Tính:
\(\frac{2}{-x-3}+\frac{3}{-x+3}+\frac{x^2+11x+12}{x^2-9}\)
A ) \(\frac{11x+1}{86}-\frac{11x-1}{88}+\frac{11x+2}{85}=\frac{11x-2}{89}\)
B ) \(\frac{-3}{x^2-5x+4}+\frac{-3}{x^2-11x+28}+\frac{-3}{x^2-17x+70}=\frac{9}{14}\)
C ) \(x^4+3x^3-7x^2-27x-18=0\)
Giải Phương Trình Sau (Nhớ ghi cách làm nha mình k đúng cho)
Giải Phương Trình Sau (Nhớ ghi cách làm nha mình ĐÁNH DẤU ĐÚNG cho)
Giải phương trình
a, \(\frac{1}{4x^2-12x+9}-\frac{3}{9-4x^2}=\frac{4}{4x^2+12x+9}\)
b, \(\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}=\frac{1}{8}\)
ai giúp mình câu (a) với ạ
ĐKXĐ: \(x\ne\pm\frac{3}{2}\)
\(\frac{1}{\left(2x-3\right)^2}+\frac{3}{\left(2x-3\right)\left(2x+3\right)}-\frac{4}{\left(2x+3\right)^2}=0\)
\(\Leftrightarrow\frac{1}{\left(2x-3\right)^2}-\frac{1}{\left(2x-3\right)\left(2x+3\right)}+\frac{4}{\left(2x-3\right)\left(2x+3\right)}-\frac{4}{\left(2x-3\right)^2}=0\)
\(\Leftrightarrow\frac{1}{2x-3}\left(\frac{1}{2x-3}-\frac{1}{2x+3}\right)-\frac{4}{2x-3}\left(\frac{1}{2x-3}-\frac{1}{2x+3}\right)=0\)
\(\Leftrightarrow\left(\frac{1}{2x-3}-\frac{4}{2x+3}\right)\left(\frac{1}{2x-3}-\frac{1}{2x+3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=2x-3\left(vn\right)\\2x+3=4\left(2x-3\right)\Rightarrow x=\frac{5}{2}\end{matrix}\right.\)
Giai pt:
1/ \(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+20}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)
2/ \(\frac{1}{x^2+5x+4}+\frac{1}{x^2+11x+28}+\frac{1}{x^2+17x+70}+\frac{1}{x^2+23x+130}=\frac{4}{13}\)
3/ \(x^2+\frac{4x^2}{\left(x+2\right)^2}=12\)
4/ \(x^3-x^2-\frac{8}{x^3-x^2}=2\)
bài 1+2: phân tích mẫu thành nhân tử r` áp dụng
1/ab=1/a-1/b
bài 3+4: quy đồng rút gọn blah...
Tìm số TB cộng của x, biết: \(\frac{x^3+x^2-11x+9}{x^3+3x^2-4}=\frac{2}{3}\)
giải phương trình
\(\frac{1}{x^2-7x+12}+\frac{1}{x^2-9x+20}+\frac{1}{x^2-11x+30}=\frac{3}{10}\)
Giải PT:
a) \(\frac{2}{x^2+4x+3}+\frac{5}{x^2+11x+24}+\frac{2}{x^2+18x+80}=\frac{9}{52}\)
b) \(x^2+\left(\frac{x}{x+1}\right)^2=\frac{5}{4}\)
c) \(x^4-3x^3+2x^2-9x+9=0\)
Câu c : \(x^4-3x^3+2x^2-9x+9=0\)
<=>\(x^4-x^3-2x^3+2x^2-9x+9=0\)
<=>\(x^3\left(x-1\right)-2x^2\left(x-1\right)-9\left(x-1\right)=0\)
<=>\(\left(x-1\right)\left(x^3-2x^2-9\right)=0\)
<=> \(x-1=0\) hoặc \(x^3-2x^2-9=0\)
Nếu x-1=0 <=> x=1
Nếu \(x^3-2x^2-9=0\)
<=> \(x^3-3x^2+x^2-9=0\)
<=>\(x^2\left(x-3\right)+\left(x-3\right)\left(x+3\right)=0\)
<=>\(\left(x-3\right)\left(x^2+x+3\right)=0\)
Vì \(x^2+x+3=\left(x+\frac{1}{2}\right)^2+\frac{11}{4}\) >0 nên x-3=0 <=> x=3
Vậy \(S=\left\{1;3\right\}\)
Câu b : \(x^2+\left(\frac{x}{x+1}\right)^2=\frac{5}{4}\)
<=> \(4x^2\left(x^2+2x+2\right)=5\left(x^2+2x+1\right)\)
<=> \(4x^4+8x^3+8x^2=5x^2+10x+5\)
<=>\(4x^4+8x^3+3x^2-10x-5=0\)
<=>\(4x^4-4x^3+12x^3-12x^2+15x^2-15x+5x-5=0\)
<=>\(\left(x-1\right)\left(4x^3+12x^2+15x+5\right)=0\)
<=>\(\left(x-1\right)\left(2x+1\right)\left(2x^2+5x+5\right)=0\)
<=>x=1 hoặc \(x=\frac{-1}{2}\)
Phương trình \(2x^2+5x+5=0\) Vô nghiệm
Let P = \(\frac{2x}{x+3}-\frac{x+1}{x-3}+\frac{3-11x}{9-x^2}\). Find the smallest integer x such that P is also an integer.
