Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
tớ ko bt lm abc , tớ lm d thôi nha , thứ lỗi
\(\frac{5}{2x-3}-\frac{1}{x+2}=\frac{5}{x-6}-\frac{7}{2x-1}\)
\(\frac{3x+13}{2x^2+x-6}=\frac{5}{x-6}+\frac{7}{1-2x}\)
\(\frac{3x+13}{\left(x+2\right)\left(2x-3\right)}=\frac{3x+37}{\left(x-6\right)\left(2x-1\right)}\)
\(\frac{10-9x}{-4x^3+32x^2-51x+18}=0\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{10}{9}\end{cases}}\)
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 \(x=\frac{4015}{2}\)
Bài 3:
a) \(\left(x-6\right).\left(2x-5\right).\left(3x+9\right)=0\)
\(\Leftrightarrow\left(x-6\right).\left(2x-5\right).3.\left(x+3\right)=0\)
Vì \(3\ne0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\2x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\2x=5\\x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\frac{5}{2}\\x=-3\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{6;\frac{5}{2};-3\right\}.\)
b) \(2x.\left(x-3\right)+5.\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right).\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\frac{5}{2}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{3;-\frac{5}{2}\right\}.\)
c) \(\left(x^2-4\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x^2-2^2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2-3+2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2;\frac{1}{3}\right\}.\)
Chúc bạn học tốt!
a)\(\frac{1}{x-1}\)-\(\frac{3x2}{x3-1}\)=\(\frac{2x}{x2+x+1}\)
<=> \(\frac{1}{x-1}\)-\(\frac{3x2}{\left(x-1\right)\left(x2+x+1\right)}\)=\(\frac{2x}{x2+x+1}\) ĐKXĐ: x khác 1
<=> x2+x+1 - 3x2 = 2x(x-1)
<=>x2+x+1 - 3x2 = 2x2-2x
<=>x2-3x-1=0( đoạn này làm nhanh nhé)
<=>x2-2*\(\frac{3}{2}\)x +\(\frac{9}{4}\)-\(\frac{9}{4}\)-1=0
<=>(x-\(\frac{3}{2}\))2-\(\frac{13}{4}\)=0
<=>(x-\(\frac{3-\sqrt{13}}{2}\))(x-\(\frac{3+\sqrt{13}}{2}\))=0
\(\begin{cases}x=\frac{3+\sqrt{13}}{2}\\x=\frac{3-\sqrt{13}}{2}\end{cases}\)
b) pt... đkxđ x khác 1;2;3
<=> 3(x-3) +2(x-2)=x-1
<=> 3x-9 +2x-4 = x-1
<=> 4x= 12
<=> x=3 ( ko thỏa đk)
vậy pt vô nghiệm
a, \(\frac{1-x}{x+1}+3=\frac{2x+3}{x+1}\)
\(=>\frac{1-x+x+1}{x+1}+2=\frac{1}{x+1}+2\)
\(=>\frac{2}{x+1}=\frac{1}{x+1}\)
\(=>2x+2=x+1\)
\(=>2x-x=1-2=-1\)
\(=>x=-1\)
vậy nghiệm của phương trình trên là {-1}
À quên ĐKXĐ của câu a là \(x\ne-1\)
Nên \(x\in\varnothing\)nhé :v
a) <=> \(6x^2-5x+3-2x+3x\left(3-2x\right)=0\)
<=> \(6x^2-5x+3-2x+9x-6x^2=0\)
<=> \(2x+3=0\)
<=> \(x=\frac{-3}{2}\)
b) <=> \(10\left(x-4\right)-2\left(3+2x\right)=20x+4\left(1-x\right)\)
<=> \(10x-40-6-4x=20x+4-4x\)
<=> \(6x-46-16x-4=0\)
<=> \(-10x-50=0\)
<=> \(-10\left(x+5\right)=0\)
<=> \(x+5=0\)
<=> \(x=-5\)
c) <=> \(8x+3\left(3x-5\right)=18\left(2x-1\right)-14\)
<=> \(8x+9x-15=36x-18-14\)
<=> \(8x+9x-36x=+15-18-14\)
<=> \(-19x=-14\)
<=> \(x=\frac{14}{19}\)
d) <=>\(2\left(6x+5\right)-10x-3=8x+2\left(2x+1\right)\)
<=> \(12x+10-10x-3=8x+4x+2\)
<=> \(2x-7=12x+2\)
<=> \(2x-12x=7+2\)
<=> \(-10x=9\)
<=> \(x=\frac{-9}{10}\)
e) <=> \(x^2-16-6x+4=\left(x-4\right)^2\)
<=> \(x^2-6x-12-\left(x-4^2\right)=0\)
<=> \(x^2-6x-12-\left(x^2-8x+16\right)=0\)
<=> \(x^2-6x-12-x^2+8x-16=0\)
<=> \(2x-28=0\)
<=> \(2\left(x-14\right)=0\)
<=> x-14=0
<=> x=14
\(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\)
\(=>x^2+x+1-3x^2=2x\left(x-1\right)\)
\(=>-2x^2+x+1=2x^2-2x\)
\(=>-4x^2+3x+1=0\)
\(=>\left(x-1\right)\left(x+\frac{1}{4}\right)=0\)'
\(=>\orbr{\begin{cases}x-1=0\\x+\frac{1}{4}\end{cases}=>\orbr{\begin{cases}x=1\\x=-\frac{1}{4}\end{cases}}}\)
a)\(\frac{2-x}{2007}-1=\frac{1-x}{2008}-\frac{x}{2009}\)
\(\Leftrightarrow\frac{2-x}{2007}-1+2=\frac{1-x}{2008}+1-\frac{x}{2009}+1\)
\(\Leftrightarrow\frac{2-x}{2007}+\frac{2007}{2007}=\frac{1-x}{2008}+\frac{2008}{2008}-\frac{x}{2009}+\frac{2009}{2009}\)
\(\Leftrightarrow\frac{2009-x}{2007}=\frac{2009-x}{2008}-\frac{2009-x}{2009}\)
\(\Leftrightarrow\frac{2009-x}{2007}-\frac{2009-x}{2008}+\frac{2009-x}{2009}=0\)
\(\Leftrightarrow\left(2009-x\right)\left(\frac{1}{2007}-\frac{1}{2008}+\frac{1}{2009}\right)=0\)
\(\Leftrightarrow2009-x=0\).Do \(\frac{1}{2007}-\frac{1}{2008}+\frac{1}{2009}\ne0\)
\(\Leftrightarrow x=2009\)
b)\(\left(12x+7\right)^2\left(3x+2\right)\left(2x+1\right)=3\)
\(\Leftrightarrow\left(12^2x^2+2\cdot12\cdot7x+7^2\right)\left(6x^2+7x+2\right)-3=0\)
\(\Leftrightarrow\left[24\left(6x^2+7x+2\right)+1\right]\left(6x^2+7x+2\right)-3=0\)
Đặt \(t=6x^2+7x+2\) ta có:
\(\left(24t+1\right)t-3=0\)\(\Leftrightarrow12t^2+t-3=0\)
Suy ra t rồi tìm đc x
VD:
INPUT: 4
OUTPUT:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1