Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 2:
\(A=\dfrac{x\left(x^3+1\right)}{x^2-x+1}-\dfrac{x\left(x^3-1\right)}{x^2+x+1}\)
\(=x\left(x+1\right)-x\left(x-1\right)\)
=x^2+x-x^2+x
=2x
\(a.\dfrac{3x-2}{5}+\dfrac{x-1}{9}=\dfrac{14x-3}{15}-\dfrac{2x+1}{9}\\ \Leftrightarrow\dfrac{27x-18}{45}+\dfrac{5x-5}{45}=\dfrac{42x-9}{45}-\dfrac{10x+5}{45}\\ \Rightarrow27x-18+5x-5=42x-9-10x-5\\ \Leftrightarrow32x-23=32x-14\\ \Leftrightarrow0x=9\\ \Rightarrow Phươngtrìnhvônghiệm\\ \Rightarrow S=\phi\)
\(b.\dfrac{x+3}{2}-\dfrac{2-x}{3}-1=\dfrac{x+5}{6}\\ \Leftrightarrow\dfrac{3x-9}{6}-\dfrac{4-2x}{6}-\dfrac{6}{6}=\dfrac{x+5}{6}\\ \Rightarrow3x-9-4+2x-6=x+5\\ \Leftrightarrow5x-19=x+5\\ \Leftrightarrow4x=24\\ \Rightarrow x=6\\ \Rightarrow S=\left\{6\right\}\)
\(c.\dfrac{x+5}{2010}+\dfrac{x+4}{2011}+\dfrac{x+3}{2012}+\dfrac{x+2}{2013}=-4\\ \Leftrightarrow\dfrac{x+5}{2010}+1+\dfrac{x+4}{2011}+1+\dfrac{x+3}{2012}+1+\dfrac{x+2}{2013}+1=-4+4\\ \Rightarrow\dfrac{2015+x}{2010}+\dfrac{2015+x}{2011}+\dfrac{2015+x}{2012}+\dfrac{2015+x}{2013}=0\\ \Leftrightarrow\left(2015+x\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}\right)=0\)
Do \(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}>0\)
nên \(2015+x=0\Rightarrow x=-2015\)
Câu d tương tự...thêm rồi chuyển vế sang :v
a) \(\dfrac{x+5}{2010}+\dfrac{x+4}{2011}+\dfrac{x+3}{2012}+\dfrac{x+2}{2013}=-4\)
\(\Rightarrow\dfrac{x+5}{2010}+\dfrac{x+4}{2011}+\dfrac{x+3}{2012}+\dfrac{x+2}{2013}+4=0\)
\(\Rightarrow\left(\dfrac{x+5}{2010}+1\right)+\left(\dfrac{x+4}{2011}+1\right)+\left(\dfrac{x+3}{2012}+1\right)+\left(\dfrac{x+2}{2013}+1\right)=0\)
\(\Rightarrow\left(\dfrac{x+5}{2010}+\dfrac{2010}{2010}\right)+\left(\dfrac{x+4}{2011}+\dfrac{2011}{2011}\right)+\left(\dfrac{x+3}{2012}+\dfrac{2012}{2012}\right)+\left(\dfrac{x+2}{2013}+\dfrac{2013}{2013}\right)=0\)
\(\Rightarrow\dfrac{x+5+2010}{2010}+\dfrac{x+4+2011}{2011}+\dfrac{x+3+2012}{2012}+\dfrac{x+2+2013}{2013}=0\)
\(\Rightarrow\dfrac{x+2015}{2010}+\dfrac{x+2015}{2011}+\dfrac{x+2015}{2012}+\dfrac{x+2015}{2013}=0\)
\(\Rightarrow\left(x+2015\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}\right)=0\)
Mà \(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}\ne0\)
\(\Rightarrow x+2015=0\)
\(\Rightarrow x=-2015\)
b) \(\dfrac{x-22}{77}+\dfrac{x-11}{78}=\dfrac{x-74}{15}+\dfrac{x-73}{16}\)
\(\Rightarrow\dfrac{x-22}{77}+\dfrac{x-11}{78}-2=\dfrac{x-74}{15}+\dfrac{x-73}{16}-2\)
