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Mỗi số hạng của vế trái cộng thêm 1, vế phải = 5. Mỗi số hạng vế trái có mẫu số giống nhau, bạn đặt x+ 2020 làm nhân tử chung, phần còn lại tự làm nhé.
mấy bài còn lại bạn đăng cx làm tương tự
\(\frac{x+24}{1996}+\frac{x+25}{1995}+\frac{x+26}{1994}+\frac{x+27}{1993}+\frac{x+2036}{4}=0\)
\(\Leftrightarrow\left(\frac{x+24}{1996}+1\right)+\left(\frac{x+25}{1995}+1\right)+\left(\frac{x+26}{1994}+1\right)+\left(\frac{x+27}{1993}+1\right)+\left(\frac{x+2036}{4}-4\right)=0\)
\(\Leftrightarrow\frac{x+2020}{1996}+\frac{x+2020}{1995}+\frac{x+2020}{1994}+\frac{x+2020}{1993}+\frac{x+2020}{4}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{1996}+\frac{1}{1995}+\frac{1}{1994}+\frac{1}{1993}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow x+2020=0\)
\(\Leftrightarrow x=-2020\)
Vậy ....
Lời giải:
PT đã cho tương đương với:
\(\frac{x+24}{1996}+1+\frac{x+25}{1995}+1+\frac{x+26}{1994}+1+\frac{x+27}{1993}+1+\frac{x+2036}{4}-4=0\)
\(\Leftrightarrow \frac{x+2020}{1996}+\frac{x+2020}{1995}+\frac{x+2020}{1994}+\frac{x+2020}{1993}+\frac{x+2020}{4}=0\)
\(\Leftrightarrow (x+2020)\left(\frac{1}{1996}+\frac{1}{1995}+\frac{1}{1994}+\frac{1}{1993}+\frac{1}{4}\right)=0\)
Dễ thấy \(\frac{1}{1996}+\frac{1}{1995}+\frac{1}{1994}+\frac{1}{1993}+\frac{1}{4}\neq 0\) nên \(x+2020=0\Rightarrow x=-2020\) là nghiệm của pt.
Vậy............
a: \(\Rightarrow\left(\dfrac{x+1}{35}+1\right)+\left(\dfrac{x+3}{33}+1\right)=\left(\dfrac{x+5}{31}+1\right)+\left(\dfrac{x+7}{29}+1\right)\)
=>x+36=0
=>x=-36
b: \(\Leftrightarrow\left(\dfrac{x-10}{1994}-1\right)+\left(\dfrac{x-8}{1996}-1\right)+\left(\dfrac{x-6}{1998}-1\right)+\left(\dfrac{x-4}{2000}-1\right)+\left(\dfrac{x-2}{2002}-1\right)=\left(\dfrac{x-2002}{2}-1\right)+\left(\dfrac{x-2000}{4}-1\right)+\left(\dfrac{x-1998}{6}-1\right)+\left(\dfrac{x-1996}{8}-1\right)+\left(\dfrac{x-1994}{10}-1\right)\)
=>x-2004=0
=>x=2004
a) \(\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)=\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+5}{95}+1\right)\)
\(\Rightarrow\dfrac{x+100}{98}+\dfrac{x+100}{97}=\dfrac{x+100}{96}+\dfrac{x+100}{95}\)
\(\Rightarrow\dfrac{x+100}{98}+\dfrac{x+100}{97}-\dfrac{x+100}{96}-\dfrac{x+100}{95}=0\)
\(\Rightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\right)=0\)
Vì \(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\ne0\) nên \(x+100=0\Leftrightarrow x=-100\)
b) \(\dfrac{x+1}{1998}+\dfrac{x+2}{1997}=\dfrac{x+3}{1996}+\dfrac{x+4}{1995}\)
\(\Rightarrow\dfrac{x+1}{1998}+1+\dfrac{x+2}{1997}+1=\dfrac{x+3}{1996}+1+\dfrac{x+4}{1995}+1\)
\(\Rightarrow\dfrac{x+1999}{1998}+\dfrac{x+1999}{1997}=\dfrac{x+1999}{1996}+\dfrac{x+1999}{1995}\)
\(\Rightarrow\dfrac{x+1999}{1998}+\dfrac{x+1999}{1997}-\dfrac{x+1999}{1996}-\dfrac{x+1999}{1995}=0\)
