Tìm \(x\), biết :
\((x+\frac{7x}{9})\left(x+\frac{7x}{33}\right)(x+\frac{7x}{33})...\left(x+\frac{7x}{9200}\right)=\frac{186}{25}\)
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4 tháng 5 2019
Đặt \(\hept{\begin{cases}\sqrt{7x+11}=a\\\sqrt{9-7x}=b\end{cases}}\)
\(\Rightarrow a^2-b^2=14x+2\)
\(\Rightarrow\frac{2}{a^2-b^2}+\frac{1}{ab}=\frac{7}{24}\)
\(\Leftrightarrow\left(b+7a\right)\left(7b-a\right)=0\)
4 tháng 5 2019
Làm nhầm phần phân tích nhân tử giờ làm lại cách khác.
Đặt \(7x+11=a\)
\(\Rightarrow7x=a-11\)
\(\Rightarrow\frac{1}{a-10}+\frac{1}{\sqrt{a\left(20-a\right)}}=\frac{7}{24}\)
\(\Leftrightarrow\frac{1}{\sqrt{a\left(20-a\right)}}=\frac{7}{24}-\frac{1}{a-10}\)
\(\Leftrightarrow\frac{1}{a\left(20-a\right)}=\left(\frac{7}{24}-\frac{1}{a-10}\right)^2\)
\(\Leftrightarrow\left(a-18\right)\left(a-16\right)\left(49a^2-630a+200\right)=0\)
PS: Bài giải trên bỏ đi nha
CN
1
28 tháng 1 2017
a) \(\frac{x+\frac{2\left(3-x\right)}{5}}{14}-\frac{5x-4\left(x-1\right)}{24}=\frac{7x+2+\frac{9-3x}{5}}{12}+\frac{2}{3}\)
<=> \(\frac{5x+2\left(3-x\right)}{70}-\frac{5x-4\left(x-1\right)}{24}=\frac{35x+10+9-3x}{60}+\frac{2}{3}\)
<=> \(12\left(5x+6-2x\right)-35\left(5x-4x+4\right)\)
<=> \(14\left(35x+10+9-3x\right)+280.2\) <=> \(12\left(3x+6\right)-35\left(x+4\right)\)
<=> \(14\left(32x+19\right)+560\)
<=> \(36x+72-35x-140=448x+226+560\)
<=> \(-447x=894\)
<=> x = -2
11 tháng 3 2020
1) \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)
<=> \(\frac{21x}{24}-\frac{100\left(x-9\right)}{24}=\frac{80x+6}{24}\)
<=> 21x - 100x + 900 = 80x + 6
<=> -79x - 80x = 6 - 900
<=> -159x = -894
<=> x = 258/53
Vậy S = {258/53}
2) \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x+1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
<=> \(\frac{3\left(4x^2+4x+1\right)}{15}-\frac{5\left(x^2+2x+1\right)}{15}=\frac{7x^2-14x-5}{15}\)
<=> 12x2 + 12x + 3 - 5x2 - 10x - 5 = 7x2 - 14x - 5
<=> 7x2 + 2x - 7x2 + 14x = -5 + 2
<=> 16x = 3
<=> x = 3/16
Vậy S = {3/16}
11 tháng 3 2020
3) 4(3x - 2) - 3(x - 4) = 7x+ 10
<=> 12x - 8 - 3x + 12 = 7x + 10
<=> 9x - 7x = 10 - 4
<=> 2x = 6
<=> x = 3
Vậy S = {3}
4) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)
<=> \(\frac{x^2+14x+40}{12}+\frac{3\left(x^2+2x-8\right)}{12}=\frac{4\left(x^2+8x-20\right)}{12}\)
<=> x2 + 14x + 40 + 3x2 + 6x - 24 = 4x2 + 32x - 80
<=> 4x2 + 20x - 4x2 - 32x = -80 - 16
<=> -12x = -96
<=> x = 8
Vậy S = {8}
8 tháng 1 2020
1.
\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
\(MC:12\)
Quy đồng :
\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)
\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)
\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)
\(\Leftrightarrow6x+9-3x=-4-9+16\)
\(\Leftrightarrow-7x=3\)
\(\Leftrightarrow x=\frac{-3}{7}\)
2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
\(MC:20\)
Quy đồng :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
4 tháng 3 2020
b) \(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\)
<=> \(\frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{1\left(x-2\right)}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)
<=> x2+2x-x+2=2
<=> x2+x=2-2
<=> x2+x=0
<=>x(x+1)=0
<=>x=0 hoặc x+1=0
<=>x=0 hoặc x = -1
4 tháng 3 2020
a) \(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)
<=>\(\frac{1.x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{5\left(2x-3\right)}{x\left(2x-3\right)}\)
<=> x-3 =10x-15
<=> x-10x= -15+3
<=> -9x = -12
<=> x = \(\frac{-12}{-9}\)
<=> x = \(\frac{4}{3}\)
Tớ biết làm đúng 100%:
\((x\cdot1+x\cdot\frac{7}{9})\left(x\cdot1+x\cdot\frac{7}{20}\right)...\left(x\cdot1+x\cdot\frac{7}{9200}\right)=\frac{186}{25}\)
\(x\cdot\left(1+\frac{7}{9}\right)\cdot x\left(1+\frac{7}{20}\right)\cdot...\cdot x\left(1+\frac{7}{9200}\right)=\frac{186}{25}\)
\(\left(x\cdot x\cdot...\cdot x\right)(\frac{16}{9}+\frac{27}{20}+...+\frac{9207}{9200})=\frac{186}{25}\)
\(\left(x\cdot x\cdot...\cdot x\right)\left(\frac{2\cdot8}{1\cdot9}+\frac{3\cdot9}{2\cdot10}+...+\frac{93\cdot99}{92\cdot100}\right)=\frac{186}{25}\)
\(x^{92}\cdot\frac{2\cdot8\cdot3\cdot9\cdot...\cdot93\cdot99}{1\cdot9\cdot2\cdot10\cdot...\cdot92\cdot100}=\frac{186}{25}\)
\(x^{92}\cdot\frac{\left(2\cdot3\cdot...\cdot93\right)\cdot\left(8\cdot9\cdot...\cdot99\right)}{\left(1\cdot2\cdot...\cdot92\right)\cdot\left(9\cdot10\cdot...\cdot100\right)}=\frac{186}{25}\)
\(x^{92}\cdot\frac{93\cdot8}{100}=\frac{186}{25}\)
\(x^{92}\cdot\frac{186}{25}=\frac{186}{25}\)
\(x^{92}=\frac{186}{25}:\frac{186}{25}\)
\(x^{92}=1\Rightarrow x=1\)
cô tớ giải rồi . x=1 (đúng 100%)
CÂU NÀY X=1 ĐÓ !
CẬU KIA ĐÚNG RỒI !