Bài 3: Tìm x biết:
a) \(|x-2|+9y^2+12xy+4x^2=0\)
b) \(3x^2+y^2+10x-2xy+26=0\)
Bài 2: Tính giá trị biểu thức 1 cách hợp lí:
\(A=263^2+74.263+37^2\)
\(B=136^2-92.136+46^2\)
\(C=-1^2+2^2-3^2+4^2-5^2+6^2-...-99^2+100^2\)
\(D=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
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
\(C=-1^2+2^2-3^2+...-99^2+100^2\)
\(C=\left(2^2-1^2\right)+\left(4^2-3^2\right)+...+\left(100^2-99^2\right)\)
\(C=\left(2-1\right)\left(1+2\right)+\left(4-3\right)\left(3+4\right)+...+\left(100-99\right)\left(99+100\right)\)
\(C=1+2+3+4+...+99+100\)
\(C=\frac{\left(100+1\right)\cdot100}{2}=5050\)
\(D=\left(3+1\right)\left(3^2+1\right)...\left(3^{32}+1\right)\)
\(2D=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)...\left(3^{32}+1\right)\)
\(2D=\left(3^2-1\right)\left(3^2+1\right)...\left(3^{32}+1\right)\)
\(2D=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{32}+1\right)\)
\(2D=\left(3^8-1\right)\left(3^8+1\right)...\left(3^{32}+1\right)\)
\(2D=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(2D=\left(3^{32}-1\right)\left(3^{32}+1\right)\)
\(2D=3^{64}-1\)
\(D=\frac{3^{64}-1}{2}\)
