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Ta có :
\(\frac{x+4}{2014}+\frac{x+3}{2015}=\frac{x+8}{2010}+\frac{x+7}{2011}\)
\(\Leftrightarrow\)\(\left(\frac{x+4}{2014}+1\right)+\left(\frac{x+3}{2015}+1\right)=\left(\frac{x+8}{2010}+1\right)+\left(\frac{x+7}{2011}+1\right)\)
\(\Leftrightarrow\)\(\frac{x+4+2014}{2014}+\frac{x+3+2015}{2015}=\frac{x+8+2010}{2010}+\frac{x+7+2011}{2011}\)
\(\Leftrightarrow\)\(\frac{x+2018}{2014}+\frac{x+2018}{2015}=\frac{x+2018}{2010}+\frac{x+2018}{2011}\)
\(\Leftrightarrow\)\(\frac{x+2018}{2014}+\frac{x+2018}{2015}-\frac{x+2018}{2010}-\frac{x+2018}{2011}=0\)
\(\Leftrightarrow\)\(\left(x-2018\right)\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2010}-\frac{1}{2011}\right)=0\)
Vì \(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2010}-\frac{1}{2011}\ne0\)
Nên \(x-2018=0\)
\(\Leftrightarrow\)\(x=2018\)
Vậy \(x=2018\)
Chúc bạn học tốt ~
Ta có: \(\frac{x+4}{2014}+\frac{x+3}{2015}=\frac{x+7}{2011}+\frac{x+8}{2010}\)
\(\Rightarrow\left(\frac{x+4}{2014}+1\right)+\left(\frac{x+3}{2015}+1\right)=\left(\frac{x+7}{2011}+1\right)+\left(\frac{x+8}{2010}+1\right)\)
\(\Rightarrow\frac{x+2018}{2014}+\frac{x+2018}{2013}=\frac{x+2018}{2011}+\frac{x+2018}{2010}\)
\(\Rightarrow\frac{x+2018}{2014}+\frac{x+2018}{2013}-\frac{x+2018}{2011}-\frac{x+2018}{2010}=0\)
\(\Rightarrow\left(x+2018\right)\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2011}-\frac{1}{2010}\right)=0\)
\(\Rightarrow x+2018=0\Rightarrow x=-2018\)
Chúc bn hc tốt! ^_^
c) <=> \(\frac{x+1}{2016}+1+\frac{x+2}{2015}+1\)\(+\frac{x+3}{2014}+1\)= \(\frac{x+4}{2013}+1+\frac{x+5}{2012}+1\)\(+\frac{x+6}{2011}\)
<=> \(\frac{x+1+2016}{2016}+\frac{x+2+2015}{2015}+\frac{x+3+2014}{2014}\) \(=\frac{x+4+2013}{2013}+\frac{x+5+2012}{2012}+\frac{x+6+2011}{2011}\)
<=> \(\frac{x+2017}{2016}+\frac{x+2017}{2015}+\frac{x+2017}{2014}-\frac{x+2017}{2013}-\frac{x+2017}{2012}-\frac{x+2017}{2011}=0\)
<=> \(\left(x+2017\right)\left(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\right)=0\)
vì \(\left(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\right)\)khác 0
=> \(x+2017=0\) => \(x=-2017\)
Vậy \(S=\left\{-2017\right\}\)
A=5-3(2x+1)^2
Ta có : (2x+1)^2\(\ge\)0
\(\Rightarrow\)-3(2x-1)^2\(\le\)0
\(\Rightarrow\)5+(-3(2x-1)^2)\(\le\)5
Dấu = xảy ra khi : (2x-1)^2=0
=> 2x-1=0 =>x=\(\frac{1}{2}\)
Vậy : A=5 tại x=\(\frac{1}{2}\)
Ta có : (x-1)^2 \(\ge\)0
=> 2(x-1)^2\(\ge\)0
=>2(x-1)^2+3 \(\ge\)3
=>\(\frac{1}{2\left(x-1\right)^2+3}\)\(\le\)\(\frac{1}{3}\)
Dấu = xảy ra khi : (x-1)^2 =0
=> x = 1
Vậy : B = \(\frac{1}{3}\)khi x = 1
\(\frac{x^2+8}{x^2+2}\)= \(\frac{x^2+2+6}{x^2+2}=1+\frac{6}{x^2+2}\)
Làm như câu B GTNN = 4 khi x =0
k vs nha
Bai 2: gtnn cua |x-2015|+|x-1| (A)
Ta thay : | x - 2015| va |x-1| \(\ge\) 0
De (A) nho nhat thi suy ra: |x-2015| = 0 => x =2015 hay |x-1| = 0 => x=1
Suy ra: A = 0+|2015-1| = 2014
Hay: A = |1-2015| + 0 = 2014
Vay A nho nhat bang 2014
Lê Chí Công!Bạn nói dễ thì làm giúp đi, sao câu nào bạn cũng nói vậy mà có khi nào giúp đâu!
