tìm x: (x+1)(x+2)(x+4)(x+5)=4
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a, (x + 9) + (x - 2) + (x + 7) + (x - 4) + (x + 5) + (x - 6) + (x + 3) + (x - 8) + (x + 1) = 95
x + 9 + x - 2 + x + 7 + x - 4 + x + 5 + x - 6 + x + 3 + x - 8 + x + 1 = 95
9x + 9 - 2 + 7 - 4 + 5 - 6 + 3 - 8 + 1 = 95
9x + 5 = 95
9x = 95 - 5
9x = 90
x = 90 : 9
x = 10
Vậy x = 10
b, {[(x : 2) - 1/2 ] : 4 - 1/4 } : 5 = 5 + 1/5
{[(x : 2) - 1/2 ] : 4 - 1/4 } : 5 = 26/5
{[(x : 2) - 1/2 ] : 4 - 1/4 } = 26/5 x 5
[(x : 2) - 1/2 ] : 4 - 1/4 = 26
[(x : 2) - 1/2 ] : 4 = 26 + 1/4
[(x : 2) - 1/2 ] : 4 = 105/4
(x : 2) - 1/2 = 105/4 x 4
(x : 2) - 1/2 = 105
x : 2 = 105 + 1/2
x : 2 = 211/2
x = 211/2 x 2
x = 211
Vậy x = 211
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`@` ` \text {Ans}`
`\downarrow`
`a,`
`1/4+3/4*x=3/2-x`
`=> 1/4 + 3/4x - 3/2 + x = 0`
`=> (1/4 - 3/2) + (3/4x + x) = 0`
`=> -5/4 + 7/4x = 0`
`=> 7/4x = 5/4`
`=> x = 5/4 \div 7/4`
`=> x = 5/7`
Vậy, `x=5/7`
`b,`
`3/5*x-1/4=1/10*x-1/2`
`=> 3/5x - 1/4 - 1/10x + 1/2 = 0`
`=> (3/5x - 1/10x) + (-1/4 + 1/2)=0`
`=> 1/2x + 1/4 = 0`
`=> 1/2x = -1/4`
`=> x = -1/4 \div 1/2`
`=> x = -1/2`
Vậy, `x=-1/2`
`c,`
`3x-3/5=x-1/4`
`=> 3x - 3/5 - x + 1/4 = 0`
`=> (3x - x) - (3/5 - 1/4) = 0`
`=> 2x - 7/20 = 0`
`=> 2x = 0,35`
`=> x = 0,35 \div 2`
`=> x = 7/40`
Vậy, `x=7/40`
`d,`
`3/2*x-2/5=1/3*x-1/4`
`=> 3/2x - 2/5 - 1/3x + 1/4 = 0`
`=> (3/2x - 1/3x) - (2/5 - 1/4) = 0`
`=> 7/6x - 3/20 = 0`
`=> 7/6x = 3/20`
`=> x = 3/20 \div 7/6`
`=> x = 9/70`
Vậy, `x=9/70`
`@` `\text {Kaizuu lv uuu}`
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (x-2)3+6(x+1)2-x3+12=0
\(\Rightarrow\)x3-6x2+12x-8+6(x2+2x+1)-x3+12=0
\(\Rightarrow\)x3-6x2+12x-8+6x2+12x+6-x3+12=0
\(\Rightarrow\)24x+10=0
\(\Rightarrow\)24x=-10
\(\Rightarrow\)x=\(\dfrac{-10}{24}=\dfrac{-5}{12}\)
b)(x-5)(x+5)-(x+3)2+3(x-2)2=(x+1)2-(x-4)(x+4)+3x2
\(\Rightarrow\)x2-25-(x2+6x+9)+3(x2-4x+4)=x2+2x+1-(x2-16)+3x2
\(\Rightarrow\)x2-25-x2-6x-9+3x2-12x+12=x2+2x+1-x2+16+3x2
\(\Rightarrow\)3x2-18x-22=3x2+2x+17
\(\Rightarrow\)3x2-18x-22-3x2-2x-17=0
\(\Rightarrow\)-20x-39=0
\(\Rightarrow\)-20x=39
\(\Rightarrow\)x=\(-\dfrac{39}{20}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b: =>4x^2+8x-8x^2+5x-10=0
=>-4x^2+13x-10=0
=>x=2 hoặc x=5/4
c: =>2x^2-5x+6x-15=2x^2+8x
=>x-15=8x
=>-7x=15
=>x=-15/7
d: =>3x^2+15x-2x-10-3x^2-12x=5
=>x-10=5
=>x=15
e: =>x^2-3x+2x^2+2x=3x^2-12
=>-x=-12
=>x=12
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn nên viết lại đề bài cho sáng sủa, rõ ràng để người đọc dễ hiểu hơn.
