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HN
16 tháng 12 2018
\(a,\frac{2x+4}{10}+\frac{2-x}{15}=\frac{\left(2x+4\right).3}{10.3}+\frac{\left(2-x\right).2}{15.2}\)
\(=\frac{6x+12}{30}+\frac{4-2x}{30}=\frac{6x+12+4-2x}{30}=\frac{4x+16}{30}\)
\(=\frac{4.\left(x+4\right)}{30}=\frac{2\left(x+4\right)}{15}\)
\(b,\frac{3x}{10}+\frac{2x-1}{15}+\frac{2-x}{20}=\frac{3x.6}{10.6}+\frac{\left(2x-1\right).4}{15.4}+\frac{\left(2-x\right).3}{20.3}\)
\(=\frac{18x}{60}+\frac{8x-4}{60}+\frac{6-3x}{60}=\frac{18x+8x-4+6-3x}{60}=\frac{23x+2}{60}\)
\(c,\frac{x+1}{2x-2}+\frac{x^2+3}{2-2x^2}=\frac{x+1}{2\left(x-1\right)}+\frac{x^2+3}{2\left(1-x^2\right)}=\frac{x+1}{2\left(x-1\right)}+\frac{-x^2-3}{2\left(x^2-1\right)}\)
\(=\frac{x+1}{2\left(x-1\right)}+\frac{-x^2-3}{2\left(x-1\right)\left(x+1\right)}\)\(=\frac{\left(x+1\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}+\frac{-x^2-3}{2\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2+2x+1-x^2-3}{2\left(x-1\right)\left(x+1\right)}=\frac{2x-2}{2\left(x-1\right)\left(x+1\right)}=\frac{2\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\)\(=\frac{1}{x+1}\)
21 tháng 10 2022
a: \(=\dfrac{6x+12+4-2x}{30}=\dfrac{4x+16}{30}=\dfrac{2x+8}{15}\)
b: \(=\dfrac{18x}{60}+\dfrac{8x-4}{60}+\dfrac{6-3x}{60}\)
\(=\dfrac{18x+8x-4+6-3x}{60}=\dfrac{23x+2}{60}\)
c: \(=\dfrac{x+1}{2\left(x-1\right)}-\dfrac{x^2+3}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2+2x+1-x^2-3}{2\left(x-1\right)\left(x+1\right)}=\dfrac{2x-2}{2\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x+1}\)
d: \(=\dfrac{x}{y\left(x-y\right)}+\dfrac{2x-y}{x\left(y-x\right)}\)
\(=\dfrac{x^2-2xy+y^2}{xy\left(x-y\right)}=\dfrac{x-y}{xy}\)
e: \(=\dfrac{x^2+2xy+y^2+x^2+y^2}{x+y}=\dfrac{2x^2+2xy+2y^2}{x+y}\)
C
22 tháng 9 2019
a) \(x^2-5x+10\)
\(=x^2-2.\frac{5}{2}x+\frac{25}{4}+\frac{15}{4}\)
\(=\left(x-\frac{5}{2}\right)^2+\frac{15}{4}\ge\frac{15}{4}>0\)
b) \(2x^2+8x+10\)
\(=2\left(x^2+4x+4+1\right)\)
\(=2\left[\left(x+2\right)^2+1\right]\ge2>0\)
20 tháng 7 2020
\(\frac{x^2-36}{2x+10}\cdot\frac{3}{6-x}=\frac{\left(x-6\right)\left(x+6\right)}{2x+10}\cdot\frac{3}{6-x}=-\frac{3\left(x+6\right)}{2x+10}=-\frac{3x+18}{2x+10}\)
\(\frac{x^2-4}{x^2-9}\cdot\frac{3x+9}{x+2}=\frac{\left(x-2\right)\left(x+2\right)}{\left(x+3\right)\left(x-3\right)}\cdot\frac{3\left(x+3\right)}{x+2}=\frac{3\left(x-2\right)}{x-3}\)
\(\frac{x^3-8}{5x+20}\cdot\frac{x^2+4x}{x^2+2x+4}=\frac{\left(x-2\right)\left(x^2+2x+4\right)}{5\left(x+4\right)}\cdot\frac{x\left(x+4\right)}{x^2+2x+4}=\frac{x\left(x-2\right)}{5}\)
\(\frac{4x+12}{\left(x+4\right)^2}:\frac{3x+9}{x+4}=\frac{4\left(x+3\right)}{\left(x+4\right)^2}\cdot\frac{x+4}{3\left(x+3\right)}=\frac{4}{3\left(x+4\right)}\)
8 tháng 8 2017
1.
A =\(2x^2-8x+10=\left(x^2-2x+1\right)+\left(x^2-6x+9\right)\)
\(=\left(x-1\right)^2+\left(x-3\right)^2=\left(x-1\right)^2+\left(3-x\right)^2\)
Có: \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\\\left(3-x\right)^2\ge0\end{matrix}\right.\forall x\)
<=> \(\left|x-1\right|+\left|x-3\right|\)
Áp dụng bđt |a| + |b| \(\ge\) |a + b| có:
\(\left|x-1\right|+\left|3-x\right|\ge\left|x-1+3-x\right|=2\)
đẳng thức xảy ra khi \(1\le x\le3\)
Vậy ................
