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\(pt\Leftrightarrow\frac{6\left(x+1\right)+3\left(x+3\right)}{4.3}=\frac{3.4.3-4\left(x+2\right)}{4.3}\)
\(\Leftrightarrow6x+6+3x+9=36-4x-8\)
\(\Leftrightarrow13x=13\)
\(\Leftrightarrow x=1\)
\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\left(x\ne-4;-5;-6;-7;-8\right)\)
\(\Leftrightarrow\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{x}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Rightarrow x^2+11x+28=54\)
\(\Leftrightarrow x^2+11x-26=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+13=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\left(tm\right)\\x=-13\left(tm\right)\end{cases}}}\)
vậy x=2; x=-13
Bài làm:
đkxđ: \(x\ne\left\{-4;-5;-6;-7\right\}\)
Ta có: \(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow x^2+11x+28=54\)
\(\Leftrightarrow x^2+11x-26=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+13=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-13\end{cases}}\)
Vậy tập nghiệm của PT \(S=\left\{-13;2\right\}\)
Gợi ý :
Bài 1 : Cộng thêm 1 vào 3 phân thức đầu, trừ cho 3 ở phân thức thứ 4, có nhân tử chung là (x+2020)
Bài 2 : Trừ mỗi phân thức cho 1, chuyển vế và có nhân tử chung là (x-2021)
Bài 3 : Phân thức thứ nhất trừ đi 1, phân thức hai trù đi 2, phân thức ba trừ đi 3, phân thức bốn trừ cho 4, phân thức 5 trừ cho 5. Có nhân tử chung là (x-100)
bài 3
\(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15.\)
=>\(\frac{x-90}{10}-1+\frac{x-76}{12}-2+\frac{x-58}{14}-3+\frac{x-36}{16}-4+\frac{x-15}{17}-5=0\)
=>\(\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)
=>\(\left(x-100\right).\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)
=>(x-100)=0 do \(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\)
=> x=100
b, \(B=\frac{\frac{x}{x+3}-\frac{9}{x^2+6x+9}}{\frac{3}{x+3}}=\frac{\frac{x}{x+3}-\frac{3^2}{x^2+2\cdot3\cdot x+3^2}}{\frac{3}{x+3}}\)
\(=\frac{\frac{x}{x+3}-\left(\frac{3}{x+3}\right)^2}{\frac{3}{x+3}}=1-\frac{3}{x+3}\)
a, Vậy điều kiện là \(x\ne3\)
c, \(B=\frac{1}{3}\Leftrightarrow1-\frac{3}{x+3}=\frac{1}{3}\)
\(\Rightarrow\frac{3}{x+3}=\frac{2}{3}\Leftrightarrow x=\frac{3}{2}\)
\(a)\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}=\frac{-3}{4}\left(x\ne-3;x\ne2\right)\)
\(\Leftrightarrow\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{1}{x-2}=\frac{-3}{4}\)
\(\Leftrightarrow\frac{x^2-4}{\left(x-2\right)\left(x+3\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{x+3}{\left(x-2\right)\left(x+3\right)}=\frac{-3}{4}\)
\(\Leftrightarrow\frac{x^2-4-5-x-3}{\left(x-2\right)\left(x+3\right)}=\frac{-3}{4}\)
\(\Leftrightarrow\frac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}=\frac{-3}{4}\)
\(\Leftrightarrow\frac{\left(x-4\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=\frac{-3}{4}\)
\(\Leftrightarrow\frac{x-4}{x-2}=\frac{-3}{4}\)
<=> 4x-16=-3x+6
<=> 4x-16+3x-6=0
<=> 7x-22=0
<=> 7x=22
<=> \(x=\frac{22}{7}\)(TMĐK)
a) Ta thấy:
\(\left(x+4\right)\left(x-4\right)=x\left(x-\frac{2}{3}\right)\)
\(\Rightarrow\left(x^2-4x\right)+\left(4x-16\right)=x^2-\frac{2}{3}x\)
\(\Rightarrow\left(x^2-16\right)-\left(4x-4x\right)=x^2-\frac{2}{3}x\)
\(\Rightarrow x^2-16-0=x^2-\frac{2}{3}x\)
\(\Rightarrow x^2-16=x^2-\frac{2}{3}x\)
\(\Rightarrow16=\frac{2}{3}x\) ( do có cùng hiệu và cùng số bị trừ )
