Giải phương trình :
\(\frac{2010x+2010}{x^2+x+1}+\frac{2010x-2010}{x^2-x+1}=\frac{2011}{x.\left(x^4+x^2+1\right)}\)
Giải phương trình sau:
\(\frac{2010x+2010}{x^2+x+1}-\frac{2010x-2010}{x^2-x+1}=\frac{2011}{x.x^6+x^2+1}\)
Giải phương trình
\(\frac{1}{2}\left(\frac{2x-2}{2009}+\frac{2x}{2010}+\frac{2x+2}{2011}\right)=\frac{33}{10}-\left(\frac{x+1}{2011}+\frac{x-1}{2009}+\frac{x}{2010}\right)\)
giải phương trình sau:
a) \(\frac{15-x}{2000}+\frac{14-x}{2001}=\frac{13-x}{2002}+\frac{12-x}{2003}\)
b) \(\frac{x-5}{2010}+\frac{x-4}{2011}=\frac{x-2010}{5}+\frac{x-2011}{4}\)
c) \(\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2=0\)
ai bít thì giúp mình với nhé
\(a,\frac{15-x}{2000}+\frac{14-x}{2001}=\frac{13-x}{2002}+\frac{12-x}{2003}\)
\(\Leftrightarrow\frac{15-x}{2000}+1+\frac{14-x}{2001}+1=\frac{13-x}{2002}+1+\frac{12-x}{2003}+1\)
\(\Leftrightarrow\frac{15-x+2000}{2000}+\frac{14-x+2001}{2001}=\frac{13-x+2002}{2002}+\frac{12-x+2003}{2003}\)
\(\Leftrightarrow\frac{2015-x}{2000}+\frac{2015-x}{2001}=\frac{2015}{2002}+\frac{2015-x}{2003}\)
\(\Leftrightarrow\left(2015-x\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
mà \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}>0\)
\(\Leftrightarrow2015-x=0\)
\(\Leftrightarrow x=2015\)
KL : PT có nghiệm \(S=\left\{2015\right\}\)
a)\(\frac{15-x}{2000}+\frac{14-x}{2001}=\frac{13-x}{2002}+\frac{12-x}{2003}\)
\(\Leftrightarrow\frac{15-x}{2000}+1+\frac{14-x}{2001}+1=\frac{13-x}{2002}+1+\frac{12-x}{2003}+1\)
\(\Leftrightarrow\frac{15-x+2000}{2000}+\frac{14-x+2001}{2001}=\frac{13-x+2002}{2002}+\frac{12-x+2003}{2003}\)
\(\Leftrightarrow\frac{2015-x}{2000}+\frac{2015-x}{2001}-\frac{2015-x}{2002}-\frac{2015-x}{2003}=0\)
\(\Leftrightarrow\left(2015-x\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right)=0\)
Vì \(\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right)\ne0\)
\(\Leftrightarrow2015-x=0\)
\(\Leftrightarrow x=2015\)
Vậy x=2015
Giải phương trình:
\(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{^{x^2}}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2=\left(x+4\right)^2\)
\(\frac{\left(2009-x\right)^2+\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}{\left(2009-x\right)^2-\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}=\frac{19}{49}\)
Bài 1: Tìm x biết: \(\frac{\left(2009-x\right)^2+\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}{\left(2009-x\right)^2-\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}=\frac{19}{49}\)
Bài 2: Tìm giá trị nhỏ nhất của biểu thức A=\(\frac{2010x+2680}{x^2+1}\)
==>Giúp mình 2 bài này nhé. Mai mình phải nộp bài rùi. Hihi <==
Bài 1:
Đặt x-2009=y. Khi đó phương trình đã cho trở thành:
\(\frac{y^2-y\left(y-1\right)+\left(y-1\right)^2}{y^2+y\left(y-1\right)+\left(y-1\right)^2}=\frac{19}{49}\)
\(\Leftrightarrow4y^2-4y-15=0\)
\(\Leftrightarrow\)(2y-5).(2y+3)=0
\(\Leftrightarrow\left[\begin{matrix}y=2,5\\y=-1,5\end{matrix}\right.\)
Thay y=x-2009. Ta được: \(\left[\begin{matrix}x=2009+2,5=2011,5\\x=2009-1,5=2007,5\end{matrix}\right.\)
Vậy x=2011,5 hoặc x=2007,5
Giải các phương trình sau:
a) \(x^3-6x^2-9x+14=0\)
b) \(\frac{\left(2010-x\right)^2-\left(2010-x\right)\left(x-2011\right)+\left(x-2011\right)^2}{\left(2010-x\right)^2+\left(2010+x\right)\left(x-2011\right)+\left(x-2011\right)^2}\)
a) \(x^3-6x^2-9x+14=0\)
\(\Leftrightarrow x^3-8x^2+2x^2+7x-16x+14=0\)
\(\Leftrightarrow\left(x^3-8x^2+7x\right)+\left(2x^2-16x+14\right)=0\)
\(\Leftrightarrow x\left(x^2-8x+7\right)+2\left(x^2-8x+7\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-8x+7\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-7x-x+7\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left[x\left(x-7\right)-\left(x-7\right)\right]=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)\left(x-7\right)=0\)
\(\Leftrightarrow x\in\left\{-2;1;7\right\}\)
giải phương trình: \(\frac{x+1}{2010}+\frac{x+2}{2009}+\frac{x+3}{2008}+...+\frac{x+2009}{2}+\frac{x+2010}{1}\)\(=\left(-2010\right)\)
\(\frac{x+1}{2010}+\frac{x+2}{2009}+\frac{x+3}{2008}+...+\frac{x+2010}{1}=\left(-2010\right)\)
\(\Rightarrow\left(\frac{x+1}{2010}+1\right)+\left(\frac{x+2}{2009}+1\right)+...+\left(\frac{x+2010}{1}+1\right)=-2010+2010\)
\(\Rightarrow\frac{x+2011}{2010}+\frac{x+2011}{2009}+...+\frac{x+2011}{1}=0\)
\(\Rightarrow\left(x+2011\right)\left(1+\frac{1}{2}+...+\frac{1}{2009}+\frac{1}{2010}\right)=0\)
\(\Rightarrow x+2011=0\Leftrightarrow x=-2011\)
cách bạn giải bài này là gì vậy ?