\(x\ne\pm3\)
\(P=\frac{2x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{11x-3}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x^2+x-6}{\left(x-3\right)\left(x+3\right)}=\frac{\left(x+3\right)\left(x-2\right)}{\left(x+3\right)\left(x-3\right)}\)
\(=\frac{x-2}{x-3}=1+\frac{1}{x-3}\)
P is an integer if and only if 1 is divisible by \(x-3\)
Therefore \(x-3=\left\{-1;1\right\}\Rightarrow x=\left\{2;4\right\}\)
\(\Rightarrow x_{min}=2\)
Giải phương trình sau
\(\frac{2}{x^2+4x+3}+\frac{5}{x^2+11x+24}+\frac{2}{x^2+18x+80}=\frac{9}{52}\)
ĐKXĐ: \(x\ne\left\{-10;-8;-3;-1\right\}\)
\(\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{5}{\left(x+3\right)\left(x+8\right)}+\frac{2}{\left(x+8\right)\left(x+10\right)}=\frac{9}{52}\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+10}=\frac{9}{52}\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+10}=\frac{9}{52}\)
\(\Leftrightarrow\frac{9}{\left(x+1\right)\left(x+10\right)}=\frac{9}{52}\)
\(\Leftrightarrow\left(x+1\right)\left(x+10\right)=52\)
\(\Leftrightarrow x^2+11x-42=0\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-14\end{matrix}\right.\)
Giải PT:
a) \(\frac{2}{x^2+4x+3}+\frac{5}{x^2+11x+24}+\frac{2}{x^2+18x+80}=\frac{9}{52}\)
b) \(x^2+\left(\frac{x}{x+1}\right)^2=\frac{5}{4}\)
c) \(x^4-3x^3+2x^2-9x+9=0\)
a) \(ĐKXĐ:x\ne-1;x\ne-3;x\ne-8;x\ne-10\)
\(\frac{2}{x^2+4x+3}+\frac{5}{x^2+11x+24}+\frac{2}{x^2+18x+8x}=\frac{9}{52}\)
\(\Leftrightarrow\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{5}{\left(x+3\right)\left(x+8\right)}+\frac{2}{\left(x+10\right)\left(x+8\right)}-\frac{9}{52}=0\)
\(\Leftrightarrow\frac{104\left(x+10\right)\left(x+8\right)+260\left(x+1\right)\left(x+10\right)+104\left(x+1\right)\left(x+3\right)-9\left(x+1\right)\left(x+3\right)\left(x+8\right)\left(x+10\right)}{52\left(x+1\right)\left(x+3\right)\left(x+8\right)\left(x+10\right)}=0\)
Đoạn này cậu tự phân tích tử rồi rút gọn nhé :D Vì hơi dài nên viết ra đây sẽ rối, k đẹp mắt cho lắm :>
\(\Leftrightarrow\frac{-927x^2+1782x+9072-9x^4-198x^3}{52\left(x+1\right)\left(x+3\right)\left(x+8\right)\left(x+10\right)}=0\)
\(\Leftrightarrow\frac{-9\left(x^4+22x^3+103x^2-198x-1008\right)}{52\left(x+1\right)\left(x+3\right)\left(x+8\right)\left(x+10\right)}=0\)
\(\Leftrightarrow\frac{-9\left(x^4-3x^3+25x^3-75x^{^2}+178x^2-534x+336x-1008\right)}{52\left(x+1\right)\left(x+3\right)\left(x+8\right)\left(x+10\right)}=0\)
\(\Leftrightarrow\frac{-9\left[x^3\left(x-3\right)+25x^2\left(x-3\right)+178x\left(x-3\right)+336\left(x-3\right)\right]}{52\left(x+1\right)\left(x+3\right)\left(x+8\right)\left(x+10\right)}=0\)
\(\Leftrightarrow\frac{-9\left(x-3\right)\left(x^3+25x^2+178x+336\right)}{52\left(x+1\right)\left(x+3\right)\left(x+8\right)\left(x+10\right)}=0\)
\(\Leftrightarrow\frac{-9\left(x-3\right)\left(x^3+14x^2+11x^2+154x+24x+336\right)}{52\left(x+1\right)\left(x+3\right)\left(x+8\right)\left(x+10\right)}=0\)
\(\Leftrightarrow\frac{-9\left(x-3\right)\left[x^2\left(x+14\right)+11x\left(x+14\right)+24\left(x+14\right)\right]}{52\left(x+1\right)\left(x+3\right)\left(x+8\right)\left(x+10\right)}=0\)
\(\Leftrightarrow\frac{-9\left(x-3\right)\left(x+14\right)\left(x^2+11x+24\right)}{52\left(x+1\right)\left(x+3\right)\left(x+8\right)\left(x+10\right)=0}\)