\(\Rightarrow\left(x+1999\right)\left(\dfrac{1}{1998}+\dfrac{1}{1997}-\dfrac{1}{1996}-\dfrac{1}{1995}\right)=0\)
Vì \(\dfrac{1}{1998}+\dfrac{1}{1997}-\dfrac{1}{1996}-\dfrac{1}{1995}\ne0\) nên \(x+1999=0\Leftrightarrow x=-1999\)
c) \(\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0\)
\(\Rightarrow\dfrac{201-x}{99}+1+\dfrac{203-x}{97}+1+\dfrac{205-x}{95}+1=0\)
\(\Rightarrow\dfrac{300-x}{99}+\dfrac{300-x}{97}+\dfrac{300-x}{95}=0\)
\(\Rightarrow\left(300-x\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\right)=0\)
Vì \(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\ne0\) nên \(300-x=0\Leftrightarrow x=300\)
Vào trang cá nhân của t mà xem.T vừa làm r.Lười gõ lại lắm T^T
\(\dfrac{x+1}{1998}+\dfrac{x+2}{1997}=\dfrac{x+3}{1996}+\dfrac{x+4}{1995}\)
\(=\dfrac{x+1}{1998}+\dfrac{x+2}{1997}-\dfrac{x+3}{1996}-\dfrac{x+4}{1995}=0\)
\(=\dfrac{x+1}{1998}+1+\dfrac{x+2}{1997}+1-\dfrac{x+3}{1996}-1-\dfrac{x+4}{1995}-1=0\)
\(=\dfrac{x+1999}{1998}+\dfrac{x+1999}{1998}-\left(\dfrac{x+3}{1996}+1\right)-\left(\dfrac{x+4}{1995}+1\right)=0\)
\(=\dfrac{x+1999}{1998}+\dfrac{x+1999}{1997}-\dfrac{x+1999}{1996}-\dfrac{x+1999}{1995}=0\)
\(=\left(x+1999\right)\left(\dfrac{1}{1998}+\dfrac{1}{1997}-\dfrac{1}{1996}-\dfrac{1}{1995}\right)=0\)
⇔\(x+1999=0\)
Vậy \(x=-1999\)
ta có (x+1/2009 +1) + ( x+3/2007 + 1)- (x+5/2005 +1) - (x+7/1993 + 1) = 0
=>(x +100/ 2009) + (x+100/2007) - (x+100/2005)-(x+100/1993)
=> (x +100) * (1/2009 + 1/2007+ 1/2005 + 1/1993) = 0
=> x = -100
Bạn cứ tinh ý để ý đến phần tử và mẫu cộng lại bằng 100. Khi bạn bỏ phần x + 100 ra thì còn lại như trên. Sau đó lược bỏ còn lại x = -100
Mạn phép mk không chép đề , mk làm luôn nhé
\(\dfrac{x+1}{2009}+1+\dfrac{x+3}{2007}+1=\dfrac{x+5}{2005}+1+\dfrac{x+7}{1993}+1\)
⇔ \(\dfrac{x+2010}{2009}+\dfrac{x+2010}{2007}-\dfrac{x+2010}{2005}-\dfrac{x+2010}{1993}=0\)
⇔( x + 2010 )\(\left(\dfrac{1}{2009}+\dfrac{1}{2007}-\dfrac{1}{2005}-\dfrac{1}{1993}\right)=0\)
Ta thấy : \(\dfrac{1}{2009}< \dfrac{1}{2007}< \dfrac{1}{2005}< \dfrac{1}{1993}\)
⇒ \(\dfrac{1}{2009}+\dfrac{1}{2007}-\dfrac{1}{2005}-\dfrac{1}{1993}< 0\)
⇒ x + 2010 = 0
⇒ x = -2010
KL....
bài 1:
\(\dfrac{x-10}{1994}+\dfrac{x-8}{1996}+\dfrac{x-6}{1998}=\dfrac{x-2002}{2}+\dfrac{x-2000}{4}+\dfrac{x-1998}{6}\)
<=>\(\left(\dfrac{x-10}{1994}-1\right)+\left(\dfrac{x-8}{1996}+-1\right)+\left(\dfrac{x-6}{1998}-1\right)=\left(\dfrac{x-2002}{2}-1\right)+\left(\dfrac{x-2000}{4}-1\right)+\left(\dfrac{x-1998}{6}-1\right)\)
<=>\(\dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}=\dfrac{x-2004}{2}+\dfrac{x-2004}{4}+\dfrac{x-2004}{6}\)
<=>\(\dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}-\dfrac{x-2004}{2}-\dfrac{x-2004}{4}-\dfrac{x-2004}{6}=0\)
<=>(x-2004)\(\left(\dfrac{1}{1994}+\dfrac{1}{1996}+\dfrac{1}{1998}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{6}\right)\)
vì 1/1994+1/1996+1/1998-1/2-1/4-1/6 khác 0
nên x-2004=0=>x=2004
vyaj.......
