Tìm x biết
(X-1)^x+2=(x-1)^x+4
(x−1)x+2=(x−1)x+4⇔(x−1)x+4−(x−1)x+2=0⇔(x−1)x+2[(x−1)2−1]=0⇔x(x−1)x+2(x−2)⇒[x=0x=1x=2\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\\ \Leftrightarrow\left(x-1\right)^{x+4}-\left(x-1\right)^{x+2}=0\\ \Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\\\Leftrightarrow x\left(x-1\right)^{x+2}\left(x-2\right)\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=2\end{matrix}\right.(x−1)x+2=(x−1)x+4⇔(x−1)x+4−(x−1)x+2=0⇔(x−1)x+2[(x−1)2−1]=0⇔x(x−1)x+2(x−2)⇒⎣⎡x=0x=1x=2
Giải bất phương trình sau
4x+14(2−x)≥x+2\dfrac{4x+1}{4\left(2-x\right)}\ge x+24(2−x)4x+1≥x+2
giải bất phương trình sau
b,x+23x+1≤x−22x−1b,\dfrac{x+2}{3x+1}\le\dfrac{x-2}{2x-1}b,3x+1x+2≤2x−1x−2
Giải pt x+1+2x+3=5\sqrt{x+1}+\sqrt{2x+3}=5x+1+2x+3=5 ???
cho pt (m−2)x4−2(m+1)x2+2m−1=0\left(m-2\right)x^4-2\left(m+1\right)x^2+2m-1=0(m−2)x4−2(m+1)x2+2m−1=0 tìm m để pt có
a,vô nghiệm b,1 nghiệm c,2 nghiệm d,3 nghiệm e,4 nghiệm
15x2−35x+23=x2−x−3\sqrt{15x^2-35x+23}=x^2-x-315x2−35x+23=x2−x−3 giải phương trình trên??
Chứng minh BĐT:
a2+4b2+4c2≥2(ab−ac+2bc)a^2+4b^2+4c^2\ge2\left(ab-ac+2bc\right)a2+4b2+4c2≥2(ab−ac+2bc)
Cho a,b,c ≥0\ge0≥0. CMR:
a3ba4+a2b2+b4+b3cb4+b2c2+c4+c3ac4+c2a2+a4≤1\dfrac{a^3b}{a^4+a^2b^2+b^4}+\dfrac{b^3c}{b^4+b^2c^2+c^4}+\dfrac{c^3a}{c^4+c^2a^2+a^4}\le1a4+a2b2+b4a3b+b4+b2c2+c4b3c+c4+c2a2+a4c3a≤1
cho a,b,c >0, và a2+b2+c2=3a^2+b^2+c^2=3a2+b2+c2=3:
CMR: a2+3ab+b26a2+8ba+11b2+a2+3ab+c26a2+8ca+11c2+c2+3cb+b26c2+8ca+11b2\dfrac{a^2+3ab+b^2}{\sqrt{6a^2+8ba+11b^2}}+\dfrac{a^2+3ab+c^2}{\sqrt{6a^2+8ca+11c^2}}+\dfrac{c^2+3cb+b^2}{\sqrt{6c^2+8ca+11b^2}}6a2+8ba+11b2a2+3ab+b2+6a2+8ca+11c2a2+3ab+c2+6c2+8ca+11b2c2+3cb+b2 ≤\leq≤ 3
{x2+y2=xy+x+yx2−y2=3\left\{{}\begin{matrix}x^2+y^2=xy+x+y\\x^2-y^2=3\end{matrix}\right.{x2+y2=xy+x+yx2−y2=3
{4xy+4(x2+y2)+3(x+y)2=72x+1x+y=1\begin{cases} 4xy +4(x^2+y^2)+\frac{3}{(x+y)^2}=7\\2x+\frac{1}{x+y}=1\end{cases}{4xy+4(x2+y2)+(x+y)23=72x+x+y1=1
