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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 ...
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...........
rút gọn biểu thức
a)2x(2x−1)2−3x(x+3)(x−3)−4x(x+1)2
=2x(4x2-4x+1)-3x.(x2-9)-4x(x2+2x+1)
=8x3-8x2+2x-3x3-27x-4x3-8x2-4x
=8x3-16x2-7x3-29x
\(B1\\ a,2x+10y=2\left(x+5y\right)\\ b,x^2+4x+4=x^2+2.2x+2^2=\left(x+2\right)^2\\ c,x^2-y^2+10y-25\\ =\left(x^2-y^2\right)+5\left(2y-5\right)\\ =\left(x-y\right)\left(x+y\right)+5\left(2y-5\right)\\ B2\)
\(a,x^2-3x+x-3=0\\ =>x\left(x-3\right)+\left(x-3\right)=0\\ =>\left(x+1\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+1=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\\ b,2x\left(x-3\right)-\dfrac{1}{2}\left(4x^2-3\right)=0\\ =>2x^2-6x-2x^2+\dfrac{3}{2}=0\\ =>-6x=-\dfrac{3}{2}\\ =>x=\left(-\dfrac{3}{2}\right):\left(-6\right)\\ =>x=\dfrac{1}{4}\\ c,x^2-\left(x-3\right)\left(2x-5\right)=9\\ =>x^2-2x^2+6x+5x-15=9\\ =>-x^2+11-15-9=0\\ =>-x^2+11x-24=0\\ =>-x^2+8x+3x-24=0\\ =>-x\left(x-8\right)+3\left(x-8\right)=0\\ =>\left(3-x\right)\left(x-8\right)=0\\ =>\left[{}\begin{matrix}3-x=0\\x-8=0\end{matrix}\right.=>\left[{}\begin{matrix}x=3\\x=8\end{matrix}\right.\)
a) \(\Rightarrow\dfrac{1}{3}x\left(x-2\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
b) \(\Rightarrow\left(x+5\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)
c) \(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
e) \(\Rightarrow\left(x+2\right)\left(x+2-x+2\right)=0\Rightarrow\left(x+2\right).4=0\Rightarrow x=-2\)
f) \(\Rightarrow x\left(2x-3\right)+2\left(2x-3\right)=0\Rightarrow\left(2x-3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)
g) \(\Rightarrow2\left(3x-2\right)^2-\left(3x-2\right)\left(3x+2\right)=0\Rightarrow\left(3x-2\right)\left(3x-6\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)
h) \(\Rightarrow x\left(x+1\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)
i) \(\Rightarrow4x\left(x+1\right)+5\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(4x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{5}{4}\end{matrix}\right.\)
a, (3x - 5)(2x - 1) - (x + 2)(6x - 1) = 0
=> 6x^2 - 3x - 10x + 5 - (6x^2 - x + 12x - 2) = 0
=> 6x^2 - 13x + 5 - 6x^2 - 11x + 2 = 0
=> -24x + 7 = 0
=> - 24x = -7
=> x = 7/24
b, (3x - 2)(3x + 2) - (3x - 1)^2 = -5
=> 9x^2 - 4 - 9x^2 + 6x - 1 = -5
=> 6x - 5 = -5
=> 6x = 0
=> x = 0
c, x^2 = -6x - 8
=> x^2 + 6x + 8 = 0
=> x^2 + 2.x.3 + 9 - 1 = 0
=> (x + 3)^2 = 1
=> x + 3 = 1 hoặc x + 3 = -1
=> x = -2 hoặc x = -4
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
\(C=-1^2+2^2-3^2+...-99^2+100^2\)
\(C=\left(2^2-1^2\right)+\left(4^2-3^2\right)+...+\left(100^2-99^2\right)\)
\(C=\left(2-1\right)\left(1+2\right)+\left(4-3\right)\left(3+4\right)+...+\left(100-99\right)\left(99+100\right)\)
\(C=1+2+3+4+...+99+100\)
\(C=\frac{\left(100+1\right)\cdot100}{2}=5050\)
\(D=\left(3+1\right)\left(3^2+1\right)...\left(3^{32}+1\right)\)
\(2D=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)...\left(3^{32}+1\right)\)
\(2D=\left(3^2-1\right)\left(3^2+1\right)...\left(3^{32}+1\right)\)
\(2D=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{32}+1\right)\)
\(2D=\left(3^8-1\right)\left(3^8+1\right)...\left(3^{32}+1\right)\)
\(2D=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(2D=\left(3^{32}-1\right)\left(3^{32}+1\right)\)
\(2D=3^{64}-1\)
\(D=\frac{3^{64}-1}{2}\)