Giải Phương trình
\(\left(2x-1\right)^3+\left(x+2\right)^3=\left(3x+1\right)^3\)
\(\frac{x-1988}{15}+\frac{x-1969}{17}+\frac{x-1946}{19}+\frac{x-1919}{21}=10\)
\(\frac{\left(2009-x\right)^2+\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}{\left(2009-x^2\right)-\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}=\frac{19}{49}\)
1. \(\left(2x-1\right)^3+\left(x+2\right)^3=\left(3x+1\right)^3\)
\(\Rightarrow8x^3-12x^2+6x-1+x^3+6x^2+12x+8=27x^3+27x^2+9x+1\)
\(\Rightarrow-18x^3-33x^2+9x+6=0\)\(\Rightarrow\left(x+2\right)\left(-18x^2+3x+3\right)=0\)
\(\Rightarrow\left(x+2\right)\left(2x-1\right)\left(-9x-3\right)=0\Rightarrow\orbr{\begin{cases}x=-2\\x=\frac{1}{2};x=-\frac{1}{3}\end{cases}}\)
Vậy \(x=-2;x=\frac{1}{2};x=-\frac{1}{3}\)
2. \(\frac{x-1988}{15}+\frac{x-1969}{17}+\frac{x-1946}{19}+\frac{x-1919}{21}=10\)
\(\Rightarrow\left(\frac{x-1988}{15}-1\right)+\left(\frac{x-1969}{17}-2\right)+\left(\frac{x-1946}{19}-3\right)+\left(\frac{x-1919}{21}-4\right)=0\)
\(\Rightarrow\frac{x-2003}{15}+\frac{x-2003}{17}+\frac{x-2003}{19}+\frac{x-2003}{21}=0\)
\(\Rightarrow x-2003=0\)do \(\frac{1}{15}+\frac{1}{17}+\frac{1}{19}+\frac{1}{21}\ne0\)
Vậy \(x=2003\)
3. Đặt \(\hept{\begin{cases}2009-x=a\\x-2010=b\end{cases}}\)
\(\Rightarrow\frac{a^2+ab+b^2}{a^2-ab+b^2}=\frac{19}{49}\Rightarrow49a^2+49ab+49b^2=19a^2-19ab+19b^2\)
\(\Rightarrow30a^2+68ab+30b^2=0\Rightarrow\left(5a+3b\right)\left(3a+5b\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5a=-3b\\3a=-5b\end{cases}}\)
Với \(5a=-3b\Rightarrow5\left(2009-x\right)=-3\left(x-2010\right)\)
\(\Rightarrow-2x=-4015\Rightarrow x=\frac{4015}{2}\)
Với \(3a=-5b\Rightarrow3\left(2009-x\right)=-5\left(x-2010\right)\)
\(\Rightarrow2x=4023\Rightarrow x=\frac{4023}{2}\)
Vậy \(x=\frac{4023}{2}\)hoặc \(x=\frac{4015}{2}\)
Giải mấy bài toán này giúp tớ đy các cậu ==" <3
Nếu được thì các cậu giải thích ra giùm mình ==" Thanks nhiều ạ <3
1) Rút gọn \(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{\left(3n+2\right)\left(3n+5\right)}\)
2) Tìm các số a,b sao cho đa thức \(x^2+ax+b\) chia hết cho \(x^2-4\)
3) Chứng minh rằng trong một tam giác tâm G, trực tâm H, tâm đường tròn ngoại tiếp tam giác là O. Thì H,G,O thẳng hàng.
4) Phân tích các đa thức sau thành nhân tử:
a) \(\left(x+y+z\right)^3-x^3-y^3-z^3\)
b) \(x^4+2010x^2+2009x+2010\)
5) Tìm x biết:
\(\frac{\left(2009-x\right)^2+\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}{\left(2009-x\right)^2-\left(2009-x\right)2010+\left(x-2010\right)^2}=\frac{19}{49}\)
==> Đang cần gấp