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\(\frac{60}{x}=\frac{30}{x-6}+\frac{30}{x+10}\)
\(\Leftrightarrow\frac{60}{x}=\frac{30}{x-6}+\frac{30}{x+10},Đkxđ:x\ne0,6,-10\)
\(\Leftrightarrow\frac{60}{x}-\frac{30}{x-6}-\frac{30}{x+10}=0\)
\(\Leftrightarrow\frac{60\left(x-6\right)\left(x+10\right)-30x\left(x+10\right)=30\left(x-6\right)}{x\left(x-6\right)\left(x+10\right)}\)
\(\Leftrightarrow\frac{\left(60x-360\right)\left(x+10\right)-30x^2-300x-30x^2+180x}{x\left(x-6\right)\left(x+10\right)}\)
\(\Leftrightarrow\frac{60x^2+600x-360x-3600-30x^2-300x-30x^2+180}{x\left(x-6\right)\left(x=10\right)}=0\)
\(\Leftrightarrow\frac{120x-3600}{x\left(x-6\right)\left(x+10\right)}=0\)
\(\Leftrightarrow120x-3600=0\)
\(\Leftrightarrow120x=3600\)
\(\Leftrightarrow x=30;x\ne0;x\ne6,x\ne-10\)
Từ đề bài, ta có:
\(2+\frac{x-30}{10}+2+\frac{x-28}{9}+2+\frac{x-26}{8}=0\)
\(\Leftrightarrow\frac{x-10}{10}+\frac{x-10}{9}+\frac{x-10}{8}=0\)
\(\Leftrightarrow\left(x-10\right)\left(\frac{1}{10}+\frac{1}{9}+\frac{1}{8}\right)=0\)
Do \(\left(\frac{1}{10}+\frac{1}{9}+\frac{1}{8}\right)>0\)nên x-10=0
<=> x=10
Vậy phương trình có nghiệm duy nhất x=10
pt <=> 1/(x+2).(x+3) + 1/(x+3).(x+4) + 1/(x+4).(x+5) + 1/(x+5).(x+6) = 1/8
<=> 1/x+2 - 1/x+3 + 1/x+3 - 1/x+4 + 1/x+4 - 1/x+5 + 1/x+5 - 1/x+6 = 1/8
<=> 1/x+2 - 1/x+6 = 1/8
<=> (x+6-x-2)/(x+2).(x+6) = 1/8
<=> 4/(x+2).(x+6) = 1/8
<=>(x+2).(x+6) = 4 : 1/8 = 32
<=>x^2 + 8x + 12 = 32
<=> x^2+8x+12-32=0
<=>x^2+8x-20=0
<=>(x-2).(x+10)=0
<=> x-2 =0 hoặc x+10 = 0
<=> x=2 hoặc x=-10
giang sinh an lanh $%###Xuyen gam cu chuoi###%$
\(\frac{x-30}{10}+\frac{x-28}{9}+\frac{x-26}{8}=-6\)
<=> \(\frac{36.\left(x-30\right)}{360}+\frac{40\left(x-28\right)}{360}+\frac{45\left(x-26\right)}{360}=\frac{-2160}{360}\)
=> \(36x-1080+40x-1120+45x-1170=-2160\)
\(< =>36x+40x+45x=-2160+1080+1120+1170\)
<=> \(121x=1210\)
<=> x = 10
phân tích mẫu thành nhân tử r` tách ra rút gọn như kiểu bài tính của lớp 5 ấy
bài tương tự : Câu hỏi của Lê Phương Oanh - Toán lớp 8 | Học trực tuyến (https://h-o-c-24.vn/hoi-dap/question/179719.html)
\(\frac{x-15}{2014}+\frac{x-20}{2019}=\frac{x-5}{2004}+\frac{x+30}{1969}\)
\(\Leftrightarrow\frac{x-15}{2014}+1+\frac{x-20}{2019}+1=\frac{x-5}{2004}+1+\frac{x+30}{1969}+1\)
\(\Leftrightarrow\frac{x-15+2014}{2014}+\frac{x-20+2019}{2019}-\frac{x-5+2004}{2004}-\frac{x+30+1969}{1969}=0\)
\(\Leftrightarrow\frac{x-1999}{2014}+\frac{x+1999}{2019}-\frac{x+1999}{2004}-\frac{x+1999}{1969}=0\)
\(\Leftrightarrow\left(x-1999\right)\left(\frac{1}{2014}+\frac{1}{2019}-\frac{1}{2004}-\frac{1}{1969}\right)=0\)
Vì \(\left(\frac{1}{2014}+\frac{1}{2019}-\frac{1}{2004}-\frac{1}{1969}\right)\ne0\)
nên \(x-1999=0\)
\(\Leftrightarrow x=1999\)
\(easy!\)(sai đề + sửa đề)
\(\frac{x-5}{2014}+\frac{x-20}{2019}-\frac{x-5}{2004}-\frac{x+3}{1969}=0\)
\(\Leftrightarrow\left(\frac{x-15}{2014}-1\right)+\left(\frac{x-20}{2019}-1\right)-\left(\frac{x-5}{2004}-1\right)-\left(\frac{x-30}{1969}-1\right)=0\)
\(\Leftrightarrow\frac{x-1999}{2014}+\frac{x-1999}{2019}-\frac{x-1999}{2004}-\frac{x-1999}{1969}=0\)
\(\Leftrightarrow\left(x-1999\right)\left(\frac{1}{2014}+\frac{1}{2019}-\frac{1}{2004}-\frac{1}{1969}\right)=0\)
dễ dàng cm được \(x-1999=0\)
\(\Leftrightarrow x=1999\)
ĐKXĐ : x khác -3;-4;-5;-6
pt <=> 1/(x+3).(x+4) + 1/(x+4).(x+5) + 1/(x+5).(x+6) = 1/18
<=> 1/x+3 - 1/x+4 + 1/x+4 - 1/x+5 + 1/x+5 - 1/x+6 = 1/18
<=> 1/x+3 - 1/x+6 = 1/18
<=> x+6-x-3/(x+3).(x+6) = 1/18
<=> 3/x^2+9x+18 = 1/18
<=> x^2+9x+18 = 3 : 1/18 = 48
<=> x^2+9x+18-48 = 0
<=> x^2+9x-30 = 0
<=>(x+9/2)^2 = 201/4
<=> x = \(\frac{+-\sqrt{201}-9}{2}\)(tm)
k mk nha
\(\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}=\frac{1}{18}\)
\(\Leftrightarrow\)\(\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}=\frac{1}{18}\)
\(\Leftrightarrow\)\(\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}=\frac{1}{18}\)
\(\Leftrightarrow\)\(\frac{1}{x+3}-\frac{1}{x+6}=\frac{1}{18}\)
\(\Leftrightarrow\)\(\frac{3}{\left(x+3\right)\left(x+6\right)}=\frac{1}{18}\)
\(\Leftrightarrow\)(x + 3)(x + 6) = 54 = 6.9 = (-6).(-9)
Đến đây giải tiếp nha
30 theo tui nghĩ là đúng
=>\(\frac{30x\left(x-6\right)}{x\left(x+10\right)\left(x-6\right)}+\frac{30x\left(x+10\right)}{x\left(x+10\right)\left(x-6\right)}=\frac{60\left(x+10\right)\left(x-6\right)}{x\left(x-6\left(x+10\right)\right)}\)
=>30x2-180x+30x2+300x=60x2-360x+600x-3600
=>60x2+120x=60x2+240x-3600
=>-120x=-3600
=>x=30
nhớ k mk........Đúng 100%