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10x + 3/12 = 1 + 6 + 8x/9
<=> 10x + 3/2 = 7 + 3x/9
<=> 180x + 27 = 126 + 16x
<=> 180x - 16x = 126 - 27
<=> 164x = 99
<=> x = 99/164
![](https://rs.olm.vn/images/avt/0.png?1311)
Xin lỗi bạn,mk ms học đến phân tích đa thức thành nhân tử nhóm nhiều hạng tử,còn phần này mk ms học còn yếu lắm.
1. \(-10x^2+11x+6\)
\(=-10x^2+15x-4x+6\)
\(=-5x\left(2x-3\right)-2\left(2x-3\right)\)
\(=\left(-5x-2\right)\left(2x-3\right)\)
2.\(10x^2-4x-6\)
\(=2\left(5x^2-2x-3\right)\)
\(=2\left(5x^2+3x-5x-3\right)\)
\(=2\left[x\left(5x+3\right)-\left(5x+3\right)\right]\)
\(=2\left(x-1\right)\left(5x+3\right)\)
3. \(10x^2+7x-6\)
\(=10x^2+12x-5x-6\)
\(=2x\left(5x+6\right)-\left(5x+6\right)\)
\(=\left(2x-1\right)\left(5x+6\right)\)
4. \(10x^2-14x-12\)
\(=2\left(5x^2-7x-6\right)\)
\(=2\left(5x^2+3x-10x-6\right)\)
\(=2\left[x\left(5x+3\right)-2\left(5x+3\right)\right]\)
\(=2\left(x-2\right)\left(5x+3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`d,(10x+3)/12=1+(6+8x)/9`
`<=>(10x+3)/12=(8x+15)/9`
`<=>30x+9=32x+60`
`<=>2x=-51`
`<=>x=-51/2`
![](https://rs.olm.vn/images/avt/0.png?1311)
(10x+3)/12= 1+ (6+8x)/9
(10x+3)*9/36= 36/36+ (6+8x)*4/36
(90x+27)/36= 36/36 + (24+32x)/36
khử mẫu ta được
90x+27 = 36+24+32x
chuyển các hạng tử chứa ẩn sang 1 vế
90x-32x=-27+36+24
58x= 33
x= 33/58
vậy nghiệm của pt đã cho là S=(33/58)
mn k cho mk nha
arigato
\(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)
\(\Rightarrow\frac{\left(10x+3\right).3}{12.3}=\frac{36}{36}+\frac{\left(6+8x\right).4}{9.4}\)
\(\Rightarrow3\left(10x+3\right)=36+4\left(6+8x\right)\)
\(\Rightarrow30x+9=36+24+48x\)
\(\Rightarrow30x-48x=36+24-9\)
\(\Rightarrow-18x=51\)
\(\Rightarrow x=-\frac{17}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>10x-14=15-9x
=>19x=29
hay x=29/19
b: \(\Leftrightarrow3\left(10x+3\right)=36+4\left(8x+6\right)\)
=>30x+9=36+32x+24
=>30x+9=32x+60
=>-2x=51
hay x=-51/2
c: \(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)
=>35x-5+60x=96-6x
=>101x=101
hay x=1
d: \(\Leftrightarrow12\left(\dfrac{1}{2}-\dfrac{3}{2}x\right)=-5x+6\)
\(\Leftrightarrow6-18x+5x-6=0\)
=>-13x=0
hay x=0
\(a,\dfrac{5x-7}{3}=\dfrac{5-3x}{2}\\ \Leftrightarrow2\left(5x-7\right)=3\left(5-3x\right)\\ \Leftrightarrow10x-14=15-9x\\ \Leftrightarrow10x-14-15+9x=0\\ \Leftrightarrow19x-19=0\\ \Leftrightarrow x=1\)
\(b,\dfrac{10x+3}{12}=1+\dfrac{6+8x}{9}\\ \Leftrightarrow\dfrac{3\left(10x+3\right)}{36}=\dfrac{36}{36}+\dfrac{4\left(6+8x\right)}{36}\\ \Leftrightarrow30x+9=36+24+32x\\ \Leftrightarrow36+24+32x-30x-9=0\\ \Leftrightarrow2x+51=0\\ \Leftrightarrow x=-\dfrac{51}{2}\)
\(c,\dfrac{7x-1}{6}+2x=\dfrac{16-x}{5}\\ \Leftrightarrow\dfrac{7x-1+12x}{6}=\dfrac{16-x}{5}\\ \Leftrightarrow5\left(19x-1\right)=6\left(16-x\right)\\ \Leftrightarrow95x-5=96-6x\\ \Leftrightarrow95x-5-96+6x=0\\ \Leftrightarrow101x-101=0\\ \Leftrightarrow x=1\)
\(d,4\left(0,5-1,5x\right)=-\dfrac{5x-6}{3}\\ \Leftrightarrow12\left(0,5-1,5x\right)=6-5x\\ \Leftrightarrow6-18x=6-5x\\ \Leftrightarrow6-5x-6+18x=0\\ \Leftrightarrow13x=0\\ \Leftrightarrow x=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{10x+3}{12}=1+\dfrac{6+8x}{9}\)
\(\Leftrightarrow\dfrac{30x+9}{36}=\dfrac{36}{36}+\dfrac{24+32x}{36}\)
\(\Leftrightarrow30x+9=36+24+32x\)
\(\Leftrightarrow-2x=51\)
\(\Leftrightarrow x=-\dfrac{51}{2}\)
\(\dfrac{10x+3}{12}=\dfrac{1+6+8x}{9}=\dfrac{7+8x}{9}\)
\(\Leftrightarrow90x+27=84+96x\)
\(\Leftrightarrow-6x=57\)
\(\Leftrightarrow x=-\dfrac{57}{6}\)
\(10x+\dfrac{3}{12}=1+6+\dfrac{8x}{9}\)
\(\dfrac{10x}{1}+\dfrac{3}{12}=\dfrac{1}{1}+\dfrac{6}{1}+\dfrac{8x}{9}\)
\(\dfrac{10x.36}{1.36}+\dfrac{3.3}{12.3}=\dfrac{1.36}{1.36}+\dfrac{6.36}{1.36}+\dfrac{8x.4}{9.4}\)
khử mẫu :
=>360x+9=36+216+32x
=>360x+9-36-216-32x=0
=>328x-243=0
=>328x=243
=>x=243/328