5\)
Giải phương trình: \(-2\left(\sqrt{1+x}+\sqrt{1-x}\right)+7=\sqrt{\left(5-2x\right)\left(5+2x\right)}-2\sqrt{1-x^2}\)
Tim x
a) \(\left(x+3\right)^3-x.\left(3x+1\right)^2+\left(2x+1\right).\left(4x^2-2x+1-3x^2\right)=54\)
b) \(\left(x-3\right)^3-\left(x-3\right).\left(x^2+3x+9\right)+6.\left(x+1\right)^2+3x^2=-33\)
Giải các phương trình:
a) \(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\)
b) \(\frac{3}{\left(x-1\right)\left(x-2\right)}+\frac{2}{\left(x-3\right)\left(x-1\right)}=\frac{1}{\left(x-2\right)\left(x-3\right)}\)
c) \(1+\frac{1}{x+2}=\frac{12}{8+x^3}\)
d) \(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{\left(x-3\right)\left(x+3\right)}\)
Phân tích đa thức thành nhân tử :
\(a,\left(x+y\right)^5-x^5-y^5\)
\(b,\left(x^2+y^2\right)^3+\left(z^2-x^2\right)^3+\left(y^2+z^2\right)^3\)
\(c,x^9-x^7-x^6-x^5+x^4+x^3+x^2+1\).
Rút gọn : \(\frac{1}{\left(x+y\right)^3}.\left(\frac{1}{x^3}+\frac{1}{y^3}\right)+\frac{3}{\left(x+y\right)^5}\left(\frac{1}{x^2}+\frac{1}{y^2}\right)+\frac{6}{\left(x+y\right)^5}\left(\frac{1}{x}+\frac{1}{y}\right)\)
giải hộ mk phương trình, thanks:
a) \(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\)
b) \(\frac{\left(x-2\right)^2}{3}-\frac{\left(2x-3\right)\left(2x+3\right)}{8}+\frac{\left(x-4\right)^2}{6}=0\)
Tìm x :
a ) \(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6.\)
b ) \(4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)
