cố lên nha các bạn:
C/M:
\(\frac{X-X^2+1}{X-X^2-1}< 1\)
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a: =5/48+10/18=95/144
b: =3/4(7/5-1/2)=3/4x9/10=27/40
c: =2/3x35/24=70/72=35/36
\(M=\left(\frac{1+x}{1-x}-\frac{1-x}{1+x}-\frac{4x^2}{x^2-1}\right):\frac{4\left(x^2-3\right)}{x\left(1-x\right)}\)
\(=\left(\frac{1+x}{1-x}-\frac{1-x}{1+x}+\frac{4x^2}{1-x^2}\right).\frac{x\left(1-x\right)}{4\left(x^3-3\right)}\)
\(=\left(\frac{\left(1+x\right)^2}{\left(1-x\right)\left(1+x\right)}-\frac{\left(1-x\right)^2}{\left(1+x\right)\left(1-x\right)}+\frac{4x^2}{\left(1+x\right)\left(1-x\right)}\right).\frac{x\left(1-x\right)}{4\left(x^3-3\right)}\)
\(=\left(\frac{\left(1+x\right)^2-\left(1-x\right)^2+4x^2}{\left(1-x\right)\left(1+x\right)}\right).\frac{x\left(1-x\right)}{4\left(x^3-3\right)}\)
\(=\frac{\left(1+x+1-x\right)\left(1+x-1+x\right)+4x^2}{\left(1-x\right)\left(1+x\right)}.\frac{x\left(1-x\right)}{4\left(x^3-3\right)}\)
\(=\frac{2.2x+4x^2}{\left(1+x\right)}.\frac{x}{4\left(x^3-3\right)}\)
\(=\frac{4x+4x^2}{\left(1+x\right)}.\frac{x}{4\left(x^3-3\right)}\)
\(=\frac{4x\left(1+x\right)}{\left(1+x\right)}.\frac{x}{4\left(x^3-3\right)}\)
\(=\frac{x}{1}.\frac{x}{\left(x^3-3\right)}\)
\(=\frac{x^2}{x^3-3}\)
a, 10+15+20+....+295+x.300+x=67
10+15+20+...+295+x(300+1)=67
10+15+20+...+295+x.301=67
8845+x.301=67
67-8845=x.301
-8878=x.301
x=-29/149/301
b,
\(\frac{1}{7.6}+\frac{1}{6.5}+\frac{1}{5.4}+\frac{1}{4.3}+\frac{1}{3.2}+\frac{1}{2.1}-\frac{1}{x+1}=\frac{59}{77}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}-\frac{1}{x+1}=\frac{59}{77}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}-\frac{1}{x+1}=\frac{59}{77}\)\(1-\frac{1}{7}-\frac{1}{x+1}=\frac{59}{77}\)
\(\frac{6}{7}-\frac{1}{x+1}=\frac{59}{77}\)
\(\frac{1}{x+1}=\frac{6}{7}-\frac{59}{77}\)
\(\frac{1}{x+1}=\frac{1}{11}\)
suy ra x+1=11
suy ra x=10
a) (x-3)+(x-2)+(x-1)+....+10+11=11
(x-3)+(x-2)+(x-1)+....+10 =0
gọi số hạng của tổng vế trái là n
(x-3+10).\(\frac{n}{2}\)=0
(x+7).n:2=0
(x+7) =0
\(\Rightarrow\)x+7=0 (do n\(\ne\)0)
x=0-7
x=-7
b) \(\frac{2}{3}\left[\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right]<=x<=4\frac{1}{3}.\left[\frac{1}{2}-\frac{1}{6}\right]\)
\(\frac{2}{3}.\frac{11}{12}<=x<=\frac{13}{3}.\frac{1}{3}\)
\(\frac{11}{18}<=x<=\frac{13}{9}\)
do x\(\in\)z nên x=1
vậy x=1
Nhận xét: \(x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\) do đó \(\frac{2}{x^2-x+1}>0\)
Ta có \(\frac{x-x^2+1}{x-x^2-1}=1+\frac{2}{x-x^2-1}=1-\frac{2}{x^2-x+1}< 1\)