CM đẳng thức
\(\dfrac{x^{24}+x^{18}+x^{12}+x^6+1}{x^{27}+x^{24}+x^{21}+x^{18}+x^{15}+x^{12}+x^9+x^6+x^3+1}\)=\(\dfrac{1}{x^3+1}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{x^{24}+x^{18}+x^{12}+x^6+1}{x^{27}+x^{24}+x^{21}+x^{18}+x^{15}+x^{12}+x^9+x^6+x^3+1}=\frac{x^{24}+x^{18}+x^{12}+x^6+1}{x^{24}\left(x^3+1\right)+x^{18}\left(x^3+1\right)+x^{12}\left(x^3+1\right)+x^6\left(x^3+1\right)+\left(x^3+1\right)}\)
=\(\frac{x^{24}+x^{18}+x^{12}+x^6+1}{\left(x^3+1\right)\left(x^{24}+x^{18}+x^{12}+x^6+1\right)}=\frac{1}{x^3+1}\)
Giải:
\(9-3\times\left(x-9\right)=6\)
\(3\times\left(x-9\right)=9-6\)
\(3\times\left(x-9\right)=3\)
\(x-9=3:3\)
\(x-9=1\)
\(x=1+9\)
\(x=10\)
\(4+6\times\left(x+1\right)=70\)
\(6\times\left(x+1\right)=70-4\)
\(6\times\left(x+1\right)=66\)
\(x+1=66:6\)
\(x+1=11\)
\(x=11-1\)
\(x=10\)
\(\dfrac{x}{13}+\dfrac{15}{26}=\dfrac{46}{52}\)
\(\dfrac{x}{13}=\dfrac{23}{26}-\dfrac{15}{26}\)
\(\dfrac{x}{13}=\dfrac{4}{13}\)
\(\Rightarrow x=4\)
\(\dfrac{11}{14}-\dfrac{3}{x}=\dfrac{5}{14}\)
\(\dfrac{3}{x}=\dfrac{11}{14}-\dfrac{5}{14}\)
\(\dfrac{3}{x}=\dfrac{3}{7}\)
\(\Rightarrow x=7\)
\(5\times\left(3+7\times x\right)=40\)
\(3+7\times x=40:5\)
\(3+7\times x=8\)
\(7\times x=8-3\)
\(7\times x=5\)
\(x=5:7\)
\(x=\dfrac{5}{7}\)
\(x\times6+12:3=120\)
\(x\times6+4=120\)
\(x\times6=120-4\)
\(x\times6=116\)
\(x=116:6\)
\(x=\dfrac{58}{3}\)
\(x\times3,7+x\times6,3=120\)
\(x\times\left(3,7+6,3\right)=120\)
\(x\times10=120\)
\(x=120:10\)
\(x=12\)
\(\left(15\times24-x\right):0,25=100:\dfrac{1}{4}\)
\(\left(360-x\right):0,25=400\)
\(360-x=400.0,25\)
\(360-x=100\)
\(x=360-100\)
\(x=260\)
\(71+65\times4=\dfrac{x+140}{x}+260\)
\(\left(x+140\right):x+260=71+260\)
\(x:x+140:x+260=331\)
\(1+140:x+260=331\)
\(140:x=331-1-260\)
\(140:x=70\)
\(x=140:70\)
\(x=2\)
\(\left(x+1\right)+\left(x+4\right)+\left(x+7\right)+...+\left(x+28\right)=155\)
\(10\times x+\left(1+4+7+...+28\right)=155\)
Số số hạng \(\left(1+4+7+...+28\right)\) :
\(\left(28-1\right):3+1=10\)
Tổng dãy \(\left(1+4+7+...+28\right)\) :
\(\left(1+28\right).10:2=145\)
\(\Rightarrow10\times x+145=155\)
\(10\times x=155-145\)
\(10\times x=10\)
\(x=10:10\)
\(x=1\)
Đều theo cách lớp 5 nha em!
1, 2x - 35 = 15
2x = 15 + 35
2x = 50
x = 50 : 2
x = 25.
2, 3x + 18 = 12
3x = 12 - 18
3x = -6
x = -6 : 3
x = -2.
3, / x - 1 / = 0
=> x \(\in\varnothing\).
4, -13 /x/ = - 26
/x/ = -26 : -13
=> \(\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)
Vậy x \(\in\){ 2 ; -2}.
5,4 - ( 27 - 3 ) = x - ( 13 - 4 )
4 - 24 = x - 9
-20 = x - 9
-x = 9 + 20
-x = 29
x = -29.
6, 47 - ( x + 15 ) = 21
47 - x - 15 = 21
-x - 15 = 21 - 47
-x - 15 = -26
-x = -26 + 15
-x = - 11
x = 11.
7, -5 -( 24 - x) = - 11
-5 - 24 + x = -11
-24 + x = -11 + 5
-24 + x = -6
x = -6 + 24
x = 18.
8, 6 - /x/ = 2
/x/ = 6 - 2
\(\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
Vậy x \(\in\left\{3;-3\right\}.\)
9, 6 + /x/ = 2
/x/ = 2 - 6
=> x = -4.
