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Lời giải:
\(a^{200}+b^{200}=a^{201}+b^{201}\)
\(\Rightarrow a^{200}(a-1)+b^{200}(b-1)=0(1)\)
\(a^{201}+b^{201}=a^{202}+b^{202}\)
\(\Rightarrow a^{201}(a-1)+b^{201}(b-1)=0(2)\)
Lấy $(2)-(1)$ suy ra:
\((a-1)(a^{201}-a^{200})+(b-1)(b^{201}-b^{200})=0\)
\(\Leftrightarrow a^{200}(a-1)^2+b^{200}(b-1)^2=0\)
Ta thấy $a^{200}(a-1)^2\geq 0; b^{200}(b-1)^2\geq 0$ với mọi $a,b$
Do đó để tổng của chúng bằng $0$ thì:
\(a^{200}(a-1)^2=b^{200}(b-1)^2=0\)
$\Rightarrow a=0$ hoặc $a=1$; $b=0$ hoặc $b=1$
Suy ra $(a,b)=(1,1); (0,0); (1,0); (0,1)$
$\Rightarrow B=a^{2019}+b^{2020}$ có thể nhận những giá trị là $0; 2; 1$
<=> (2-x/201 + 1) + (x/203 - 1) = (1-x/202 + 1) + (1-1)
<=> 203-x/201 + x-203/203 = 203-x/202
<=> 203-x/201 - 203-x/203 - 203-x/202 = 0
<=> (203-x).(1/201-1/203-1/202) = 0
<=> 203-x = 0 ( vì 1/201-1/203-1/202 khác 0 )
<=> x=203
Vậy x=203
k mk nha
Ta có \(\left(a^{201}+b^{201}\right)^2=\left(a^{200}+b^{200}\right)\left(a^{202}+b^{202}\right)\Leftrightarrow2a^{201}b^{201}=a^{200}b^{202}+a^{202}b^{200}\Leftrightarrow2ab=a^2+b^2\Leftrightarrow\left(a-b\right)^2=0\Leftrightarrow a=b\).
Khi đó \(a^{200}=a^{201}\Leftrightarrow a=1\).
Do đó P = 2.
a) \(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\)
\(\Leftrightarrow\frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\)
\(\Leftrightarrow\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
\(\Leftrightarrow x+2005=0\)
\(\Leftrightarrow x=-2005\)
b) Sửa đề :
\(\frac{201-x}{99}+\frac{203-x}{97}+\frac{205-x}{95}+3=0\)
\(\Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\)
\(\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\)
\(\Leftrightarrow x=300\)
c) \(\frac{2-x}{2002}-1=\frac{1-x}{2003}-\frac{x}{2004}\)
\(\Leftrightarrow\frac{2-x}{2002}+1=\frac{1-x}{2003}+1-\frac{x}{2004}+1\)
\(\Leftrightarrow\frac{2004-x}{2002}=\frac{2004-x}{2003}-\frac{2004-x}{2004}\)
\(\Leftrightarrow\left(2004-x\right)\left(\frac{1}{2002}-\frac{1}{2003}+\frac{1}{2004}\right)=0\)
\(\Leftrightarrow x=2004\)
Vậy....
a)\(\frac{201-x}{99}+1+\frac{203-x}{97}+1+\frac{205-x}{95}+1=0\)
\(\Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\)
\(\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\)
Mà \(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\ne0\Rightarrow300-x=0\Rightarrow x=300\)
b)\(\frac{2-x}{2002}+1=\frac{1-x}{2003}+2-\frac{x}{2004}\)
\(\Leftrightarrow\frac{2004-x}{2002}=\frac{1-x}{2003}+1+1-\frac{x}{2004}\)
\(\Leftrightarrow\frac{2004-x}{2002}=\frac{2004-x}{2003}+\frac{2004-x}{2004}\)
\(\Leftrightarrow\left(2004-x\right)\left(\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}\right)=0\)
Mà \(\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}\ne0\Rightarrow2004-x=0\Rightarrow x=2004\)
c)\(\frac{x^2-10x-29}{1971}+\frac{x^2-10x-27}{1973}-2=\frac{x^2-10x-1971}{29}+\frac{x^2-10x-1973}{27}-2\)
\(\Leftrightarrow\frac{x^2-10x-2000}{1971}+\frac{x^2-10x-2000}{1973}=\frac{x^2-10x-2000}{29}+\frac{x^2-10x-2000}{27}\)
\(\Leftrightarrow\left(x^2-10x-2000\right)\left(\frac{1}{1971}+\frac{1}{1973}-\frac{1}{29}-\frac{1}{27}\right)=0\)
Mà\(\frac{1}{1971}+\frac{1}{1973}-\frac{1}{29}-\frac{1}{27}\ne0\)
\(\Rightarrow x^2-10x-2000=0\Leftrightarrow\left(x+40\right)\left(x-50\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+40=0\\x-50=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-40\\x=50\end{matrix}\right.\)
a) Ta có: \(37^2+2\cdot37\cdot13+13^2\)
\(=\left(37+13\right)^2=50^2=2500\)
b) Ta có: \(201^2=\left(200+1\right)^2\)
\(=200^2+2\cdot200+1\)
\(=40000+200+1=40201\)
c) Ta có: \(37\cdot43=\left(40+3\right)\cdot\left(40-3\right)\)
\(=40^2-3^2=1600-9=1591\)
A = 1/401
B = 201 x 201 - 200 x 200 / 201 x 201 + 200 x 200
B = 201 x 200 + 201 - 200 x 200 / 201 x 200 + 201 + 201
B = (201 - 200) x 200 + 201 / ( 201 + 200) x 200 + 201
B = 1 x 200 + 201 / 401 x 200 + 201
B = 401 / 401 x 200 + 201
ta có 401 / 200 x 401 + 201 > 401 / 201 x 401
ta có : 401 / 201 x 401 = 1/201 > 1/401
mà : 401/401 x 200 + 201 > 401 / 201 x 401 > 1/401