\(\Rightarrow\left(\dfrac{x-22}{77}-1\right)+\left(\dfrac{x-11}{78}-1\right)=\left(\dfrac{x-74}{15}-1\right)+\left(\dfrac{x-73}{16}-1\right)\)
\(\Rightarrow\left(\dfrac{x-22}{77}-\dfrac{77}{77}\right)+\left(\dfrac{x-11}{78}-\dfrac{78}{78}\right)=\left(\dfrac{x-74}{15}-\dfrac{15}{15}\right)+\left(\dfrac{x-73}{16}-\dfrac{16}{16}\right)\)
\(\Rightarrow\dfrac{x-22-77}{77}+\dfrac{x-11-78}{78}=\dfrac{x-74-15}{15}+\dfrac{x-73-16}{16}\)
\(\Rightarrow\dfrac{x-99}{77}+\dfrac{x-99}{78}=\dfrac{x-99}{15}+\dfrac{x-99}{16}\)
\(\Rightarrow\left(x-99\right)\left(\dfrac{1}{77}+\dfrac{1}{78}\right)=\left(x-99\right)\left(\dfrac{1}{15}+\dfrac{1}{16}\right)\)
\(\Rightarrow\left(x-99\right)\left(\dfrac{1}{77}+\dfrac{1}{78}\right)-\left(x-99\right)\left(\dfrac{1}{15}+\dfrac{1}{16}\right)=0\)
\(\Rightarrow\left(x-99\right)\left(\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}\right)=0\)
Mà \(\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}\ne0\)
\(\Rightarrow x-99=0\)
\(\Rightarrow x=99\)
a/\(\dfrac{8}{x-8}+1+\dfrac{11}{x-11}+1=\dfrac{9}{x-9}+1+\dfrac{10}{x-10}+1\)
=>\(\dfrac{8+x-8}{x-8}+\dfrac{11+x-11}{x-11}=\dfrac{9+x-9}{x-9}+\dfrac{10+x-10}{x-10}\)
=>\(\dfrac{x}{x-8}+\dfrac{x}{x-11}-\dfrac{x}{x-9}-\dfrac{x}{x-10}=0\)
=>x.\(\left(\dfrac{1}{x-8}+\dfrac{1}{x-11}+\dfrac{1}{x-9}+\dfrac{1}{x-10}\right)=0\)
=>x=0
b/\(\dfrac{x}{x-3}-1+\dfrac{x}{x-5}-1=\dfrac{x}{x-4}-1+\dfrac{x}{x-6}-1\)
=>\(\dfrac{x-x+3}{x-3}+\dfrac{x-x+5}{x-5}-\dfrac{x-x+4}{x-4}-\dfrac{x-6+6}{x-6}=0\)
=>\(\dfrac{3}{x-3}+\dfrac{5}{x-5}-\dfrac{4}{x-4}-\dfrac{6}{x-6}=0\)
Đến đây thì bạn giải giống câu a
a) \(\frac{4x + 3}{5} - \frac{6x - 2}{7}=\frac{5x + 4}{3} + 3\)
\(\Leftrightarrow \frac{21(4x +3) - 15(6x - 2)}{105}=\frac{35(5x + 4) + 315}{105}\)
<=> 21(4x + 3) - 15(6x - 2) = 35(5x + 4) + 315
<=> 84x + 63 - 90x + 30 = 175x + 140 + 315
<=> - 6x + 93 = 175x + 455
<=> - 6x - 175x = 455 - 93
<=> - 181x = 362
<=> x = - 2
Vậy ................................
b) \(\frac{3x - 11}{11} - \frac{x}{3} = \frac{3x - 5}{7} - \frac{5x - 3}{9}\)
\(\Leftrightarrow \frac{63(3x - 11) - 231x}{693}=\frac{99(3x - 5) - 77(5x - 3)}{693}\)
<=> 63(3x - 11) - 231x = 99(3x - 5) - 77(5x - 3)
<=> 189x - 693 - 231x = 297x - 495 - 385x + 231
<=> - 42x - 693 = - 88x - 264
<=> - 42x + 88x = -264 + 693
<=> 46x = 429
<=> x = \(\frac{429}{46}\)
Vậy ........
