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TT
21 tháng 12 2017
\(\dfrac{5x-1+x+1}{3x^2y}=\dfrac{6x}{3x^2y}=\dfrac{2}{xy}\)
\(\dfrac{21x^2+22y}{36x^3y^2}\)
\(\dfrac{x\left(4x-7\right)+7x-16}{\left(x+2\right)\left(4x-7\right)}=\dfrac{4x^2-16}{\left(x+2\right)\left(4x-7\right)}=\dfrac{4\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(4x-7\right)}=\dfrac{4x-8}{4x-7}=1-\dfrac{1}{4x-7}\)
NN
3 tháng 12 2017
a) \(\dfrac{3x-2}{2xy}+\dfrac{7x+2}{2xy}\)
\(=\dfrac{\left(3x-2\right)+\left(7x+2\right)}{2xy}\)
\(=\dfrac{3x-2+7x+2}{2xy}\)
\(=\dfrac{10x}{2xy}\)
\(=\dfrac{5}{y}\)
b) \(\dfrac{5x+y^2}{x^2y}+\dfrac{x^2-5y}{xy^2}\) MTC: \(x^2y^2\)
\(=\dfrac{y\left(5x+y^2\right)}{x^2y^2}+\dfrac{x\left(x^2-5y\right)}{x^2y^2}\)
\(=\dfrac{y\left(5x+y^2\right)+x\left(x^2-5y\right)}{x^2y^2}\)
\(=\dfrac{5xy+y^3+x^3-5xy}{x^2y^2}\)
\(=\dfrac{y^3+x^3}{x^2y^2}\)
c) \(\dfrac{3x-2}{2xy}-\dfrac{7x-y}{2xy}\)
\(=\dfrac{\left(3x-2\right)-\left(7x-y\right)}{2xy}\)
\(=\dfrac{3x-2-7x+y}{2xy}\)
\(=\dfrac{-2-4x+y}{2xy}\)
d) \(\dfrac{5x+y^2}{x^2y}-\dfrac{5y-x^2}{xy^2}\) MTC: \(x^2y^2\)
\(=\dfrac{y\left(5x+y^2\right)}{x^2y^2}-\dfrac{x\left(5y-x^2\right)}{x^2y^2}\)
\(=\dfrac{y\left(5x+y^2\right)-x\left(5y-x^2\right)}{x^2y^2}\)
\(=\dfrac{5xy+y^3-5xy+x^3}{x^2y^2}\)
\(=\dfrac{y^3+x^3}{x^2y^2}\)
e) \(\dfrac{16xy}{3x-1}.\dfrac{3-9x}{12xy^3}\)
\(=\dfrac{16xy\left(3-9x\right)}{12xy^3\left(3x-1\right)}\)
\(=\dfrac{4\left(3-9x\right)}{3y^2\left(3x-1\right)}\)
\(=\dfrac{-4\left(9x-3\right)}{3y^2\left(3x-1\right)}\)
\(=\dfrac{-4.3\left(3x-1\right)}{3y^2\left(3x-1\right)}\)
\(=\dfrac{-12}{3y^2}\)
\(=\dfrac{-4}{y^2}\)
f) \(\dfrac{8xy}{3x-1}:\dfrac{12xy^3}{5-15x}\)
\(=\dfrac{8xy}{3x-1}.\dfrac{5-15x}{12xy^3}\)
\(=\dfrac{8xy\left(5-15x\right)}{12xy^3\left(3x-1\right)}\)
\(=\dfrac{2\left(5-15x\right)}{3y^2\left(3x-1\right)}\)
\(=\dfrac{-2\left(15x-5\right)}{3y^2\left(3x-1\right)}\)
\(=\dfrac{-2.5\left(3x-1\right)}{3y^2\left(3x-1\right)}\)
\(=\dfrac{-10}{3y^2}\)
HH
19 tháng 6 2018
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 ...