f: =>4(x^2+4x-5)-x^2-7x-10=3(x^2+x-2)
=>4x^2+16x-20-x^2-7x-10-3x^2-3x+6=0
=>6x-24=0
=>x=4
e: =>8x+16-5x^2-10x+4(x^2-x-2)=4-x^2
=>-5x^2-2x+16+4x^2-4x-8=4-x^2
=>-6x+8=4
=>-6x=-4
=>x=2/3
d: =>2x^2+3x^2-3=5x^2+5x
=>5x=-3
=>x=-3/5
b: =>2x^2-8x+3x-12+x^2-7x+10=3x^2-12x-5x+20
=>-12x-2=-17x+20
=>5x=22
=>x=22/5
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (2x + 1)(1 - 2x) + (1 - 2x)2 = 18
= ( 1 - 2x) \(\left[\left(2x+1+1-2x\right)\right]\) = 18
= 2(1 - 2x) - 18 = 0
= 2 - 4x - 18 = 0
= -16 - 4x = 0
= -4x = 16
= x = \(\dfrac{16}{-4}=-4\)
b) 2(x + 1)2 -(x - 3)(x + 3) - (x - 4)2 = 0
= 2 (x2 + 2x + 1) - (x2 - 9) - (x2 - 8x + 16) = 0
= 2x2 + 4x + 2 - x2 + 9 - x2 + 8x - 16 = 0
= 12x - 5 = 0
= 12x = 5
= x = \(\dfrac{5}{12}\)
c) (x - 5)2 - x(x - 4) = 9
= x2 - 10x + 25 - x2 + 4x - 9 = 0
= -6x + 16 = 0
= -6x = -16
= x = \(\dfrac{-16}{-6}=\dfrac{8}{3}\)
d) (x - 5)2 + (x - 4)(1 - x)
= x2 - 10x + 25 + 5x - x2 - 4 = 0
= -5x + 21 = 0
= -5x = -21
= x = \(\dfrac{-21}{-5}=\dfrac{21}{5}\)
Chúc bạn học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
a: x*3/4=1/5
=>x=1/5:3/4=1/5*4/3=4/15
b: x*3/7=2/5
=>x=2/5:3/7=2/5*7/3=14/15
c: 1/3+2/9=2/12x
=>1/6x=3/9+2/9=5/9
=>x=5/9*6=30/9=10/3
d: 4/15*x-2/3=1/5
=>4/15*x=2/3+1/5=10/15+3/15=13/15
=>4x=13
=>x=13/4
e: x:1/7=2/3
=>x=2/3*1/7=2/21
f: 1/9:x=7/3
=>x=1/9:7/3=1/9*3/7=3/63=1/21
j: 1/4+5/12=8/3:x
=>8/3:x=3/12+5/12=8/12=2/3
=>x=4
h: =>7/4:x=1/5+1/2=7/10
=>x=7/4:7/10=10/4=5/2
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Bài 1 :
\(\frac{x-1}{x-5}=\frac{6}{7}\Leftrightarrow7x-7=6x-30\)
\(\Leftrightarrow x=-23\)
\(\frac{x-2}{x-1}=\frac{x+4}{x+7}\)ĐK : \(x\ne1;-7\)
\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=\left(x+4\right)\left(x-1\right)\)
\(\Leftrightarrow x^2+5x-14=x^2+3x-4\)
\(\Leftrightarrow2x-10=0\Leftrightarrow x=5\)
![](https://rs.olm.vn/images/avt/0.png?1311)