3/ ta để ý thấy ở số mũ sẽ có thừa số 1000-103=0
nên số mũ chắc chắn bằng 0
mà số nào mũ 0 cũng bằng 1 nên A=1
5/ vì |2/3x-1/6|> hoặc = 0
nên A nhỏ nhất khi |2/3x-6|=0
=>A=-1/3
6/ =>14x=10y=>x=10/14y
23x:2y=23x-y=256=28
=>3x-y=8
=>3.10/4y-y=8
=>6,5y=8
=>y=16/13
=>x=10/14y=10/14.16/13=80/91
8/106-57=56.26-56.5=56(26-5)=59.56
có chứa thừa số 59 nên chia hết 59
4/ tính x
sau đó thế vào tinh y,z
PT đã cho suy ra thành
\(\left(\frac{x^{2010}}{a^2+b^2+c^2+d^2}-\frac{x^{2010}}{a^2}\right)+\left(\frac{y^{2010}}{a^2+b^2+c^2+d^2}-\frac{y^{2010}}{b^2}\right)+\left(\frac{z^{2010}}{a^2+b^2+c^2+d^2}-\frac{z^{2010}}{c^2}\right)\)
\(+\left(\frac{t^{2010}}{a^2+b^2+c^2+d^2}-\frac{t^{2010}}{d^2}\right)=0\)
\(=>x^{2010}\left(\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{a^2}\right)+\left(tương\right)Tựnha=0\)
Do
\(\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{a^2}\ne0\)
máy cái bạn tự suy ra cx thế
\(=>x^{2010}=y^{2010}=z^{2010}=t^{2010}=0=>x=y=z=t=0\)
ta có
\(T=x^{2011}+y^{2011}+z^{2011}+t^{2011}=0+0+0+0=0\)
Ta có:
\(\frac{x^{2010}+y^{2010}+z^{2010}+t^{2010}}{a^2+b^2+c^2+d^2}=\frac{x^{2010}}{a^2}+\frac{y^{2010}}{b^2}+\frac{z^{2010}}{c^2}+\frac{t^{2010}}{d^2}\)
<=> \(x^{2010}\left(\frac{1}{a^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)+y^{2010}\left(\frac{1}{b^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)\)
\(+z^{2010}\left(\frac{1}{c^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)+t^{2010}\left(\frac{1}{d^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)=0\)(1)
Lại có: \(x^{2010};y^{2010};z^{2010};t^{2010}\ge0;\forall x,y,z,t\)
và với mọi a; b ; c ; d khác 0 có:
\(\frac{1}{a^2}-\frac{1}{a^2+b^2+c^2+d^2}>0\)
\(\frac{1}{b^2}-\frac{1}{a^2+b^2+c^2+d^2}>0\);
\(\frac{1}{c^2}-\frac{1}{a^2+b^2+c^2+d^2}>0\);
\(\frac{1}{d^2}-\frac{1}{a^2+b^2+c^2+d^2}>0\)
=> \(x^{2010}\left(\frac{1}{a^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)\ge0\)
\(y^{2010}\left(\frac{1}{b^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)\ge0\)
\(z^{2010}\left(\frac{1}{c^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)\ge0\)
\(t^{2010}\left(\frac{1}{d^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)\ge0\)
=> \(x^{2010}\left(\frac{1}{a^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)+y^{2010}\left(\frac{1}{b^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)\)
\(+z^{2010}\left(\frac{1}{c^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)+t^{2010}\left(\frac{1}{d^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)\ge0\)
Như vậy (1) xảy ra<=> \(x^{2010}=y^{2010}=z^{2010}=t^{2010}=0\)
<=> x = y = z = t = 0
Thay vào T ta có : T = 0
a,\(\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}=\frac{x+1}{8}+\frac{x+1}{9}\) (1)
