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Đề bài lạ thế!
\(A=-\frac{8}{5}x^3+\frac{36}{5}x^2y-\frac{54}{5}xy^2+\frac{27}{5}y^3\)
\(=-\frac{1}{5}\left(8x^3-36x^2y+54xy^2-27y^3\right)\)
=\(-\frac{1}{5}\left(\left(2x\right)^3-3.\left(2x\right)^2.3y+3.2x.\left(3y\right)^2-\left(3y\right)^3\right)\)
\(=-\frac{1}{5}\left(2x-3y\right)^3=-\frac{1}{5}.4^3=-\frac{64}{5}\)
a) \(\frac{x+1}{94}+\frac{x+2}{93}+\frac{x+3}{92}=\frac{x+4}{91}+\frac{x+5}{90}+\frac{x+6}{89}\)
\(\Leftrightarrow\left(\frac{x+1}{94}+1\right)+\left(\frac{x+2}{93}+1\right)+\left(\frac{x+3}{92}+1\right)=\left(\frac{x+4}{91}+1\right)+\left(\frac{x+5}{90}+1\right)+\left(\frac{x+6}{89}+1\right)\)
\(\Leftrightarrow\frac{x+95}{94}+\frac{x+95}{93}+\frac{x+95}{92}-\frac{x+95}{91}-\frac{x+95}{90}-\frac{x+95}{89}=0\)
\(\Leftrightarrow\) \(\left(x+95\right)\left(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}\right)=0\)
Vì \(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}\ne0\)
\(\Rightarrow x+95=0\)
\(\Leftrightarrow x=-95\)
Vậy phương trình có một nghiệm x = -95
b) \(\frac{x-1}{59}+\frac{x-2}{58}+\frac{x-3}{57}=\frac{x-4}{56}+\frac{x-5}{55}+\frac{x-6}{54}\)
\(\Leftrightarrow\left(\frac{x-1}{59}-1\right)+\left(\frac{x-2}{58}-1\right)+\left(\frac{x-3}{57}-1\right)=\left(\frac{x-4}{56}-1\right)+\left(\frac{x-5}{55}-1\right)+\left(\frac{x-6}{54}-1\right)\)
\(\Leftrightarrow\frac{x-60}{59}+\frac{x-60}{58}+\frac{x-60}{57}-\frac{x-60}{56}-\frac{x-60}{55}-\frac{x-60}{54}=0\)
\(\Leftrightarrow\left(x-60\right)\left(\frac{1}{59}+\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}-\frac{1}{54}\right)=0\)
Vì \(\frac{1}{59}+\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}-\frac{1}{54}\ne0\)
\(\Rightarrow x-60=0\)
\(\Leftrightarrow x=60\)
Vậy phương trình có một nghiệm x = 60
a) \(\frac{x+1}{94}+\frac{x+2}{93}+\frac{x+3}{92}=\frac{x+4}{91}+\frac{x+5}{90}+\frac{x+6}{89}\)
\(\Rightarrow\left(\frac{x+1}{94}+1\right)+\left(\frac{x+2}{93}+1\right)+\left(\frac{x+3}{92}+1\right)=\left(\frac{x+4}{91}+1\right)+\left(\frac{x+5}{90}+1\right)+\left(\frac{x+6}{89}+1\right)\)
\(\Rightarrow\frac{x+95}{94}+\frac{x+95}{93}+\frac{x+95}{92}=\frac{x+95}{91}+\frac{x+95}{90}+\frac{x+95}{89}\)
\(\Rightarrow\frac{x+95}{94}+\frac{x+95}{93}+\frac{x+95}{92}-\frac{x+95}{91}-\frac{x+95}{90}-\frac{x+95}{89}=0\)
\(\Rightarrow\left(x+95\right)\left(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}\right)=0\)
Mà \(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}\ne0\)
\(\Rightarrow x+95=0\)
\(\Rightarrow x=-95\)
Vậy x = -95
b) \(\frac{x-1}{59}+\frac{x-2}{58}+\frac{x-3}{57}=\frac{x-4}{56}+\frac{x-5}{55}+\frac{x-6}{54}\)
\(\Rightarrow\left(\frac{x-1}{59}-1\right)+\left(\frac{x-2}{58}-1\right)+\left(\frac{x-3}{57}-1\right)=\left(\frac{x-4}{56}-1\right)+\left(\frac{x-5}{55}-1\right)+\left(\frac{x-6}{54}-1\right)\)
\(\Rightarrow\frac{x-60}{59}+\frac{x-60}{58}+\frac{x-60}{57}-\frac{x-60}{56}-\frac{x-5}{55}-\frac{x-6}{54}=0\)
\(\Rightarrow\left(x-60\right)\left(\frac{1}{59}+\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}-\frac{1}{54}\right)=0\)
Mà \(\frac{1}{59}+\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}-\frac{1}{54}\ne0\)
\(\Rightarrow x-60=0\)
\(\Rightarrow x=60\)
Vậy x = 60
Xin lỗi mình làm hơi tắt nha !!!Còn 1 cách nữa ,nếu bạn muốn thì nói với mình nha !!
