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\(a\text{) }pt\Leftrightarrow\left(y^2+2y+1\right)+\left[\left(2^x\right)^2-2.2^x+1\right]=0\)
\(\Leftrightarrow\left(y+1\right)^2+\left(2^x-1\right)^2=0\)
\(\Leftrightarrow y+1=0\text{ và }2^x-1=0\)
\(\Leftrightarrow y=-1\text{ và }x=0\)
\(b\text{) }pt\Leftrightarrow\left(4x^2+4y^2+8xy\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\)
\(\Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
\(\Leftrightarrow x+y=0\text{ và }x-1=0\text{ và }y+1=0\)
\(\Leftrightarrow x=1\text{ và }y=-1\)
1. Đặt \(t=x^2,t\ge0\)
\(3x^4+4x^2-2\ge3.0+4.0-2=-2\)
=> MIN = -2 khi x = 0
2. \(\left(x^2+2\right)\left(x+1\right)=0\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x^2+2=0\\x+1=0\end{array}\right.\)
Vì \(x^2+2\ge2>0\) => Vô nghiệm
Vậy x+1 = 0 => x = -1
3. Kết quả là 10
4. Ko rõ đề
\(x^2+2xy+6x+6y+2y^2+8=0\)
\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(6x+6y\right)+9+y^2-1=0\)
\(\Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)+9=1-y^2\)
\(\left(x+y+3\right)^2=1-y^2\)
Do \(VP=1-y^2\le1\forall x\) \(\Rightarrow VT=\left(x+y+3\right)^2\le1\)
\(\Leftrightarrow-1\le x+y+3\le1\)
\(\Leftrightarrow-1+2013\le x+y+3+2013\le1+2013\)
\(\Leftrightarrow2012\le x+y+2016\le2014\) hay \(2012\le B\le2014\)
B đạt MIN là 2012 \(\Leftrightarrow\hept{\begin{cases}y=0\\x+y+3=-1\end{cases}\Rightarrow\hept{\begin{cases}y=0\\x=-4\end{cases}}}\)
B đạt MAX là 2014 \(\Leftrightarrow\hept{\begin{cases}y=0\\x+y+3=1\end{cases}\Leftrightarrow\hept{\begin{cases}y=0\\x=-2\end{cases}}}\)
Câu 1: xin sửa đề :D
CM: \(n\left(n+1\right)\left(n+2\right)\left(n+3\right)+1\)là 1 scp
\(n\left(n+1\right)\left(n+2\right)\left(n+3\right)+1\)
\(=\left(n^2+3n\right)\left(n^2+3n+2\right)+1\)
\(=\left(n^2+3n\right)^2+2\left(n^2+3n\right)+1\)
\(=\left(n^2+3n+1\right)^2\)là scp
\(M=3x^4-8x^3-6x^2+8x+3\)
\(=3x^4-12x^3+4x^3+9x^2+x^2-16x^2+12x-4x+3\)
\(=\left(3x^4-12x^3+9x^2\right)+\left(4x^3-16x^2+12x\right)+\left(x^2-4x+3\right)\)
\(=3x^2\left(x^2-4x+3\right)+4x\left(x^2-4x+3\right)+\left(x^2-4x+3\right)\)
\(=\left(3x^2+4x+1\right)\left(x^2-4x+3\right)\)
\(=\left(3x^2+3x+x+1\right)\left(x^2-3x-x+3\right)\)
\(=\left[3x\left(x+1\right)+\left(x+1\right)\right]\left[x\left(x-3\right)-\left(x-3\right)\right]\)
\(=\left(3x+1\right)\left(x+1\right)\left(x-1\right)\left(x-3\right)\)
M = 0\(\Leftrightarrow\left(3x+1\right)\left(x+1\right)\left(x-1\right)\left(x-3\right)=0\)
\(\Rightarrow x\in\left\{\frac{-1}{3};-1;1;3\right\}\)
a: Đặt a=6x+7
=>a-1=6x+6; a+1=6x+8
\(\left(6x+6\right)\cdot\left(6x+7\right)^2\cdot\left(6x+8\right)=12\)
=>\(\left(a-1\right)\cdot a^2\cdot\left(a+1\right)=12\)
=>\(a^2\left(a^2-1\right)=12\)
=>\(a^4-a^2-12=0\)
=>\(\left(a^2-4\right)\left(a^2+3\right)=0\)
=>\(a^2-4=0\)
=>\(a^2=4\)
=>\(\left[\begin{array}{l}a=2\\ a=-2\end{array}\right.\Rightarrow\left[\begin{array}{l}6x+7=2\\ 6x+7=-2\end{array}\right.\Rightarrow\left[\begin{array}{l}6x=-5\\ 6x=-9\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac56\\ x=-\frac96=-\frac32\end{array}\right.\)
b:
ĐKXĐ: x<>2y
\(x^2-3xy+2y^2=0\)
=>\(x^2-xy-2xy+2y^2=0\)
=>x(x-y)-2y(x-y)=0
=>(x-y)(x-2y)=0
mà x<>2y
nên x-y=0
=>x=y
\(\frac{1}{\left|x-2y\right|}+\left|x+2y\right|=4\)
=>\(\left|\frac{1}{y-2y}\right|+\left|y+2y\right|=4\)
=>\(\frac{1}{\left|-y\right|}+3\left|y\right|=4\)
=>\(\frac{1}{\left|y\right|}+3\cdot\left|y\right|=4\)
Đặt a=|y|
=>\(\frac{1}{a}+3a=4\)
=>\(\frac{3a^2+1}{a}=4\)
=>\(3a^2+1-4a=0\)
=>(3a-1)(a-1)=0
TH1: 3a-1=0
=>3a=1
=>\(a=\frac13\)
=>\(\left|y\right|=\frac13\)
=>\(\left[\begin{array}{l}y=\frac13\\ y=-\frac13\end{array}\right.\)
Nếu \(y=\frac13\) thì \(x=y=\frac13\)
Nếu \(y=-\frac13\) thì \(x=y=-\frac13\)
TH2: a-1=0
=>a=1
=>|y|=1
=>\(\left[\begin{array}{l}y=1\\ y=-1\end{array}\right.\)
Nếu y=1 thì x=y=1
Nếu y=-1 thì x=y=-1