Tìm x
a) 4.(x-5)-3.(4-x)=-4
b) 12.(x+6)-9.(x-2)=63
c) 2.x.(x-5).x.(2.x-3)=4
d) (x-2).(x+3)=0
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a,3/4 . (-5/12)+3/4.(-7/12)
` 3/4 . [ - ( 5/12 + 7/12 ) ] `
`3/4 . (-1) = -3/4 `
`2/3 . x - 0,5 = 3/4 `
` x - 0,5 = 3/4 - 2/3 `
` x-0,5 = 1/12 `
` x = 1/12 + 0,5 `
` x= 7/12 `
Bài 2:
a: =>2/3x=3/4+1/2=3/4+2/4=5/4
=>x=5/4:2/3=5/4*3/2=15/8
b:=>-2x+4=3x-12
=>-5x=-16
=>x=16/5
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:
a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)
\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)
\(\Leftrightarrow-12x^2+14x+13=0\)
\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)
b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)
\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)
hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
a: \(=\dfrac{x^2+3x+2-x^2+2x+8}{\left(x-2\right)\left(x+2\right)}=\dfrac{5x+10}{\left(x-2\right)\left(x+2\right)}=\dfrac{5}{x-2}\)
b: \(=\dfrac{x^2-4x+3-x^2-3x-2+8x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x-1}\)
c: \(=\dfrac{x+2}{x\left(x-2\right)}+\dfrac{2}{x\left(x+2\right)}+\dfrac{3x+2}{\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{x^2+2x+2x-4+3x+2}{x\left(x-2\right)\left(x+2\right)}=\dfrac{x^2+7x-2}{x\left(x-2\right)\left(x+2\right)}\)
a,
\(\dfrac{x+1}{x-2}-\dfrac{x}{x+2}+\dfrac{8}{x^2-4}\\ =\dfrac{x^2+3x+2-x^2+2x+8}{\left(x-2\right)\left(x+2\right)}=\dfrac{5x+10}{\left(x-2\right)\left(x+2\right)}=\dfrac{5\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{5}{x-2}\)
b,
\(\dfrac{x-3}{x+1}-\dfrac{x+2}{x-1}+\dfrac{8x}{x^2-1}\\ =\dfrac{x^2-4x+3-x^2-3x-2+8x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{1}{x-1}\)
Bài 3:
Gọi số nhóm là x
Theo đề, ta có: \(x\in\left\{1;2;3;4;6;9;12;18;36\right\}\)
mà 2<x<6
nên \(x\in\left\{3;4\right\}\)
Vậy: Có 2 cách chia nhóm
a) 25 - x = 12 + 6 =18
x=25-18=7 Vậy x=7
b) 7 + 2 x ( x -3 ) = 11
2.(x-3)=11-7=4
x-3=4:2=2
x=3+2=5
c) 102 : ( 2.x + 13) : 4) = 6
(2.x+13):4=102:6=17
2.x+13=17.4=68
2.x=68-13=55
x=27,5 Vậy x=27,5
Bài 3:
Gọi số nhóm là x
Theo đề, ta có: x∈{1;2;3;4;6;9;12;18;36}x∈{1;2;3;4;6;9;12;18;36}
mà 2<x<6
nên x∈{3;4}x∈{3;4}
Vậy: Có 2 cách chia nhóm
còn bài 1 chắc bn làm đc nha tick mk nha
a)\(\dfrac{x}{60}=-\dfrac{3}{4}\)
\(\Rightarrow x\cdot4=60\cdot\left(-3\right)\)
\(x\cdot4=-180\)
x=45
b)\(\dfrac{2}{5}=\dfrac{12}{x}\)
\(\Rightarrow2x=5\cdot12\)
\(2x=60\)
x=30
c)\(x-\dfrac{5}{7}=\dfrac{6}{21}\)
\(x=\dfrac{2}{7}+\dfrac{5}{7}\)
x=1
d)\(x+\dfrac{7}{8}=\dfrac{63}{24}\)
\(x=\dfrac{21}{8}-\dfrac{7}{8}\)
\(\dfrac{14}{8}\)
1/(2.x-5)+17=6
=> 2x - 5 = -11
=> 2x = -6
=> x = 3
vậy_
2/10-2.(4-3x)=-4
=> 2(4 - 3x) = 14
=> 4 - 3x = 7
=> 3x = -3
=> x = -1
3/-12+3.(-x+7)=-18
=> 3(-x+7) = -6
=> -x+7 = -2
=> -x = -9
=> x = 9
4/24:(3.x-2)=-3
=> 3x - 2 = -8
=> 3x = -6
=> x = -2
5/-45:5.(-3-2.x)=3
=> 5(-3 - 2x) = -15
=> -3 - 2x = -3
=> - 2x = 0
=> x = 0
6/x.(x+7)=0
=> x = 0 hoặc x + 7 = 0
=> x = 0 hoặc x = -7
7/(x+12).(x-3)=0
=> x + 12 = 0 hoặc x - 3 = 0
=> x = -12 hoặc x = 3
8/(-x+5).(3-x)=0
=> -x + 5 = 0 hoặc 3 - x = 0
=> x = 5 hoặc x = 3
9/x.(2+x).(7-x)=0
=> x = 0 hoặc 2 + x = 0 hoặc 7 - x = 0
=> x = 0 hoặc x = -2 hoặc x = 7
10/(x-1).(x+2).(-x-3)=0
=> x - 1 = 0 hoặc x + 2 = 0 hoặc -x-3 = 0
=> x = 1 hoặc x = -2 hoặc x = -3
a) \(\left(x+1\right)^3-\left(x-1\right)^3-6\cdot\left(x-1\right)^2=10\)
\(\Rightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\cdot\left(x^2-2x+1\right)=10\)
\(\Rightarrow6x^2+2-6x^2+12x-6=10\)
\(\Rightarrow12x-4=10\)
\(\Rightarrow12x=14\)
\(\Rightarrow x=\dfrac{7}{6}\)
b) \(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=42\)
\(\Rightarrow x\left(x^2-25\right)-\left(x^3+8\right)=42\)
\(\Rightarrow x^3-25x-x^3-8=42\)
\(\Rightarrow-25x-8=42\)
\(\Rightarrow-25x=50\)
\(\Rightarrow x=\dfrac{50}{-25}=-2\)
c) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)
\(\Rightarrow x^3-6x^2+12x-8-\left(x^3-27\right)+6\left(x^2+2x+1\right)=49\)
\(\Rightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)
\(\Rightarrow24x+25=49\)
\(\Rightarrow24x=24\)
\(\Rightarrow x=\dfrac{24}{24}=1\)
a) 4.(x-5)-3.(4-x)=-4 => 4x - 20 - 12 + 3x = -4 => 7x = 28 => x=4
b) 12.(x+6)-9.(x-2)=63 => 12x + 72 - 9x + 18 = 63 => 3x = -27 => x= -9
d) (x-2).(x+3)=0 => [ x-2 = 0 => x = 2
x+3 = 0 => x = -3
còn câu c thì bn coi lại cái đề nha , coi thử nó có lộn ko nha ^^