2x - 35 = 15
=> 2x = 15 + 35
=> x = 50 : 2
=> x = 25
3x + 18 = 12
=> 3x = 12 - 18
=> x = ( -6 ) : 3
=> x = -2
| x - 1 | = 0
=> x - 1 = 0
=> x = 0 + 1
=> x = 1
-13 * | x | = -26
=> | x | = -26 : ( -13 )
=> | x | = 2
1: Ta có: \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)
\(\Leftrightarrow5x+20+12x-28=7x+2\)
\(\Leftrightarrow17x-7x=2+8=10\)
hay x=1
2: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)
\(\Leftrightarrow\dfrac{6x}{36}+\dfrac{4\left(1-3x\right)}{36}=\dfrac{3\left(-x+1\right)}{36}\)
\(\Leftrightarrow6x+4-12x=-3x+3\)
\(\Leftrightarrow-6x+3x=3-4\)
hay \(x=\dfrac{1}{3}\)
3: Ta có: \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)
\(\Leftrightarrow4x-12-x-2=6x-3\)
\(\Leftrightarrow3x-14-6x+3=0\)
\(\Leftrightarrow-3x=11\)
hay \(x=-\dfrac{11}{3}\)
4: Ta có: \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)
\(\Leftrightarrow3x-6-8x-12=x+6\)
\(\Leftrightarrow-5x-x=6+18\)
hay x=-4
5: Ta có: \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)
\(\Leftrightarrow6x-3+2x-6=-1\)
\(\Leftrightarrow8x=8\)
hay x=1
Mk làm luôn nhé , không chép lại đề đâu
Q = \(\dfrac{x^6\left(x^4-x^2+1\right)-x^3\left(x^4-x^2+1\right)+x^4-x^2+1}{x^{18}\left(x^{12}+x^6+1\right)+x^{12}+x^6+1}\)
\(Q=\dfrac{\left(x^4-x^2+1\right)\left(x^6-x^3+1\right)}{\left(x^{12}+x^6+1\right)\left(x^{18}+1\right)}\)
\(Q=\dfrac{\left(x^4-x^2+1\right)\left(x^6-x^3+1\right)}{\left(x^{12}+x^6+1\right)\left[\left(x^6\right)^3+1\right]}\)
\(Q=\dfrac{\left(x^4-x^2+1\right)\left(x^6-x^3+1\right)}{\left(x^{12}+2x^6+1-x^6\right)\left[\left(x^2\right)^3+1\right]\left(x^{12}-x^6+1\right)}\)
\(Q=\dfrac{\left(x^4-x^2+1\right)\left(x^6-x^3+1\right)}{\left[\left(x^6+1\right)-\left(x^3\right)^2\right]\left(x^2+1\right)\left(x^4-x^2+1\right)\left(x^{12}-x^6+1\right)}\)
\(Q=\dfrac{\left(x^6-x^3+1\right)}{\left(x^6-x^3+1\right)\left(x^6+1+x^3\right)\left(x^2+1\right)\left(x^{12}-x^6+1\right)}\)
\(Q=\dfrac{1}{\left(x^6+1+x^3\right)\left(x^2+1\right)\left(x^{12}-x^6+1\right)}\)
\(\frac{x^{10}-x^8-x^7+x^6+x^6+x^4-x^3-x^2+1}{x^{30}+x^{24}+x^{18}+x^{12}+x^6+1}=\frac{(x^{10}-x^8+x^6)-(x^7-x^5+x^3)+(x^4-x^2+1)}{ (x^{30}+x^{18}+x^{24})+(x^{12}+x^6+1)} \)
=\(\frac{(x^4-x^2+1)(x^6-x^3+1)}{(x^{12}+x^6+1)(x^{18}+1 )}=\frac{(x^4-x^2+1)(x^6-x^3+1)}{(x^{12}+2x^6+1-x^6) (x^6+1)(x^{12}-x^6+1)}=\frac{(x^4-x^2+1)(x^6-x^3+1)}{ (x^6-x^3+1)(x^6+x^3+1)(x^2+1)(x^4-x^2+1)(x^12-x^6+1 )} \)
=\(\frac{1}{(x^6+x^2+1)(x^2+1)(x^{12}-x^6+1)}\)
Giải:
a) \(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\)
\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\)
\(\Rightarrow5x\in\left\{0;\pm5;10\right\}\)
\(\Rightarrow x\in\left\{0;\pm1;2\right\}\)
b) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow\dfrac{4}{x-6}=\dfrac{-12}{18}\)
\(\Rightarrow-12.\left(x-6\right)=4.18\)
\(\Rightarrow-12x+72=72\)
\(\Rightarrow-12x=72-72\)
\(\Rightarrow-12x=0\)
\(\Rightarrow x=0:-12\)
\(\Rightarrow x=0\)
\(\Rightarrow\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow y=\dfrac{-12.24}{18}=-16\)
c) \(\dfrac{x+46}{20}=x.\dfrac{2}{5}\)
\(\dfrac{x+46}{20}=\dfrac{2x}{5}\)
\(\Rightarrow5.\left(x+46\right)=2x.20\)
\(\Rightarrow5x+230=40x\)
\(\Rightarrow5x-40x=-230\)
\(\Rightarrow-35x=-230\)
\(\Rightarrow x=-230:-35\)
\(\Rightarrow x=\dfrac{46}{7}\)
Chúc bạn học tốt!
Xét \(x^{27}+x^{24}+x^{21}+x^{18}+x^{15}+x^{12}+x^9+x^6+x^3+1\)
\(=\left(x^{27}+x^{21}+x^{15}+x^9+x^3\right)+\left(x^{24}+x^{18}+x^{12}+x^6+1\right)\)
\(=x^3\left(x^{24}+x^{18}+x^{12}+x^6+1\right)+\left(x^{24}+x^{18}+x^{12}+x^6+1\right)\)
\(=\left(x^3+1\right)\left(x^{24}+x^{18}+x^{12}+x^6+1\right)\)
Vậy ta có
\(VT=\dfrac{x^{24}+x^{18}+x^{12}+x^6+1}{\left(x^3+1\right)\left(x^{24}+x^{18}+x^{12}+x^6+1\right)}=\dfrac{1}{x^3+1}\) (đpcm)