Có: \(\left(-5\right)^{39}=\left[\left(-5\right)^3\right]^{13}=\left(-125\right)^{13}\)
\(\left(-2\right)^{91}=\left[\left(-2\right)^7\right]^{13}=\left(-128\right)^{13}\)
Vì \(\left(-125\right)^{13}>\left(-128\right)^{13}\Rightarrow\left(-5\right)^{39}>\left(-2\right)^{91}\)
a) Rút gọn:
P = \(\left(2-\frac{x-1}{2x-3}\right)\div\left(\frac{6x+1}{2x^2-x-3}+\frac{x}{x+1}\right)\)
b) So sánh P với \(\frac{3}{2}\)
Tim x
a) \(\left(4x+3\right).\left(4x-3\right)-\left(4x-5\right)^2=46\)
b) \(\left(x+1\right)^3+2x-\left(x-1\right)^3-3.\left[\left(x+1\right)^2+\left(x-1\right)^2\right]+5=0\)
cho \(Q\left(x\right)=x^4+mx^3+nx^2+px+q\) biết \(Q\left(1\right)=5,Q\left(2\right)=7,Q\left(3\right)=9,Q\left(4\right)=11\).
Tìm m,n,p,q và tính \(Q\left(10\right),Q\left(11\right),Q\left(12\right),Q\left(13\right)\)
HELP ME.... mình cảm ơn
rút gọn
a)\(\left(7x-8\right).\left(7x+8\right)-10.\left(2x+3\right)^2+5x.\left(3x-2\right)^24x.\left(x-5\right)^2\)
b) \(\left(3x+7\right)^3-\left(5x-y\right).\left(25x^2+5xy+y^2\right)+\left(x+2y\right)^3\)
Tìm x
a) \(\left(2x-1\right).\left(2x+1\right)-4x^2=3\)
b) \(5x.\left(x-3\right)^2-5.\left(x-1\right)^3+15.\left(x+2\right).\left(x-2\right)=5\)
B1 ( kết quả thôi ko cần lời giải)
a) \(\left(4x-3\right)\left(3x+2\right)-\left(6x+1\right)\left(2x-5\right)+1\)
b) \(\left(3x+4\right)^2+\left(4x-1\right)^2+\left(2+5x\right)\left(2-5x\right)\)
c) \(\left(2x+1\right)\left(4x^2-2x+1\right)+\left(2-3x\right)\left(4+6x+9x^2\right)-9\)
B2 tìm x(kết quả)
a) \(3x\left(x-4\right)-x\left(5+3x\right)=-34\)
b) \(\left(3x+1\right)^2+\left(5x-2\right)^2=34\left(x+2\right)\left(x-2\right)\)
c) \(x^3+3x^2+3x+28=0\)
Rút gọn : \(\frac{1}{\left(x+y\right)^3}.\left(\frac{1}{x^3}+\frac{1}{y^3}\right)+\frac{3}{\left(x+y\right)^5}\left(\frac{1}{x^2}+\frac{1}{y^2}\right)+\frac{6}{\left(x+y\right)^5}\left(\frac{1}{x}+\frac{1}{y}\right)\)
Rút gon
\(\left(x-3\right)\left(x+3\right)-\left(x+5\right)\left(x-1\right)-\left(x-4\right)^2\)
Phân tích đa thức thành nhân tử :
a ) \(\left(x+5\right)^5-x^5-y^5\)
b ) \(A=3abc+a^2\left(a-b-c\right)+b^2\left(b-a-c\right)+c^2\left(c-a-b\right)-c\left(b-c\right)\left(a-c\right)\).