<=> \(\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}-\frac{x+1}{8}-\frac{x+1}{9}=0\)
<=> \(\left(x+1\right)\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}\right)=0\)
=> x+1=0 (vì \(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}\ne0\))
<=> x=-1
Vậy pt (1) có tập nghiệm S\(=\left\{-1\right\}\)
b, \(\frac{x+6}{2015}+\frac{x+5}{2016}+\frac{x+4}{2017}=\frac{x+3}{2018}+\frac{x+2}{2019}+\frac{x+1}{2010}\)(2)
<=> \(\frac{x+6}{2015}+1+\frac{x+5}{2016}+1+\frac{x+4}{2017}+1=\frac{x+3}{2018}+1+\frac{x+2}{2019}+1+\frac{x+1}{2020}+1\)
<=> \(\frac{x+2021}{2015}+\frac{x+2021}{2016}+\frac{x+2021}{2017}-\frac{x+2021}{2018}-\frac{x+2021}{2019}-\frac{x+2021}{2020}=0\)
<=> \(\left(x+2021\right)\left(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)
=> x+2021=0(vì \(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\ne0\))
<=> x=-2021
Vậy pt (2) có tập nghiệm S=\(\left\{-2021\right\}\)
c,\(\frac{x+6}{2016}+\frac{x+7}{2017}+\frac{x+8}{2018}=\frac{x+9}{2019}+\frac{x+10}{2020}+1\) (3)
<=> \(\frac{x+6}{2016}-1+\frac{x+7}{2017}-1+\frac{x+8}{2018}-1=\frac{x+9}{2019}-1+\frac{x+10}{2020}-1+1-1\)
<=> \(\frac{x-2010}{2016}+\frac{x-2010}{2017}+\frac{x-2010}{2018}=\frac{x-2010}{2019}+\frac{x-2010}{2020}\)
<=> \(\frac{x-2010}{2016}+\frac{x-2010}{2017}+\frac{x-2010}{2018}-\frac{x-2010}{2019}-\frac{x-2010}{2020}=0\)
<=> \(\left(x-2010\right)\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)
=> x-2010=0 (vì \(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\ne0\))
<=> x=2010
Vậy pt (3) có tập nghiệm S=\(\left\{2010\right\}\)
d, \(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15\) (4)
<=>\(\frac{x-90}{10}-1+\frac{x-76}{12}-2+\frac{x-58}{14}-3+\frac{x-36}{16}-4+\frac{x-15}{17}-5=15-1-2-3-4-5\)
<=> \(\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)
<=> (x-100)(\(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\))=0
=> x -100=0(vì \(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\))
<=> x=100
Vậy pt (4) có tập nghiệm S=\(\left\{100\right\}\)
a) \(\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}=\frac{x+1}{8}+\frac{x+1}{9}\)
\(\Leftrightarrow\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}-\frac{x+1}{8}-\frac{x+1}{9}=0\)
\(\Leftrightarrow\left(x+1\right).\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}\right)=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=0-1\)
\(\Rightarrow x=-1\)
Vậy \(x=-1.\)
Mình chỉ làm câu a) thôi nhé.
Chúc bạn học tốt!
bạn bui le anh kia. người ta ko biết làm thì kệ người ta chứ. tự nhiên đi bảo người ta là bị chập mạch. nếu bạn là tôi, bạn bị người khác nói là bị chập mạnh thì bạn thấy thế nào?