bài 2:
\(\dfrac{x-85}{15}+\dfrac{x-74}{13}+\dfrac{x-67}{11}+\dfrac{x-64}{9}=10\)
<=>\(\left(\dfrac{x-85}{15}-1\right)+\left(\dfrac{x-74}{13}-2\right)+\left(\dfrac{x-67}{11}-3\right)+\left(\dfrac{x-64}{9}-4\right)=0\)
<=>\(\dfrac{x-100}{15}+\dfrac{x-100}{13}+\dfrac{x-100}{11}+\dfrac{x-100}{9}=0\)
<=>\(\left(x-100\right)\left(\dfrac{1}{15}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{9}\right)=0\)
vì 1/15+1/13+1/11+1/9 khác 0
=>x-100=0<=>x=100
4)a)\(\dfrac{x+5}{x-5}-\dfrac{x-5}{x+5}=\dfrac{20}{x^2-25}\)(1)
ĐKXĐ:\(\left\{{}\begin{matrix}x-5\ne0\\x+5\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne5\\x\ne-5\end{matrix}\right.\)
(1)\(\Rightarrow\left(x+5\right)\left(x+5\right)-\left(x-5\right)\left(x-5\right)=20\)
\(\Leftrightarrow x^2+10x+25-\left(x^2-10x+25\right)=20\)
\(\Leftrightarrow x^2+10x+25-x^2+10x-25=20\)
\(\Leftrightarrow x^2-x^2+10x+10x=-25+25=20\)
\(\Leftrightarrow20x=20\)
\(\Leftrightarrow x=1\left(nh\text{ậ}n\right)\)
S=\(\left\{1\right\}\)
Câu 1:
Tìm max:
Áp dụng BĐT Bunhiacopxky ta có:
\(y^2=(3\sqrt{x-1}+4\sqrt{5-x})^2\leq (3^2+4^2)(x-1+5-x)\)
\(\Rightarrow y^2\leq 100\Rightarrow y\leq 10\)
Vậy \(y_{\max}=10\)
Dấu đẳng thức xảy ra khi \(\frac{\sqrt{x-1}}{3}=\frac{\sqrt{5-x}}{4}\Leftrightarrow x=\frac{61}{25}\)
Tìm min:
Ta có bổ đề sau: Với $a,b\geq 0$ thì \(\sqrt{a}+\sqrt{b}\geq \sqrt{a+b}\)
Chứng minh:
\(\sqrt{a}+\sqrt{b}\geq \sqrt{a+b}\)
\(\Leftrightarrow (\sqrt{a}+\sqrt{b})^2\geq a+b\)
\(\Leftrightarrow \sqrt{ab}\geq 0\) (luôn đúng).
Dấu "=" xảy ra khi $ab=0$
--------------------
Áp dụng bổ đề trên vào bài toán ta có:
\(\sqrt{x-1}+\sqrt{5-x}\geq \sqrt{(x-1)+(5-x)}=2\)
\(\sqrt{5-x}\geq 0\)
\(\Rightarrow y=3(\sqrt{x-1}+\sqrt{5-x})+\sqrt{5-x}\geq 3.2+0=6\)
Vậy $y_{\min}=6$
Dấu "=" xảy ra khi \(\left\{\begin{matrix} (x-1)(5-x)=0\\ 5-x=0\end{matrix}\right.\Leftrightarrow x=5\)
Bài 2:
\(A=\sqrt{(x-1994)^2}+\sqrt{(x+1995)^2}=|x-1994|+|x+1995|\)
Áp dụng BĐT dạng \(|a|+|b|\geq |a+b|\) ta có:
\(A=|x-1994|+|x+1995|=|1994-x|+|x+1995|\geq |1994-x+x+1995|=3989\)
Vậy \(A_{\min}=3989\)
Đẳng thức xảy ra khi \((1994-x)(x+1995)\geq 0\Leftrightarrow -1995\leq x\leq 1994\)
\(\dfrac{x-1}{1992}+\dfrac{x-2}{1993}=\dfrac{x-3}{1994}+\dfrac{x-4}{1995}\)
\(\Rightarrow\left(\dfrac{x-1}{1992}+1\right)+\left(\dfrac{x-2}{1993}+1\right)=\left(\dfrac{x-3}{1994}+1\right)+\left(\dfrac{x-4}{1995}+1\right)\)
\(\Rightarrow\left(\dfrac{x-1+1992}{1992}\right)+\left(\dfrac{x-2+1993}{1993}\right)=\left(\dfrac{x-3+1994}{1994}\right)+\left(\dfrac{x-4+1995}{1995}\right)\)
\(\Rightarrow\dfrac{x+1991}{1992}+\dfrac{x+1991}{1993}=\dfrac{x+1991}{1994}+\dfrac{x+1991}{1995}\)
\(\Rightarrow\dfrac{x+1991}{1992}+\dfrac{x+1991}{1993}-\dfrac{x+1991}{1994}-\dfrac{x+1991}{1995}=0\)
\(\Rightarrow\left(x+1991\right)\left(\dfrac{1}{1992}+\dfrac{1}{1993}-\dfrac{1}{1994}-\dfrac{1}{1995}\right)=0\)
\(\Rightarrow\left(x+1991\right)=0\) ( vì \(\left(\dfrac{1}{1992}+\dfrac{1}{1993}-\dfrac{1}{1994}-\dfrac{1}{1995}\right)\ne0\)
\(\Rightarrow x=-1991\)
Lời giải hay😊😉