Tìm x, biết: \(\frac{-1}{2}.\left(3x+5\right)-\frac{2}{9}.\left(9-6x\right)=\frac{3}{5}.\left(x-10\right)-3\)
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26 tháng 2 2018
\(\frac{2x-3}{\left(7-6x\right)^2}+\frac{x-2}{\left(7-6x\right)^2}=\frac{6x-3}{\left(3x-5\right)^2}-\frac{12x-10}{\left(3x-5\right)^2}\)
\(\Leftrightarrow\frac{2x-3+x-2}{\left(7-6x\right)^2}=\frac{6x-3-12x+10}{\left(3x-5\right)^2}\)
\(\Leftrightarrow\frac{3x-5}{\left(7-6x\right)^2}=\frac{7-6x}{\left(3x-5\right)^2}\)
\(\Leftrightarrow\left(7-6x\right)^3=\left(3x-5\right)^3\)
\(\Leftrightarrow7-6x=3x-5\)
\(\Leftrightarrow7+5=3x+6x\)
\(\Leftrightarrow12=9x\)
\(\Leftrightarrow x=\frac{4}{3}\)
Vậy \(x=\frac{4}{3}\)
21 tháng 7 2019
\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)
= \(\frac{3x\left(x-y\right)}{5.2.\left(x+y\right)\left(x-y\right)}-\frac{x\left(x+y\right)}{10\left(x^2-y^2\right)}\)
= \(\frac{3x^2-3xy-x^2-xy}{10\left(x^2-y^2\right)}\)
= \(\frac{3x\left(x-y\right)}{10\left(x^2-y^2\right)}\)
= \(\frac{3x}{10\left(x+y\right)}\)
JK
10 tháng 10 2019
a, (3x - 5)(2x - 1) - (x + 2)(6x - 1) = 0
=> 6x^2 - 3x - 10x + 5 - (6x^2 - x + 12x - 2) = 0
=> 6x^2 - 13x + 5 - 6x^2 - 11x + 2 = 0
=> -24x + 7 = 0
=> - 24x = -7
=> x = 7/24
b, (3x - 2)(3x + 2) - (3x - 1)^2 = -5
=> 9x^2 - 4 - 9x^2 + 6x - 1 = -5
=> 6x - 5 = -5
=> 6x = 0
=> x = 0
c, x^2 = -6x - 8
=> x^2 + 6x + 8 = 0
=> x^2 + 2.x.3 + 9 - 1 = 0
=> (x + 3)^2 = 1
=> x + 3 = 1 hoặc x + 3 = -1
=> x = -2 hoặc x = -4
11 tháng 3 2020
1) \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)
<=> \(\frac{21x}{24}-\frac{100\left(x-9\right)}{24}=\frac{80x+6}{24}\)
<=> 21x - 100x + 900 = 80x + 6
<=> -79x - 80x = 6 - 900
<=> -159x = -894
<=> x = 258/53
Vậy S = {258/53}
2) \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x+1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
<=> \(\frac{3\left(4x^2+4x+1\right)}{15}-\frac{5\left(x^2+2x+1\right)}{15}=\frac{7x^2-14x-5}{15}\)
<=> 12x2 + 12x + 3 - 5x2 - 10x - 5 = 7x2 - 14x - 5
<=> 7x2 + 2x - 7x2 + 14x = -5 + 2
<=> 16x = 3
<=> x = 3/16
Vậy S = {3/16}
11 tháng 3 2020
3) 4(3x - 2) - 3(x - 4) = 7x+ 10
<=> 12x - 8 - 3x + 12 = 7x + 10
<=> 9x - 7x = 10 - 4
<=> 2x = 6
<=> x = 3
Vậy S = {3}
4) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)
<=> \(\frac{x^2+14x+40}{12}+\frac{3\left(x^2+2x-8\right)}{12}=\frac{4\left(x^2+8x-20\right)}{12}\)
<=> x2 + 14x + 40 + 3x2 + 6x - 24 = 4x2 + 32x - 80
<=> 4x2 + 20x - 4x2 - 32x = -80 - 16
<=> -12x = -96
<=> x = 8
Vậy S = {8}
13 tháng 2 2020
a) \(\frac{7x}{8}-5\left(x-9\right)=\frac{20x+1,5}{6}\)
\(\Leftrightarrow\frac{7x}{8}-\frac{40\left(x-9\right)}{8}=\frac{20x+1,5}{6}\)
\(\Leftrightarrow\frac{7x}{8}-\frac{40x-360}{8}=\frac{20x+1,5}{6}\)
\(\Leftrightarrow\frac{360-33x}{8}=\frac{20x+1,5}{6}\)
\(\Leftrightarrow2160-198x=160x+12\)
\(\Leftrightarrow358x=2148\)
\(\Leftrightarrow x=6\)
Vậy nghiệm của pt x=6
b) \(\frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x+1\right)}{7}-5\)
\(\Leftrightarrow\frac{10\left(x-1\right)+4}{12}-\frac{21x-3}{12}=\frac{4x+2}{7}-\frac{35}{7}\)
\(\Leftrightarrow\frac{-11x-3}{12}=\frac{4x-33}{7}\)
\(\Leftrightarrow-77x-21=48x-396\)
\(\Leftrightarrow125x=375\)
\(\Leftrightarrow3\)
Vậy nghiệm của pt x=3
16 tháng 6 2016
a) \(\left|3x-\frac{1}{2}\right|+\left|\frac{1}{2}y+\frac{3}{5}\right|=0\)
=>\(3x-\frac{1}{2}=0;\frac{1}{2}y+\frac{3}{5}=0\left(\left|3x-\frac{1}{2}\right|;\left|\frac{1}{2}y+\frac{3}{5}\right|\ge0\right)\)
=>\(x=\frac{1}{6};y=\frac{-6}{5}\)
b)\(\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|\le0\)
Ta lại có:
\(\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|\ge0\)
=>\(\frac{3}{2}x+\frac{1}{9}=0;\frac{1}{5}y-\frac{1}{2}=0\Rightarrow x=-\frac{2}{27};y=\frac{5}{2}\)