\(\Leftrightarrow\frac{-9\left(x+14\right)\left(x-3\right)\left(x+3\right)\left(x+8\right)}{52\left(x+1\right)\left(x+3\right)\left(x+8\right)\left(x+10\right)}=0\)
\(\Leftrightarrow\frac{-9\left(x+14\right)\left(x-3\right)}{52\left(x+1\right)\left(x+10\right)}=0\)
\(\Leftrightarrow-9x^2-99x+378=0\)
\(\Leftrightarrow x^2+11x-42=0\)
\(\Leftrightarrow\left(x+14\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+14=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-14\\x=3\end{cases}}}\)
Vậy tập nghiệm của phương trình là : \(S=\left\{-14;3\right\}\)
b) \(ĐKXĐ:x\ne-1\)
\(x^2+\left(\frac{x}{x+1}\right)^2=\frac{5}{4}\)
\(\Leftrightarrow x^2+\frac{x^2}{\left(x+1\right)^2}-\frac{5}{4}=0\)
\(\Leftrightarrow\frac{4x^2\left(x^2+2x+1\right)+4x^2-5\left(x^2+2x+1\right)}{\left(x+1\right)^2}=0\)
\(\Leftrightarrow4x^4+8x^3+4x^2+4x^2-5x^2-10x-5=0\)
\(\Leftrightarrow4x^2+8x^3+3x^2-10x-5=0\)
\(\Leftrightarrow4x^4+2x^3+6x^3+3x^2-10x-5=0\)
\(\Leftrightarrow2x^3\left(2x+1\right)+3x^2\left(2x+1\right)-5\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x^3+3x^2-5\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x^3-2x^2+5x^2-5x+5x-5\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left[2x^2\left(x-1\right)+5x\left(x-1\right)+5\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x-1\right)\left(2x^2+5x+5\right)=0\)
\(\Leftrightarrow2x+1=0\)
hoặc \(x-1=0\)
hoặc \(2x^2+5x+5=0\)
\(\Leftrightarrow\) \(x=-\frac{1}{2}\left(tm\right)\)
hoặc \(x=1\left(tm\right)\)
hoặc \(\left(x+\frac{5}{4}\right)^2+\frac{55}{16}=0\left(ktm\right)\)
Vậy tập nghiệm của phương trình là : \(S=\left\{-\frac{1}{2};1\right\}\)
c) \(x^4-3x^3+2x^2-9x+9=0\)
\(\Leftrightarrow x^4-x^3-2x^3+2x^2-9x+9=0\)
\(\Leftrightarrow x^3\left(x-1\right)-2x^2\left(x-1\right)-9\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3-2x^2-9\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[\left(x^3-3x^2\right)+\left(x^2-9\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x-3\right)+\left(x-3\right)\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(x^2+x+3\right)=0\)
\(\Leftrightarrow\)\(x-1=0\)
hoặc \(x-3=0\)
hoặc \(x^2+x+3=0\)
\(\Leftrightarrow\)\(x=1\left(tm\right)\)
hoặc \(x=3\left(tm\right)\)
hoặc \(\left(x-\frac{1}{2}\right)^2+\frac{11}{4}=0\left(ktm\right)\)
Vậy tập nghiệm của phương trình là :\(S=\left\{1;3\right\}\)
\(ĐKXĐ:x\ne-1;x\ne-3;x\ne-8;x\ne-10\)
\(pt\Leftrightarrow\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{5}{\left(x+3\right)\left(x+8\right)}+\frac{2}{\left(x+8\right)\left(x+10\right)}=\frac{9}{52}\)
\(\Leftrightarrow\frac{\left(x+3\right)-\left(x+1\right)}{\left(x+1\right)\left(x+3\right)}+\frac{\left(x+8\right)-\left(x+3\right)}{\left(x+3\right)\left(x+8\right)}+\frac{\left(x+10\right)-\left(x+8\right)}{\left(x+8\right)\left(x+10\right)}\)
\(=\frac{9}{52}\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+10}=\frac{9}{52}\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+10}=\frac{9}{52}\)
\(\Leftrightarrow\frac{9}{\left(x+1\right)\left(x+10\right)}=\frac{9}{52}\)
\(\Leftrightarrow\left(x+1\right)\left(x+10\right)=52\)
\(\Leftrightarrow x^2+11x+10=52\)
\(\Leftrightarrow x^2+11x-42=0\)
\(\Delta=11^2+4.42=289,\sqrt{289}=17\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-11+17}{2}=3\\x=\frac{-11-17}{2}=-14\end{cases}}\)
Vậy pt có 2 nghiệm là 3 và -14
战哥làm dài và khó hiểu))):????