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Bài 3:
a) \(\left|x-2\right|+9y^2+12xy+4x^2=0\)
\(\Leftrightarrow\left|x-2\right|+\left(3y+2x\right)^2=0\)
Dễ thấy \(VT\ge0\forall x;y\)
\(\Rightarrow\left\{{}\begin{matrix}x-2=0\\3y+2x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\frac{-4}{3}\end{matrix}\right.\)
Vậy...
b) \(3x^2+y^2+10x-2xy+26=0\)
\(\Leftrightarrow x^2-2xy+y^2+2x^2+10x+26=0\)
\(\Leftrightarrow\left(x-y\right)^2+2\left(x^2+5x+\frac{25}{4}\right)+\frac{27}{2}=0\)
\(\Leftrightarrow\left(x-y\right)^2+2\left(x+\frac{5}{2}\right)^2=\frac{-27}{2}\)
Dễ thấy \(VT\ge0\forall x;y\) mặt khác \(VP< 0\)
Do đó pt vô nghiệm
Bài 2:
\(A=263^2+74\cdot263+37^2\)
\(A=263^2+2\cdot263\cdot37+37^2\)
\(A=\left(263+37\right)^2\)
\(A=300^2\)
\(A=90000\)
b) tương tự
\(C=-1^2+2^2-3^2+...-99^2+100^2\)
\(C=\left(2^2-1^2\right)+\left(4^2-3^2\right)+...+\left(100^2-99^2\right)\)
\(C=\left(2-1\right)\left(1+2\right)+\left(4-3\right)\left(3+4\right)+...+\left(100-99\right)\left(99+100\right)\)
\(C=1+2+3+4+...+99+100\)
\(C=\frac{\left(100+1\right)\cdot100}{2}=5050\)
\(D=\left(3+1\right)\left(3^2+1\right)...\left(3^{32}+1\right)\)
\(2D=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)...\left(3^{32}+1\right)\)
\(2D=\left(3^2-1\right)\left(3^2+1\right)...\left(3^{32}+1\right)\)
\(2D=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{32}+1\right)\)
\(2D=\left(3^8-1\right)\left(3^8+1\right)...\left(3^{32}+1\right)\)
\(2D=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(2D=\left(3^{32}-1\right)\left(3^{32}+1\right)\)
\(2D=3^{64}-1\)
\(D=\frac{3^{64}-1}{2}\)
Bài 1 :
a ) \(2x\left(x+1\right)+2\left(x+1\right)=\left(x+1\right)\left(2x+2\right)=2\left(x+1\right)^2\)
b ) \(y^2\left(x^2+y\right)-zx^2-zy=y^2\left(x^2+y\right)-z\left(x^2+y\right)=\left(x^2+y\right)\left(y^2-z\right)\)
c ) \(4x\left(x-2y\right)+8y\left(2y-x\right)=4x\left(x-2y\right)-8y\left(x-2y\right)=4\left(x-2y\right)^2\)
d ) \(3x\left(x+1\right)^2-5x^2\left(x+1\right)+7\left(x+1\right)=\left(x+1\right)\left(3x^2+3x-5x^2+7\right)=\left(x+1\right)\left(3x-2x^2+7\right)\)
e ) \(x^2-6xy+9y^2=\left(x-3x\right)^2\)
Bài 1 :
f ) \(x^3+6x^2y+12xy^2+8y^3=\left(x+2y\right)^3\)
g ) \(x^3-64=\left(x-4\right)\left(x^2+4x+16\right)\)
h ) \(125x^3+y^6=\left(5x+y^2\right)\left(25x^2-5xy^2+y^4\right)\)
Câu 3 kiểm tra lại đề lại với , nếu đúng thì phức tạp lắm, còn sửa lại đề thì là :
\(y^2+2y+4^x-2^{x+1}+2=0\)
\(=>\left(y^2+2y+1\right)+2^{2x}-2^x.2+1=0\)
\(=>\left(y+1\right)^2+\left(\left(2^x\right)^2-2^x.2.1+1^2\right)=0\)
\(=>\left(y+1\right)^2+\left(2^x-1\right)^2=0\)
Dấu = xảy ra khi :
\(\hept{\begin{cases}y+1=0\\2^x-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}y=-1\\x=0\end{cases}}}\)
CHÚC BẠN HỌC TỐT...........
\(A=x^2+3x+7\)
\(=x^2+2.1,5x+2,25+4,75\)
\(=\left(x+1,5\right)^2+4,75\ge4,75\)
Vậy \(A_{min}=4,75\Leftrightarrow x=-1,5\)
\(B=2x^2-8x\)
\(=2\left(x^2-4x\right)\)
\(=2\left(x^2-4x+4-4\right)\)
\(=2\left[\left(x-2\right)^2-4\right]\)
\(=2\left(x-2\right)^2-8\ge-8\)
Vậy \(B_{min}=-8\Leftrightarrow x=2\)
B1: a) \(\left|x-2\right|+9y^2+12xy+4x^2=0\)
=> \(\left|x-2\right|+\left(3y+2x\right)^2=0\)
Ta có: \(\left|x-2\right|\ge0\forall x\)
\(\left(3y+2x\right)^2\ge0\forall x;y\)
=> \(\left|x-2\right|+\left(3y+2x\right)^2\ge0\forall x;y\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}x-2=0\\3y+2x=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\3y=-2x\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\3y=-2.2=-4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=-\frac{4}{3}\end{cases}}\)
Vậy ...
\(A=263^2+74.263+37^2\)
\(=263^2+2.263.37+37^2\)
\(=\left(263+37\right)^2\)
\(=300^2=90000\)
\(B=136^2-92.136+46^2\)
\(=136^2-2.136.46+46^2\)
\(=\left(136-46\right)^2\)
\(=90^2=8100\)