Ta có : \(\frac{x-1}{59}+\frac{x-2}{58}+\frac{x-3}{57}=\frac{x-4}{56}+\frac{x-5}{55}+\frac{x-6}{54}\)
\(\Leftrightarrow\frac{x}{59}+\frac{x}{58}+\frac{x}{57}-\frac{x}{56}-\frac{x}{55}-\frac{x}{54}=\frac{1}{59}+\frac{2}{58}+\frac{3}{57}-\frac{4}{56}-\frac{5}{55}-\frac{6}{54}\)
<=> x = 60
Vậy x = 60
Bạn kiểm tra lại đề nhé. Chỗ
\(.....=\frac{x-4}{56}+\frac{x-5}{56}+\frac{x-6}{54}\)
A=\(\frac{13-x}{x+3}+\frac{6x^2+6}{x^4-8x^2-9}-\frac{3x+6}{\left(x+2\right)\left(x+3\right)}-\frac{2}{x-3}=0\)\(\Leftrightarrow\frac{13-x}{x+3}+\frac{6\left(x^2+1\right)}{\left(x-3\right)\left(x+3\right)\left(x^2+1\right)}-\frac{3\left(x+2\right)}{\left(x+2\right)\left(x+3\right)}-\frac{2}{x-3}=0\) ( với \(x^4-8x^2-9=x^4-9x^2+x^2-9=x^2\left(x^2-9\right)+\left(x^2-9\right)=\left(x^2-9\right)\left(x^2+1\right)=\left(x-3\right)\left(x+3\right)\left(x^2+1\right)\)
A= \(\frac{13-x}{x+3}+\frac{6}{\left(x-3\right)\left(x+3\right)}-\frac{3}{x+3}-\frac{2}{x-3}=0\) \(\Leftrightarrow\frac{10-x}{x+3}+\frac{6}{\left(x-3\right)\left(x+3\right)}-\frac{2}{x-3}=0\) \(\Leftrightarrow\left(10x-30\right)\left(x-3\right)+6-2\left(x+3\right)=0\Leftrightarrow-x^2+11x-30=0\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=6\\x=5\end{array}\right.\)
(x -1)/59 -1 +(x-2)/58 -1 +(x-3)/57 -1 = (x-3)/56 -1 +(x-4)/55 -1 +(x-5)/54 -1
<=> (x-60)/59 +(x-60)/58 + (X-60)/57 -(x-60)/56 - (X-60)/55 -(X-60)/54 =0
<=> (x-60).(1/59 +1/58 +1/57 -1/56 -1/55 - 1/54)=0
vì 1/59 +1/58 +1/57 -1/56 -1/55 -1/54 <0
nên x-60 =0 <=> x=60
đề bài của bạn bi sai vì vế trái không thể bằng vế phải nếu đề đúng thì phải là :
(x-1)/59 +(x-2)/58 +(x-3)/57 =(x-4)/56 +(x-5)/55 +(x-6)/54
khí đó bạn giải cách như trên ,chúc bạn học toán tốt