T
8 tháng 8 2017
1.
a)
\(A=2x^2-8x+10=2\left(x^2-4x+4\right)+2\ge=2\left(x-2\right)^2+2\ge2\)
Đẳng thức xảy ra \(\Leftrightarrow x=2\)
b)
\(B=3x^2-x+20=3\left(x^2-\dfrac{1}{3}x+\dfrac{1}{36}\right)+\dfrac{239}{12}=3\left(x-\dfrac{1}{6}\right)^2+\dfrac{239}{12}\ge\dfrac{239}{12}\)
Đẳng thức xảy ra \(\Leftrightarrow x=\dfrac{1}{6}\)
c) ĐK: \(x\ne-1\)
\(C=\dfrac{x^2+x+1}{x^2+2x+1}=\dfrac{4x^2+4x+4}{4x^2+8x+4}\)
\(=\dfrac{3x^2+6x+3}{4x^2+8x+4}+\dfrac{x^2-2x+1}{4x^2+8x+4}\)
\(=\dfrac{3\left(x^2+2x+1\right)}{4\left(x^2+2x+1\right)}+\dfrac{\left(x-1\right)^2}{4x^2+8x+4}=\dfrac{3}{4}+\dfrac{\left(x-1\right)^2}{4x^2+8x+4}\ge\dfrac{3}{4}\)
Đẳng thức xảy ra \(\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)
4 tháng 5 2018
\(a)x^3-2x^2-3x+10=0\)
\(\Rightarrow(x-1)^3=0\)
\(\Rightarrow x-1=0\)
\(x=1\)
10 tháng 4 2020
\(a, 2x^2 + 5x + 10 = x^2 + 5x - 11\)
\(<=> x^2 + 21 = 0 \)
\(Do x^2 + 21 > 0\)
=> Pt vô nghiệm
\(b, 2x^2 - 6x + 7 = 0\)
\(<=> 2(x^2 - 3x+7/2)=0\)
\(<=> (x-3/2)^2 +7/4 = 0 \)
Tương tự như trên thì pt vô nghiệm
\(c, |x^2 + 3x+20| + |x-3| = 0\)
Ta có : \(|x^2 + 3x+20| = |(x+3/2)^2 + 17,75| > 0\)
\(=> |x^2 + 3x+20| + |x-3| > 0\)
=> Pt vô nghiệm
20 tháng 7 2022
a: \(A=-\left(x^2-4x-3\right)\)
\(=-\left(x^2-4x+4-7\right)\)
\(=-\left(x-2\right)^2+7< =7\)
Dấu '=' xảy ra khi x=2
b: \(B=-\left(x^2-x+\dfrac{1}{4}-\dfrac{1}{4}\right)=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}< =\dfrac{1}{4}\)
Dấu '=' xảy ra khi x=1/2
c: \(C=-2\left(x^2-x+\dfrac{5}{2}\right)\)
\(=-2\left(x^2-x+\dfrac{1}{4}+\dfrac{9}{4}\right)\)
\(=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}< =-\dfrac{9}{2}\)
Dấu '=' xảy ra khi x=1/2
e: \(E=-\left(x^2+6x+9+1\right)=-\left(x+3\right)^2-1< =-1\)
Dấu = xảy ra khi x=-3
13 tháng 11 2017
Bài 2 .
a) \(\dfrac{2x}{x^2+2xy}+\dfrac{y}{xy-2y^2}+\dfrac{4}{x^2-4y^2}\)
\(=\dfrac{2x}{x\left(x+2y\right)}+\dfrac{y}{y\left(x-2y\right)}+\dfrac{4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{2xy\left(x-2y\right)+xy\left(x+2y\right)+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{2x^2y-2xy^2+x^2y+2xy^2+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{3x^2y+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
b) Sai đề hay sao ý
c) \(\dfrac{2x+y}{2x^2-xy}+\dfrac{16x}{y^2-4x^2}+\dfrac{2x-y}{2x^2+xy}\)
\(=\dfrac{2x+y}{x\left(2x-y\right)}+\dfrac{-16x}{\left(2x-y\right)\left(2x+y\right)}+\dfrac{2x-y}{x\left(2x+y\right)}\)
\(=\dfrac{\left(2x+y\right)^2-16x^2+\left(2x-y\right)^2}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\dfrac{4x^2+4xy+y^2-16x^2+4x^2-4xy+y^2}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\dfrac{-8x^2}{x\left(2x-y\right)\left(2x+y\right)}\)
d) \(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
.....
\(=\dfrac{16}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{32}{1-x^{32}}\)
\(2x+10=20\)
\(\Rightarrow2x=10\)
\(\Rightarrow x=5\)
trả lời
2x +10=20
<=> 2x= 20-10=10
<=>x=10:2=5
học tốt
DNMT :))