\(\Rightarrow x=16:\frac{2}{3}\)
\(\Rightarrow x=24\)
Vậy x = 24
b.) x^3-x^2-2x=0
x(x^2-x-2)=0
x(x^2-2x+x-2)=0
x(x(x-2)+x-2)=0
x(x-2)(x+1)=0
suy ra x=0 hoặc x-2=0 hoặc x+1=0
vậy x=0 hoặc x=2 hoặc x=-1
hình như câu c đề phải là (x+4)/120 thì phải đó bạn
c.)(x+4)/120+(x+8)/116=(x+5)/119+(x+7)/117
(x+4)/120+(x+8)/116-(x+5)/119-(x+7)/117=0
(x+4)/120+1+(x+8)/116+1-(x+5)/119-1-(x+7)/117-1=0
(x+4)/120+1+(x+8)/116+1-((x+5)/119+1)-((x+7)/117+1)=0
(x+124)/120+(x+124)/116-(x+124)/119-(x+124)/117=0
(x+124)(1/120+1/116-1/119-1/117)=0
suy ra x+124=0
x=-124
sai đề rồi nha...bạn thay dấu suy ra thành dấu tương đương giùm mik..mik bị nhầm
\(\frac{x+5}{65}+\frac{x+10}{60}=\frac{x+15}{35}+\frac{x+20}{50}\)
\(\Rightarrow\frac{x+5}{65}+\frac{x+10}{60}-\frac{x+15}{55}-\frac{x+20}{50}+2-2=0\)
\(\Rightarrow\left(\frac{x+5}{65}+1\right)+\left(\frac{x+10}{60}+1\right)-\left(\frac{x+15}{55}+1\right)-\left(\frac{x+20}{50}+1\right)=0\\ \)
\(\Rightarrow\left(\frac{x+5}{65}+\frac{65}{65}\right)+\left(\frac{x+10}{60}+\frac{60}{60}\right)-\left(\frac{x+15}{55}+\frac{55}{55}\right)-\left(\frac{x+20}{50}+\frac{50}{50}\right)=0\)
\(\Rightarrow\frac{x+70}{65}+\frac{x+70}{60}-\frac{x+70}{55}-\frac{x+70}{50}=0\)
\(\Rightarrow\left(x+70\right)\left(\frac{1}{65}+\frac{1}{60}-\frac{1}{55}-\frac{1}{50}\right)=0\)
\(\Rightarrow x+70=0\left(\frac{1}{65}+\frac{1}{60}-\frac{1}{55}-\frac{1}{50}\nè0\right)\)
\(\Leftrightarrow x=-70\)
học tốt...............nhớ k cho mik nha
từ trên ta có (x+2)/13+(2x+45)/15-(3x+8)/37-(4x+69)/9=0
(x+2)/13+1+(2x+45)/15-1-(3x+8)/37-1-(4x+69)/9+1=0
(x+15)/13+(2x+30)/15-((3x+8)/37+1)-((4x+69)/9-1)=0
(x+15)/13+2(x+15)/15-3(x+15)/37-4(x+15)/9=0
(x+15)(1/13+2/15-3/37-4/9)=0
suy ra x+15=0
x=-15
\(\frac{x+2}{13}+\frac{2x+45}{15}=\frac{3x+8}{37}+\frac{4x+69}{9}\)
<=> \(\left(\frac{x+2}{13}+1\right)+\left(\frac{2x+45}{15}-1\right)=\left(\frac{3x+8}{37}+1\right)+\left(\frac{4x+69}{9}-1\right)\)
<=> \(\frac{x+2+13}{13}+\frac{2x+45-15}{15}=\frac{3x+8+37}{37}+\frac{4x+69-9}{9}\)
<=> \(\frac{x+15}{13}+\frac{2\left(x+15\right)}{13}=\frac{3\left(x+15\right)}{37}+\frac{4\left(x+15\right)}{9}\)
<=> \(\frac{x+15}{13}+\frac{2\left(x+15\right)}{13}-\frac{3\left(x+15\right)}{37}-\frac{4\left(x+15\right)}{9}=0\)
<=> \(\left(x+15\right)\left(\frac{1}{13}+\frac{2}{13}-\frac{3}{37}-\frac{4}{9}\right)=0\)
Vì \(\frac{1}{13}+\frac{2}{13}-\frac{3}{37}-\frac{4}{9}\ne0\)
<=> x + 15 = 0
<=> x = -15
1) Ta có pt : \(4x^2+\frac{1}{x^2}=8x+\frac{4}{x}\)
\(\Leftrightarrow4x^2+4+\frac{1}{x^2}=8x+4+\frac{4}{x}\)
\(\Leftrightarrow\left(2x+\frac{1}{x}\right)^2=4\left(2x+\frac{1}{x}\right)+4\)
\(\Leftrightarrow\left(2x+\frac{1}{x}\right)^2-4\left(2x+\frac{1}{x}\right)+4=8\)
\(\Leftrightarrow\left(2x+\frac{1}{x}-2\right)^2=8\)
Đến đây dễ rồi nhé, chia 2 TH.
Bài làm
\(\frac{50}{\left(20-x\right)}-\frac{50}{\left(x+20\right)}=\frac{4}{3}\)
\(\Leftrightarrow\frac{3.50\left(x+20\right)}{3.\left(20-x\right)\left(x+20\right)}-\frac{3.50\left(20-x\right)}{3\left(20-x\right)\left(x+20\right)}=\frac{4\left(20-x\right)\left(x+20\right)}{3\left(20-x\right)\left(x+20\right)}\)
\(\Rightarrow150\left(x+20\right)-150\left(20-x\right)=4\left(20^2-x^2\right)\)
\(\Leftrightarrow150x+3000-3000+150x=1600-4x^2\)
\(\Leftrightarrow4x^2+300x-1600=0\)
Đoạn đó còn mà bạn, chưa giải hết hay sao kia