\(I=3\left(x^2-\dfrac{5}{3}x+1\right)\)
\(I=3\left(x^2-2.x.\dfrac{5}{6}+\left(\dfrac{5}{6}\right)^2-\left(\dfrac{5}{6}\right)^2+1\right)\)
\(I=3\left[\left(x-\dfrac{5}{6}\right)^2+\dfrac{11}{36}\right]\)
\(I=3\left(x-\dfrac{5}{6}\right)^2+\dfrac{11}{12}\)
mình ra là \(\dfrac{11}{36}\)mà bn
bn coi lại đi
I=3x2-5x+3
I=3(x2-\(\dfrac{5}{3}\)x+1)
I=3[x2-2.x.\(\dfrac{5}{3}\)+\(\left(\dfrac{5}{6}\right)^2\)-\(\left(\dfrac{5}{6}\right)^2\)+1]
I=3(x-\(\dfrac{5}{3}\))2+\(\dfrac{11}{36}\)
I=3(x-\(\dfrac{5}{3}\))2+\(\dfrac{11}{36}\)≥\(\dfrac{11}{36}\)
vậy Min I= \(\dfrac{11}{36}\)khi x =\(\dfrac{5}{3}\)
Theo mik nghĩ là vậy á
CHÚC BN HỌC TỐT
Giải:
1) \(\dfrac{-1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)\)
\(=\dfrac{-1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)\)
\(=\dfrac{-1}{12}-\dfrac{55}{24}\)
\(=\dfrac{-19}{8}\)
2) \(-1,75-\left(\dfrac{-1}{9}-2\dfrac{1}{18}\right)\)
\(=-\dfrac{7}{4}+\dfrac{1}{9}+2\dfrac{1}{18}\)
\(=-\dfrac{7}{4}+\dfrac{1}{9}+\dfrac{37}{18}\)
\(=\dfrac{5}{12}\)
3) \(-\dfrac{5}{6}-\left(-\dfrac{3}{8}+\dfrac{1}{10}\right)\)
\(=-\dfrac{5}{6}+\dfrac{3}{8}-\dfrac{1}{10}\)
\(=-\dfrac{67}{120}\)
4) \(\dfrac{2}{5}+\left(-\dfrac{4}{3}\right)+\left(-\dfrac{1}{2}\right)\)
\(=\dfrac{2}{5}-\dfrac{4}{3}-\dfrac{1}{2}\)
\(=-\dfrac{43}{30}\)
5) \(\dfrac{3}{12}-\left(\dfrac{6}{15}-\dfrac{3}{10}\right)\)
\(=\dfrac{3}{12}-\dfrac{6}{15}+\dfrac{3}{10}\)
\(=\dfrac{3}{20}\)
6) \(\left(8\dfrac{5}{11}+3\dfrac{5}{8}\right)-3\dfrac{5}{11}\)
\(=8\dfrac{5}{11}+3\dfrac{5}{8}-3\dfrac{5}{11}\)
\(=8+\dfrac{5}{11}+3+\dfrac{5}{8}-3-\dfrac{5}{11}\)
\(=8+\dfrac{5}{8}\)
\(=\dfrac{69}{8}\)
7) \(-\dfrac{1}{4}.13\dfrac{9}{11}-0,25.6\dfrac{2}{11}\)
\(=-\dfrac{1}{4}.13\dfrac{9}{11}-\dfrac{1}{4}.6\dfrac{2}{11}\)
\(=-\dfrac{1}{4}\left(13\dfrac{9}{11}+6\dfrac{2}{11}\right)\)
\(=-\dfrac{1}{4}\left(13+\dfrac{9}{11}+6+\dfrac{2}{11}\right)\)
\(=-\dfrac{1}{4}\left(13+6+1\right)\)
\(=-\dfrac{1}{4}.20=-5\)
8) \(\dfrac{4}{9}:\left(-\dfrac{1}{7}\right)+6\dfrac{5}{9}:\left(-\dfrac{1}{7}\right)\)
\(=\dfrac{4}{9}\left(-7\right)+6\dfrac{5}{9}\left(-7\right)\)
\(=-7\left(\dfrac{4}{9}+6\dfrac{5}{9}\right)\)
\(=-7\left(\dfrac{4}{9}+6+\dfrac{5}{9}\right)\)
\(=-7\left(6+1\right)\)
\(=-7.7=-49\)
Vậy ...
\(2-\dfrac{2}{3}x>\dfrac{1}{4}\left(x-2\right)+11\\ \Leftrightarrow24-8x>3\left(x-2\right)+132\\ \Leftrightarrow24-8x>3x-6+132\\ \Leftrightarrow-8x-3x>126-24\\ \Leftrightarrow-11x>102\\ \Leftrightarrow x< -\dfrac{102}{11}\)
Vậy bất phương trình có tạp nghiệm \(S=\left\{x/x< -\dfrac{102}{11}\right\}\)
2 - 2/3x = 1/4 (x - 2 ) + 11
<=> -2/3x - 1/4x = 11 - 1/2 - 2
<=> -11/12x = 17/2
<=> x = 102/11
\(S=\dfrac{11}{2^2}+\dfrac{11}{2^3}+\dfrac{11}{2^4}+...+\dfrac{11}{2^{2011}}\)
\(\Rightarrow2S=\dfrac{11}{2}+\dfrac{11}{2^2}+\dfrac{11}{2^3}+...+\dfrac{11}{2^{2010}}\)
\(\Rightarrow2S-S=\left(\dfrac{11}{2}+\dfrac{11}{2^2}+\dfrac{11}{2^3}+...+\dfrac{11}{2^{2010}}\right)\)
\(-\left(\dfrac{11}{2^2}+\dfrac{11}{2^3}+\dfrac{11}{2^4}+...+\dfrac{11}{2^{2011}}\right)\)
\(\Rightarrow S=\dfrac{11}{2}-\dfrac{11}{2^{2010}}=\dfrac{11.2^{2009}}{2^{2010}}-\dfrac{11}{2^{2010}}=\dfrac{11.\left(2^{2009}-1\right)}{2^{2010}}\)
đáp án này k giống với sách của mình