a)5^(x-2)(x+3)=1

=>5^(x-2)(x+3)=5⁰

=>(x-2)(x+3)=0

=>x-2=0=>x=2.

Và x+3=0=>x=-3.

Vậy x=2 và x=-3.

b)Câu này mik ko làm dc.

c)x.(6-x)²⁰⁰³=(6-x)²⁰⁰³​

=>x=(6-x)²⁰⁰³​:(6-x)²⁰⁰³​

=>x=1.

Vậy x=1.

d)2.3^x=10.3¹²+8.3¹²

=>2.3^x=2.

1. Ta có: \(x\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)

=> \(x\left(6-x\right)^{2003}-\left(6-x\right)^{2003}=0\)

=> \(\left(6-x\right)^{2003}\left(x-1\right)=0\)

=> \(\orbr{\begin{cases}\left(6-x\right)^{2003}=0\\x-1=0\end{cases}}\)

=> \(\orbr{\begin{cases}6-x=0\\x=1\end{cases}}\)

=> \(\orbr{\begin{cases}x=6\\x=1\end{cases}}\)

Bài 2. Ta có: (3x - 5)¹⁰⁰ \(\ge\)0 \(\forall\)x

       (2y + 1)¹⁰⁰ \(\ge\)0 \(\forall\)y

=> (3x - 5)¹⁰⁰ + (2y + 1)¹⁰⁰ \(\ge\)0 \(\forall\)x;y

Dấu "=" xảy ra khi: \(\hept{\begin{cases}3x-5=0\\2y+1=0\end{cases}}\) => \(\hept{\begin{cases}3x=5\\2y=-1\end{cases}}\) => \(\hept{\begin{cases}x=\frac{5}{3}\\y=-\frac{1}{2}\end{cases}}\)

Vậy ...

1. x( 6 - x )²⁰⁰³ = ( 6 - x )²⁰⁰³

<=> x( 6 - x )²⁰⁰³ - ( 6 - x )²⁰⁰³ = 0

<=> ( x - 1 )( 6 - x )²⁰⁰³ = 0

<=> \(\orbr{\begin{cases}x-1=0\\6-x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=6\end{cases}}\)

2. \(\left(3x-5\right)^{100}+\left(2y+1\right)^{100}\le0\)

\(\hept{\begin{cases}\left(3x-5\right)^{100}\ge0\forall x\\\left(2y+1\right)^{100}\ge0\forall y\end{cases}\Rightarrow}\left(3x-5\right)^{100}+\left(2y+1\right)^{100}\ge0\forall x,y\)

Dấu " = " xảy ra <=>  \(\hept{\begin{cases}3x-5=0\\2y+1=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{5}{3}\\y=-\frac{1}{2}\end{cases}}\)

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)

\(2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)

\(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)

\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2001}{2003}\)

\(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2001}{2003}\)

\(\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2001}{2003}:2\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)

\(-\frac{1}{x+1}=\frac{2001}{4006}-\frac{1}{2}\)

\(-\frac{1}{x+1}=-\frac{1}{2003}\)

\(\Rightarrow x+1=2003\)

\(\Rightarrow x=2012\)

Ta có: \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+..+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)

\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)

\(\Rightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)

\(\Rightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{2003}:2\)

\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)

\(\Rightarrow\frac{2003}{4006}-\frac{1}{x+1}=\frac{2001}{4006}\)

\(\Rightarrow\frac{1}{x+1}=\frac{2003}{4006}-\frac{2001}{4006}\)

\(\Rightarrow\frac{1}{x+1}=\frac{2}{4006}=\frac{1}{2003}\)

=> x + 1 = 2003

=> x = 2002

Vậy x = 2002

Duyệt nha !!!

chúc hk tốt!!!

Ta có \(\left(x+\sqrt{x^2+2003}\right).\left(y+\sqrt{y^2+2003}\right)=2003\)

\(\Rightarrow\frac{-2003}{x-\sqrt{x^2+2003}}.\frac{-2003}{y-\sqrt{y^2+2003}}=2003\)

\(\Leftrightarrow\left(x-\sqrt{x^2+2003}\right)\left(y-\sqrt{y^2+2003}\right)=2003\)

\(\Rightarrow\left(x+\sqrt{x^2+2003}\right).\left(y+\sqrt{y^2+2003}\right)=\left(x-\sqrt{x^2+2003}\right).\left(y-\sqrt{y^2+2003}\right)\)

\(\Leftrightarrow xy+x\sqrt{y^2+2003}+y\sqrt{x^2+2003}+\sqrt{\left(x^2+2003\right)\left(y^2+2003\right)}=xy-x\sqrt{y^2+2003}-y\sqrt{x^2+2003}+\sqrt{\left(x^2+2003\right)\left(y^2+2003\right)}\)

\(\Leftrightarrow x\sqrt{y^2+2003}=-y\sqrt{x^2+2003}\left(1\right)\)

Ta thấy pt (1)có 1 nghiệm \(x=y=0\)

\(\left(1\right)\Rightarrow\hept{\begin{cases}x^2\left(y^2+2003\right)=y^2\left(x^2+2003\right)\\x>0;y< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x^2=y^2\\x>0;y< 0\end{cases}\Leftrightarrow}x=-y}\)

Vậy \(x+y=0\)

Áp dụng \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\) rút gọn rồi quy đồng làm nốt

\(2003-\left|x-2003\right|=x\)

\(\Leftrightarrow\left|x-2003\right|=2003-x\left(1\right)\)

+ ) Nếu : \(x\ge2003\) thì ( 1 ) \(\Leftrightarrow x-2003=2003-x\)

\(\Leftrightarrow2x=2.2003\)

\(\Leftrightarrow x=2003\left(nhận\right)\)

+ ) Nếu \(x< 2003\) thì ( 1 ) \(\Leftrightarrow2003-x=2003-x\)

\(\Leftrightarrow0.x=0\)

Vậy pt có vô số nghiệm với \(x< 2003\)

mk chỉ biết đáp số là \(x=2003^{ }\) thôi à

13 +16 +110 +....+1x(x+1):2 =20012003 

26 +212 +220 +....+2x(x+1) =20012003 

2(12.3 +13.4 +14.5 +....+1x(x+1) )=20012003 

12 −13 +13 −14 +14 −15 +....+1x −1x+1 =20012003 :2=20014006 

12 −1x+1 =20014006 

1x+1 =12 −20014006 =12003 

=> x+1 = 2003

=> x = 2003 - 1

=> x = 2002

 Xin 1 tích đúng 

\(\Rightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+.....+\frac{2}{x.\left(x+1\right)}=\frac{2001}{2003}\)

\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2001}{2003}\)

\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2001}{2003}\)

\(\Rightarrow\frac{x-1}{x+1}=\frac{2001}{2003}\)

\(\Rightarrow2x=4004\)

\(\Rightarrow x=2002\)

\(PT\Leftrightarrow\frac{x+4+2000}{2000}+\frac{x+3+2001}{2001}=\frac{x+2+2002}{2002}+\frac{x+1+2003}{2003}\)

<=> \(\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)

<=> \(\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)

<=> x + 2004 = 0

<=> x = -2004.

\(\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}+1\right)\)

\(\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)

\(\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)

\(x+2004=0\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\right)\)

\(\Rightarrow x=-2004\)

     \(\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}+1\right)\)  

\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+20004}{2003}\)

\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)

\(\Leftrightarrow x+2004=0:\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\)

\(\Leftrightarrow x+2004=0\)

\(\Leftrightarrow x=0-2004=-2004\)

Vậy \(x=-2004\)