7 tháng 6 2017
giải pt sau
g) 11+8x-3=5x-3+x
\(\Leftrightarrow\) 8x + 8 = 6x - 3
<=> 8x-6x = -3 - 8
<=> 2x = -11
=> x=-\(\dfrac{11}{2}\)
Vậy tập nghiệm của PT là : S={\(-\dfrac{11}{2}\)}
h)4-2x+15=9x+4-2x
<=> 19 - 2x = 7x + 4
<=> -2x - 7x = 4 - 19
<=> -9x = -15
=> x=\(\dfrac{15}{9}=\dfrac{5}{3}\)
Vậy tập nghiệm của pt là : S={\(\dfrac{5}{3}\)}
g)\(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=\dfrac{5}{3}+2x\)
<=> \(\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{5.2+6.2x}{6}\)
<=> 9x + 6 - 3x + 1 = 10 + 12x
<=> 6x + 7 = 10 + 12x
<=> 6x -12x = 10-7
<=> -6x = 3
=> x= \(-\dfrac{1}{2}\)
Vậy tập nghiệm của PT là : S={\(-\dfrac{1}{2}\)}
\(h,\dfrac{x+4}{5}-x+4=\dfrac{4x+2}{5}-5\)
<=> \(\dfrac{x+4-5\left(x+4\right)}{5}=\dfrac{4x+2-5.5}{5}\)
<=> x + 4 - 5x - 20 = 4x + 2 - 25
<=> x - 5x - 4x = 2-25-4+20
<=> -8x = -7
=> x= \(\dfrac{7}{8}\)
Vậy tập nghiệm của PT là S={\(\dfrac{7}{8}\)}
\(i,\dfrac{4x+3}{5}-\dfrac{6x-2}{7}=\dfrac{5x+4}{3}+3\)
<=> \(\dfrac{21\left(4x+3\right)}{105}\)-\(\dfrac{15\left(6x-2\right)}{105}\)=\(\dfrac{35\left(5x+4\right)+3.105}{105}\)
<=> 84x + 63 - 90x + 30 = 175x + 140 + 315
<=> 84x - 90x - 175x = 140 + 315 - 63 - 30
<=> -181x = 362
=> x = -2
Vậy tập nghiệm của PT là : S={-2}
K) \(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)
<=> \(\dfrac{5\left(5x+2\right)}{30}-\dfrac{10\left(8x-1\right)}{30}=\dfrac{6\left(4x+2\right)-150}{30}\)
<=> 25x + 10 - 80x - 10 = 24x + 12 - 150
<=> -55x = 24x - 138
<=> -55x - 24x = -138
=> -79x = -138
=> x=\(\dfrac{138}{79}\)
Vậy tập nghiệm của PT là S={\(\dfrac{138}{79}\)}
m) \(\dfrac{2x-1}{5}-\dfrac{x-2}{3}=\dfrac{x+7}{15}\)
<=> \(\dfrac{3\left(2x-1\right)-5\left(x-2\right)}{15}=\dfrac{x+7}{15}\)
<=> 6x - 3 - 5x + 10 = x+7
<=> x + 7 = x+7
<=> 0x = 0
=> PT vô nghiệm
Vậy S=\(\varnothing\)
n)\(\dfrac{1}{4}\left(x+3\right)=3-\dfrac{1}{2}\left(x+1\right)-\dfrac{1}{3}\left(x+2\right)\)
<=> \(\dfrac{1}{4}x+\dfrac{3}{4}=3-\dfrac{1}{2}x-\dfrac{1}{2}-\dfrac{1}{3}x-\dfrac{2}{3}\)
<=> \(\dfrac{1}{4}x+\dfrac{1}{2}x+\dfrac{1}{3}x=3-\dfrac{1}{2}-\dfrac{2}{3}-\dfrac{3}{4}\)
<=> \(\dfrac{13}{12}x=\dfrac{13}{12}\)
=> x= 1
Vậy S={1}
p) \(\dfrac{x}{3}-\dfrac{2x+1}{6}=\dfrac{x}{6}-6\)
<=> \(\dfrac{2x-2x+1}{6}=\dfrac{x-36}{6}\)
<=> 2x -2x + 1= x-36
<=> 2x-2x-x = -37
=> x = 37
Vậy S={37}
q) \(\dfrac{2+x}{5}-0,5x=\dfrac{1-2x}{4}+0,25\)
<=> \(\dfrac{4\left(2+x\right)-20.0,5x}{20}=\dfrac{5\left(1-2x\right)+20.0,25}{20}\)
<=> 8 + 4x - 10x = 5 - 10x + 5
<=> 4x-10x + 10x = 5+5-8
<=> 4x = 2
=> x= \(\dfrac{1}{2}\)
Vậy S={\(\dfrac{1}{2}\)}
7 tháng 6 2017
g) \(11+8x-3=5x-3+x\)
\(\Leftrightarrow8+8x=6x-3\)
\(\Leftrightarrow8x-6x=-3-8\)
\(\Leftrightarrow2x=-11\)
\(\Leftrightarrow x=-\dfrac{11}{2}\)
h, \(4-2x+15=9x+4-2x\)
\(\Leftrightarrow-2x-9x+2x=4-4-15\)
\(\Leftrightarrow-9x=-15\)
\(\Leftrightarrow x=\dfrac{-15}{-9}=\dfrac{5}{3}\)
28 tháng 10 2018
\(I=3\left(x^2-\dfrac{5}{3}x+1\right)\)
\(I=3\left(x^2-2.x.\dfrac{5}{6}+\left(\dfrac{5}{6}\right)^2-\left(\dfrac{5}{6}\right)^2+1\right)\)
\(I=3\left[\left(x-\dfrac{5}{6}\right)^2+\dfrac{11}{36}\right]\)
\(I=3\left(x-\dfrac{5}{6}\right)^2+\dfrac{11}{12}\)
T
28 tháng 10 2018
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
11 tháng 12 2018
\(a,\frac{4x^3}{10x^2y}=\frac{2x}{5y}\)
\(b,\frac{10xy^5\left(2x-3y\right)}{12xy\left(2x-3y\right)}=\frac{5y^4}{6}\)
Hok Tốt~~
11 tháng 12 2018
\(\frac{4x^3}{10x^2y}=\frac{2x}{5y}\)
\(\frac{10xy^5\left(2x-3y\right)}{12xy\left(2x-3y\right)}=\frac{5y^4}{4}\)
Tham khảo nhé~